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# PHYSICS 555 PHYSICS 555

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This 189 page Class Notes was uploaded by Bryana Greenholt on Monday September 21, 2015. The Class Notes belongs to PHYSICS 555 at Ohio State University taught by Staff in Fall. Since its upload, it has received 48 views. For similar materials see /class/209996/physics-555-ohio-state-university in Physics 2 at Ohio State University.

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53 THE DIVERGENCEAND CURL OFB 221 53 The Divergence and Curl of B 531 StraightLine Currents The magnetic eld of an in nite straight Wire is shown in Fig 527 the current is coming out of the page At a glance it is clear that this eld has a nonzero curl something you ll never see in an electrostatic eld let s calculate it B Figure 527 According to Eq 536 the integral of B around a circular path of radius 5 centered at the wire is 1 1 yiBdhyi idm i dlzaol 271s 2715 Notice that the answer is independent of s that s because B decreases at the same rate as the circumference increases In fact it doesn t have to be a circle any old loop that encloses the wire would give the same answer For if we use cylindrical coordinates 5 db z with the current owing along the z axis B L0quot 541 275 anddlds sd dziso I 1 I 2 yiedlz i sd i d a0L 71 s 271 0 This assumes the loop encircles the wire exactly once if it went around twice the d would run from 0 to 471 and if it didn t enclose the wire at all then d would go from 1 to 2 and back again with f dd 2 0 Fig 528 222 CHAPTER 5 MAGNETOSTATICS Loop Wire A l Figure 528 Figure 529 Now suppose we have a bundle of straight wires Each wire that passes through our loop contributes uo I and those outside contribute nothing Fig 529 The line integral will then be yiB d1 norm 542 where Ienc stands for the total current enclosed by the integration path If the ow of charge is represented by a volume current density J the enclosed current is Ienc 39 d3 543 with the integral taken over the surface bounded by the loop Applying Stokes theorem to Eq 542 then VXBdaM0Jda and hence V szqu 544 With minimal labor we have actually obtained the general formula for the curl of B But our derivation is seriously awed by the restriction to in nite straight line currents and combinations thereof Most current con gurations cannot be constructed out of in nite straight wires and we have no right to assume that Eq 544 applies to them So the next section is devoted to the formal derivation of the divergence and curl of B starting from the Biot Savart law itself 532 The Divergence and Curl of B The Biot Savart law for the general case of a volume current reads Br zxadr 545 471 a 53 THE DIVERGENCE AND CURL OFB 223 x9y2 Figure 530 This formula gives the magnetic eld at a point r x y z in terms of an integral over the current distribution J x y z Fig 530 It is best to be absolutely explicit at this stage B is a function of x y z J is a function of x y Z a x x iy y z z i dr 2 dxdydZ The integration is over the primed coordinates the divergence and the curl are to be taken with respect to the unprimed coordinates Applying the divergence to Eq 545 we obtain VB vJx12gt dr 546 471 7L Invoking product rule number 6 But V X J 0 because J doesn t depend on the unprimed variables x y z whereas v x 1242 0 Prob 162 So 54s Evidently the divergence of the magnetic eld is zero Applying the curl to Eq 545 we obtain VXBz Vx Jxl dr 549 471 42 Again our strategy is to expand the integrand using the appropriate product rule in this case number 8 VXJXJv JV 550 224 CHAPTER 5 MAGNETOSTATICS I have dropped terms involving derivatives of J because J does not depend on x y z The second term integrates to zero as we ll see in the next paragraph The rst term involves the divergence we were at pains to calculate in Chapter 1 Eq 1100 V 4215300 551 1 Thus M0 3 V X B E Jr 4718 r rdr u0Jr which con rms that Eq 544 is not restricted to straight line currents but holds quite generally in magnetostatics To complete the argument however we must check that the second term in Eq 550 integrates to zero Because the derivative acts only on IALdz we can switch from V to V at the cost of a minus sign9 JV 2 JV 2 552 L L The x component in particular is x x x x x x 3 gtv 3 Ji T using product lule 5 Now for steady currents the divergence of J is zero Eq 531 so I2 x x lU39Vu zlivl 43 J and therefore this contribution to the integral 549 can be written V MJ 0112 MJda 553 v 3 S 3 The reason for switching from V to V was precisely to permit this integration by parts But what region are we integrating over Well it s the volume that appears in the Biot Savart law 545 large enough that is to include all the current You can make it bigger than that if you like J 0 out there anyway so it will add nothing to the integral The essential point is that on the boundary the current is zero all current is safely inside and hence the surface integral 553 vanishes10 9The point here is that It depends only on the difference between the coordinates and B Bx f x x39 83x fx x gt 10H J itself extends to in nity as in the case of an in nite straight wire the Surface integral is still typicall zero though the analysis calls for greater care 53 THE DI VERGENCE AND CURL OF B 225 533 Applications of Ampere s Law The equation for the curl of B m is called Ampere s law in differential form It can be converted to integral form by the usual device of applying one of the fundamental theorems in this case Stokes theorem V xBday Bd1u0Jda Now f J da is the total current passing through the surface Fig 531 which we call Ienc the current enclosed by the amperian loop Thus 9913611 norm 555 This is the integral version of Ampere s law it generalizes Eq 542 to arbitrary steady currents Notice that Eq 555 inherits the sign ambiguity of Stokes theorem Sect 135 Which way around the 100p am I supposed to go And which direction through the surface corresponds to a positive current The resolution as always is the righthand rule If the ngers of your right hand indicate the direction of integration around the boundary then your thumb de nes the direction of a positive current Boundary line Figure 531 Just as the BiotSavart law plays a role in magnetostatics that Coulomb s law assumed in electrostatics so Ampere s plays the role of Gauss s Electrostatics Coulomb gt Gauss Magnetostatics Biot Savart gt Ampere In particular for currents with appropriate symmetry Ampere s law in integral form offers a lovely and extraordinarily ef cient means for calculating the magnetic eld 226 CHAPTER 5 MAGNETOSTATICS Example 57 Find the magnetic eld a distance s from a long straight wire Fig 532 carrying a steady current I the same problem we solved in Ex 55 using the BiotSavart law Solution We know the direction of B is circumferential circling around the wire as indicated by the right hand rule By symmetry the magnitude of B is constant around an amperian loop of radius s centered on the wire So Ampere s law gives Bdl8 dl 8271s zaolenczaol or 1 0 B 2713 This is the same answer we got before Eq 536 but it was obtained this time with far less effort z Sheet of current Amperian loop S Amperlan loop B x 1 Figure 532 Figure 533 Example 58 Find the magnetic eld of an in nite uniform surface current K K it owing over the xy plane Fig 533 Solution First of all what is the direction of B Could it have any xcomponent No A glance at the BiotSavart law 539 reveals that Bis perpendicularto K Could it have a zcomponent No again You could con rm this by noting that any vertical contribution from a lament at y is canceled by the corresponding lament at y But there is a nicer argument Suppose the eld pointed away from the plane By reversing the direction of the current I could make it point toward the plane in the Biot Savart law changing the sign of the current switches the sign of the eld But the z component of B cannot possibly depend on the direction of the current in the xy plane Think about it So B can only have a y component and a quick check with your right hand should convince you that it points to the left above the plane and to the right below it 53 THE DIVERGENCE AND CURL OF B 227 With this in mind we draw a rectangular amperian loop as shown in Fig 533 parallel to the yz plane and extending an equal distance above and below the surface Applying Ampere s law we nd 8 dl 231 016m 2 MOKI one Bl comes from the top segment and the other from the bottom so B uOZK or more precisely u02K I forzlt0 i uo2Ky for z gt o 556 Notice that the eld is independent of the distance from the plane just like the electric eld of a uniform surface charge Ex 24 Example 59 Find the magnetic eld of a very long solenoid consisting of n closely wound turns per unit length on a cylinder of radius R and carrying a steady current I Fig 534 The point of making the windings so close is that one can then pretend each turn is circular If this troubles you after all there is a net current I in the direction of the solenoid s axis no matter how tight the winding picture instead a sheet of aluminum foil wrapped around the cylinder carrying the equivalent uniform surface current K n Fig 535 Or make a double winding going up to one end and then always in the same sense going back down again thereby eliminating the net longitudinal current But in truth this is all unnecessary fastidiousness for the eld inside a solenoid is huge relatively Speaking and the eld of the longitudinal current is at most a tiny re nement Solution First of all what is the direction of B Could it have a radial component No For suppose BS were positive if we reversed the direction of the current BS would then be negative But switching I is physically equivalent to turning the solenoid upside down and Figure 534 Figure 535 CHAPTER 5 MAGNETOSTATICS Amperian loop Amperian loops Figure 536 Figure 537 that certainly should not alter the radial eld How about a circumferential component No For B would be constant around an amperian loop concentric with the solenoid Fig 536 and hence B dl B 27ts hole C 0 since the loop encloses no current So the magnetic eld of an in nite closely wound solenoid runs parallel to the axis From the right hand rule we expect that it points upward inside the solenoid and downward outside Moreover it certainly approaches zero as you go very far away With this in mind let s apply Ampere s law to the two rectangular loops in Fig 537 Loop 1 lies entirely outside the solenoid with its sides at distances a and b from the axis f3 0 3a BbL Molenc 0 Evidently the eld outside does not depend on the distance from the axis But we know that it goes to zero for large s It must therefore be zero everywhere This astonishing result can also be derived from the BiotSavart law of course but it s much more dif cult See Prob 544 As for loop 2 which is half inside and half outside Ampere s law gives fB dl 2 BL uoIenC HonIL where B is the eld inside the solenoid The right side of the loop contributes nothing since 8 0 out there Conclusion B uOnIi inside the solenoid 0 outside the solenoid 557 Notice that the eld inside is uniform in this sense the solenoid is to magnetostatics what the parallel plate capacitor is to electrostatics a simple device for producing strong uniform elds 53 THE DI VERGENCE AND CURL OF B 229 Like Gauss s law Ampere s law is always true for steady currents but it is not always useful Only when the symmetry of the problem enables you to pull B outside the integral 99 B dl can you calculate the magnetic eld from Amp re s law When it does work it s by far the fastest method when it doesn t you have to fall back on the BiotSavart law The current con gurations that can be handled by Amp re s law are 1 In nite straight lines prototype Ex 57 2 In nite planes prototype Ex 58 3 In nite solenoids prototype Ex 59 4 Toroids prototype Ex 510 The last of these is a surprising and elegant application of Ampere s law it is treated in the following example As in Exs 58 and 59 the hard part is guring out the direction of the eld which we will now have done once and for all for each of the four geometries the actual application of Ampere s law takes only one line Example 510 A toroidal coil c0nsists of a circular ring or donut around which a long wire is wrapped Fig 538 The winding is uniform and tight enough so that each turn can be considered a closed loop The crosssectional shape of the coil is immaterial I made it rectangular in Fig 538 for the sake of simplicity but it could just as well be circular or even some weird asymmetn39cal form as in Fig 539 just as long as the shape remains the same all the way around the ring In that case it follows that the magnetic eld of the toroid is circumferential at all points both inside and outside the coil quotII 1 In Figure 538 Proof According to the Biot Savart law the eld at r due to the current element at r is I X L dl 47 43 We may as well put r in the xz plane Fig 539 so its Cartesian components are x 0 z while the source coordinates are dB 1quot scos W s39 sin d z Then IL x s cos s s1n z z Since the current has no gt component I IS s Z i or in Cartesian coordinates I 1 cos 1 sin 1 CHAPTER 5 MAGNETOSTATICS x r x Figure 539 Accordingly 1 y 2 I X4 2 IS cos gt IS sinrp Z x s cos d s sin gt Z Z sin Isz z s Iz 2 Iz x s cos Is cos z Z 3 st Sin W 2 But there is a symmetrically situated current element at r with the same s the same d the same dl the same 5 and the same IZ but negative gt Fig 539 Because sin gt change sign the I and 2 contributions from Iquot and r cancel leaving only a 9 term Thus the eld at r is in the y direction and in general the eld points in the is direction qed Now that we know the eld is circumferential determining its magnitude is ridiculousl easy Just apply Ampere s law to a circle of radius s about the axis of the toroid BZ s 016m and hence NI 20 V 3 for points inside the coil B m 558 0 for points outside the coil where N is the total number of turns 53 THE DI VERGENCE AND CURL OF B 231 Problem 513 A steady current I ows down a long cylindrical wire of radius a Fig 540 Find the magnetic eld both inside and outside the wire if a The current is uniformly distributed over the outside surface of the wire b The current is distributed in such a way that J is pr0portional to s the distance from the axis Figure 540 Figure 541 Problem 514 A thick slab extending from z a to z a carries a uniform volume current J J a Fig 541 Find the magnetic eld as a function of z both inside and outside the slab Problem 515 Two long coaxial solenoids each carry current I but in opposite directions as shown in Fig 542 The inner solenoid radius a has ni turns per unit length and the outer one radius 7 has n2 Find B in each of the three regions i inside the inner solenoid ii between them and iii outside both Figure 542 Figure 543 Problem 516 A large parallel plate capacitor with uniform surface charge a on the upper plate and a on the lower is moving with a constant speed 12 as shown in Fig 543 a Find the magnetic eld between the plates and also above and below them b Find the magnetic force per unit area on the upper plate including its direction c At what speed v would the magnetic force balance the electrical force11 11See footnote 8 232 CHAPTER 5 MAGNETOSTATICS Problem 517 Show that the magnetic eld of an in nite solenoid runs parallel to the axis regardless of the crosssectional shape of the coil as long as that shape is constant along the length of the solenoid What is the magnitude of the eld inside and outside of such a coil Show that the toroid eld 558 reduces to the solenoid eld when the radius of the donut is so large that a segment can be considered essentially straight Problem 518 In calculating the current enclosed by an amperian loop one must in general evaluate an integral of the form Ienc J da 8 The trouble is there are in nitely many surfaces that share the same boundary line Which one are we supposed to use 534 Comparison of Magnetostatics and Electrostatics The divergence and curl of the electrostatic eld are l V E p Gauss s law 60 V X E 0 no name These are Maxwell s equations for electrostatics Together with the boundary condition E gt 0 far from all charges Maxwell s equations determine the eld if the source charge density p is given they contain essentially the same information as Coulomb s law plus the principle of superposition The divergence and curl of the magnetostatic eld are V B 0 no name V X B MOJ Ampere s law These are Maxwell s equations for magnetostatiCS Again together with the boundary condition B gt 0 far from all currents Maxwell s equations determine the magnetic eld they are equivalent to the Biot Savart law plus superpOSition Maxwell s equations and the force law F QE v x B constitute the most elegant formulation of electrostatics and magnetostatics The electric eld diverges away from a positive charge the magnetic eld line curls around a current Fig 544 Electric eld lines originate on pOSitive charges and terminate on negative ones magnetic eld lines do not begin or end anywhere to do so would require a nonzero divergence They either form closed loops or extend out to in nity To put it another way there are no point sources for B as there are for E there exists no magnetic analog to electric charge This is the physical content of the statement V B O Coulomb and others believed that magnetism was produced by magnetic charges magnetic monopoles as we would now call them and in some older books you will still 53 THE DIVERGENCE AND CURL OFB 233 a Electrostatic eld b Magnetostatic eld of a point charge of a long wire Figure 544 nd references to a magnetic version of Coulomb s law giving the force of attraction or repulsion between them It was Amp re who rst speculated that all magnetic effects are attributable to electric charges in motion currents As far as we know Ampere was right nevertheleSs it remains an Open experimental question whether magnetic monopoles exist in nature they are obviously pretty rare or somebody would have found one and in fact some recent elementary particle theories require them For our purpOses though B is divergenceless and there are no magnetic monopoles It takes a moving electric charge to produce a magnetic eld and it takes another moving electric charge to feel a magnetic eld Typically electric forces are enormously larger than magnetic ones That s not some thing you can tell from the theory as such it has to do with the sizes of the fundamental constants 60 and MO In general it is only when both the source charges and the test charge are moving at velocities comparable to the speed of light that the magnetic force approaches the electric fOrce in strength Problems 512 and 516 illustrate this rule How is it then that we ever notice magnetic effects at all The answer is that both in the production of a magnetic eld Biot Savart and in its detection Lorentz it is the current charge times velocity that enters and we can compensate for a smallish velocity by pouring huge quan tities of charge down the wire Ordinarily this charge would simultaneously generate so large an electric force as to swamp the magnetic one But if we arrange to keep the wire neutral by embedding in it an equal amount of Opposite charge at rest the electric eld cancels out leaving the magnetic eld to stand alone It sounds very elaborate but of course this is precisely what happens in an ordinary current carrying wire 12Arr apparent detection B Cabrera Phys Rev Lett 48 1378 1982 has never been reproducedand not for want of trying For a delightful brief history of ideas about magnetism see Chapter 1 in D C Mattis The Theory of Magnetism New York Harper and Row 1965 234 CHAPTER 5 MAGNETOSTATICS Problem 519 a Find the density p of mobile charges in a piece of copper assuming each atom contributes one free electron Look up the necessary physical constants b Calculate the average electron velocity in a copper wire 1 mm in diameter carrying a current of l A Note this is literally a snail s pace How then can you carry On a long distance telephone conversation c What is the force of attraction between two Such wires 1 cm apart 1 If you could somehow rem0ve the stationary positive ions what would the electrical repulsion force be How many times greater than the magnetic force is it Problem 5201s Ampere s law consistent with the general rule Eq 146 that divergence of curl is always zero Show that Ampere s law cannot be valid in general outside magneto statics Is there any such defect in the other three Maxwell equatiOns Problem 521 Suppose there did exist magnetic monOpoles How would you modify Maxwell s equations and the force law to accommodate them If you think there are several plausible options list them and suggest how you might decide experimentally which one is right 54 Magnetic Vector Potential 541 The Vector Potential Just as V X E O permitted us to introduce a scalar potential V in electrostatics E V V so V B O invites the introduction of a vector potential A in magnetostatics m The former is authorized by Theorem 1 of Sect 162 the latter by Theorem 2 the proof of Theorem 2 is developed in Prob 530 The potential formulation automatically takes care of V B 0 since the divergence of a curl is always zero there remains Amp re s law VxBVxVxAVVA V2AM0J 560 Now the electric potential had a builtin ambiguity you can add to V any function whose gradient is zero which is to say any constant without altering the physical quantity E Likewise you can add to the magnetic potential any function whose curl vanishes which is to say the gradient of any scalar with no effect on B We can exploit this freedom to eliminate the divergence of A V A O 561 54 MAGNETIC VECTOR POTENTIAL 235 To prove that this is always possible suppose that our original potential A0 is not divergenceless If we add to it the gradient of A A A0 VA the new divergence is VAVA0V2x We can accommodate Eq 561 then if a function A can be found that satis es V2 V A0 But this is mathematically identical to Poisson s equation 224 sz 60 with V A0 in place of peo as the source And we know how to solve Poisson s equation that s what electrostatics is all about given the charge distribution nd the potential In particular if 0 goes to zero at in nity the solution is Eq 229 l p d T 47T 0 2 and by the same token if V A0 goes to zero at in nity then 1 V 39 A0 A 471 a dt If V A0 does not go to zero at in nity we ll have to use other means to discover the appropriate A just as we get the electric potential by other means when the charge distribu tion extends to in nity But the essential point remains It is always possible to make the vector potential divergenceless To put it the other way around The de nition B V X A speci es the curl of A but it doesn t say anything about the divergence we are at liberty to pick that as we see t and zero is ordinarily the simplest choice With this condition on A Ampere s law 560 becomes This again is nothing but Poisson s equation or rather it is three Poisson s equations one for each Cartesian13 component Assuming J goes to zero at in nity we can read off the solution Ar E ail 563 471 a l3In Cartesian coordinates V2A 2 VzAx VZAy VZAz so Eq 562 reduces to VZAX pOJx VZAy 2 10 Jy and VZAZ p0 JZ In curvilinear coordinates the unit vectors themselves are functions of position and must be differentiated so it is not the case for example that V2 Ar 2 p0 1 The safest way to calculate the Laplacian of a vector in terms of its curvilinear components is to use VZA V V A V x V x A Remember also that even if you calculate integrals such as 563 using curvilinear coordinates you must rst express J in terms of its Cartesian components see Sect 141 236 CHAPTER 5 MAGNETOSTATICS For line and surface currents A O Ecit5 O Ildle A Edd 564 471 a 471 a 471 L If the current does not go to zero at in nity we have to nd other ways to get A some of these are explored in Ex 512 and in the problems at the end of the section It must be said that A is not as useful as V For one thing it s still a vector and although Eqs 563 and 564 are somewhat easier to work with than the Biot Savart law you still have to fuss with components It would be nice if we could get away with a scalar potential B VU 565 but this is incompatible with Ampere s law since the curl of a gradient is always zero A magnetostatic scalar potential can be used if you stick scrupulously to simplyconnected current free regions but as a theoretical tool it is of limited interest See Prob 528 More over since magnetic forces do no work A does not admit a simple physical interpretation in terms of potential energy per unit charge In some contexts it can be interpreted as momentum per unit charge Nevertheless the vector potential has substantial theoretical importance as we shall see in Chapter 10 Example 511 A spherical shell of radius R carrying a uniform surface charge a is set spinning at angular velocity 0 Find the vector potential it produces at point r Fig 545 Solution It might seem natural to align the polar axis along a but in fact the integration is easier if we let I lie on the z axis so that a is tilted at an angle 11 We may as well orient the x axis so that a lies in the xz plane as shown in Fig 546 According to Eq 564 I Ar KI daly 4n L Figure 545 Figure 546 14M D Semon and 1 R Taylor Am J Phys 64 1361 1996 54 MAGNETIC VECTOR POTENTIAL 237 where K 2 av o 1 R2 r2 2Rr cos 6 and da R2 sin6 d6 dqb Now the velocity of a point r in a rotating rigid body is given by a X r in this case 2 y 2 vzwxr z usin1 0 wcoszl R sin 49 cos 15 R sin 49 sin 15 R cos 49 Ra cos 1 sin 19 sin 12 cos 1 sin 19 cos sin 1 cos 6 y sin 1 sino sin 2 Notice that each of these terms save one involves either sin 15 or cos 15 Since Zn 271 sinqbdqb cos dqb 0 0 0 such terms contribute nothing There remains A u0R3aasin1 cos6 sin6 d6 1 2 0 R2r2 2chos6 Letting u E cos 6 the integral becomes 1 u R2 2 R A du wR2 72 2Rru R2 r2 2Rru 3R2quot2 1 1 l 3R2r2 R2 r2 RrR r R2 r2 RrR m If the point r lies inside the sphere then R gt r and this expression reduces to 2r 3R2 if r lies outside the sphere so that R lt r it reduces to 2R3r2 Noting that w x r w sin 1 y we have nally R M0 1 0 X 1 for points inside the sphere AG 2 R4 566 m 3 0 0 X 1 for points outside the sphere r Having evaluated the integral I revert to the natural coordinates of Fig 545 in which a coincides with the z axis and the point r is at r 49 b R 1 Mrsin gb r 5 R Ar 49 b 567 u0R4wa si116 A 739 2 R 3 r2 Curiously the eld inside this spherical shell is uniform 2uoRaa BVxA 3 A 2 A 2 cos6 i s1n69 guoasz guoaRw 568 238 CHAPTER 5 MAGNETOSTATICS Example 512 Find the vector potential of an in nite solenoid with n turns per unit length radius R and current I Solution This time we cannot use Eq 564 since the current itself extends to in nity But here s a cute method that does the job Notice that ffAdlVXAdaBda 569 where I is the ux of B through the loop in question This is reminiscent of Ampere s law in the integral form 555 dB d1 Moleno In fact it s the same equation with B gt A and uolenc gt I If symmetry permits we can determine A from I in the same way we got B from lane in Sect 533 The present problem with a uniform longitudinal magnetic eld uOnI inside the solenoid and no eld outside is analogous to the Ampere s law problem of a fat wire carrying a uniformly distributed current The vector potential is circumferential it mimics the magnetic eld of the wire using a circular amperian loop at radius s inside the solenoid we have yiA d1 A27rs B da uOnI7rs2 Ago quot1 2 For an amperian loop outside the solenoid the ux is s43 fors lt R 570 B da uOnI7rR2 since the eld only extends out to R Thus 1 R2 A A quot forsgtR 571 2 s If you have any doubts about this answer check it Does V x A B Does V A 0 If so we re done Typically the direction of A will mimic the direction of the current For instance both were azimuthal in Exs 511 and 512 Indeed if all the current ows in one direction then Eq 563 suggests that A must point that way too Thus the potential of a nite segment of straight wire Prob 522 is in the direction of the current Of course if the current extends to infinity you can t use Eq 563 in the rst place see Probs 525 and 526 Moreover you can always add an arbitrary constant vector to A this is analogous to changing the reference point for V and it won t affect the divergence or curl of A which is all that matters in Eq 563 we have chosen the constant so that A goes to zero at in nity In principle you could even use a vector potential that is not divergenceless in which case all bets are off Despite all these caveats the essential point remains Ordinarily the direction of A will match the direction of the current 54 MAGNETIC VECTOR POTENTIAL 239 Problem 522 Find the magnetic vector potential of a nite segment of straight Wire carrying a current I Put the wire on the z axis from z to 22 and use Eq 564 Check that your answer is consistent with Eq 5 35 Problem 523 What current density would produce the vector potential A kc where k is a constant in cylindrical coordinates Problem 524 If B is uniform show that Ar r x B works That is check that V A 0 and V X A B Is this result unique or are there other functions with the same divergence and curl Problem 525 a By whatever means you can think of short of looking it up nd the vector potential a distance s from an in nite straight wire carrying a current 1 Check that V A 0 and V X A B b Find the magnetic potential inside the wire if it has radius R and the current is uniformly distributed Problem 526 Find the vector potential above and below the plane surface current in Ex 58 Problem 527 a Check that Eq 563 is consistent with Eq 561 by applying the divergence b Check that Eq 563 is consistent with Eq 545 by applying the curl c Check that Eq563 is consistent with Eq 562 by applying the Laplacian Problem 528 Suppose you want to de ne a magnetic scalar potential U Eq 565 in the vicinity of a currentcarrying wire First of all you must stay away from the wire itself there V x B 7395 0 but that s not enough Show by applying Ampere s law to a path that starts at a and circles the wire returning to b Fig 547 that the scalar potential cannot be single valued that is U a 7E U b even if they represent the same physical point As an example nd the scalar potential for an in nite straight wire To avoid a multivalued potential you must restrict yourself to simply connected regions that remain on one side or the other of every wire never allowing you to go all the way around Amperian loop 1p I Figure 547 240 CHAPTER 5 MAGNETOSTATICS Problem 529 Use the results of Ex 511 to nd the eld inside a uniformly charged sphere of total charge Q and radius R which is rotating at a constant angular velocity a Problem 530 a Complete the proof of Theorem 2 Sect 162 That is show that any divergenceless vector eld F can be written as the curl of a vector potential A What you have to do is nd A x A y and AZ such that i aAzay BAyBz Fx ii anaz BAZax Fy and iii BAyBx BAX8y Fz Here s one way to do it Pick Ax 0 and solve ii and iii for Ay and AZ Note that the constants of integration here are themselves functions of y and z they re constant only with respect to x Now plug these expressions into i and use the fact that V F 0 to obtain x y x Ay 20 Fzx yzdx AZ f0 Fx0yzdy 0 Fyx yzdx b By direct differentiation check that the A you obtained in part a satis es V x A F Is A divergenceless This was a very asymmetrical construction and it would be surprising if it were although we know that there exists a vector whose curl is F and whose divergence is zero c As an example let F y z y x 2 Calculate A and con rm that V x A F For further discussion see Prob 551 542 Summary Magnetostatic Boundary Conditions In Chapter 2 I drew a triangular diagram to summarize the relations among the three fundamental quantities of electrostatics the charge density p the electric eld E and the potential V A similar diagram can be constructed for magnetostatics Fig 548 relating Figure 548 54 MAGNETIC VECTOR POTENTIAL 241 the current density J the eld B and the potential A There is one missing link in the diagram the equation for A in terms of B It s unlikely you would ever need such a formula but in case you are interested see Probs 550 and 551 Just as the electric eld suffers a discontinuity at a surface charge so the magnetic eld is discontinuous at a surface current Only this time it is the tangential component that changes For if we apply Eq 548 in the integral form v Bda0 to a wafer thin pillbox straddling the surface Fig 549 we get J J Babove Bbelow 53972 As for the tangential components an amperian loop running perpendicular to the current Fig 550 yields B d1 Elbow Bl elow MOIeHC MOKZ 01 B B elow MK 573 above Thus the component of B that is parallel to the surface but perpendicular to the current is discontinuous in the amount uoK A similar amperian loop running parallel to the current reveals that the parallel component is continuous These results can be summarized in a single formula Babove Bbelow M0 K X f1 574 where fl is a unit vector perpendicular to the surface pointing upward Figure 549 242 CHAPTER 5 MAGNETOS TATI C S Figure 550 Like the scalar potential in electrostatics the vector potential is continuous across an boundary Aabove AbelOWa 575 for V A 0 guarantees15 that the normal component is continuous and V x A B in theform ffAdlBdadgt means that the tangential components are continuous the ux through an amperian loop of vanishing thickness is zero But the derivative of A inherits the discontinuity of B aAabove aAbelow K 576 an an M0 Problem 531 a Check Eq 574 for the con guration in Ex 59 b Check Eqs 575 and 576 for the con guration in Ex 51 Problem 532 Prove Eq 576 using Eqs 561 574 and 575 S uggestion I d set up Cartesian coordinates at the surface with z perpendicular to the surface and x parallel to the current 543 Multipole Expansion of the Vector Potential If you want an approximate formula for the vector potential of a localized current distri bution valid at distant points a multipole expansion is in order Remember the idea of a multipole expansion is to write the potential in the form of a power series in 1 r where r is the distance to the point in question Fig 551 if r is sufficiently large the series will be 15Note that Eqs 575 and 576 presuppose that A is divergenceless 54 MAGNETIC VECTOR POTENTIAL 243 Figure 551 dominated by the lowest nonvanishing contribution and the higher terms can be ignored As we found in Sect 341 Eq 394 1 1 quot 39 P cos9 577 5 Vrz r 2 2rr cos 9 r g r n Accordingly the vector potential of a current loop can be written M011M01001 Ar gdl zgz mfm Pncos9dl 578 n or more explicitly I 1 1 AI 2 rC089dll 471 r r l 3 l r 3 r2 cos2 9 dl 39 As in the multipole expansion of V we call the rst term which goes like 1 r the monopole term the second which goes like 1 r2 dipole the third quadrupole and so on Now it happens that the magnetic monopole term is always zero for the integral is just the total vector displacement around a closed loop 579 8 d1 0 580 This re ects the fact that there are apparently no magnetic monopoles in nature an as sumption contained in Maxwell s equation V B 0 on which the entire theory of vector potential is predicated 244 CHAPTER 5 MAGNETOS TATI CS In the absence of any monopole contribution the dominant term is the dipole except in the rare case where it too vanishes M01 M01 A Adipr m r COS 9 d1 4an ltr 39 r This integral can be rewritten in a more illuminating way if we invoke Eq 1108 with c f f39r dl f39 X da 582 Then 583 where m is the magnetic dipole moment Here a is the vector area of the loop Prob 161 if the loop is at a is the ordinary area enclosed with the direction assigned by the usual right hand rule ngers in the direction of the current Example 513 Find the magnetic dipole moment of the bookend shaped loop shown in Fig 552 All sides have length w and it carries a current I Solution This wire could be considered the superposition of two plane Square loops Fig 55 3 I The extra sides A B cancel when the two are put together since the currents ow in opposite directions The net magnetic dipole moment is mIw2yIw22 its magnitude is x21w2 and it points along the 450 line 2 y Figure 552 54 MAGNETIC VECTOR POTENTIAL 245 Figure 553 It is clear from Eq 584 that the magnetic dipole moment is independent of the choice of origin You may remember that the electric dipole moment is independent of the origin only when the total charge vanishes Sect 343 Since the magnetic monopole moment is always zero it is not really surprising that the magnetic dipole moment is always independent of ongln Although the dipole term dominates the multipole expansion unless m 0 and thus offers a good approximation to the true potential it is not ordinarily the exact potential there will be quadrupole octopole and higher contributions You might ask is it possible to devise a current distribution whose potential is pure dipole for which Eq 583 is exact Well yes and no like the electrical analog it can be done but the model is a bit contrived To begin with you must take an in nitesimally small loop at the origin but then in order to keep the dipole moment nite you have to crank the current up to in nity with the product m Ia held xed In practice the dipole potential is a suitable approximation whenever the distance r greatly exceeds the size of the loop The magnetic eld of a pure dipole is easiest to calculate if we put m at the origin and let it point in the z direction Fig 554 According to Eq 583 the potential at point Figure 554 246 CHAPTER 5 MAGNETOSTATICS Z Z IIIIIIMQ y E39llllllllllly y a Field of a quotpurequot dipole a Field of a quotphysicalquot dipole Figure 555 r 6 1 is 10 m sin 6 A A r 585 d1p 4g r2 and hence Mom A Bdipr V gtltA 4 32cos6f sin619 586 Jrr Surprisingly this is identical in structure to the eld of an electric dipole Eq 3103 Up close however the eld of a physical magnetic dipole a small current loop looks quite different from the eld of a physical electric dipole plus and minus charges a short distance apart Compare Fig 555 with Fig 337 Problem 533 Show that the magnetic eld of a dipole can be written in coordinatefree form I A A Bdipr g mmm ml 58 Problem 534 A circular loop of wire with radius R lies in the x y plane centered at the origin and carries a current 1 running counterclockwise as viewed from the positive 2 axis a What is its magnetic dipole moment b What is the approximate magnetic eld at points far from the origin c Show that for points on the z axis your an3wer is consistent with the exact eld Ex 56 I when z gtgt R Problem 535 A phonograph record of radius R carrying a uniform surface charge a is rotating at constant angular velocity u Find its magnetic dipole moment Problem 536 Find the magnetic dipole moment of the spinning spherical shell in Ex 51 1 Show that for points r gt R the potential is that of a perfect dipole Problem 537 Find the exact magnetic eld a distance 2 above the center of a square loop 01 side w carrying a current I Verify that it reduces to the eld of a dipole with the appropriate dipole moment when 1 gtgt w 54 MAGNETIC VECTOR POTENTIAL 247 More Problems on Chapter 5 Problem 538 It may have occurred to you that since parallel currents attract the current within a single wire should contract into a tiny concentrated stream along the axis Yet in practice the current typically distributes itself quite uniformly over the wire How do you account for this If the positive charges density p are at rest and the negative charges density p move at speed 1 and none of these depends on the distance from the axis show that p py2 where y E 1 1 1262 and c2 laoeo If the wire as a whole is neutral where is the compensating charge located16 Notice that for typical velocities see Prob 519 the two charge densities are essentially unchanged by the current since y z 1 In plasmas however where the positive charges are also free to move this socalled pinch effect can be very signi cant Problem 539 A current I ows to the right through a rectangular bar of conducting material in the presence of a uniform magnetic eld B pointing out of the page Fig 556 a If the moving charges are positive in which direction are they de ected by the magnetic eld This de ection results in an accumulation of charge on the upper and lower surfaces of the bar which in turn produces an electric force to counteract the magnetic one Equilibrium occurs when the two exactly cancel This phenomenon is knoWn as the Hall effect b Find the resulting potential difference the Hall voltage between the top and bottom of the bar in terms of B u the speed of the charges and the relevant dimensions of the bar17 0 How would your analysis change if the moving charges were negative The Hall effect is the classic way of determining the sign of the mobile charge carriers in a material Figure 556 Figure 557 Problem 540 A plane wire loop of irregular shape is situated so that part of it is in a uniform magnetic eld B in Fig 557 the eld occupies the shaded region and points perpendicular to the plane of the 100p The loop carries a current I Show that the net magnetic force on the loop is F I B w where w is the chord subtended Generalize this result to the case where the magnetic eld region itself has an irregular shape What is the direction of the force 16For further discussion see D C Gabuzda Am J Phys 61 360 1993 17The potential within the bar makes an interesting boundaryvalue problem See M J Moelter J Evans and G Elliot Am J Phys 66 6681998 CHAPTER 5 IVIAGNETOS TATI CS Field region Particle trajectory Figure 558 Problem 541 A circularly symmetrical magnetic eld B depends only on the distance from the axis pointing perpendicular to the page occupies the shaded region in Fig 558 If the total ux f B da is zero show that a charged particle that starts out at the center will emerge from the eld region on a radial path provided it escapes at all if the initial velocity is too great it may simply circle around forever On the reverse trajectory a particle red at the center from outside will hit its target though it may follow a weird route getting there Hints Calculate the total angular momentum acquired by the particle using the Lorentz force 1aw Problem 542 Calculate the magnetic force of attraction between the northern and southern hemispheres of a spinning charged spherical shell Ex 511 Answer n4u002w2 R4 Problem 543 Consider the motion of a particle with mass m and electric charge qe in the eld of a hypothetical stationary magnetic monopole qm at the origin qmf 471 r2 a Find the acceleration of qe expressing your answer in terms of q qm m r the position of the particle and v its velocity b Show that the speed 1 v is a constant of the motion c Show that the vector quantity MOqeq m A E r r Q m X V 4H is a constant of the motion Hint39 differentiate it with respect to time and prove using the equation of motion from a that the derivative is zero d Choosing spherical coordinates r 6 b With the polar z axis along Q i calculate Q 43 and show that 6 is a constant of the motion so qe moves on the surface of a cone something Poincare rst discovered in 1896 18In point of fact the charge follows a geodesic on the cone The original paper is H Poincare Comptes rendus de l Academie des Sciences 123 530 1896 for a more modern treatment see B Rossi and S Olbert Introduction to the Physics of Space New York McGrawHill 1970 54 MAGNETIC VECTOR POTENTIAL 249 ii calculate Q rquot and show that the magnitude of Q is A iii calculate Q 0 show that and determine the constant k e By expressing v2 in spherical coordinates obtain the equation for the trajectory in the form dr 39 a f F that is determine the function f r f Solve this equation for r Problem 544 Use the Biot Savart law most conveniently in the form 539 appropriate to surface currents to nd the eld inside and outside an in nitely long solenoid of radius R with n turns per unit length carrying a steady current I Problem 545 A semicircular wire carries a steady current I itmust be hooked up to some other wires to complete the circuit but we re not concerned with them here Find the magnetic eld atapoint P on the other semicircle Fig 559 Answer MOI871 R lntan 641 tan I z 0 1 C3 1 Figure 559 Figure 560 Problem 546 The magnetic eld on the axis of a circular current loop Eq 538 is far from uniform it falls off sharply with increasing 2 You can produce a more nearly uniform eld by using two such loops a distance d apart Fig 560 a Find the eld B as a function of z and show that BBBz is zero at the point midway between them z 0 Now if you pick 1 just right the seCond derivative of B will also vanish at the midpoint This arrangement is known as a Helmholtz coil it s a convenient way of producing relatively uniform elds in the laboratory b Determine 1 such that BZBBz2 0 at the midpoint and nd the resulting magnetic eld at the center Answer 8M015x5 R CHAPTER 5 MAGNETOS TATICS Problem 547 Find the magnetic eld at a point z gt R on the axis of a the rotating disk and b the rotating sphere in Prob 56 Problem 548 SuppOse you wanted to nd the eld of a circular loop Ex 56 at a point r that is not directly above the center Fig 561 You might as well choose your axes so that r lies in the yz plane at 0 y z The source point is R cos b39 R sin b39 0 and runs from 0 to 2n Set up the integrals from which you could calculate Bx By and Bz and evaluate BI explicitly 7 dll Figure 561 Figure 562 Problem 549 Magnetostati39cs treats the source current the one that sets up the eld and the recipient current the one that experiences the force so asymmetrically that it is by no means obvious that the magnetic force betWeen two current loops is consistent with Newton s third law Show starting with the BiotSavart law 532 and the Lorentz force law 516 that the force on loop 2 due to loop 1 Fig 562 can be written as 0 392 F I I dl dl 588 2 4 1 2 jg jg 42 1 2 In this form it is clear thath F1 since Ii changes direction when the roles of 1 and 2 are interchanged If you seem to be getting an extra term it will help to note that 1 12 do Problem 550 a One way to ll in the missing link in Fig 548 is to exploit the analogy between the de ning equations for A V A 0 V X A B and Maxwell s equations for B V B 0 V x B MOJ Evidently A depends on B in exactly the same way that B depends on qu to wit the BiotSavart law Use this observation to write down the formula for A in terms of B b The electrical analog to your result in a is 1 Er J2 4n 42 Vr dr39 Derive it by exploiting the appropriate analogy 252 54 MAGNETIC VECTOR POTENTIAL 251 Problem 551 Another way to ll in the missing link in Fig 548 is to look for a magnetostatic analog to Eq 221 The obvious candidate would be r Ar B x dl O a Test this formula for the simplest possible case uniform B use the origin as your reference point Is the result consistent with Prob 524 You could cure this problem by throwing in a factor of but the aw in this equation runs deeper b Show that f B x dl is not independent of path by calculating f B x dl around the rectangular loop shown in Fig 563 As far as I know19 the best one can do along these lines is the pair of equations i Vr rltf01Etrdt ii Ar r x fol ABArdi Equation i amounts to selecting a radial path for the integral in Eq 221 equation ii constitutes a more symmetrical solution to Prob 530 c Use ii to nd the vector potential for uniform B d Use ii to nd the Vector potential of an in nite straight wire carrying a steady current I Does ii automatically satisfy V A 0 Answer MOI2719z s s 2 Figure 563 Problem 552 a Construct the scalar potential U r for a pure magnetic dipole m b Construct a scalar potential for the spinning spherical shell Ex 511 Hintz for r gt R this is a pure dipole eld as you can see by comparing Eqs 567 and 585 c Try doing the same for the interior of a solid spinning sphere Hint if you solved Prob 529 you already know the eld set it equal to VU and solve for U What s the trouble 399R L Bishop and 51 Goldberg Tensor Analysis on Manifolds Section 45 New York Macmillan 1968 CHAPTER 5 MAGNETOS TATI CS Problem 553 Just as V B 0 allows us to express B as the curl of a vector potential B V x A so V A 0 permits us to write A itself as the curl of a higher potential A V x W And this hierarchy can be extended ad in nitum a Find the general formula for W as an integral over B which holds when B gt 0 at 00 b Determine W for the case of a uniform magnetic eld B Hint see Prob 524 c Find W inside and outside an in nite solenoid Hint see Ex 512 Problem 554 Prove the following uniqueness theorem If the current density J is speci ed throughout a volume V and either the potential A or the magnetic eld B is speci ed on the surface 8 bounding V then the magnetic eld itself is uniquely determined throughout 1 Hint First use the divergence theorem to show that V xUV xV UV x V xVdry UxV xVda f0r arbitrary vector functions U and V Problem 555 A magnetic dipole m 2 m0 2 is situated at the origin in an otherwise uniform magnetic eld B Bo 2 Show that there exists a spherical surface centered at the origin through which no magnetic eld lines pass Find the radius of this sphere and sketch the eld lines inside and out Problem 556 A thin uniform donut carrying charge Q and mass M rotates about its axis as shown in Fig 564 a Find the ratio of its magnetic dipole moment to its angular momentum This is called the gyromagnetic ratio or magnetomechanical ratio b What is the gyromagnetic ratio for a uniform spinning sphere This requires no new calculation simply decompose the sphere into in nitesimal rings and apply the result of part all c According to quantum mechanics the angular momentum of a spinning electron is yr where n is Planck s constant What then is the electron s magnetic dipole moment in A m2 397 This semiclassical value is actually off by a factor of almost exactly 2 Dirac s relativistic electron theory got the 2 right and Feynman Schwinger and Tomonaga later calculated tin further corrections The determination of the electron s magnetic dipole moment remains the nest achievement of quantum electrodynamics and exhibits perhaps the most stunningly precise agreement between theory and experiment in all of physics Incidentally the quantity eh 2m where e is the charge of the electron and m is its mass is called the Bohr magneton Figure 564 54 MAGNETIC VECTOR POTENTIAL 253 Problem 557 a Prove that the average magnetic eld over a sphere of radius R due to steady currents within the sphere is 0 2m 5 F where m is the total dipole moment of the sphere Contrast the electrostatic result Eq 3105 This is tough so I ll give you a start 1 Bave 371R Write B as V X A and apply Prob 160b Now put in Eq 563 and do the surface integral rst showing that l 4 f d3 gTEF see Fig 565 Use Eq 591 if you like b Show that the average magnetic eld due to steady currents outside the sphere is the same as the eld they produce at the center 589 Bave Figure 565 Problem 558 A uniformly charged solid sphere of radius R carries a total charge Q and is set spinning with angular velocity to about the z axis a What is the magnetic dipole moment of the sphere b Find the average magnetic eld within the sphere see Prob 557 c Find the approximate vector potential at a point r 6 where r gtgt R d Find the exact potential at a point r 6 outside the sphere and check that it is consistent with c Hint refer to Ex 511 e Find the magnetic eld at a point r 6 inside the sphere and check that it is consistent with b Problem 559 Using Eq 586 calculate the average magnetic eld of a dipole over a sphere of radius R centered at the origin Do the angular integrals rst Compare your answer with the general theorem in Prob 557 Explain the discrepancy and indicate how Eq 587 can be corrected to resolve the ambiguity at r 0 If you get stuck refer to Prob 342 254 CHAPTER 5 MAGNETOS TATI CS Evidently the true eld of a magnetic dipole is Bd r i3m m m m83r 5 90 p 471 r3 3 Compare the electrostatic analog Eq 3106 Incidentally the delta function term is respon sible for the hyper ne splitting in atomic spectra see for example D J Grif ths Am J Phys 50 698 1982 Problem 560 I worked out the multipole expansion for the vector potential of a line current because that s the most common type and in some respects the easiest to handle For a volume current J a Write down the multipole expansion analogous to Eq 578 b Write down the monopole potential and prove that it vanishes c Using Eqs 1107 and 584 sh0w that the dipole moment can be written mrXJdr 591 Problem 561 A thin glass rod of radius R and length L carries a uniform surface charge a It is set spinning about its axis at an angular velocity a Find the magnetic eld at a distance s gtgt R from the center of the rod Fig 566 Hint treat it as a stack of magnetic dipoles Answws MowaLR3452 L 2232 Figure 566 Chapter 6 Magnetic Fields in Matter 61 Magnetization 611 Diamagnets Paramagnets Ferromagnets If you ask the average person what magnetism is you will probably be told about horse shoe magnets compass needles and the North Pole none of which has any obvious connection with moving charges or current carrying wires Yet all magnetic phenomena are due to electric charges in motion and in fact if you could examine a piece of magnetic material on an atomic scale you would nd tiny currents electrons orbiting around nuclei and electrons spinning about their axes For macroscopic purposes these current loops are so small that we may treat them as magnetic dipoles Ordinarily they cancel each other out because of the random orientation of the atoms But when a magnetic eld is applied a net alignment of these magnetic dipoles occurs and the medium becomes magnetically polarized or magnetized Unlike electric polarization which is almost always in the same direction as E some materials acquire a magnetization parallel to B paramagnets and some opposite to B diamagnets A few substances called ferromagnets in deference to the most common example iron retain their magnetization even after the external eld has been removed for these the magnetization is not determined by the present eld but by the whole magnetic history of the object Permanent magnets made of iron are the most familiar examples of magnetism though from a theoretical point of view they are the most complicated I ll save ferromagnetism for the end of the chapter and begin with qualitative models of paramagnetism and diamagnetism 612 Torques and Forces on Magnetic Dipoles A magnetic dipole experiences a torque in a magnetic eld just as an electric dipole does in an electric eld Let s calculate the torque on a rectangular current loop in a uniform eld B Since any current loop could be built up from in nitesimal rectangles with all 255 256 CHAPTER 6 MAGNETIC FIELDS IN MATTER Figure 61 the internal sides canceling as indicated in Fig 61 there is no actual loss of generality in using this shape but if you prefer to start from scratch with an arbitrary shape see Prob 62 Center the loop at the origin and tilt it an angle 6 from the z axis towards the y axis Fig 62 Let B point in the z direction The forces on the two sloping sides cancel they tend to stretch the loop but they don t rotate it The forces on the horizontal sides are likewise equal and opposite so the net force on the loop is zero but they do generate a torque N aF sin 6 2 The magnitude of the force on each of these segments is F I 3 and therefore N IabB sine 2 m3 sinex Figure 62 61 MAGNETIZATION 257 0139 lt6 where m I ah is the magnetic dipole moment of the loop Equation 61 gives the exact torque on any localized current distribution in the presence of a uniform eld in a nonuniform eld it is the exact torque about the center for a perfect dipole of in nitesimal size Notice that Eq 61 is identical in form to the electrical analog Eq 44 N p X E In particular the torque is again in such a direction as to line the dipole up parallel to the eld It is this torque that accounts for paramagnetism Since every electron constitutes a magnetic dipole picture it if you wish as a tiny spinning sphere of charge you might expect paramagnetism to be a universal phenomenon Actually the laws of quantum mechanics speci cally the Pauli exclusion principle dictate that the electrons within a given atom lock together in pairs with opposing spins and this effectively neutralizes the torque on the combination As a result paramagnetism normally occurs in atoms or molecules with an odd number of electrons where the extra unpaired member is subject to the magnetic torque Even here the alignment is far from complete since random thermal collisions tend to destroy the order In a uniform eld the net force on a current loop is zero FI leBIlt dlXB0 the constant B comes outside the integral and the net displacement 55 d1 around a closed loop vanishes In a nonuniform eld this is no longer the case For example suppose a circular wire of radius R carrying a current I is suspended above a short solenoid in the fringing region Fig 63 Here B has a radial component and there is a net downward force on the loop Fig 64 F 27rIRBcos6 62 NaDI Figure 63 Figure 64 258 CHAPTER 6 MAGNETIC FIELDS IN MATTER For an in nitesimal loop with dipole moment m in a eld B the force is F VmB 63 see Prob 64 Once again the magnetic formula is identical to its electrical twin provided we agree to write the latter in the form F Vp E If you re starting to get a sense of deja vu perhaps you will have more respect for those early physicists who thought magnetic dipoles consisted of positive and negative magnetic charges north and south poles they called them separated by a small distance just like electric dipoles Fig 65a They wrote down a Coulomb s law for the attraction and repulsion of these poles and developed the whole of magnetostatics in exact analogy to electrostatics It s not a bad model for many purposes it gives the correct eld of a dipole at least away from the origin the right torque on a dipole at least on a stationary dipole and the proper force on a dipole at least in the absence of external currents But it s bad physics because there s no such thing as a single magnetic north pole or south pole If you break a bar magnet in half you don t get a north pole in one hand and a south pole in the other you get two complete magnets Magnetism is not due to magnetic monopoles but rather to moving electric charges magnetic dipoles are tiny current loops Fig 65C and it s an extraordinary thing really that the formulas involving m bear any resemblance at all to the corresponding formulas for p Sometimes it is easier to think in terms of the Gilbert model of a magnetic dipole separated monopoles instead of the physical correct Ampere model current loop Indeed this picture occasionally offers a quick and clever solution to an otherwise cumbersome problem you just copy the corresponding result from electrostatics changing p to m l 60 to uo and E to B But whenever the closeup features of the dipole come into play the two models can yield strikingly different answers My advice is to use the Gilbert model if you like to get an intuitive feel for a problem but never rely on it for quantitative results O O m p m I S a Magnetic dipole b Electric dipole a Magnetic dipole Gilbert model Ampere model 2 Figure 65 6 I MAGNETIZATION 259 Problem 61 Calculate the torque exerted on the square loop shown in Fig 66 due to the circular loop assume r is much larger than a or b If the square loop is free to rotate what will its equilibrium orientation be Figure 66 Problem 62 Starting from the Lorentz force law in the form of Eq 516 show that the torque on any steady current distribution not just a square loop in a uniform eld B is m x B Problem 63 Find the force of attraction between two magnetic dipoles m1 and m2 oriented as shown in Fig 67 a distance r apart a using Eq 62 and b using Eq 63 m1 m2 O gt O gt E y r X Figure 67 Figure 68 Problem 64 Derive Eq 63 Here s one way to do it Assume the dipole is an in nitesimal square of side 6 if it s not chop it up into squares and apply the argument to each one Choose axes as shown in Fig 68 and calculate F I fdl X B along each of the four sides Expand B in a Taylor series on the right side for instance 3B BB0ez B00ze 3y 00z For a more sophisticated method see Prob 622 260 CHAPTER 6 MAGNETIC FIELDS IN MATTER Problem 65 A uniform current density J JO 2 lls a slab straddling the yz plane from x a to x o A magnetic dipole m m0 2 is situated at the origin a Find the force on the dipole using Eq 63 b Do the same for a dipole pointing in the y direction m moy c In the electrostatic case the expressions F Vp E and F p VE are equivalent prove it but this is not the case for the magnetic analogs explain Why As an example calculate m VB for the con gurations in a and b 613 Effect of 3 Magnetic Field on Atomic Orbits Electrons not only spin they also revolve around the nucleus for simplicity let s assume the orbit is a circle of radius R Fig 69 Although technically this orbital motion does not constitute a steady current in practice the period T 27 R v is so short that unless you blink awfully fast it s going to look like a steady current e w I T 27TR Accordingly the orbital dipole moment 7 R2 is 1 A m esz 64 The minus sign accounts for the negative charge of the electron Like any other magnetic dipole this one is subject to a torque m x B when the atom is placed in a magnetic eld But it s a lot harder to tilt the entire orbit than it is the spin so the orbital contribution to paramagnetism is small There is however a more signi cant effect on the orbital motion llm Figure 69 6 I MAGNE TIZATI ON 261 Figure 610 The electron speeds up or slows down depending on the orientation of B For whereas the centripetal acceleration vz R is ordinarily sustained by electrical forces alone1 65 in the presence of a magnetic eld there is an additional force ev X B For the sake of argument let s say that B is perpendicular to the plane of the orbit as shown in Fig 610 then 2 2 1 e t7 quotB m 66 4713960 R2 e e R Under these conditions the new speed i is greater than 1 me 2 2 me B v v v vv v ev R gt R gt or assuming the change A1 t7 v is small eRB v 67 2mg When B is turned on then the electron speeds up2 A change in orbital speed means a change in the dipole moment 64 l e2R2 A A R A B 68 m 2e 1 z 4me Notice that the change in m is opposite to the direction of B An electron circling the other way would have a dipole moment pointing upward but such an orbit would be slowed 1To avoid confusion with the magnetic dipole moment m I ll write the electron mass with subscript me 2I said earlier Eq 51 1 that magnetic elds do no work and are incapable of speeding a particle up I stand by that However as we shall see in Chapter 7 a changing magnetic eld induces an electric eld and it is the latter that accelerates the electrons in this instance 262 CHAPTER 6 MAGNETIC FIELDS IN MATTER down by the eld so the change is still opposite to B Ordinarily the electron orbits are randomly oriented and the orbital dipole moments cancel out But in the presence of a magnetic eld each atom picks up a little extra dipole moment and these increments are all antiparallel to the eld This is the mechanism responsible for diamagnetism It is a universal phenomenon affecting all atoms However it is typically much weaker than paramagnetism and is therefore observed mainly in atoms with even numbers of electrons where paramagnetism is usually absent In deriving Eq 68 I assumed that the orbit remains circular with its original radius R 1 cannot offer a justi cation for this at the present stage If the atom is stationary while the eld is turned on then my assumption can be proved this is not magnetostatics however and the details will have to await Chapter 7 see Prob 749 If the atom is moved into the eld the situation is enormously more complicated But never mind I m only trying to give you a qualitative account of diamagnetism Assume if you prefer that the velocity remains the same while the radius changes the formula 68 is altered by a factor of 2 but the conclusion is unaffected The truth is that this classical model is fundamentally awed diamagnetism is really a quantum phenomenon so there s not much point in re ning the details3 What is important is the empirical fact that in diamagnetic materials the induced dipole moments point opposite to the magnetic eld 614 Magnetization In the presence of a magnetic eld matter becomes magnetized that is upon microscopic examination it will be found to contain many tiny dipoles with a net alignment along some direction We have discussed two mechanisms that account for this magnetic polarization 1 paramagnetism the dipoles associated with the spins of unpaired electrons experience a torque tending to line them up parallel to the eld and 2 diamagnetism the orbital speed of the electrons is altered in such a way as to change the orbital dipole moment in a direction opposite to the eld Whatever the cause we describe the state of magnetic polarization by the vector quantity M E magnetic dipole moment per unit volume 69 M is called the magnetization it plays a role analogous to the polarization P in elec trostatics In the following section we will not worry about how the magnetization gm there it could be paramagnetism diamagnetism or even ferromagnetism we shall take M as given and calculate the eld this magnetization itself produces Incidentally it may have surprised you to learn that materials other than the famou ferromagnetic trio iron nickel and cobalt are affected by a magnetic eld at all You cannot of course pick up a piece of wood or aluminum with a magnet The reason is that diamagnetism and paramagnetism are extremely weak It takes a delicate experiment and a powerful magnet to detect them at all If you were to suspend a piece of paramagnetic 3S L O Dell and R K P Zia Am J Phys 54 32 1986 R Peierls Surprises in Theoretical Physiz39x Section 43 Princeton NJ Princeton University Press 1979 R P Feynman R B Leighton and M Sands The Feynman Lectures on Physics Vol 2 Sec 34 36 New York AddisonWesley 1966 62 THE FIELD OFA MAGNETIZEi OBJECT 263 material above a solenoid as in Fig 63 the induced magnetization would be upward and hence the force downward By contrast the magnetization of a diamagnetic object would be downward and the force upward In general when a sample is placed in a region of nonuni form eld the paramagnet is attracted into the eld whereas the diamagnet is repelled away But the actual forces are pitifully weak in a typical experimental arrangement the force on a comparable sample of iron would be 104 or 105 times as great That s why it was reasonable for us to calculate the eld inside a piece of copper wire say in Chapter 5 without worrying about the effects of magnetization Problem 66 Of the following materials which would you expect to be paramagnetic and which diamagnetic Aluminum copper copper chloride CuClz carbon lead nitrogen N2 salt NaCl sodium sulfur water Actually copper is slightly diamagnetic otherwise they re all what you d expect 62 The Field of a Magnetized Object 621 Bound Currents Suppose we have a piece of magnetized material the magnetic dipole moment per unit volume M is given What eld does this object produce Well the vector potential of a single dipole m is given by Eq 583 nomxi Ar 47 42 610 In the magnetized object each volume element d 17 carries a dipole moment M d t so the total vector potential is Fig 61 1 Ar f M as 611 47 42 Figure 6 11 264 CHAPTER 6 MAGNETIC FIELDS IN MATTER That does it in principle But as in the electrical case Sect 421 the integral can be cast in a more illuminating form by exploiting the identity owl V1 4 Ar 5 MM x 6 dt Integrating by parts using product rule 7 gives Ar 5 72 V X Mr dr V x dt Problem l60b invites us to express the latter as a surface integral With this 1 1 Ar E f V x Mr dt E yi Mr x da 612 47 4 47 4 The rst term looks just like the potential of a volume current 613 while the second looks like the potential of a surface current where is the normal unit vector With these de nitions Ar L dt K r da 615 47 V 4 47 s 4 What this means is that the potential and hence also the eld of a magnetized object is the same as would be produced by a volume current J V x M throughout the material plus a surface currenth M x on the boundary Instead of integrating the contributions of all the in nitesimal dipoles as in Eq 61 1 we rst determine these bound currents and then nd the eld they produce in the same way we would calculate the eld of any other volume and surface currents Notice the striking parallel with the electrical case there the eld of a polarized object was the same as that of a bound volume charge p1 V P plus a bound surface charge ab P Example 61 Find the magnetic eld of a uniformly magnetized sphere Solution Choosing the z axis along the direction of M Fig 612 we have JbVXM0 KbMx Msin043 62 THE FIELD OF A MAGNETIZED OBJECT 265 Figure 612 Now a rotating spherical shell of uniform surface charge 0 corresponds to a surface current density K UV 2 aszine It follows therefore that the eld of a uniformly magnetized sphere is identical to the eld of a Spinning spherical shell with the identi cation oRa gt M Referring back to Ex 511 I conclude that 2 B gpoM 616 inside the sphere whereas the eld outside is the same as that of a pure dipole 4 m 7r R3M 3 Notice that the internal eld is uniform like the electric eld inside a uniformly polarized sphere Eq 414 although the actual formulas for the two cases are curiously different 3 in place of The external elds are also analogous pure dipole in both instances Problem 67 An in nitely long circular cylinder carries a uniform magnetization M parallel to its axis Find the magnetic eld due to M inside and outside the cylinder Problem 68 A long circular cylinder of radius R carries a magnetization M ks2 13 where k is a constant s is the distance from the axis and d is the usual azimuthal unit vector Fig 613 Find the magnetic eld due to M for points inside and outside the cylinder Problem 69 A short circular cylinder of radius a and length L carries a frozenin uniform magnetization M parallel to its axis Find the bound current and sketch the magnetic eld of the cylinder Make three sketches one for L gtgt a one for L ltlt a and one for L m a Compare this bar magnet with the bar electret of Prob 411 266 CHAPTER 6 MAGNETIC FIELDS IN MATTER S a l Figure 613 Figure 614 Problem 610 An iron rod of length L and square cross section side a is given a uniform longitudinal magnetization M and then bent around into a circle with a narrow gap width w as shown in Fig 614 Find the magnetic eld at the center of the gap assuming w ltlt a ltlt L Hints treat it as the superposition of a complete torus plus a square loop with reversed current 622 Physical Interpretation of Bound Currents In the last section we found that the eld of a magnetized object is identical to the eld that would be produced by a certain distribution of bound currents J1 and K1 I want to show you how these bound currents arise physically This will be a heuristic argument the rigorous derivation has already been given Figure 615 depicts a thin slab of uniformly magnetized material with the dipoles represented by tiny current loops Notice that all the internal currents cancel every time there is one going to the right a contiguous one is going to the left However at the edge there is no adjacent loop to do the canceling The whole thing then is equivalent to a single ribbon of current I owing around the boundary Fig 616 62 THE FIELD OF A MAGNETIZED OBJECT 267 M Figure 615 Figure 616 What is this current in terms of M Say that each of the tiny loops has area a and thickness t Fig 617 In terms of the magnetization M its dipole moment is m 2 Mat In terms of the cirCulating current I hoWever m Ia Therefore I 2 Mt so the surface current is K I t M Using the outwarddrawn unit vector Fig 616 the direction of K is conveniently indicated by the cross product KMX This expression also records the fact that there is no current on the top or bottom surfaCe of the slab here M is parallel to n so the cross product vanishes M a 11 Figure 617 This bound surface current is exactly what we obtained in Sect 621 It is a peculiar kind of current in the sense that no single charge makes the whole trip on the contrary each charge moves only in a tiny little loop within a single atom Nevertheless the net effect is a macroscopic current owing over the surface of the magnetized object We call it a bound current to remind ourselves that every charge is attached to a particular atom but it s a perfectly genuine current and it produces a magnetic eld in the same way any other current does When the magnetization is nonuniform the internal currents no longer cancel Figure 618a shows two adjacent chunks of magnetized material with a larger arrow on the one to the right suggesting greater magnetization at that point On the surface where they join there is a net current in the x direction given by 8M 1 1My dy 1140le dy dz 268 CHAPTER 6 MAGNETIC FIELDS IN MATTER Z My an MZ y Myz dz dz p i i Myz a dy b Figure 618 The correSponding volume current density is therefore 3MZ By By the same token a nonuniform magnetization in the ydirection would contribute an amount 3My3z Fig 618b so 1bx 3Mz 3My J bx 3y az In general then J1 V X M consistent again with the result of Sect 621 Incidentally like any other steady current J1 should obey the conservation law 531 VJbzo Does it Yes for the divergence of a curl is always zero 623 The Magnetic Field Inside Matter Like the electric eld the actual microscopic magnetic eld inside matter uctuates wildly from point to point and instant to instant When we speak of the magnetic eld in matter we mean the macroscopic eld the average over regions large enough to contain many atoms The magnetization M is smoothed out in the same sense It is this macrosc0pic eld one obtains when the methods of Sect 621 are applied to points inside magnetized material as you can prove for yourself in the following problem Problem 611 In Sect 621 we began with the potential of a perfect dipole Eq 610 whereas in fact we are dealing with physical dipoles Show by the method of Sect 423 that we nonetheless get the correct macrosc0pic eld I IVIATTER 63 THE AUXILIARY FIELD H 269 270 CHAPTER 6 MAGNETIC FIELDS N 63 The Auxiliary H this results in bound currents we cannot turn them on oroff independently as we can free currents In applying Eq 620 all we need to worry about 1s the free current wh1ch we know about because we put it there In particular when symmetry permits we can calculate H immediately from Eq 620 by the usual Ampere s law methods For example Probs 67 In Sect 62 we found that the effect of magnetization is to establish bound currents J1 and 68 can be done in one line by noting that H 0 V X M within the material and Kb 2 M X n on the surface The eld due to magnetization of the medium is just the eld produced by these bound currents We are now ready to put everything together the eld attributable to bound currents plus the eld due to everything else which I shall call the free current The free current might ow through wires imbedded in the magnetized substance or if the latter is a conductor through the 631 Amp re s law in Magnetized Materials Example 62 A long copper rod of radius R carries a uniformly distributed free current I Fig 619 Find H inside and outside the rod material itself In any eVent the total current can be written as Solution C0pper is weakly diamagnetic so the dipoles will line up opposite to the eld This results in a bound current running antiparallel to I within the wire and parallel to I along the J Jb J f 617 surface see Fig 620 Just how great these bound currents will be we are not yet in a pos1tion There is no new physics in Eq 617 it is simply a convenience to separate the current into these two parts because they got there by quite different means the free current is there because somebody hooked up a wire to a battery it involves actual transport of charge the bound current is there because of magnetization it results from the conspiracy of many aligned atomic dipoles In view of Eqs 613 and 617 Ampere s law can be written 1 u V XB JJfJb JfV X M 0 or collecting together the two curls l V X B M J f 0 The quantity in parentheses is designated by the letter H 1 0 H E B M 618 Amperian 100p In terms of H then Ampere s law reads m f H d Ifmc 620 or in integral form where Ifenc is the total free current passing through the Amperian loop H plays a role in magnetostatics analogous to D in electrostatics Just as D allowed us to write Gauss s law in terms of the free charge alone H permits us to eXpress Ampere 39s law in terms of the free current alone and free current is what we control directly Bound current like bound charge comes along for the ride the material gets magnetized and Figure 619 Figure 620 6 3 THE AUXILIARY FIELD H 271 to say but in order to calculate H it is suf cient to realize that all the currents are longitudinal so B M and therefore also H are circumferential Applying Eq 620 to an Amperian loop of radius s lt R H27rs I 1 2 fenc m so H L s lt R 62 271R2 T within the wire Meanwhile outside the wire I A H E d s 3 R 622 In the latter region as always in empty space M 0 so I x BM0H 43 BR 2715 the same as for a nonmagnetized wire Ex 57 Inside the wire B cannot be determined at this stage since we have no way of knowing M though in practice the magnetization in copper is so slight that for most purposes we can ignore it altogether As it turns out H is a more useful quantity than D In the laboratory you will frequently hear people talking about H more often even than B but you will never hear anyone Speak of D only E The reason is this To build an electromagnet you run a certain free current through a coil The current is the thing you read on the dial and this determines H or at any rate the line integral of H B depends on the speci c materials you used and even if iron is present on the history of your magnet On the other hand if you want to set up an electric eld you do not plaster a known free charge on the plates of a parallel plate capacitor rather you connect them to a battery of known voltage It s the potential di ference you read on your dial and that determines E or at any rate the line integral of E D depends on the details of the dielectric you re using If it were easy to measure charge and hard to measure potential then you d nd experimentalists talking about D instead of E So the relative familiarity of H as contrasted with D derives from purely practical considerations theoretically they re all on equal footing Many authors call H not B the magnetic eld Then they have to invent a new word for B the ux density or magnetic induction an absurd choice since that term already has at least two other meanings in electrodynamics Anyway B is indisputably the fundamental quantity so I shall continue to call it the magnetic eld as everyone does in the spoken language H has no sensible name just call it H 4 4For those who disagree I quote A Sommerfeld39s Electrodynamics New York Academic Press 1952 p 45 The unhappy term magnetic eld for H should be avoided as far as possible It seems to us that this term has led into error none less than Maxwell himself 272 CHAPTER 6 MAGNETIC FIELDS IN MATTER Problem 612 An in nitely long cylinder of radius R carries a frozen in magnetization parallel to the axis Mksz where k is a constant and s is the distance from the axis there is no free current anywhere Find the magnetic eld inside and outside the cylinder by two different methods a As in Sect 62 locate all the bound currents and calculate the eld they produce b Use Ampere s law in the form of Eq 620 to nd H and then get B from Eq 618 Notice that the second method is much faster and avoids any explicit reference to the bound currents Problem 613 Suppose the eld inside a large piece of magnetic material is B0 so that H0 2 1 M0Bo M a Now a small spherical cavity is hollowed out of the material Fig 621 Find the eld at the center of the cavity in terms of B0 and M Also nd H at the center of the cavity in terms of H0 and M b Do the same for a long needleshaped cavity running parallel to M c Do the same for a thin wafershaped cavity perpendicular to M a Sphere b Needle c Wafer Figure 621 63 THE AUXILIARY FIELD H 273 Assume the cavities are small enough so that M B0 and H0 are essentially constant Com pare Prob 416 Hints Carving out a cavity is the same as superimposing an object of the same shape but opposite magnetization 632 A Deceptive Parallel Equation 619 looks just like Am re s original law 554 only the total current is replaced by the free current and B is replaced by uOH As in the case of D however I must warn you against reading too much into this correspondence It does not say that uOH is just like B only its source is J f instead of J For the curl alone does not determine a vector eld you must know the divergence as well And whereas V B 0 the divergence of H is not in general zero In fact from Eq 618 V H V M 623 Only when the divergence of M vanishes is the parallel between B and uOH faithful If you think I m being pedantic consider the example of the bar magnet a short cylinder of iron that carries a permanent uniform magnetization M parallel to its axis See Probs 69 and 614 In this case there is no free current anywhere and a naive application of Eq 620 might lead you to suppose that H 0 and hence that B uOM inside the magnet and B 0 outside which is nonsense It is quite true that the curl of H vanishes everywhere but the divergence does not Can you see where V M 0 Advice When you are asked to nd B or H in a problem involving magnetic materials rst look for symmetry If the problem exhibits cylindrical plane solenoidal or toroidal symmetry then you can get H directly from Eq 620 by the usual Ampere s law methods Evidently in such cases V M is automatically zero since the free current alone determines the answer If the requisite symmetry is absent you ll have to think of another approach and in particular you must not assume that H is zero just because you see no free current 633 Boundary Conditions The magnetostatic boundary conditions of Sect 542 can be rewritten in terms of H and the free current From Eq 623 it follows that J J J J Habove Hbelow Mab0ve Mbelow 63924 while Eq 619 says HII above H Kf x n 625 below 274 CHAPTER 6 MAGNETIC FIELDS IN MATTER In the presence of materials these are sometimes more useful than the corresponding bound ary conditions on B Eqs 572 and 573 1 L Babove Bbelow 0 626 and Bu below Bu above p0K X n 627 You might want to check them for Ex 62 or Prob 614 Problem 614 For the bar magnet of Prob 69 make careful sketches of M B and H assuming L is about 2a Compare Prob 417 Problem 615 If J f 0 everywhere the curl of H vanishes Eq 619 and we can express H as the gradient of a scalar potential W H VW According to Eq 623 then VZW V M so W obeys Poisson s equation with V M as the source This opens up all the machiner of Chapter 3 As an example nd the eld inside a uniformly magnetized sphere Ex 61 b separation of variables Hints V M 0 everywhere except at the surface r R so 1139 satis es Laplace s equation in the regions r lt R and r gt R use Eq 365 and from Eq 624 gure out the appropriate boundary condition on W 64 Linear and Nonlinear Media 641 Magnetic Susceptibility and Permeability In paramagnetic and diamagnetic materials the magnetization is sustained by the eld when B is removed M disappears In fact for most substances the magnetization is proportional to the eld provided the eld is not too strong For notational consistency with the electrical case Eq 430 I should express the prOportionality thus M imB incorrectl 628i 0 But custom dictates that it be written in terms of H instead of B M XmH 629 The constant of prOportionality m is called the magnetic susceptibility it is a dimen sionless quantity that varies from one substance to another pOSitive for paramagnets and negative for diamagnets Typical values are around 10 5 see Table 61 64 LINEAR AND NONLINEAR MEDIA 275 Material Susceptibility Material Susceptibility Diamagnetic Paramagnetic Bismuth l6 x 10 4 Oxygen 19 x 10 6 Gold 34 x 105 Sodium 85 x 106 Silver 24 X 105 Aluminum 21 X 10 5 Copper 97 x 10 6 Tungsten 78 X 10 5 Water 90 x 10 6 Platinum 28 x 10 4 Carbon Dioxide l2 x 10 8 Liquid Oxygen 200 C 39 x 10 3 Hydrogen 22 x 10 9 Gadolinium 48 x 10 1 Table 61 Magnetic Susceptibilities unless otherwise speci ed values are for 1 atm 20 C Source Handbook of Chemistry and Physics 67th ed Boca Raton CRC Press Inc 1986 Materials that obey Eq 629 are called linear media In view of Eq 618 B M0H M M00 XmH 630 for linear media Thus B is also proportional to H5 B uH 631 where M E 1100 Xm 632 p is called the permeability of the material6 In a vacuum where there is no matter to magnetize the susceptibility Xm vanishes and the permeability is no That s why no is called the permeability of free space Example 63 An in nite solenoid n turns per unit length current I is lled with linear material of suscep tibility Xm Find the magnetic eld inside the solenoid Solution Since B is due in part to bound currents which we don t yet know we cannot compute it directly However this is one of those symmetrical cases in which we can get H from the free current alone using Ampere s law in the form of Eq 620 Hnli 5 Physically therefore Eq 628 would say exactly the same as Eq 629 only the constant Xm would have a different value Equation 629 is a little more convenient because experimentalists find it handier to work with H than B 6If you factor out no what s left is called the relative permeability tr 2 1 Xm llllo By the way formulas for H in terms of B Eq 631 in the case of linear media are called constitutive relations just like those for D in terms of E 276 CHAPTER 6 MAGNETIC FIELDS IN MATTER Figure 622 Fig 622 According to Eq 631 then B 2 lion xmn12 If the medium is paramagnetic the eld is slightly enhanced if it s diamagnetic the eld ix somewhat reduced This re ects the fact that the bound surface current szMx XmHgtlt anI is in the same direction as I in the former case xm gt 0 and opposite in the latter xm lt 0 You might suppose that linear media avoid the defect in the parallel between B and H since M and H are now proportional to B does it not follow that their divergence like B39s must always vanish Unfortunately it does not at the boundary between two materials of different permeability the divergence of M can actually be in nite For instance at the end of a cylinder of linear paramagnetic material M is zero on one side but not on the other For the Gaussian pillbox shown in Fig 623 f M da 75 0 and hence by the divergence theorem V M cannot vanish everywhere within Gaussian pillbox M 0 Vacuum Figure 623 64 LINEAR AND NONLINEAR MEDIA 277 Incidentally the volume bound current density in a homogeneous linear material is proportional to the free current density J V X M V X XmH Xme 633 In particular unless free current actually ows through the material all bound current will be at the surface Problem 616 A coaxial cable consists of two very long cylindrical tubes separated by linear insulating material of magnetic susceptibility Xm A current I ows down the inner conductor and returns along the outer one in each case the current distributes itself uniformly over the surface Fig 624 Find the magnetic eld in the region between the tubes As a check calculate the magnetization and the bound currents and con rm that together of course with the free currents they generate the correct eld Figure 624 Problem 617 A current I ows down a long straight wire of radius a If the wire is made of linear material copper say or aluminum with susceptibility Xm and the current is distributed uniformly what is the magnetic eld a distance s from the axis Find all the bound currents What is the net bound current owing down the wire Problem 618 A sphere of linear magnetic material is placed in an otherwise uniform magnetic eld B0 Find the new eld inside the sphere Hint See Prob 615 or Prob 423 Problem 619 On the basis of the naive model presented in Sect 613 estimate the magnetic susceptibility of a diamagnetic metal such as copper Compare your answer with the empirical value in Table 61 and comment on any discrepancy 278 CHAPTER 6 MAGNETIC FIELDS IN MATTER 642 Ferromagnetism In a linear medium the alignment of atomic dipoles is maintained by a magnetic eld im posed from the outside Ferromagnets which are emphatically not linear7 require no external elds to sustain the magnetization the alignment is frozen in Like paramag netism ferromagnetism involves the magnetic dipoles associated with the spins of unpaired electrons The new feature which makes ferromagnetism so different from paramagnetism is the interaction between nearby dipoles In a ferromagnet each dipole likes to point in the same direction as its neighbors The reason for this preference is essentially quantum mechanical and I shall not endeavor to explain it here it is enough to know that the cor relation is so strong as to align virtually 100 of the unpaired electron spins If you could somehow magnify a piece of iron and see the individual dipoles as tiny arrows it would look something like Fig 625 with all the spins pointing the same way Figure 625 But if that is true why isn t every wrench and nail a powerful magnet The answer is that the alignment occurs in relatively small patches called domains Each domain contains billions of dipoles all lined up these domains are actually visible under a microscope using suitable etching techniques see Fig 626 but the domains themselves are randoml oriented The household wrench contains an enormous number of domains and their magnetic elds cancel so the wrench as a whole is not magnetized Actually the orientation of domains is not completely random within a given crystal there may be some preferential alignment along the crystal axes But there will be just as many domains pointing one wa as the other so there is still no large scale magnetization Moreover the crystals themselves are randomly oriented within any sizable chunk of metal How then would you produce a permanent magnet such as they sell in toy stores If you put a piece of iron into a strong magnetic eld the torque N m X B tends to align the dipoles parallel to the eld Since they like to stay parallel to their neighbors most of the dipoles will resist this torque However at the boundary between two domains there 7In this sense it is misleading to speak of the susceptibility or permeability of a ferromagnet The terms are used for such materials but they refer to the proportionality factor between a di ferential increase in H and the resulting di ferentinl change in M or B moreover they are not constants but functions of H 64 LINEAR AND NONLINEAR MEDIA 279 Ferromagnetic domains Photo courtesy of R W DeBlois Figure 626 are competing neighbors and the torque will throw its weight on the side of the domain most nearly parallel to the eld this domain will win over some converts at the expense of the less favorably oriented one The net effect of the magnetic eld then is to move the domain boundaries Domains parallel to the eld grow and the others shrink If the eld is strong enough one domain takes over entirely and the iron is said to be saturated It turns out that this process the shifting of domain boundaries in response to an external eld is not entirely reversible When the eld is switched off there will be some return to randomly oriented domains but it is far from complete there remains a preponderance of domains in the original direction The object is now a permanent magnet A simple way to accomplish this in practice is to wrap a coil of wire around the object to be magnetized Fig 627 Run a current I through the coil this provides the external magnetic eld pointing to the left in the diagram As you increase the current the eld increases the domain boundaries move and the magnetization grows Eventually you reach the saturation point with all the dipoles aligned and a further increase in current has no effect on M Fig 628 point b Now suppose you reduce the current Instead of retracing the path back to M 0 there is only a partial return to randomly oriented domains M decreases but even with the current off there is some residual magnetization point c The wrench is now a permanent 280 CHAPTER 6 MAGNETIC FIELDS IN MATTER Figure 627 magnet If you want to eliminate the remaining magnetization you ll have to run a current backwards through the coil a negative I Now the external eld points to the right and as you increase I negatively M drops down to zero point d If you turn I still higher you soon reach saturation in the other direction all the dipoles now pointing to the right e AI this stage switching off the current will leave the wrench with a permanent magnetization 10 the right point f To complete the story turn I on again in the positive sense M returns to zero point g and eventually to the forward saturation point b The path we have traced out is called a hysteresis loop Notice that the magnetization of the wrench depends not only on the applied eld that is on I but also on its prev10us magnetic history 8 For instance at three different times in our experiment the current was zero a c and f yet the magnetization was different for each of them Actually it is customary to draw hysteresis loops as plots of B against H rather than M against I If our coil is approximated by a long solenoid with it turns per unit length then H nI so H and I are proportional Meanwhile B n0H M but in practice M is huge compared to H so to all intents and purposes B is proportional to M M Permanent c b Saturatron Magnet d a g I o e f Permanent Saturation Magnet Figure 628 8Etymologically the word hysteresis has nothing to do with the word history nor with the word hysteria 1t derives from a Greek verb meaning to lag behind 64 LINEAR AND NONLINEAR MEDIA 281 To make the units consistent teslas I have plotted aoH horizontally Fig 629 notice however that the vertical scale is 104 times greater than the horizontal one Roughly speaking aoH is the eld our coil would have produced in the absence of any iron B is what we actually got and compared to OH it is gigantic A little current goes a long way when you have ferromagnetic materials around That s why anyone who wants to make a powerful electromagnet will wrap the coil around an iron core It doesn t take much of an external eld to move the domain boundaries and as soon as you ve done that you have all the dipoles in the iron working with you 15 13 2390 0HX104 Figure 629 One nal point concerning ferromagnetism It all follows remember from the fact that the dipoles within a given domain line up parallel to one another Random theimal motions compete with this ordering but as long as the temperature doesn t get too high they cannot budge the dipoles out of line It s not surprising though that very high temperatures do destroy the alignment What is surprising is that this occurs at a precise temperature 770 C for iron Below this temperature called the Curie point iron is ferromagnetic above it is paramagnetic The Curie point is rather like the boiling point or the freezing point in that there is no gradual transition from ferro to paramagnetic behavior any more than there is between water and ice These abrupt changes in the properties of a substance occurring at sharply de ned temperatures are known in statistical mechanics as phase transitions Problem 620 How would you go about demagnetizing a permanent magnet such as the wrench we have been discussing at point c in the hysteresis loop That is how could you restore it to its original state with M 0 at I 0 Problem 621 a Show that the energy of a magnetic dipole in a magnetic eld B is given by U m B 634 282 CHAPTER 6 MAGNETIC FIELDS IN MATTER 71361 r 3 2 62 V Figure 630 Assume that the magnitude of the dipole moment is xed and all you have to do is move it into place and rotate it into its nal orientation The energy required to keep the current owing is a different problem which we will confront in Chapter 7 Compare Eq 46 b Show that the interaction energy of two magnetic dipoles separated by a displacement r i given by M0 1 A A 7 U m1 m2 3m1rm2rl 6339 47f r3 Compare Eq 47 c Express your answer to b in terms of the angles 61 and 62 in Fig 630 and use the result to nd the stable con guration two dipoles Would adopt if held a xed distance apart but left free to rotate d Suppose you had a large collection of compass needles mounted on pins at regular interval along a straight line How would they point assuming the earth s magnetic eld can be neglected A rectangular array of compass needles also aligns itself spontaneously and this is sometimes used as a demonstration of ferromagnetic behavior on a large scale It s a bit of a fraud however since the mechanism here is purely classical and much weaker than the quantum mechanical exchange forces that are actually responsible for ferromagnetism More Problems on Chapter 6 Problem 622 In Prob 64 you calculated the force on a dipole by brute force Here s a more elegant approach First writ39e Br as a Taylor expansion about the center of the loop 30 E Bro r r0 VolBr0 where 10 is the position of the dipole and V0 denotes differentiation with respect to r0 Put this into the Lorentz force law Eq 516 to obtain F 1 if d1 X r V0Br0 Or numbering the Cartesian coordinates from 1 to 3 3 Fl 1 ijk r1 dlj V0Bkro jkl1 64 LINEAR AND NONLINEAR MEDIA 283 where eijk is the LeviCivita symbol 1 if ijk 123 231 or 312 1 if ijk 132 213 or 321 0 otherwise in terms of which the cross product can be written A X Bi Zikzl eijkAJBk Use Eq 1108 to evaluate the integral Note that 3 2 61716ij 5115km 5im5kl 391 where 5 is the Kronecker delta Prob 345 Problem 623 Notice the following parallel VD VB 0 V X E 0 60E D P no free charge 0 V X H 0 uQH B uQM no free current Thus the transcription D gt B E gt H P gt MOM 60 gt uo turns an electrostatic problem into an analogous magnetostatic one Use this observation together with your knowledge of the electrostatic results to rederive a the magnetic eld inside a uniformly magnetized sphere Eq 616 b the magnetic eld inside a sphere of linear magnetic material in an otherwise uniform magnetic eld Prob 618 c the average magnetic eld over a sphere due to steady currents within the sphere Eq 589 Problem 624 Compare Eqs 215 49 and 611 Notice that if 0 P and M are uniform the same integral is involved in all three r 22 Therefore if you happen to know the electric eld of a uniformly charged object you can immediately write down the scalar potential of a uniformly polarized object and the vector potential of a uniformly magnetized object of the same shape Use this observation to obtain V inside and outside a uniformly polarized sphere Ex 42 and A inside and outside a uniformly magnetized sphere Ex 61 Problem 625 A familiar toy consists of donut shaped permanent magnets magnetization parallel to the axis which slide frictionlessly on a vertical rod Fig 631 Treat the magnets as dipoles with mass md and dipole moment m a If you put two backtoback magnets on the rod the upper one will oat the magnetic force upward balancing the gravitational force downward At what height z does it oat b If you now add a third magnet parallel to the bottom one what is the ratio of the two heights Determine the actual number to three signi cant digits Answer39 a 3uom227rmdg14 b 08501 Problem 626 At the interface between one linear magnetic material and another the magnetic eld lines bend see Fig 632 Show that tan 62 tan 01 LgM assuming there is no free current at the boundary Compare Eq 468 284 CHAPTER 6 MAGNETIC FIELDS IN MATTER a b Figure 631 Figure 632 Problem 627 A magnetic dipole m is imbedded at the center of a sphere radius R of linear magnetic material permeability u Show that the magnetic eldinside the sphere 0 lt r 5 R is L 47f What is the eld outside the sphere 2Mo mm i WWW n 2m MR3 Problem 628 You are asked to referee a grant application which proposes to determine Whether the magnetization of iron is due to Ampere dipoles current loops or Gilbert dipoles separated magnetic monopoles The experiment will involve a cylinder of iron radius R and length L 10R uniformly magnetized along the direction of the axis If the dipoles are Amperetype the magnetization is equivalent to a surface bound current Kb M 3 if they are Gilberttype the magnetization is equivalent to surface monopole densities ab lM at the two ends Unfortunately these two con gurations produce identical magnetic elds at exterior points However the interior elds are radically different in the rst case B is in the same general direction as M whereas in the second it is roughly opposite to M The applicant proposes to measure this internal eld by carving out a small cavity and nding the torque on a tiny compass needle placed inside Assuming that the obvious technical dif culties can be overcome and that the question itself is worthy of study would you advise funding this experiment If so what shape cavity would you recommend If not what is wrong with the proposal Hint refer to Probs 411 416 69 and 613 Chapter 7 Electrodynamics 71 Electromotive Force 711 Ohm s Law To make a current ow you have to push on the charges How fast they move in response to a given push depends on the nature of the material For most substances the current density J is proportional to the force per unit charge f J of 71 The proportionality factor a not to be confused with surface charge is an empirical con stant that varies from one material to another it s called the conductivity of the medium Actually the handbooks usually list the reciprocal of 0 called the resistivity p 1 0 not to be confused with charge density I m sorry but we re running out of Greek 1et ters and this is the standard notation Some typical values are listed in Table 71 Notice that even insulators conduct slightly though the conductivity of a metal is astronomically greater by a factor of 1022 or so In fact for most purposes metals can be regarded as perfect conductors with a 00 In principle the force that drives the charges to produce the current could be anything chemical gravitational or trained ants with tiny harnesses For our purposes though it s usually an electromagnetic force that does the job In this case Eq 71 becomes J0Ev XB 72 Ordinarily the velocity of the charges is suf ciently small that the second term can be ignored 03gt However in plasmas for instance the magnetic contribution to f can be signi cant Equation 73 is called Ohm s law though the physics behind it is really contained in Eq 71 of which 73 is just a special case 285 286 CHAPTER 7 ELECTRODYNAMICS Material Resistivity Material Resistivity Conductors Semiconductors Silver 159 x 10 8 Salt water saturated 44 x 10 2 Copper 168 x 10 8 Germanium 46 x 10 Gold 221 x 108 Diamond 27 Aluminum 265 x 108 Silicon 25 x 103 Iron 961 X 10 8 Insulators Mercury 958 X 10 7 Water pure 25 x 105 Nichrome 100 X 10 6 Wood 108 10 Manganese 144 x 1076 Glass 1010 1014 Graphite 14 x 10 5 Quartz fused 1016 Table 71 Resistivities in ohmmeters all values are for 1 atm 200 C Source Handbook of Chemistry and Physics 78th ed Boca Raton CRC Press Inc 1997 I know you re confused because I said E 0 inside a conductor Sect 251 But that s for stationary charges J 0 Moreover for perfect conductors E J a 0 even if current is owing In practice metals are such good conductors that the electric eld required to drive current in them is negligible Thus we routinely treat the connecting wires in electric circuits for example as equipotentials Resistors by contrast are made from poorly conducting materials Example 71 A cylindrical resistor of crosssectional area A and length L is made from material with conductivity cr See Fig 71 as indicated the cross section need not be circular but I In assume it is the same all the way down If the potential is constant over each end and the potential difference between the ends is V what current ows Nv Figure 71 Solution As it turns out the electric eld is uniform within the wire 1 11 prove this in a moment It follows from Eq 73 that the current density is also uniform so crA IJAcrEA V L 71 ELECTROMOTIVE FORCE 287 Example 72 Two long cylinders radii a and b are separated by material of conductivity a Fig 72 If they are maintained at a potential difference V what current ows from one to the other in a length L Figure 72 Solution The eld between the cylinders is A A s 27160s where A is the charge per unit length on the inner cylinder The current is therefore szJdaafEdaiAL 60 The integral is over any surface enclosing the inner cylinder Meanwhile the potential difference between the cylinders is 1 V f Edllenlt 1 27160 a 271aL lnba 39 As these examples illustrate the total current owing from one electrode to the other is proportional to the potential difference between them V IR 74 This of course is the more familiar version of Ohm s law The constant of proportionality R is called the resistance it s a function of the geometry of the arrangement and the conductivity of the medium between the electrodes In Ex 71 R L 0A in Ex 72 R ln ba2noL Resistance is measured in ohms S2 an ohm is a volt per ampere Notice that the proportionality between V and I is a direct consequence of Eq 73 if you want to double V you simply double the charge everywhere but that doubles E which doubles J which doubles I 288 CHAPTER 7 ELECTRODYNAMICS For steady currents and uniform conductivity 1 VE V39J0 75 0 Eq 531 and therefore the charge density is zero any unbalanced charge resides on the surface We proved this long ago for the case of stationary charges using the fact that E 0 evidently it is still true when the charges are allowed to move It follows in particular that Laplace s equation holds within a homogeneous ohmic material carrying a steady current so all the tools and tricks of Chapter 3 are available for computing the potential Example 73 I asserted that the eld in Ex 71 is uniform Let s prove it Solution Within the cylinder V obeys Laplace s equation What are the boundary conditions At the left end the potential is constant we may as well set it equal to zero At the right end the potential is likewise constant call it V0 On the cylindrical surface J 0 or else charge would be leaking out into the surrounding space which we take to be nonconducting Therefore E 0 and hence aVan 0 With V or its normal derivative speci ed on all surfaces the potential is uniquely determined Prob 34 But it s easy to guess one potential that obeys Laplace s equation and ts these boundary conditions E L where z is measured along the axis The uniqueness theorem guarantees that this is the solution The corresponding eld is Vz VOA E VV Lz which is indeed uniform qed Figure 73 Contrast the enormously more dif cult problem that arises if the conducting material is removed leaving only a metal plate at either end Fig 73 Evidently in the present case charge arranges itself over the surface of the wire in just such a way as to produce a nice uniform eld within 1Calculating this surface charge is not easy See for example J D Jackson Am J Phys 64 855 1996 Nor is it a simple matter to determine the eld outside the wire see Prob 757 71 ELECTROMOTIVE FORCE 289 I don t suppose there is any formula in physics more widely known than Ohm s law and yet it s not really a true law in the sense of Gauss s law or Ampere s law rather it is a rule of thumb that applies pretty well to many substances You re not going to win a Nobel prize for nding an exception In fact when you stop to think about it it s a little surprising that Ohm s law eVer holds After all a given eld E produces a force qE on a charge q and according to Newton s second law the charge will accelerate But if the charges are accelerating why doesn t the current increase with time growing larger and larger the longer you leave the eld on Ohm s law implies on the contrary that a constant eld produces a constant current which suggests a constant velocity Isn t that a contradiction of Newton s law No for we are forgetting the frequent collisions electrons make as they pass down the wire It s a little like this Suppose you re driving down a street with a stop sign at every intersection so that although you accelerate constantly in between you are obliged to start all over again with each new block Your average speed is then a constant in spite of the fact that save for the periodic abrupt stops you are always accelerating If the length of a block is A and your acceleration is a the time it takes to go a block is and hence the average velocity is 1 t Aa 1 m 2quot V 2 But wait That s no good either It says that the velocity is proportional to the square root of the acceleration and therefore that the current should be proportional to the Square root of the eld There s another twist to the story The charges in practice are already moving quite fast because of their thermal energy But the thermal velocities have random directions and average to zero The net drift velocity we re concerned with is a tiny extra bit Prob 519 So the time between collisions is actually much shorter than we supposed in fact A I 1thermal and therefore 1 t aA Uave a 2 21Ithermal If there are n molecules per unit volume and f free electrons per molecule each with charge q and mass m the current density is n A F n A 2 J quotfCIVave fq q 76 zvtherrnal m 2m Uthermal I don t claim that the term in parentheses is an accurate formula for the conductivity2 but it 2This classical model due to Drude bears little resemblance to the modern quantum theory of conductivity See for instance D Park s Introduction to the Quantum Theory 3rd ed Chap 15 New York McGraw Hill 1992 290 CHAPTER 7 ELECTRODYNAMICS does indicate the basic ingredients and it correctly predicts that conductivity is proportional to the density of the moving charges and ordinarily decreases with increasing temperature As a result of all the collisions the work done by the electrical force is converted into heat in the resistOr Since the work done per unit charge is V and the charge owing per unit time is I the power delivered is P 2 VI 12R 77 This is the Joule heating law With I in amperes and R in ohms P comes out in walls joules per second Problem 71 Two concentric metal spherical shells of radius a and b respectively are separated by weakly conducting material of conductivity 0 Fig 74a a If they are maintained at a potential difference V what current ows from one to the other b What is the resistance between the shells c Notice that if 9 gtgt a the outer radius 9 is irrelevant How do you account for that Exploit this observation to determine the current owing between two metal spheres each of radiu a immersed deep in the sea and held quite far apart Fig 74b if the potential difference between them is V This arrangement can be used to measure the conductivity of sea water i a b Figure 74 Problem 72 A capacitor C has been charged up to potential V0 at time t 0 it is connected to a resistor R and begins to discharge Fig 75a a Determine the charge on the capacitor as a function of time Qt What is the current through the resistor I t 71 ELECTROMOTIVE FORCE 291 Q C Q 8 V0 Q D R D Q R a b Figure 75 b What was the original energy stored in the capacitor Eq 255 By integrating Eq 77 con rm that the heat delivered to the resistor is equal to the energy lost by the capacitor Now imagine charging up the capacitor by connecting it and the resistor to a battery of xed voltage V0 at timet 0 Fig 75b c Again determine Qt and It d Find the total energy output of the battery f V01 dt Determine the heat delivered to the resistor What is the nal energy stored in the capacitor What fraction of the work done by the battery shows up as energy in the capacitor Notice that the answer is independent of R Problem 73 a Two metal objects are embedded in weakly conducting material of conductivity at Fig 76 Show that the resistance between them is related to the capacitance of the arrangement by 60 ch b Suppose you connected a battery between 1 and 2 and charged them up to a potential difference V0 If you then disconnect the battery the charge will gradually leak off Show that Vt Voe T and nd the time constant I in terms of 50 and a G5 6 Figure 76 Problem 74 Suppose the conductivity of the material separating the cylinders in Ex 72 is not uniform speci cally as k s for some constant k Find the resistance between the cylinders Hintz Because a is a function of position Eq 75 does not hold the charge density is not zero in the resistive medium and E does not go like I s But we do know that for steady currents I is the same across each cylindrical surface Take it from there 292 CHAPTER 7 ELECTRODYNAMICS 712 Electromotive Force If you think about a typical electric circuit Fig 77 a battery hooked up to a light bulb say there arises a perplexing question In practice the current is the same all the way around the loop at any given moment why is this the case when the only obvious driving force is inside the battery Off hand you might expect this to produce a large current in the battery and none at all in the lamp Who s doing the pushing in the rest of the circuit and how does it happen that this push is exactly right to produce the same current in each segment What s more given that the charges in a typical wire move literally at a snail s pace see Prob 519 why doesn t it take half an hour for the news to reach the light bulb How do all the charges know to start moving at the same instant l Figure 77 Figure 78 AnSWer If the current is not the same all the way around for instance during the rst split second after the switch is closed then charge is piling up somewhere and here s the crucial point the electric eld of this accumulating charge is in such a direction as to even out the ow Suppose for instance that the current into the bend in Fig 78 is greater than the current out Then charge piles up at the knee and this produces a eld aiming away from the kink This eld opposes the current owing in slowing it down and promotes the current owing out speeding it up until these currents are equal at which point there is no further accumulation of charge and equilibrium is established It s a beautiful system automatically selfcorrecting to keep the current uniform and it does it all so quickly that in practice you can safely assume the current is the same all around the circuit even in systems that oscillate at radio frequencies The upshot of all this is that there are really two forces involved in driving current around a circuit the source f3 which is ordinarily con ned to one portion of the loop a battery say and the electrostatic force which serves to smooth out the ow and communicate the in uence of the source to distant parts of the circuit f f E 78 The physical agency responsible for f S can be any one of many different things in a battery it s a chemical force in a piezoelectric crystal mechanical pressure is converted into an 71 ELECTROMOTIVE FORCE 293 electrical impulse in a thermocouple it s a temperature gradient that does the job in a photoelectric cell it s light and in a Van de Graaff generator the electrons are literally loaded onto a conveyer belt and swept along Whatever the mechanism its net effect is determined by the line integral of f around the circuit 79 Because 35 E d1 0 for electrostatic elds it doesn t matter whether you use f or f5 8 is called the electromotive force or emf of the circuit It s a lousy term since this is not a force at all it s the integral of a force per unit charge Some people prefer the word electromotance but emf is so ingrained that I think we d better stick with it Within an ideal source of emf a resistanceless battery3 for instance the net force on the charges is zero Eq 71 with o 00 so E fs The potential difference between the terminals a and b is therefore I h V Edlfsdlfsdl 710 we can extend the integral to the entire loop because f5 0 outside the source The function of a battery then is to establish and maintain a voltage difference equal to the electromotive force a 6 V battery for example holds the positive terminal 6 V above the negative terminal The resulting electrostatic eld drives current around the rest of the circuit notice however that inside the battery f 5 drives current in the direction opposite to E Because it s the line integral of f5 8 can be interpreted as the work done per unit charge by the source indeed in some books electromotive force is de ned this way However as you ll see in the next section there is some subtlety involved in this interpretation so I prefer Eq 79 Problem 75 A battery of emf 5 and internal resistance r is hooked up to a variable load resistance R If you want to deliver the maximum possible power to the load what resistance R should you choose You can t change 5 and r of course Problem 76 A rectangular loop of wire is situated so that one end height h is between the plates of a parallelplate capacitor Fig 79 oriented parallel to the eld E The other end is way outside where the eld is essentially zero What is the emf in this loop If the total resistance is R what current ows EXplain Warning this is a trick question so be careful if you have invented a perpetual motion machine there s probably something wrong with it 3Real batteries have a certain internal resistance r and the potential difference between their terminals is S 7 Ir when a current I is owing For an illuminating discussion of how batteries work see D Roberts Am J Phys 51 829 1983 294 CHAPTER 7 ELECTRODYNAMICS Figure 79 713 Motional emf In the last section I listed several possible sources of electromotive force in a circuit batteries being the most familiar But I did not mention the most common one of all the generator Generators exploit motional emf s which arise when you move a wire through a magnetic eld Figure 710 shows a primitive model for a generator In the shaded region there is a uniform magnetic eld B pointing into the page and the resistor R represents whatever it is maybe a light bulb or a toaster we re trying to drive current through If the entire loop is pulled to the right with speed 1 the charges in segment ab experience a magnetic force whose vertical component qu drives current around the loop in the clockwise direction The emf is rmagalv3n 711 where h is the width of the loop The horizontal segments be and ad contribute nothing since the force here is perpendicular to the wire Notice that the integral you perform to calculate 8 Eq 79 or 711 is carried out at one instant of time take a snapshot of the loop if you like and work from that Thus d l for the segment ab in Fig 710 points straight up even though the loop is moving to the right You can t quarrel with this it s simply the way emf is de ned but it is important to be clear about it k 2 AAAA VVVV C Figure 710 7 ELECTROMOTIVE FORCE 295 N MB fpu Figure 711 In particular although the magnetic force is responsible for establishing the emf it is certainly not doing any work magnetic forces never do work Who then is supplying the energy that heats the resistor Answer The person who s pulling on the loop With the current owing charges in segment ab have a vertical velocity call it u in addition to the horizontal velocity v they inherit from the motion of the loop Accordingly the magnetic force has a component tu to the left To counteract this the person pulling on the wire must exert a force per unit charge fpull B to the right Fig 711 This force is transmitted to the charge by the structure of the wire Meanwhile the particle is actually moving in the direction of the resultant velocity w and the distance it goes is h cos 6 The work done per unit charge is therefore Emull MB h gtsin6 UBh 5 cos 6 sin 9 coming from the dot product As it turns out then the work done per unit charge is exactly equal to the emf though the integrals are taken along entirely different paths Fig 712 and completely different forces are involved To calculate the emf you integrate around the loop at one instant but to calculate the work done you follow a charge in its motion around the loop fpuu contributes nothing to the emf because it is perpendicular to the wire whereas fmag contributes nothing to work because it is perpendicular to the motion of the charge4 There is a particularly nice way of expressing the emf generated in a moving loop Let lt1 be the ux of B through the loop EBda 712 For the rectangular loop in Fig 710 I th 4For further discussion see E P Mosca Am J Phys 42 295 1974 296 CHAPTER 7 ELECTRODYNAMICS b c b c h A V hcos 0 a a V d a a39 d b Integration path for calculating work done follow the charge around the loop a Integration for computing 8 follow the wire at one instant of time Figure 712 As the loop moves the ux decreases d lt1 d Bh x Bhv dt d t The minus sign accounts for the fact that dxdt is negative But this is precisely the emf Eq 711 evidently the emf generated in the loop is minus the rate of change of ux through the loop dd 8 3 This is the ux rule for motional emf Apart from its delightful simplicity it has the virtue of applying to nonrectangular loops moving in arbitrary directions through nonuniform magnetic elds in fact the loop need not even maintain a xed shape Proof Figure 713 shows a loop of wire at time t and also a short time d t later Suppose we compute the ux at time t using surface S and the ux at time t dt using the surface consisting of 8 plus the ribbon that connects the new position of the loop to the old The change in ux then is dCDCIgttdt qgttqgtribbon Bda ribbon Focus your attention on point P in time d t it moves to P Let v be the velocity of the wire and u the velocity of a charge down the wire w v u is the resultant velocity of a charge at P The in nitesimal element of area on the ribbon can be written as da v x dl dt 71 ELECTROMOTIVE FORCE 297 Surface 8 Loop at Loop at time t time t dt Enlargement of da Figure 713 see inset in Fig 713 Therefore dltIgt EfBvxdl Since w v u and u is parallel to d we can also write this as 2 B dl dt wgtlt Now the scalar triple product can be rewritten Bwxdl wadl dd B dl dt w x But w x B is the magnetic force per unit charge fmag so dd 17 fmag 3961 and the integral of fmag is the emf 298 CHAPTER 7 ELECTRODYNAMI C S There is a sign ambiguity in the de nition of emf Eq 79 Which way around the loop are you supposed to integrate There is a compensatory ambiguity in the de nition of ux Eq 712 Which is the positive direction for da In applying the ux rule sign consistency is governed as always by your right hand If your ngers de ne the positive direction around the loop then your thumb indicates the direction of da Should the emf come out negative it means the current will ow in the negative direction around the circuit The ux rule is a nifty short cut for calculating motional emf s It does not contain any new physics Occasionally you will run across problems that cannot be handled by the ux rule for these one must go back to the Lorentz force law itself Example 74 A metal disk of radius a rotates with angular velocity 0 about a vertical axis through a uniform eld B pointing up A circuit is made by connecting one end of a resistor to the axle and the other end to a sliding contact which touches the outer edge of the disk Fig 714 Find the current in the resistor Sliding contact J 1 Figure 714 Solution The speed of a point on the disk at a distance s from the axis is v cos so the force per unit charge is fmag v x B wsBs The emf is therefore a a 82f fmagdswa sdszw a 0 0 2 7 and the current is 8 0302 I R 2R The trouble with the ux rule is that it assumes the current ows along a well de ned path whereas in this example the current spreads out over the whole disk It s not even clear what the ux through the circuit would mean in this context Even more tricky is the case of eddy currents Take a chunk of aluminum say and shake it around in a nonuniform magnetic eld Currents will be generated in the material and you will feel a kind of viscous drag as though you were pulling the block through molasses this is the force I called fpull in the discussion of motional emf Eddy currents are notoriously dif cult to calculate5 but easy and dramatic to demonstrate You may have witnessed the classic experiment in which an 5See for example W M Saslow Am J Phys 60 693 1992 71 ELECTROMOTIVE FORCE 299 a b Figure 715 aluminum disk mounted as a pendulum on a horizontal axis swings down and passes between the poles of a magnet Fig 715a When it enters the eld region it suddenly slows way down To con rm that eddy currents are responsible one repeats the process using a disk that has many slots cut in it to prevent the ow of large scale currents Fig 715b This time the disk swings freely unimpeded by the eld Problem 77 A metal bar of mass m slides frictionlesst on two parallel conducting rails a distance 1 apart Fig 716 A resistor R is connected across the rails and a uniform magnetic eld B pointing into the page lls the entire region AAAA VVVV N Figure 716 300 CHAPTER 7 ELECTRODYNAMICS a If the bar m0ves to the right at speed 12 what is the current in the resistor In what direction does it ow b What is the magnetic force on the bar In what direction c If the bar starts out with speed v0 at time t O and is left to slide what is its speed at a later time t d The initial kinetic energy of the bar was of course m 1202 Check that the energy delivered to the resistor is exactly mvoz Problem 78 A square loop of wire side a lies on a table a distance s from a very long straight wire which carries a current I as shown in Fig 717 a Find the ux of B through the loop b If someone now pulls the loop directly away from the wire at speed 1 what emf is generated In what direction clockwise or counterclockwise does the current ow c What if the loop is pulled to the right at speed 1 instead of away a Figure 717 Problem 79 An in nite number of different surfaces can be t to a given boundary line and yet in de ning the magnetic ux through a loop ltIgt f B da I never speci ed the particular surface to be used Justify this apparent Oversight Problem 710 A square loop side a is mounted on a vertical shaft and rotated at angular velocity 0 Fig 718 A uniform magnetic eld B points to the right Find the 50 for this alternating current generator Problem 711 A square loop is cut out of a thick sheet of aluminum It is then placed so that the top portion is in a uniform magnetic eld B and allowed to fall under gravity Fig 719 In the diagram shading indicates the eld region B points into the page If the magnetic eld is l T a pretty standard laboratory eld nd the terminal velocity of the loop in ms Find the velocity of the loop as a function of time How long does it take in seconds to reach S21 90 of the terminal velocity What would happen if you cut a tiny slit in the ring breaking the circuit N0te39 The dimensions of the loop cancel out determine the actual numbers in the units indicated 72 ELECTROMAGNETIC INDUCTION a gt gt B a gt I Figure 718 301 Figure 719 2 Electromagnetic Induction 721 Faraday s Law In 1831 Michael Faraday reported on a series of experiments including three that with some violence to history can be characterized as follows Experiment 1 He pulled a loop of wire to the right through a magnetic eld Fig 720a A current owed in the loop Experiment 2 He moved the magnet to the left holding the loop still Fig 720b Again a current owed in the loop Experiment 3 With both the loop and the magnet at rest Fig 720c he changed the strength of the eld he used an electromagnet and varied the current in the coil Once again current owed in the loop a b Figure 720 C changing magnetic field 302 CHAPTER 7 ELECTRODYNAMI C S The rst experiment of course is an example of motional emf conveniently expressed by the ux rule d 1 dt I don t think it will surprise you to learn that exactly the same emf arises in Experiment 2 all that really matters is the relative motion of the magnet and the loop Indeed in the light of special relativity is has to be so But Faraday knew nothing of relativity and in classical electrodynamics this simple reciprocity is a coincidence with remarkable implications For if the loop moves it s a magnetic force that sets up the emf but if the loop is stationary the force cannot be magnetic stationary charges experience no magnetic forces In that case what is responsible What sort of eld exerts a force on charges at rest Well electric elds do of course but in this case there doesn t seem to be any electric eld in sight Faraday had an ingenious inspiration A changing magnetic eld induces an electric ehi It is this induced electric eld that accounts for the emf in Experiment 26 Indeed if as Faraday found empirically the emf is again equal to the rate of change of the ux dd dt then E is related to the change in B by the equation 3B yiEdl da 715 at This is Faraday s law in integral form We can convert it to differential form by applying Stokes theorem BB VxE 7161 at Note that Faraday s law reduces to the old rule f E dl 0 or in differential form V x E 0 in the static case constant B as of course it should In Experiment 3 the magnetic eld changes for entirely different reasons but according to Faraday s law an electric eld will again be induced giving rise to an emf dlt1gtdt Indeed one can subsume all three cases and for that matter any combination of them into a kind of universal ux rule Whenever and for whatever reason the magnetic ux through a loop changes an emf dd 6 717 dt will appear in the loop 6You might argue that the magnetic eld in Experiment 2 is not really changing just moving What 1 mean is that if you sit at a xed location the eld does change as the magnet passes by 72 ELECTROMAGNETIC INDUCTION 303 Many people call this Faraday s law Maybe I m overly fastidious but I nd this confus ing There are really two totally different mechanisms underlying Eq 717 and to identify them both as Faraday s law is a little like saying that because identical twins look alike we ought to call them by the same name In Faraday s rst experiment it s the Lorentz force law at work the emf is magnetic But in the other two it s an electric eld induced by the changing magnetic eld that does the job Viewed in this light it is quite astonishing that all three processes yield the same formula for the emf In fact it was precisely this coincidence that led Einstein to the special theory of relativity he sought a deeper un derstanding of what is in classical electrodynamics a peculiar accident But that s a story for Chapter 12 In the meantime I shall reserve the term Faraday s law for electric elds induced by changing magnetic elds and I do not regard Experiment 1 as an instance of Faraday s law I Example 75 A long cylindrical magnet of length L and radius a carries a uniform magnetization M parallel to its axis It passes at constant velocity 1 through a circular wire ring of slightly larger diameter Fig 721 Graph the emf induced in the ring as a function of time Figure 721 Solution The magnetic eld is the same as that of a long solenoid with surface current Kb 2 M So the eld inside is B MOM except near the ends where it starts to spread out The ux through the ring is zero when the magnet is far away it builds up to a maximum of MOMJTQZ as the leading end passes through and it drops back to zero as the trailing end emerges Fig 722a The emf is minus the derivative of ltIgt with respect to time so it consists of two spikes as shown in Fig 722b Keeping track of the signs in Faraday s law can be a real headache For instance in Ex 75 we39 would like to know which way around the ring the induced current ows In principle the right hand rule does the job we called lt1 positive to the left in Fig 722a so the positive direction for current in the ring is counterclockwise as viewed from the left since the rst spike in Fig 722b is negatiVe the rst current pulse ows clockwise and the second counterclockwise But there s a handy rule called Lenz s law whose sole purpose is to help you get the directions right7 7Lenz s law applies to motional emf s too but for them it is usually easier to get the direction of the current from the Lorentz force law 304 CHAPTER 7 ELECTRODYNAMICS I ttoMtta2 a t 8 LI b t Figure 722 Nature abhors a change in ux The induced current will ow in such a direction that the ux it produces tends to cancel the change As the front end of the magnet in Ex 75 enters the ring the ux increases so the current in the ring must generate a eld to the right it therefore ows clockwise Notice that it is the change in ux not the ux itself that nature abhors when the tail end of the magnet exits the ring the ux drops so the induced current ows counterclockwise in an effort to restore it Faraday induction is a kind of inertial phenomenon A conducting loop likes to maintain a constant ux through it if you try to change the ux the loop responds by sending a current around in such a direction as to frustrate your efforts It doesn t succeed completely the ux produced by the induced current is typically only a tiny fraction of the original All Lenz s law tells you is the direction of the ow Example 76 The jumping ring demonstration If you wind a solenoidal coil around an iron core the iron is there to beef up the magnetic eld place a metal ring on top and plug it in the ring will jump several feet in the air Fig 723 Why Solution Before you turned on the current the ux through the ring was zero Afterward a ux appeared upward in the diagram and the emf generated in the ring led to a current in the ring which according to Lenz s law was in such a direction that its eld tended to canCel this new ux This means that the current in the loop is opposite to the current in the solenoid And opposite currents repel so the ring ies off8 8For further discussion of the jumping ring and the related oating ring see C S Schneider and J P Ertel Am J Phys 66 686 1998 72 ELECTROMAGNETIC INDUCTION 305 i ring CD lt E solenoid Figure 723 Problem 712 A long solenoid of radius a is driven by an alternating current so that the eld inside is sinusoidal Bt Bo coswt i A circular loop of wire of radius a 2 and resistance R is placed inside the solenoid and coaxial with it Find the current induced in the loop as a function of time Problem 713 A square loop of wire with sides of length a lies in the rst quadrant of the x y plane with one corner at the origin In this region there is a nonuniform time dependent magnetic eld By t ky3t2 2 where k is a constant Find the emf induced in the loop Problem 714 As a lecture demonstration a short cylindrical bar magnet is dropped down a vertical aluminum pipe of slightly larger diameter about 2 meters long It takes several seconds to emerge at the bottom whereas an otherwise identical piece of nnmagnetized iron makes the trip in a fraction of a second Explain why the magnet falls more slowly 722 The Induced Electric Field What Faraday s discovery tells us is that there are really two distinct kinds of electric elds those attributable directly to electric charges and those associated with changing magnetic elds9 The former can be calculated in the static case using Coulomb s law the latter can be found by exploiting the analogy between Faraday s law VXE at 9You could I suppose introduce an entirely new word to denote the eld generated by a changing B Electro dynamics would then involve three elds E elds produced by electric charges V vE l 60 p V X E 0 B elds produced by electric currents V B 0 V x B Mon and G elds produced by changing magnetic elds V G 0 V x G BBBt Because E and G exertforces in the same way F qE G it is tidier to regard their sum as a single entity and call the whole thing the electric eld 306 CHAPTER 7 ELECTRODYNAMICS and Ampere s law V X B 0 Of course the curl alone is not enough to determine a eld you must also specify the divergence But as long as E is a pure Faraday eld due exclusively to a changing B with 0 0 Gauss s law says V E 0 while for magnetic elds of course V B 0 always So the parallel is complete andI conclude that Faradayinduced electric elds are determined by 8B8t in exactly the same way as magnetostatic elds are determined by MOJ 39 x In particular if symmetry permits we can use all the tricks associated w1th Ampere 5 law in integral form fl 39 d1 MOIeHCv only this time it s Faraday s law in integral form Edl 718 The rate of change of magnetic ux through the Amperian loop plays the role forrnerl assigned to 016116 Example 77 A uniform magnetic eld Bt pointing straight up lls the shaded circular region of Fig 724 If B is changing with time what is the induced electric eld Solution E points in the circumferential direction just like the magnetic eld inside a long straight wire carrying a uniform current density Draw an Amperian loop of radius s and apply Faraday s law dCD d dB fr dl E27rs 17 E 71831 ns2 Therefore If B is increasing E runs clackwise as viewed from above Example 78 A line charge A is glued onto the rim of a wheel of radius b which is then suspended horizontally as shown in Fig 725 so that it is free to rotate the spokes are made of some nonconducting material wood maybe In the central region out to radius a there is a uniform magnetic eld B0 pointing up Now someone turns the eld off What happens 72 ELECTROMAGNETIC INDUCTION 307 Bt Rotation d1 I direction 7 Amperian loop of radius s Figure 724 Figure 725 Solution The changing magnetic eld will induce an electric eld curling around the axis of the wheel This electric eld exerts a force on the charges at the rim and the wheel starts to turn According to Lenz s law it will rotate in such a direction that its eld tends to restore the upward ux The motion then is counterclockwise as viewed from above Quantitatively Faraday s law says from d na2dB dt dt Now the torque on a segment of length dl is r X F or bAE dl The total torque on the wheel is therefore N of Edl bAna2 f and the angular momentum imparted to the wheel is 2 0 th Jta b dB Ana2bBo 30 It doesn t matter how fast or slow you turn off the eld the ultimate angular velocity of the wheel is the same regardless If you nd yourself wondering where this angular momentum came from you re getting ahead of the story Wait for the next chapter A nal word on this example It s the electric eld that did the rotating To convince you of this I deliberately set things up so that the magnetic eld is always zero at the location of the charge on the rim The experimenter may tell you she never put in any electric elds all she did was switch off the magnetic eld But when she did that an electric eld automatically appeared and it s this electric eld that turned the wheel 308 CHAPTER 7 ELECTRODYNAMICS I must warn you now of a small fraud that tarnishes many applications of Faraday s law Electromagnetic induction of course occurs only when the magnetic elds are changing and yet we would like to use the apparatus of magnetostatics Ampere s law the Biot Savart law and the rest to calculate those magnetic elds Technically any result derived in this way is only approximately correct But in practice the error is usually negligible unless the eld uctuates extremely rapidly or you are interested in points very far from the source Even the case of a wire snipped by a pair of scissors Prob 718 is static enough for Ampere s law to apply This regime in which magnetostatic rules can be used to calculate the magnetic eld on the right hand side of Faraday s law is called quasistatic Generall speaking it is only when we come to electromagnetic waves and radiation that we must worry seriously about the breakdown of magnetostatics itself Example 79 An in nitely long straight wire carries a slowly varying current I t Determine the induced electric eld as a function of the distance s from the wire Figure 726 Solution In the quasistatic approximation the magnetic eld is MOI271s and it circles around the wire Like the B eld of a solenoid E here runs parallel to the axis For the rectangular Amperian loop in Fig 726 Faraday s law gives from Es0l Esl Bda d S 1 ldl E9 ds E9 lns lnso 271 dt SO s 271 dt Thus d1 Em g gamsK 2 719 where K is a constant that is to say it is independent of s it might still be a function of t The actual value of K depends on the whole history of the function I t we ll see some examples in Chapter 10 10This example is arti cial and not just in the usual sense of involving in nite wires but in a more subtle respect It assumes that the current is the same at any given instant all the way down the line This is a safe assumption for the short wires in typical electric circuits but not in practice for long wires transmission lines unless you supply a distributed and synchronized driving mechanism But never mind the problem doesn t inquire how you would produce such a current it only asks what elds would result if you did Variations on this problem are discussed in M A Heald Am J Phys 54 l 142 1986 and references cited therein 72 ELECTROMAGNETIC INDUCTION 309 Equation 719 has the peculiar implication that E blOWs up as s goes to in nity That can t be true What s gone wrong Answer We have overstepped the limits of the quasistatic approximation As we shall see in Chapter 9 electromagnetic news travels at the speed of light and at large distances B depends not on the current now but on the current as it was at some earlier time indeed a whole range of earlier times since different points on the wire are different distances away If r is the time it takes I to change substantially then the quasistatic approximation should hold only for s ltlt CI 720 and hence Eq 719 simply does not apply at extremely large s Problem 715 A long solenoid with radius a and n turns per unit length carries a timedependent current I t in the 46 direction Find the electric eld magnitude and direction at a distance s from the axis both inside and outside the solenoid in the quasistatic approximation Problem 716 An alternating current I 2 IO cos wt ows down a long straight wire and returns along a coaxial conducting tube of radius a a In what direction does the induced electric eld point radial circumferential or longitu dinal b Assuming that the eld goes to zero as s gt 00 nd Es t Incidentally this is not at all the way electric elds actually behave in coaxial cables for reasons suggested in footnote 10 See Sect 953 or J G Cherveniak Am J Phys 54 946 1986 for a more realistic treatment Problem 717 A long solenoid of radius a carrying n turns per unit length is looped by a wire with resistance R as shown in Fig 727 a If the current in the solenoid is increasing at a constant rate d I d t k what current ows in the loop and which way left or right does it pass through the resistor b If the current I in the solenoid is constant but the solenoid is pulled out of the loop and reinserted in the opposite direction what total charge passes through the resistor Mill WNW R Figure 727 310 CHAPTER 7 ELECTRODYNAMICS l k Figure 728 Problem 718 A square loop side a resistance R lies a distance s from an in nite straight wire that carries current I Fig 728 Now someone cuts the wire so that I dr0ps to zero In what direction does the induced current in the square loop ow and what total charge passes a given point in the loop during the time this current ows If you don t like the scissors model turn the current down gradually l xtI f0r0 t lor 0 fort gt lor t Problem 719 A toroidal coil has a rectangular cross section with inner radius a outer radius a w and height h It carries a total of N tightly wound turns and the current is increasing at a constant rate d I dt k If 11 and h are both much less than a nd the elecln39c eld at a point z above the center of the toroid Hint exploit the analogy between Faraday elds and magnetostatic elds and refer to Ex 56 723 Inductance Suppose you have two loops of Wire at rest Fig 729 If you run a steady current 11 around loop 1 it produces a magnetic eld B1 Some of the eld lines pass through 100p 2 let 192 be the ux of B1 through 2 You might have a tough time actually calculating B1 but a glance at the BiOtSavart law LO d1 X3 I B 47 If a2 reveals one signi cant fact about this eld It is proportional to the current 11 Therefore so too is the ux through loop 2 12 2B1da2 72 ELECTROMAGNETIC INDUCTION 311 dl2 Loop 2 Loop 1 dl1 Figure 729 Figure 730 Thus 192 M2111 721 where M21 is the constant of proportionality it is known as the mutual inductance of the two loops There is a cute formula for the mutual inductance which you can derive by expressing the ux in terms of the vector potential and invoking Stokes theorem 2B1dazVXA1dag A1dlz Now according to Eq 563 A M011 dli 1 47 a and hence I dl 192 1 6112 47 a Evidently Mo dli d12 M 21 4 f jg a 722 This is the Neumann formula it involves a double line integral one integration around loop 1 the other around loop 2 Fig 730 It s not very useful for practical calculations but it does reveal two important things about mutual inductance 1 M21 is a purely geOmetrical quantity having to do with the sizes shapes and relative positions of the two loops 2 The integral in Eq 722 is unchanged if we switch the roles of loops 1 and 2 it follows that M21 M12 723 312 CHAPTER 7 ELECTRODYNAMICS This is an astonishing conclusion Whatever the shapes and positions of the loops the ux through 2 when we run a current around I is identical to the ux through I when we send the same current 1 around 2 We may as well drop the subscripts and call them both M Example 710 A short solenoid length l and radius a with n1 turns per unit length lies on the axis of a very long solenoid radius 1 n2 turns per unit length as shown in Fig 731 Current I ows in the short solenoid What is the ux through the long solenoid quot quot W r My 39 39 Kev Figure 731 Solution Since the inner solenoid is short it has a very complicated eld moreover it puts a different amount of ux through each turn of the outer solenoid It would be a miserable task to compute the total ux this way However if we exploit the equality of the mutual inductances the problem becomes very easy Just look at the reverse situation run the current I through the outer solenoid and calculate the ux through the inner one The eld inside the long solenoid is constant B uonzl Eq 557 so the ux through a single loop of the short solenoid is Bria2 uonzlfraz There are nll turns in all so the total ux through the inner solenoid is CD uorraznlnzll This is also the ux a current I in the short solenoid would put through the long one which is what we set out to nd Incidentally the mutual inductance in this case is M uorraznlnzl Suppose now that you vary the current in 100p l The flux through loop 2 will vary accordingly and Faraday s law says this changing ux will induce an emf in loop 2 d I d I 2 M1 dt dt In quoting Eq 72l which was based on the Biot Savart law I am tacitly assuming that 7 24 the currents change slowly enough for the con guration to be considered quasistatic What 52 72 ELECTROMAGNETIC INDUCTION 3 13 Figure 732 a remarkable thing Every time you change the current in loop 1 an induced current ows in loop 2 even though there are no wires connecting them Come to think of it a changing current not only induces an emf in any nearby loops it also induces an emf in the source loop itself Fig 732 Once again the eld and therefore also the ux is proportional to the current ch L 725 The constant of proportionality L is called the selfinductance or simply the inductance of the loop As with M it depends on the geometry size and shape of the loop If the current changes the emf induced in the loop is d 5 L dt 726 Inductance is measured in henries H a henry is a volt second per ampere Example 711 Find the self inductance of a toroidal coil with rectangular cross section inner radius a outer radius 7 height h which carries a total of N turns Solution The magnetic eld inside the toroid is Eq 558 B 7701 39 271s The ux through a single turn Fig 733 is N b Bda Ih 1dsLmhm 2 271 a s 271 a The total ux is N times this so the selfinductance Eq 725 is a N2 b L 0 1n 727 271 a 3 14 CHAPTER 7 ELECTRODYNAMICS Axis Figure 733 Inductance like capacitance is an intrinsically positive quantity Lenz s law which is enforced by the minus sign in Eq 726 dictates that the emf is in such a direction as to oppose any change in current For this reason it is called a back emf Whenever you try to alter the current in a wire you must ght against this back emf Thus inductance plays somewhat the same role in electric circuits that mass plays in mechanical systems The greater L is the harder it is to change the current just as the larger the mass the harder it is to change an object s velocity Example 712 Suppose a current I is owing around a loop when someone suddenly cuts the wire The current drops instantaneously to zero This generates a whopping back emf for although I may be small d I dt is enormous That s why you often draw a spark when you unplug an iron or toaster electromagnetic induction is desperately trying to keep the current going even if it has to jump the gap in the circuit Nothing so dramatic occurs when you plug in a toaster or iron In this case induction opposes the sudden increase in current prescribing instead a smooth and continuous buildup Suppose for instance that a battery which supplies a constant emf 5390 is connected to a circuit of resistance R and inductance L Fig 734 What current ows Vvvv 2 80 AAAA Figure 734 72 ELECTROMAGNETIC INDUCTION 315 SR LR 2LR 3LR Figure 735 Solution The total emf in this circuit is that provided by the battery plus that resulting from the selfinductance Ohm s law then says11 5 L 2 IR 0 dt This is a rstorder differential equation for I as a function of time The general solution as you can easily derive for yourself is where k is a constant to be determined by the initial conditions In particular if the circuit is plugged in at time t 0 so 0 0 then k has the value 0R and 9 RLz 1a R 1 e 728 This function is plotted in Fig 735 Had there been no inductance in the circuit the current would have jumped immediately to 0 R In practice every circuit has some self inductance and the current approaches 60 R asymptotically The quantity r E L R is called the time constant it tells you how long the current takes to reach a substantial fraction roughly two thirds of its nal value Problem 720 A small loop of wire radius a lies a distance z above the center of a large loop radius 7 as shown in Fig 736 The planes of the two loops are parallel and perpendicular to the common axis a Suppose current I ows in the big loop Find the ux through the little loop The little loop is so small that you may consider the eld of the big loop to be essentially constant b Suppose current I ows in the little loop Find the ux through the big loop The little loop is so small that you may treat it as a magnetic dipole c Find the mutual inductances and con rm that M12 2 M21 HNotice that Ld I Mr goes on the left side of the equationgit is part of the emf that together with 50 establishes the voltage across the resistor Eq 710 CHAPTER 7 ELECTRODYNAMICS Figure 736 Figure 737 Problem 721 A square 100p of wire of side a lies midway between two long wires 3a apart and in the same plane Actually the long wires are sides of a large rectangular loop but the short ends are so far away that they can be neglected A clockwise current I in the square loop is gradually increasing d I dt k a constant Find the emf induced in the big loop Which way will the induced current ow Problem 722 Find the selfinductance per unit length of a long solenoid of radius R carrying n turns per unit length Problem 723 Try to compute the selfinductance of the hairpin loop shown in Fig 737 Neglect the contribution from the ends most of the ux comes from the long straight section You ll run into a snag that is characteristic of many selfinductance calculations To get a de nite answer assume the wire has a tiny radius 6 and ignore any ux through the wire itself Problem 724 An alternating current 0 cosaz amplitude 05 A frequency 60 Hz ows down a straight wire which runs along the axis of a toroidal coil with rectangular cross section inner radius 1 cm outer radius 2 cm height 1 cm 1000 turns The coil is connected to a 500 2 resistor a In the quasistatic approximation what emf is induced in the toroid Find the current Ir I in the resistor b Calculate the back emf in the coil due to the current Irt What is the ratio of the amplitudes of this back emf and the direct emf in a Problem 725 A capacitor C is charged up to a potential V and connected to an inductor L as shown schematically in Fig 738 At time t 0 the switch S is closed Find the current in the circuit as a function of time How does your answer change if a resistor R is included in series with C and L 72 ELECTROMAGNETIC INDUCTION 317 S 0 L C Figure 738 724 Energy in Magnetic Fields It takes a certain amount of energy to start a current owing in a circuit I m not talking about the energy delivered to the resistors and converted into heat that is irretrievably lost as far as the circuit is concerned and can be large or small depending on how long you let the current run What I am concerned with rather is the work you must do against the back emf to get the current going This is a xed amount and it is recoverable you get it back when the current is turned off In the meantime it represents energy latent in the circuit as we ll see in a moment it can be regarded as energy stored in the magnetic eld The work done on a unit charge against the back emf in one trip around the circuit is E the minus sign records the fact that this is the work done by you against the emf not the work done by the emf The amount of charge per unit time passing down the wire is I So the total work done per unit time is dW d1 51 LI dt dt If we start with zero current and build it up to a nal value I the work done integrating the last equation over time is 1 2 W 5 L1 729 It does not depend on how long we take to crank up the current only on the geometry of the loop in the form of L and the nal current I There is a nicer way to write W which has the advantage that it is readily generalized to surface and volume currents Remember that the ux CD through the loop is equal to L1 Eq 725 On the other hand ltIgtBdaVXAday Adl S S 73 where 73 is the perimeter of the loop and S is any surface bounded by 73 Thus le Adl 318 CHAPTER 7 ELECTRODYNAMICS and therefore 1 W EIfA dl The vector sign might as well go on the I l W 5 39ltfA I dl 730 In this form the generalization to volume currents is obvious I W AJdr 731 2 1 But we can do even better and expreSs W entirely in terms of the magnetic eld Ampere s law V X B 0 lets us eliminate J 1 A V x B dr 732 2M0 Integration by parts enables us to move the derivative from B to A speci cally product rule6 states that VAXBBVXA AVXB AVXBBB VAXB W Bsz VAXBdT 2M0 Bzdr AXBda 733 2M0 1 s where S is the surface bounding the volume V Now the integration in Eq 731 is to be taken over the entire volume occupied by the current But any region larger than this will do just as well for J is zero out there anyway In Eq 733 the larger the region we pick the greater is the contribution from the volume integral and therefore the smaller is that of the surface integral this makes sense as the surface gets farther from the current both A and B decrease In particular if we agree to integrate over all space then the surface integral goes to zero and we are left with l w m B2dr 734 2M0 all space In view of this result we say the energy is stored in the magnetic eld in the amount B22u0 per unit volume This is a nice way to think of it though someone looking at Eq 731 might prefer to say that the energy is stored in the current distribution in the Consequently 72 ELECTROMAGNETIC INDUCTION 319 amount A J per unit volume The distinction is one of bookkeeping the important quantity is the total energy W and we shall not worry about Where if anywhere the energy is located You might nd it strange that is takes energy to set up a magnetic eld after all magnetic elds themselves do no work The point is that producing a magnetic eld where previously there was none requires changing the eld and a changing B eld according to Faraday induces an electric eld The latter of course can do work In the beginning there is no E and at the end there is no E but in between while B is building up there is an E and it is against this that the work is done You see Why I could not calculate the energy stored in a magnetostatic eld back in Chapter 5 In the light of this it is extraordinary how similar the magnetic energy formulas are to their electrostatic counterparts 1 Welec 5 V0dr 62313sz 243 and 245 l 1 Wmag dT f 32 df and Example 713 A long coaxial cable carries current I the current ows down the surface of the inner cylinder radius a and back along the outer cylinder radius 7 as shown in Fig 739 Find the magnetic energy stored in a section of length l Figure 739 Solution According to Ampere s law the eld between the cylinders is Mob B 271s Elsewhere the eld is zero Thus the energy per unit volume is a L01 2 012 2H0 271s 871252 The energy in a cylindrical shell of length radius s and thickness ds then is 2 2 Mo V V 10 1 ds 871252 27rls ds 4T S 320 CHAPTER 7 ELECTRODYNAMICS Integrating from a to b we have 121 9 W 0 ln lt 471 a By the way this suggests a very simple way to calculate the selfinductance of the cable According to Eq 729 the energy can also be written as L12 Comparing the two expres 12 l b L M Oln 271 a This method of calculating selfinductance is especially useful when the current is not con ned to a single path but spreads over some surface or volume In such cases different parts of the current may circle different amounts of ux and it can be very tricky to get L directly from Eq 725 sions Problem 726 Find the energy stored in a section of length l of a long solenoid radius R current I n turns per unit length a using Eq 729 you found L in Prob 722 b using Eq 730 we worked out A in Ex 512 c using Eq 734 d using Eq 733 take as your volume the cylindrical tube from radius a lt R out to radius 9 gt R Problem 727 Calculate the energy stored in the toroidal coil of Ex 7 l by applying Eq 734 Use the answer to check Eq 727 Problem 728 A long cable carries current in one direction uniformly distributed over its circular cross section The current returns along the surface there is a very thin insulating sheath separating the currents Find the self inductance per unit length Problem 729 Suppose the circuit in Fig 740 has been connected for a long time when suddenly at time t 0 switch S is thrown bypassing the battery A l S B L 80 R Figure 740 12Notice the similarity to Eq 727 in a sense the rectangular toroid is a short coaxial cable turned on its side 73 MAXWELL S EQUATIONS 321 32 Figure 741 a What is the current at any subsequent time t b What is the total energy delivered to the resistor c Show that this is equal to the energy originally stored in the inductor Problem 730 Two tiny wire loops with areas a1 and 32 are situated a displacement IL apart Fig 741 a Find their mutual inductance Hint39 Treat them as magnetic dipoles and use Eq 5 87 Is your formula consistent with Eq 723 b Suppose a current 1 is owing in loop 1 and we propose to turn on a current 2 in loop 2 How much work must be done against the mutually induced emf to keep the current 11 owing in loop 1 In light of this result comment on Eq 635 73 Maxwell s Equations 731 Electrodynamics Before Maxwell So far we have encountered the following laws specifying the divergence and curl of electric and magnetic elds 1 i V E 0 Gauss s law 60 ii V B 0 no name iii V x E E Faraday s law iv V x B 2 OJ Ampere s law These equations represent the state of electromagnetic theory over a century ago when Maxwell began his work They were not written in so compact a form in those days but their physical content was familiar Now it happens there is a fatal inconsistency in these 322 CHAPTER 7 ELECTRODYNAMICS formulas It has to do with the old rule that divergence of curl is always zero If you appl the divergence to number iii everything works out 8B VVXEV E 8 V B gt 8t The left side is zero because divergence of curl is zero the right side is zero by virtue of equation ii But when you do the same thing to number iv you get into trouble V V X B 0V 39J 735 the left side must be zero but the right side in general is not For steady currents the divergence of J is zero but evidently when we go beyond magnetOstatics Ampere s 1a cannot be right There s another way to see that Amp re s law is bound to fail for nonsteady currents Suppose we re in the process of charging up a capacitor Fig 742 In integral form Amp re s law reads I B 39 d1 HOIenc I want to apply it to the Amperian loop shown in the diagram How do I determine 16m Well it s the total current passing through the loop or more precisely the current piercing a surface that has the loop for its boundary In this case the simplest surface lies in the plane of the loop the wire punctures this surface so Ienc I Fine but what if I draw instead the balloonshaped surfaCe in Fig 742 No current passes through this surface and I conclude that 13nC 0 We never had this problem in magnetostatics because the con ict arises only when charge is piling up somewhere in this case on the capacitor plates But for nonsteady currents such as this one the current enclosed by a loop is an ill de ned notion since it depends entirely on what surface you use If this seems pedantic to you obviously one should use the planar surface remember that the Amperian loop could be some contorted shape that doesn t even lie in a plane Amperian loop HF Capacitor B attery Figure 742 73 MAX WELL S EQUATIONS 323 Of course we had no right to expect Ampere s law to hold outside of magnetostatics after all we derived it from the BiotSavart law However in Maxwell s time there was no experimental reason to doubt that Ampere s law was of wider validity The aw was a purely theoretical one and Maxwell xed it by purely theoretical arguments 732 How Maxwell Fixed Ampere s Law The problem is on the right side of Eq 735 which should be zero but isn t Applying the continuity equation 529 and Gauss s law the offending term can be rewritten 80 8 8E vJ eVE V 6 at at 0 lt 0 at It might occur to you that if we were to combine 608E8t with J in Ampere s law it would be just right to kill off the extra divergence 3E V X B 10 MOEOE 736 Maxwell himself had other reasons for wanting to add this quantity to Ampere s law To him the rescue of the continuity equation was a happy dividend rather than a primary motive But today we recognize this argument as a far more compelling one than Maxwell s which was based on a nowdiscredited model of the ether13 Such a modi cation changes nothing as far as magnetostatics is concerned when E is constant we still have V X B OJ In fact Maxwell s term is hard to detect in ordinary electromagnetic experiments where it must compete for recognition with J that s why Faraday and the others never discovered it in the laboratory However it plays a crucial role in the propagation of electromagnetic waves as we ll see in Chapter 9 Apart from curing the defect in Ampere s law Maxwell s term has a certain aesthetic appeal Just as a changing magnetic eld induces an electric eld Faraday s law so A changing electric eld induces a magnetic eld Of course theoretical convenience and aesthetic consistency are only suggestive there might after all be other ways to doctor up Ampere s law The real con rmation of Maxwell s theory came in 1888 with Hertz s experiments on electromagnetic waves Maxwell called his extra term the displacement current 8E E e 737 Jd 0 at It s a misleading name since 50aEaz has nothing to do with current except that it adds to J in Ampere s law Let s see now how the displacement current resolves the parad0x of the charging capacitor Fig 742 If the capacitor plates are very close together I didn t 13For the history of this subject see A M Bork Am J Phys 31 854 1963 324 CHAPTER 7 ELECTRODYNAMICS draw them that way but the calculation is simpler if you assume this then the electric eld between them is l E 1Q a a 60 60 A where Q is the charge on the plate and A is its area Thus between the plates 8E l d Q l I 81 60A d1 eoA39 Now Eq 736 reads in integral form 8E f B dl Molenc 060113 738 If we choose the at surface then E 0 and Ienc I If on the other hand we use the balloonshaped surface then 13nC 0 but f8E8t da 2 I 60 So we get the same answer for either surface though in the rst case it comes from the genuine current and in the second from the displacement current Problem 731 A fat wire radius a carries a constant current I uniformly distributed over its cross section A narrow gap in the wire of width w ltlt a forms a parallelplate capacitor as shown in Fig 743 Find the magnetic eld in the gap at a distance s lt a from the axis Figure 743 Problem 732 The preceding problem was an arti cial model for the charging capacitor de signed to avoid complications associated with the current spreading out over the surface of the plates For a more realistic model imagine thin wires that connect to the centers of the plates Fig 744a Again the current I is constant the radius of the capacitor is a and the separation of the plates is w ltlt a Assume that the current flows out over the plates in such a way that the surface charge is uniform at any given time and is zero at t 0 a Find the electric eld between the plates as a function of t 73 MAX WELL S EQUATIONS 325 b Figure 744 b Find the displacement current through a circle of radius s in the plane midway between the plates Using this circle as your Amperian loop and the at surface that spans it nd the magnetic eld at a distance s from the axis c Repeat part b but this time use the cylindrical surface in Fig 744b which extends to the left through the plate and terminates outside the capacitor Notice that the displacement current through this surface is zero and there are two contributions to lenc1 Problem 733 Refer to Prob 716 to which the correct answer was 01060 E z a S7 271 s sinat 1n z a Find the displacement current density J d b Integrate it to get the total displacement current IdJdda c Compare Id and I What s their ratio If the outer cylinder were say 2 mm in diameter how high would the frequency have to be for Id to be 1 of I This problem is designed to indicate why Faraday never discovered displacement currents and why it is ordinarily safe to ignore them unless the frequency is extremely high 14This problem raises an interesting quasiphilosophical question If you measure B in the laboratory have you detected the effects of displacement current as b would suggest or merely con rmed the effects of ordinary currents as c implies See D F Bartlett Am J Phys 58 1168 1990 326 CHAPTER 7 ELECTRODYNAMICS 733 Maxwell s Equations In the last section we put the nishing touches on Maxwell s equations 1 i V E p Gauss s law 60 ii V B 0 no name 8B 739 111 V x E E Faraday s law 3E 1v V X B MOJ M0603 Ampere s law With Maxwell s correction Together with the force law FqEV XB 740 they summarize the entire theoretical content of classical electrodynamics15 save for some special properties of matter which we encountered in Chapters 4 and 6 Even the continuit equation 310 at which is the mathematical expression of conservation of charge can be derived from Maxwell s equations by applying the divergence to number iv I have written Maxwell s equations in the traditional way which emphasizes that the specify the divergence and curl of E and B In this form they reinforce the notion that electric elds can be produced either by charges p or by changing magnetic elds BB8t and magnetic elds can be produced either by currents J or by changing electric elds BE8r 1 Actually this is somewhat misleading because when you come right down to it 8B 81 and BEBt are themselves due to charges and currents I think it is logically preferable to write VJ 7411 1 8B 1VE p 111VXE 0 60 at 7421 8E 11 V B 0 1V V X B quot M0605 0 with the elds E and B on the left and the sources p and J on the right This notation emphasizes that all electromagnetic elds are ultimately attributable to charges and currents MaXWell s equations tell you how charges produce elds reciprocally the force law tells you how elds affect charges 15 Like any differential equations Maxwell s must be supplemented by suitable boundary conditions Because these are typically obvious from the context eg E and B go to zero at large distances from a localized charge distribution it is easy to forget that they play an essential role 73 MAX WELL S EQUATIONS 327 Problem 734 Suppose 1 1261 of Br t 0 Em t 47160 r the theta function is de ned in Prob 145b Show that these elds satisfy all of Maxwell s equations and deteimine 0 and J Describe the physical situation that gives rise to these elds 734 Magnetic Charge There is a pleasing symmetry about Maxwell s equations it is particularly striking in free space where p and J vanish 3B VE0 VXE 3t 8E VXBZMOEOE If you replace E by B and B by aoeoE the rst pair of equations turns into the second and Vice versa This symmetry16 between E and B is spoiled though by the charge term in Gauss s law and the current term in Ampere s law You can t help wondering why the corresponding quantities are missing from V B 0 and V x E 8B8t What if we had 1 8B 1 VB pe 111 V XE MOJm a 60 at 743 3E 11 V 39B Mopm 1V V X B MoJe M0605 Then pm would represent the density of magnetic charge and pe the density of electric charge J m would be the current of magnetic charge and Je the current of electric charge Both charges would be conserved 3pm 744 at at The former follows by application of the divergence to iii the latter by taking the diver gence of iV In a sense Maxwell s equations beg for magnetic charge to exist it would t in so nicely And yet in spite of a diligent search no one has ever found any17 As far as we know pm is zero everywhere and so is Jm B is not on equal footing with E there exist 16Don t be distracted by the pesky constants m and 60 these are present Only because the SI system measures E and B in different units and would not occur for instance in the Gaussian system 17For an extensive bibliography see A S Goldhaber and W P Trower Am J Phys 58 429 1990 328 CHAPTER 7 ELECTRODYNAMICS stationary sources for E electric charges but none for B This is re ected in the fact that magnetic multipole expansions have no monopole term and magnetic dipoles consist of current loops not separated north and south poles Apparently God just didn t make any magnetic charge In the quantum theory of electrodynamics by the way it s a more than merely aesthetic shame that magnetic charge does not seem to exist Dirac showed that the existence of magnetic charge would explain why electric charge is quantized See Prob 812 Problem 735 Assuming that Coulomb s law for magnetic charges qm reads F E m 22 745 471 a2 work out the force law for a monopole qm moving with velocity v through electric and magnetic elds E and B For interesting commentary see W Rindler Am J Phys 57 993 1989 Problem 736 Suppose a magnetic monopole qm passes through a resistanceless loop of wire with self inductance L What current is induced in the loop This is one of the methods used to search for monopoles in the laboratory see B Cabrera Phys Rev Lett 48 1378 1982 735 Maxwell s Equations in Matter Maxwell s equations in the form 739 are complete and correct as they stand However when you are working with materials that are subject to electric and magnetic polarization there is a more convenient way to write them For inside polarized matter there will be accumulations of bound charge and current over which you exert no direct control It would be nice to reformulate Maxwell s equations in such a way as to make explicit reference only to those sources we control directly the free charges and currents We have already learned from the static case that an electric polarization P produces a bound charge density p V P 746 Eq 412 Likewise a magnetic polarization or magnetization M results in a bound current Jb V x M 747 Eq 613 There s just one new feature to consider in the nonstatic case Any change in the electric polarization involves a ow of bound charge call it J p which must be included in the total current For suppose we examine a tiny chunk of polarized material Fig 745 The polarization introduces a charge density ab P at one end and ab at the other Eq 411 If P now increases a bit the charge on each end increases accordingly giving a net current 73 MAXWELL S EQUATIONS 329 day 0 Figure 745 The current density therefore is 8P 8t 39 This polarization current has nothing whatever to do with the bound current J b The latter is associated with magnetization of the material and involves the spin and orbital motion of electrons J p by contrast is the result of the linear motion of charge when the electric polarization changes If P points to the right and is increasing then each plus charge moves a bit to the right and each minus charge to the left the cumulative effect is the polarization current J p In this connection we ought to check that Eq 748 is consistent with the continuity equation J p 748 an a VJsz tVP at 3t a Yes The continuity equation is satis ed in fact J p is essential to account for the con servation of bound charge Incidentally a changing magnetization does not lead to any analogous accumulation of charge or current The bound current J b V x M varies in response to changes in M to be sure but that s about it In view of all this the total charge density can be separated into two parts PpfpbPf VP 749 and the current density into three parts 8P JJfJbJszfVXM 750 Gauss s law can now be written as l VE ltpf VP 60 or V Dpf 751 where D as in the static case is given by D E 60E P 752 330 CHAPTER 7 ELECTRODYNAMI CS Meanwhile Ampere s law with Maxwell s term becomes 3P 3E VXBM0 JfVXM M0 0 3t 3t 0139 V XHJfE 753 where as before 1 E B M 754 0 Faraday s law and V B 0 are not affected by our separation of charge and current into free and bound parts since they do not involve p or J In terms of free charges and currents then Maxwell s equations read 8B i VDpf 111 VxE E 755 8D 11VB0 1vVxHJfE Some people regard these as the true Maxwell s equations but please understand that they are in no way more general than 739 they simply re ect a convenient division of charge and current into free and nonfree parts And they have the disadvantage of hybrid notation since they contain both E and D both B and H They must be supplemented therefore by appropriate constitutive relations giving D and H in terms of E and B These depend on the nature of the material for linear media P eoer and M me 756 so 1 D 2 6E and H B 757 it where e 2 601 Xe and u E uol xm Incidentally you ll remember that D is called the electric displacement that s why the second term in the AmpereMaxwell equation iv is called the displacement current generalizing Eq 737 3D E 758 Jd Problem 737 Sea water at frequency v 4 X 108 Hz has permittivity e 8160 permeability a no and resistivity 0 023 Qm What is the ratio of conduction current to displacement current Hints consider a parallelplate capacitor immersed in sea water and driven by a voltage V0 cos 271 vt 73 MAX WELL S EQUATIONS 331 736 Boundary Conditions In general the elds E B D and H will be discontinuous at a boundary between two different media or at a surface that carries charge density a or current density K The explicit form of these discontinuities can be deduced from Maxwell s equations 755 in their integral form 39 Dd 2 lt1 159 a Qf ii yiuda0 S d 39 E 111 3 dl dt 9B da for any surface S bounded by the iv f HdlIf i Dda Closedloop P 79 dt 5 over any closed surface 8 Applying i to a tiny waferthin Gaussian pillbox extending just slightly into the material on either side of the boundary we obtain Fig 746 D1 a D2aafa The positive direction for a is from 2 toward 1 The edge of the wafer contributes nothing in the limit as the thickness goes to zero nor does any volume change density Thus the component of D that is perpendicular to the interface is discontinuous in the amount Figure 746 332 CHAPTER 7 ELECTRODYNAMICS Figure 747 Identical reasoning applied to equation ii yields Turning to iii a very thin Amperian loop straddling the surface Fig 747 gives d E1 l E2l Bda dt 5 But in the limit as the width of the loop goes to zero the ux vanishes I have already dropped the contribution of the two ends to 56 E dl on the same grounds Therefore El E39239 0 761 That is the components of E parallel to the interface are continuous across the boundary By the same token iv implies Hl39l H2llfenc where I fem is the free current passing through the Amperian loop No volume current density will contribute in the limit of in nitesimal width but a surface current can In fact if n is a unit vector perpendicular to the interface pointing from 2 toward 1 so that 13 x l is normal to the Amperian loop then IfenCKf39 XlKfX l and hence H ll H39239 Kf x n 762 So the parallel components of H are discontinuous by an amount proportional to the free surface current density 73 MAX WELL S EQUATIONS 333 Equations 75962 are the general boundary conditions for electrodynamics In the case of linear media they can be expressed in terms of E and B alone i ElEfL 2EjL of iii E E o 39 1 1 763 11 Bli 3 0 w B 139 Bg Kf x n M1 M2 In particular if there is no free charge or free current at the interface then i 51EL 52E 0 iii E E 0 764 1 1 11 Bf B 0 w B B 0 M1 M2 As we shall see in Chapter 9 these equations are the basis for the theory of re ection and refraction More Problems on Chapter 7 Problem 738 Two very large metal plates are held a distance d apart one at potential zero the other at potential V0 Fig 748 A metal sphere of radius a a ltlt d is sliced in two and one hemisphere placed on the grounded plate so that its potential is likewise zero If the region between the plates is lled with weakly conducting material of uniform conductivity 0 what current ows to the hemisphere Answen39 371a20d V0 Hint study Ex 38 Problem 739 Two long straight copper pipes each of radius a are held a distance 2d apart see Fig 749 One is at potential V0 the other at V0 The space surrounding the pipes is lled with weakly conducting material of conductivity 0 Find the current per unit length which ows from one pipe to the other Hint refer to Prob 311 Problem 740 A common textbook problem asks you to calculate the resistance of a cone shaped object of resistivity p with length L radius a at one end and radius 7 at the other Fig 750 The two ends are at and are taken to be equipotentials The suggested method is to slice it into circular disks of width dz nd the resistance of each disk and integrate to get the total l m Figure 748 Figure 749 336 CHAPTER 7 ELECTRODYNAMICS c The induced current on the surface of the superconductor the xy plane can be determined from the boundary condition on the tangential component of B Eq 574 B 0K X 2 Using the eld you get from the image con guration show that 3mrh A W 4 where r is the distance from the origin Problem 744 If a magnetic dipole levitating above an in nite superconducting plane Prob 743 is free to rotate what orientation will it adopt and how high above the surface will it oat Problem 745 A perfectly conducting spherical shell of radius a rotates about the z axis with angular velocity a in a uniform magnetic eld B BO 2 Calculate the emf developed between the north pole and the equator Answer B0wa2 Problem 746 Refer to Prob 71 l and use the result of Prob 540 if it helps a Does the square ring fall faster in the orientation shown Fig 719 or when rotated 45C about an axis coming out of the page Find the ratio of the two terminal velocities If you dropped the loop which orientation would it assume in falling Answer 5 2y 1 where l is the length of a side and y is the height of the center above the edge of the magnetic eld in the rotated con guration b How long does is take a circular ring to cross the bottom of the magnetic eld at its changing terminal velocity Problem 747 a Use the analogy between Faraday s law and Ampere s law together with the BiotSavan law to show that A l 8 Br t X L dr E 765 r t 4 a 1 L2 for Faradayinduced electric elds b Referring to Prob 550a show that 8A E 766 at where A is the vector potential Check this result by taking the curl of both sides c A spherical shell of radius R carries a uniform surface charge a It spins about a xed axis at an angular velocity az that changes slowly with time Find the electric eld inside and outside the sphere Hint39 There are two contributions here the Coulomb eld due to the charge and the Faraday eld due to the changing B Refer to Ex 511 and use Eq 766 Problem 748 Electrons undergoing cyclotron motion can be speeded up by increasing the magnetic eld the accompanying electric eld will impart tangential acceleration This is the principle of the betatron One would like to keep the radius of the orbit constant during the process Show that this can be achieved by designing a magnet such that the average eld over the area of the orbit is twice the eld at the circumference Fig 752 Assume the electrons start from rest in zero eld and that the apparatus is symmetric about the center of the orbit Assume also that the electron velocity remains well below the speed of light so that nonrelativistic mechanics applies Hint39 differentiate Eq 53 with respect to time and use F 2 ma 2 qE 73 MAX WELL S EQUATIONS 335 1 Figure 751 A superconductor is a perfect conductor with the additional property that the constant B inside is in fact zero This ux exclusion is known as the Meissner effectlg C Show that the current in a superconductor is con ned to the surface 1 Superconductivity is lost above a certain critical temperature Tc which varies from one material to another Suppose you had a sphere radius a above its critical temperature and you held it in a uniform magnetic eld 802 while cooling it below To Find the induced surface current density K as a function of the polar angle 6 Problem 743 A familiar demonstration of superconductivity Prob 742 is the levitation of a magnet over a piece of superconducting material This phenomenon can be analyzed using the method of images19 Treat the magnet as a perfect dipole m a height z above the origin and constrained to point in the z direction and pretend that the superconductor occupies the entire half space below the xy plane Because of the Meissner effect B 0 for z 5 0 and since B is divergenceless the normal z component is continuous so BZ 0 just above the surface This boundary condition is met by the image con guration in which an identical dipole is placed at z as a stand in for the superconductor the two arrangements therefore produce the same magnetic eld in the region z gt 0 a Which way should the image dipole point z or z b Find the force on the magnet due to the induced currents in the superconductor which is to say the force due to the image dipole Set it equal to Mg where M is the mass of the magnet to determine the height h at which the magnet will oat Hint refer to Prob 63 CHAPTER 7 ELECTRODYNAMICS Figure 750 a Calculate R this way b Explain why this method is fundamentally awed See J D Romano and R H Price Am J Phys 64 1150 1996 c Suppose the ends are instead spherical surfaces centered at the apex of the cone Calculate the resistance in that case Let L be the distance between the centers of the circular perlmeterx of the end caps Answer p2nabb a2 L2 b a2 L Problem 741 A rare case in which the electrostatic eld E for a circuit can actually be calculated is the following M A Heald Am J Phys 52 522 1984 Imagine an in nitel long cylindrical sheet of uniform resistivity and radius a A slot corresponding to the batter t is maintained at iVoZ at q in and a steady current ows over the surface as indicated in Fig 751 According to Ohm s law then Va n lt lt 71 a Use separation of variables in cylindrical coordinates to determine Vs 4 inside ancll outside the cylinder Answer Von tan1s sin a s cos 45 s lt a Von tan a sin S a COS W S gt 0 b Find the surface charge density on the cylinder Answer 60 V0 71a tan 2 Problem 742 In a perfect conductor the conductivity is in nite so E 0 Eq 73 and an net charge resides on the surface just as it does for an imperfect conductor in electrostatics 18The Meissner effect is sometimes referred to as perfect diamagnetism in the sense that the eld inside is not merely reduced but canceled entirely However the surface currents responsible for this are free not bound so the actual mechanism is quite different 19w M Saslow Am J Phys 59 16 1991 a Show that the magnetic eld is constant BB3t 0 inside a perfect conductor b Show that the magnetic ux through a perfectly conducting loop is constant 73 MAXWELL S EQUATIONS 337 B Electron orbit b a I 8 b R13 Solenoid E R2 at lt a A Figure 752 Figure 753 Problem 749 An atomic electron charge q circles about the nucleus charge Q in an orbit of radius r the centripetal acceleration is provided of course by the Coulomb attraction of opposite charges Now a small magnetic eld dB is slowly turned on perpendicular to the plane of the orbit Show that the increase in kinetic energy d T imparted by the induced electric eld is just right to sustain circular motion at the same radius r That s why in my discussion of diamagnetism I assumed the radius is xed See Sect 613 and the references cited there Problem 750 The current in a long solenoid is increasing linearly with time so that the ux is proportional to t CI at Two voltmeters are connected to diametrically opposite points A and B together with resistors R 1 and R2 as shown in Fig 753 What is the reading on each voltmeter Assume that these are ideal voltmeters that draw negligible current they have huge internal resistance and that a voltmeter registers fab E dl between the terminals and through the meter Answer V oleRl R2 V2 ozR2R1 R2 Notice that V1 9E V2 even though they are connected to the same points See R H Romer Am J Phys 50 1089 1982 Problem 751 In the discussion of motional emf Sect 713 I assumed that the wire loop Fig 710 has a resistance R the current generated is then I vBh R But what if the wire is made out of perfectly conducting material so that R is zero In that case the current is limited only by the back emf associated with the selfinductance L of the loop which would ordinarily be negligible in comparison with I R Show that in this regime the loop mass m executes simple harmonic motion and nd its frequency20 Answer a BhM Problem 752 a Use the N eumann formula Eq 722 to calculate the mutual inductance of the con guration in Fig 736 assuming a is very small a ltlt 17 a ltlt z Compare your answer to Prob 720 b For the general case not assuming a is small sh0w that Mab 1l 85 2m 20For a collection of related problems see W M Saslow Am J Phys 55 986 1987 and R H Romer Eur J Phys 11 103 1990 338 CHAPTER 7 ELECTRODYNAMICS Secondary Primary N2 turns N turns Figure 754 where ab Problem 753 Two coils are wrapped around a cylindrical form in such a way that the same ux passes through every turn of both coils In practice this is achieved by inserting an iron core through the cylinder this has the effect of concentrating the ux The primary coil has N1 turns and the secondary has N2 Fig 754 If the current I in the primary is changing show that the emf in the secondary is given by g amp 767 51 N 1 i where 51 is the back emf of the primary This a primitive transformer a device for raising or lowering the emf of an alternating current source By choosing the appropriate number of turns any desired secondary emf can be obtained If you think this violates the conservation of energy check out Prob 754 Problem 754 A transformer Prob 753 takes an input AC voltage of amplitude V1 and delivers an output voltage of amplitude V2 which is determined by the turns ratio V2 V1 N2 N1 If N2 gt N1 the output voltage is greater than the input voltage Why doesn t thrs violate conservation of energy Answer Power is the product of voltage and current evidentl if the voltage goes up the current must come down The purpose of this problem is to see exactly how this works out in a simpli ed model a In an ideal transformer the same ux passes through all turns of the primary and of the secondary Show that in this case M 2 L1L2 where M is the mutual inductance of the coils and L1 L2 are their individual selfinductances b Suppose the primary is driven with AC voltage Vin 2 V1 cos wt and the secondary is connected to a resistor R Show that the two currents satisfy the relations d1 d I 2 d I 2 d 1 t L M IR letMdt VICOSW 2dt dt 2 25 CONDUCTORS 103 which puts out a eld 0260 on either side pointing away from the surface Fig 250 Thus a A Eabove Eother n9 260 0 A Ebelow Eother n 260 and hence l Eother E E above E below E average Averaging is really just a device for removing the contribution of the patch itself That argument applies to any surface charge in the particular case of a conductor the eld is zero inside and oeo outside Eq 248 so the average is 0260 and the force per unit area is l f 02 251 260 This amounts to an outward electrostatic pressure on the surface tending to draw the conductor into the eld regardless of the sign of a Expressing the pressure in terms of the eld just outside the surface P 630152 252 Problem 237 Two large metal plates each of area A are held a distance d apart Suppose we put a charge Q on each plate what is the electrostatic pressure on the plates Problem 238 A metal Sphere of radius R carries a totallcharge Q What is the force of repulsion between the northern hemisphere and the southern hemisphere 254 Capacitors Suppose we have two conductors and we put charge Q on one and Q on the other Fig 251 Since V is constant over a conductor we can speak unambiguously of the potential difference between them VV V Edl lt We don t know how the charge distributes itself over the two conductors and calculating the eld would be a mess if their shapes are complicated but this much we do know E is proportional to Q For E is given by Coulomb s law 1 p E 4dr 47T 0 t2 104 CHAPTER 2 ELECTROSTATICS Figure 251 so if you double 0 you double E Wait a minute How do we know that doubling Q and also Q simply doubles 0 Maybe the charge moves around into a completely different con guration quadrupling p in some places and halving it in others just so the total charge on each conductor is doubled The fact is that this concern is unwarranted doubling Q does double 0 everywhere it doesn t shift the charge around The proof of this will come in Chapter 3 for now you ll just have to believe me Since E is proportional to Q so also is V The constant of proportionality is called the capacitance of the arrangement 5 Q C V 253 Capacitance is a purely geometrical quantity determined by the sizes shapes and separation of the two conductors In SI units C is measured in farads F a farad is a coulomb per volt Actually this turns out to be inconveniently large9 more practical units are the microfarad 10 6 F and the picofarad1012 F Notice that V is by de nition the potential of the positive conductor less that of the negative one likewise Q is the charge of the positive conductor Accordingly capacitance is an intrinsically positive quantity By the way you will occasionally hear someone speak of the capacitance of a single conductor In this case the second conductor with the negative charge is an imaginary spherical shell of in nite radius surrounding the one conductor It contributes nothing to the eld so the capacitance is given by Eq 253 where V is the potential with in nity as the reference point Example 210 Find the capacitance of a parallelplate capacitor consisting of two metal surfaces of area A held a distance d apart Fig 252 Figure 252 9In the second edition I claimed you would need a forklift to carry a 1 F capacitor This is no longer the case you can now buy a 1 F capacitor that ts comfortably in a soup spoon 25 CONDUCTORS 105 Solution If we put Q on the top and Q on the bottom they will spread out uniformly over the two surfaces provided the area is reasonably large and the separation distance small10 The surface charge density then is a Q A on the top plate and so the eld according to Ex 25 is l 60 Q A The potential difference between the plates is therefore Q d7 A60 and hence C 254 If for instance the plates are square with sides 1 cm long and they are held 1 mm apart then the capacitance is 9 X 10 13 F Example 21 1 Find the capacitance of two concentric spherical metal shells with radii a and 19 Solution Place charge Q on the inner sphere and Q on the outer one The eld between the spheres is l E 99 47160 r2 so the potential difference between them is a a1 1 l V Edl i dr Q 1 47160 b r2 47160 a b As promised V is proportional to Q the capacitance is Q ab V 7160b a To charge up a capacitor you have to rem0ve electrons from the positive plate and carry them to the negative plate In doing so you ght against the electric eld which is pulling them back toward the positive conductor and pushing them away from the negative one How much work does it take then to charge the capacitor up to a nal amount Q Suppose that at some intermediate stage in the proceSs the charge on the positive plate is q so that the potential difference is qC According to Eq 238 the Work you must do to transport the next piece of charge dq is dW dq 10The exact solution is not easy even for the simpler case of circular plates See G T Carlson and B L lllman Am J Phys 62 1099 1994 106 CHAPTER 2 ELECTROSTATICS The total work necessary then to go from q 0 to q Q is Q 2 q 1 Q d W 0 C q 2 C or since Q 2 CV 1 W ECV2 255 where V is the nal potential of the capacitor Problem 239 Find the capacitance per unit length of two Coaxial metal cylindrical tubes of radii a and b Fig 253 Figure 253 Problem 240 Suppose the plates of a parallelplate capacitor move closer together by an in nitesimal distance 6 as a result of their mutual attraction 39 a Use Eq 252 to express the amount of work done by electrostatic forces in terms of the eld E and the area of the plates A b Use Eq 246 to express the energy lost by the eld in this process This problem is supposed to be easy but it contains the embryo of an alternative derivation of Eq 252 using conservation of energy More Problems on Chapter 2 Problem 241 Find the electric eld at a height z above the center of a square sheet side a carrying a uniform surface charge 7 Check your result for the limiting cases a gt 00 and z gtgt 0 Answers 0260471tan1 1 112222 1 Problem 242 If the electric eld in some region is given in spherical coordinates by the expression A Ai39 Bsin9cos r where A and B are constants what is the charge density Answers 60A B sin r2 Er 25 CONDUCTORS 107 Problem 243 Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere Express your answer in terms of the radius R and the total charge Q Answer 1471603 Q216R2 Problem 244 An inverted hemispherical bowl of radius R carries a uniform surface charge density 7 Find the potential difference between the north pole and the center Answer Ra26mm 1 Problem 245 A sphere of radius R carries a charge density p r kr where k is a constant Find the energy of the con guration Check your answer by calculating it in at least two different ways Answers 71sz7 7E Problem 246 The electric potential of some con guration is given by the expression eiAr Vr A r where A and A are constants Find the electric eld Er the charge density pr and the total charge Q Answers p 60A47183r Aze rr Problem 247 Two in nitely long wires running parallel to the x axis carry uniform charge densities A and A Fig 254 a Find the potential at any point x y 2 using the origin as your reference b Show that the equipotential surfaces are circular cylinders and locate the axis and radius of the cylinder corresponding to a given potential V0 Problem 248 In a vacuum diode electrons are boiled off a hot cathode at potential zero and accelerated across a gap to the anode which is held at positive potential V0 The cloud of moving electrons within the gap called space charge quickly builds up to the point where it reduces the eld at the surface of the cathode to zero From then on a steady current I flows between the plates Suppose the plates are large relative to the separation A gtgt d2 in Fig 255 so that edge effects can be neglected Then V p and v the speed of the electrons are all functions of x alone d 4 A Electron x Anode Cathode V0 V0 Figure 254 Figure 255 108 CHAPTER 2 ELECTROSTATICS a Write Poisson s equation for the region between the plates b Assuming the electrons start from rest at the cathode what is their speed at point x where the potential is Vx c In the steady state I is independent of x What then is the relation between p and v 1 Use these three results to obtain a differential equation for V by eliminating p and v e Solve this equation for V as a function of x V0 and d Plot Vx and compare it to the potential without spacecharge Also nd p and v as functions of x f Show that 1 KV032 256 and nd the constant K Equation 256 is called the ChildLangmuir law It holds for other geometries as well whenever spacecharge limits the current Notice that the spacecharge limited diode is nonlinear it does not obey Ohm s law Problem 249 Imagine that new and extraordinarily precise measurements have revealed an error in Coulomb s law The actual force of interaction between two point charges is found to be 1 I A F 1qu 1 84 4 47160 02 A where A is a new constant of nature it has dimensions of length obviously and is a huge number say half the radius of the kn0wn universe so that the correction is small which is why no one ever noticed the discrepancy before You are charged with the task of reformulating electrostatics to accommodate the new discovery Assume the principle of superposition still holds a What is the electric eld of a charge distribution p replacing Eq 28 b Does this electric eld admit a scalar potential Explain brie y how you reached your conclusion No formal proof necessary just a persuasive argument c Find the potential of a point charge q the analog to Eq 226 If your answer to b was no better go back and change it Use 00 as your reference point 1 For a point charge q at the origin show that I l Eda Vd C f9 A2V qu where S is the surface V the volume of any sphere centered at 1 e Show that this result generalizes 1 1 Eda Vdr Qenc 5 A2 12 60 for any charge distribution This is the next best thing to Gauss s Law in the new electro statics f Draw the triangle diagram like Fig 235 for this world putting in all the appropriate formulas Think of Poisson s equation as the formula for p in terms of V and Gauss s law differential form as an equation for p in terms of E 25 CONDUCTORS 109 Problem 250 Suppose an electric eld Ex y z has the form where a is a constant What is the charge density How do you account for the fact that the eld points in a particular direction when the charge density is uniform This is a more subtle problem than it looks and worthy of careful thought Problem 251 All of electrostatics f0110ws from the l r2 character of Coulomb s law together with the principle of superposition An analogous theory can therefore be constructed for N ewton s law of universal gravitation What is the gravitational energy of a sphere of mass M and radius R assuming the density is uniform Use your result to estimate the gravitational energy of the sun look up the relevant numbers The sun radiates at a rate of 386 x 1026 W if all this came from stored gravitational energy how long would the sun last The sun is in fact much older than that so evidently this is not the source of its power Problem 252 We know that the charge on a conductor goes to the surface but just how it distributes itself there is not easy to determine One famous example in which the surface charge density can be calculated explicitly is the ellipsoid 2 2 2 x y z a 2 7 2 1 In this case11 2 Q x2 y2 zZ 257 a 471abc a4 b4 c4 where Q is the total charge By choosing appropriate values for a b and c obtain from Eq 257 a the net both sides surface charge density ar on a circular disk of radius R b the net surface charge density 0x on an in nite conducting ribbon in the x y plane which straddles the y axis from x a to x a let A be the total charge per unit length of ribbon c the net charge per unit length Mx on a conducting needle running from x a to x a In each case sketch the graph of your result I lFor the derivation which is a real tour de force see W R Smythe Static and Dynamic Electricity 3rd ed New York Hemisphere 1989 Sect 502 Chapter 3 Special Techniques 31 Laplace s Equation 311 Introduction The primary task of electrostatiCS is to nd the electric eld of a given stationary charge distribution In principle this purpose is accomplished by Coulomb s law in the form of Eq 28 Er 3pr dr 31 47160 42 Unfortunately integrals of this type can be dif cult to calculate for any but the simplest charge con gurations Occasionally we can get around this by exploiting symmetry and using Gauss s law but ordinarily the best strategy is rst to calculate the potential V which is given by the somewhat more tractable Eq 229 Vr L1pr dr 32 41160 I Still even this integral is often too tough to handle analytically More0ver in problems involving conductors 0 itself may not be known in advance since charge is free to move around the only thing we control directly is the total charge or perhaps the potential of each conductor In such cases it is fruitful to recast the problem in differential form using Poisson s equation 224 1 V2V p 33 60 which together with appropriate boundary conditions is equivalent to Eq 32 Very often in fact we are interested in nding the potential in a region where 0 0 If 0 0 everywhere of course then V 0 and there is nothing further to say that s not what I 110 31 LAPLACE S EQUATION 11 1 mean There may be plenty of charge elsewhere but we re con ning our attention to places where there is no charge In this case Poisson s equation reduces to Laplace s equation 2 V v 0 34 or written out in Cartesian coordinates 82V 82V 32 V 52 372 35 This formula is so fundamental to the subject that one might almost say electrostatics is the study of Laplace s equation At the same time it is a ubiquitous equation appearing in such diverse branches of physics as gravitation and magnetism the theory of heat and the study of soap bubbles In mathematics it plays a major role in analytic function theory To get a feel for Laplace s equation and its solutions which are called harmonic functions we shall begin with the one and two dimensional versions which are easier to picture and illustrate all the essential properties of the three dimensional case though the one dimensional example lacks the richness of the other two 312 Laplace s Equation in One Dimension Suppose V depends on only one variable x Then Laplace s equation becomes dZV 0 dx2 39 The general solution is Vx mx b 36 the equation for a straight line It contains two undetermined constants m and b as is appropriate for a secondorder ordinary differential equation They are xed in any particular case by the boundary conditions of that problem For instance it might be speci ed that V 4 atx 1 and V Oatx 5 In thatcasem l andb 5 so V x 5 see Fig 31 OWP 123456x Figure 31 112 CHAPTER 3 SPECIAL TECHNIQUES I want to call your attention to two features of this result they may seem silly and obvious in one dimension where I can write down the general solution explicitly but the analogs in two and three dimensions are powerful and by no means obvious 1 Vx is the average of Vx a and Vx a for any a Vx Vx a Vx a Laplace s equation is a kind of averaging instruction it tells you to assign to the point x the average of the values to the left and to the right of x Solutions to Laplace s equation are in this sense as boring as they could possibly be and yet t the end points prOperly N Laplace s equation tolerates no local maxima or minima extreme values of V must occur at the end points Actually this is a consequence of 1 for if there were a local maximum V at that point would be greater than on either side and therefore could not be the average Ordinarily you expect the second derivative to be negative at a maximum and positive at a minimum Since Laplace s equation requires on the contrary that the second derivative be zero it seems reasonable that solutions should exhibit no extrema HOWever this is not a proof since there exist functions that have maxima and minima at points where the second derivative vanishes x4 for example has such a minimum at the point x 0 313 Laplace s Equation in Two Dimensions If V depends on two variables Laplace s equation becomes 82V 82V 3x2 8y2 This is no longer an ordinary differential equation that is one involving ordinary derivatives only it is a partial differential equation As a consequence some of the simple rules you may be familiar with do not apply For instance the general solution to this equation doesn t contain just two arbitrary constants or for that matter any nite number despite the fact that it s a secondorder equation Indeed one cannot write down a general solution at least not in a closed form like Eq 36 Nevertheless it is possible to deduce certain properties common to all solutions It may help to have a physical example in mind Picture a thin rubber sheet or a soap lm stretched over some support For de niteness suppose you take a cardboard box cut a wavy line all the way around and remove the top part Fig 32 Now glue a tightly stretched rubber membrane over the box so that it ts like a drum head it won t be a at drumhead of course unless you chose to cut the edges off straight Now if you lay out coordinates x y on the bottom of the box the height Vx y of the sheet above the point 31 LAPLACE S EQUATION 1 13 Figure 32 x y will satisfy Laplace s equation1 The one dimensional analog would be a rubber band stretched between two points Of course it would form a straight line Harmonic functions in two dimensions have the same properties we noted in one di mension 1 The value of V at a point x y is the average of those around the point More precisely if you draw a circle of any radius R about the point x y the average value of V on the circle is equal to the value at the center 1 Vx M f le circle This incidentally suggests the method of relaxation on which computer solutions to Laplace s equation are based Starting with speci ed values for V at the boundary and reasonable guesses for V on a grid of interior points the rst pass reassigns to each point the average of its nearest neighbors The second pass repeats the process using the corrected values and so on After a few iterations the numbers begin to settle down so that subsequent passes produce negligible changes and a numerical solution to Laplace s equation with the given boundary values has been achieved2 V has no local maxima or minima all extrema occur at the boundaries As before this follows from 1 Again Laplace s equation picks the most featureless func tion possible consistent with the boundary conditions no hills no valleys just the smoothest surface available For instance if you put a pingpong ball on the stretched rubber sheet of Fig 32 it will roll over to one side and fall off it will not nd a 1Actually the equation Satis ed by a rubber sheet is 1 2 a ava av 0 h 1av2av2 w r 8x g 3x 8y g By C e g 3x y it reduces approximately to Laplace s equation as long as the surface does not deviate too radically from a plane 2See for example E M Purcell Electricity and Magnetism 2nd ed problem 330 p 119 New York McGrawHill 1985 114 CHAPTER 3 SPECIAL TECHNIQUES pocket somewhere to settle into for Laplace s equation allows no such dents in the surface From a geometrical point of view just as a straight line is the shortest distance between two points so a harmonic function in two dimensions minimizes the surface area spanning the given boundary line 314 Laplace s Equation in Three Dimensions In three dimensions I can neither provide you with an explicit solution as in one dimension nor offer a suggestive physical example to guide your intuition as I did in two dimensions Nevertheless the same two properties remain true and this time I will sketch a proof 1 The value of V at point r is the average value of V over a spherical surface of radius 1 Cellteled at I I i da sphere 2 As a consequence V can have no local maxima or minima the extreme values of V must occur at the boundaries For if V had a local maximum at r then by the very nature of maximum 1 could draw a sphere around r Over which all values of V and a fortiori the average would be less than at r Proof Let s begin by calculating the average potential over a spherical surface of radius R due to a single point charge q located outside the sphere We may as well center the sphere at the origin and choose coordinates so that q lies on the z axis Fig 33 The potential at a point on the surface is 1 Z 4J39 07 y where ll2z2R2 2chos9 so 1 q 2 2 712 2 R 2Rcos9 R srn9d9d Vave 4nR2 4n60 Z Z 7T 2 q Lvz2R2 2chos9 411602ZR 0 q 1 1 q R z R 411602ZRZ 4760Z But this is precisely the potential due to q at the center of the sphere By the superposition principle the same goes for any collection of charges outside the sphere their average potential Over the sphere is equal to the net potential they produce at the center qed 31 LAPLACE S EQUATION 115 Figure 33 Problem 31 Find the average potential over a spherical surface of radius R due to a point charge q located inside same as above in other words only with z lt R In this case of course Laplace s equation does not hold within the sphere Show that in general QCHC 471 0R Vave Veenter where Vwmer is the potential at the center due to all the external charges and Qenc is the total enclosed charge Problem 32 In one sentence justify Earnshaw s Theorem A charged particle cannot be held in a stable equilibrium by electrostatic forces alone As an example consider the cubical arrangement of xed charges in Fig 34 It looks off hand as though a positive charge at the center would be suspended in midair since it is repelled away from each corner Where is the leak in this electrostatic bottle To harness nuclear fusion as a practical energy source it is necessary to heat a plasma soup of charges particles to fantastic temperatures so hot that contact would vaporize any ordinary pot Eamshaw s theorem says that electrostatic containment is also out of the question Fortunately it is possible to con ne a hot plasma magnetically q q I l I qt Figure 34 116 CHAPTER 3 SPECIAL TECHNIQUES Problem 33 Find the general solution to Laplace s equation in spherical coordinates for the case where V depends only on r Do the same for cylindrical coordinates assuming V depends only on s 315 Boundary Conditions and Uniqueness Theorems Laplace s equation does not by itself determine V in addition a suitable set of boundary conditions must be supplied This raises a delicate question What are appropriate boundary conditions suf cient to determine the answer and yet not so strong as to generate incon sistencies The onedimensional case is easy for here the general solution V mx b contains two arbitrary constants and we therefore require two boundary conditions We might for instance specify the value of the function at the two ends or we might give the value of the function and its derivative at one end or the value at one end and the derivative at the other and so on But we cannot get away with just the value or just the derivative at one end this is insuf cient information Nor would it do to specify the derivatives at both ends this would either be redundant if the two are equal or inconsistent if they are not In two or three dimensions we are confronted by a partial differential equation and it is not so easy to see what would constitute aceeptable boundary conditions Is the shape of a taut rubber membrane for instance uniquely determined by the frame over which it is stretched or like a canning jar lid can it snap from one stable con guration to another The answer as I think your intuition would suggest is that V is uniquely determined by its value at the boundary canning jars evidently don t obey Laplace s equation However other boundary conditions can also be used see Prob 34 The proof that a proposed set of boundary conditions will suf ce is usually presented in the form of a uniqueness theorem There are many such theorems for electrostatics all sharing the same basic format I ll show you the two most useful ones3 First uniqueness theorem The solution to Laplace s equation in some volume V is uniquely determined if V is speci ed on the boundary surface 8 Proof In Fig 35 Ihave drawn such a region and its boundary There could also be islands inside so long as V is given on all their surfaces also the outer boundary could be at in nity where V is ordinarily taken to be zero Suppose there were two solutions to Laplace s equatiOn V2V1 0 and V2V2 0 both of which assume the speci ed value on the surface I want to prove that they must be equal The trick is look at their di erence V3 EVl V2 3 do not intend to prove the existence of solutions here that s a much more dif cult job In context the existence is generally clear on physical grounds 3 LAPLACE S EQUATION 117 V speci ed on this surface 8 Figure 35 This obeys Laplace s equation V2V3 vzvl V2V2 0 and it takes the value zero on all boundaries since V1 and V2 are equal there But Laplace s equation allows no local maxima or minima all extrema oc cur on the boundaries So the maximum and minimum of V3 are both zero Therefore V3 must be zero everywhere and hence V1 2 V2 qed Example 31 Show that the potential is constant inside an enclosure completely surrounded by conducting material provided there is no charge within the enclosure Solution The potential on the cavity wall is some constant V0 that s item iv in Sect 251 so the potential inside is a functiOn that satis es Laplace s equation and has the constant value V0 at the boundary It doesn t take a genius to think of one solution to this problem V V0 everywhere The uniqueness theorem guarantees that this is the only solution It follows that the eld inside an empty cavity is zero the same result we found in Sect 252 on rather different grounds The uniqueness theorem is a license to your imagination It doesn t matter how you come by your solution if a it satis es Laplace s equation and b it has the correct value on the boundaries then it s right You ll see the power of this argument when we come to the method of images Incidentally it is easy to improve on the rst uniqueneSS theorem I assumed there was no charge inside the region in question so the potential obeyed Laplace s equation but 1 18 CHAPTER 3 SPECIAL TECHNIQUES we may as well throw in some charge in which case V obeys Poisson s equation The argument is the same only this time V2V i VZV i l 0 2 pa 60 60 so 2 2 2 1 1 V V3V V1 V V2 0 00 60 60 Once again the di erence V3 E V1 V2 satis es Laplace s equation and has the value zero on all boundaries so V3 0 and hence V V2 Corollary The potential in a volume V is uniquely determined if a the charge density throughout the region and b the value of V on all boundaries are speci ed 316 Conductors and the Second Uniqueness Theorem The simplest way to set the boundary conditions for an electrostatic problem is to specify the value of V on all surfaces surrounding the region of interest And this situation often occurs in practice In the laboratory we have conductors connected to batteries which maintain a given potential or to ground which is the experimentalist s word for V 0 However there are other circumstances in which we do not know the potential at the boundary but rather the charges on various conducting surfaces Suppose I put charge Q1 on the rst conductor Q2 on the second and so on I m not telling you how the charge distributes itself over each conducting surface because as soon as I put it on it moves around in a way I do not control And for good measure let s say there is some speci ed charge density 0 in the region between the conduCtors Is the electric eld now uniquely determined Or are there perhaps a number of different ways the charges could arrange themselves on their respective conductors each leading to a different eld Second uniqueness theorem In a volume V surrounded by conductors and containing a speci ed charge density p the electric eld is uniquely determined if the total charge on each conductor is given Fig 36 The region as a whole can be bounded by another conductor or else unbounded Proof Suppose there are two elds satisfying the conditions of the problem Both obey Gauss s law in differential form in the space between the conductors l l 60 60 And both obey Gauss s law in integral form for a Gaussian surface enclosing each conductor 1 l Elda Qi E239d3 Qi 6O 6O ith conducting ith conducting surface surface 31 LAPLACE S EQUATION 119 Integration surfaces Outer boundary could be at in nity Figure 36 Likewise for the outer boundary whether this is just inside an enclosing con ductor or at in nity 1 l i E1 da 2 Qtot i E2 da Qtot 60 60 Outer outer boundary boundary As before we examine the difference E3 E E1 E2 which obeys V E3 0 37 in the region between the conductors and ing da 0 38 over each boundary surface Now there is one nal piece of information we must exploit Although we do not know how the charge Q distributes itself over the ith conducting surface we do know that each conductor is an equipotential and hence V3 is a constant not necessarily the some constant over each conducting surface It need not be zero for the potentials V1 and V2 may not be equal all we know for sure is that both are constant over any given conductor Next comes a trick Invoking product rule number 5 we nd that V V3133 V3ltV E3 E3 W E3gt2 120 CHAPTER 3 SPECIAL TECHNIQUES Here I have used Eq 37 and E3 V V3 Integrating this over the entire region between the conductors and applying the divergence theorem to the left side V V3E3dt i V3E3 da E32dt V S V The surface integral covers all boundaries of the region in question the con ductors and outer boundary Now V3 is a constant over each surface if the outer boundary is in nity V3 2 0 there so it comes outside each integral and what remains is zero according to Eq 38 Therefore E32dt 0 V But this integrand is never negative the only way the integral can vanish is if E3 0 everywhere Consequently E1 E2 and the theorem is proved This proof was not easy and there is a real danger that the theorem itself will seem more plausible to you than the proof In case you think the second uniqueness theorem is obvious consider this example of Purcell s Figure 37 shows a comfortable electrostatic con guration consisting of four conductors with charges EQ situated so that the plusses are near the minuses It looks very stable Now what happens if we join them in pairs by tiny wires as indicated in Fig 38 Since the positive charges are very near negative charges which is where they like to be you might well guess that nothing will happen the con guration still looks stable Well that sounds reasonable but it s wrong The con guration in Fig 38 is impossible For there are now effectively two conductors and the total charge on each is zero One possible way to distribute zero charge over these conductors is to have no accumulation of charge anywhere and hence zero eld everywhere Fig 39 By the second uniqueness theorem this must be the solution The charge will ow down the tiny wires canceling itself off 9 969 Figure 37 Figure 38 32 THE METHOD OF IMAGES 121 0 00 Figure 39 Problem 34 Prove that the eld is uniquely determined when the charge density p is given and either V or the normal derivative BVBn is speci ed on each boundary surface Do not assume the boundaries are conductors or that V is constant over any given surface Problem 35 A more elegant proof of the second uniqueness theorem uses Green s identity Prob 160c with T U V3 Supply the details 32 The Method of Images 321 The Classic Image Problem Suppose a point charge q is held a distance d above an in nite grounded conducting plane Fig 310 Question What is the potential in the region above the plane It s not just 1 4n60q 1 for q will induce a certain amount of negative charge on the nearby surface of the conductor the total potential is due in part to q directly and in part to this induced charge But how can we possibly calculate the potential when we don t know how much charge is induced or how it is distributed Figure 310 122 CHAPTER 3 SPECIAL TECHNIQUES Figure 311 From a mathematical point of view our problem is to solve Poisson s equation in the region z gt 0 with a single point charge q at 0 0 d subject to the boundary conditions 1 V 0 when z 0 since the conducting plane is grounded and 2 V gt 0 far from the charge that is for x2 y2 z2 gtgt d2 The rst uniqueness theorem actually its corollary guarantees that there is only one function that meets these requirements If by trick or clever guess we can discover such a function it s got to be the right answer Trick Forget about the actual problem we re going to study a completely different situation This new problem consists of two point charges q at 0 0 d and q at 0 0 d and no conducting plane Fig 311 For this con guration I can easily write down the potential 1 q 4 V 39 x y Z 47T60 sz y2 Z d2 x2 y2 Z The denominators represent the distances from x y z to the charges q and q respec tively It follows that 1 V0whenz0and 2 V gt0forx2yzz2 gtgtd2 and the only charge in the region z gt 0 is the point charge q at 0 0 d But these are precisely the conditions of the original problem Evidently the second con guration happens to produce exactly the same potential as the rst con guration in the upper region z 3 0 The loWer region z lt 0 is completely different but who cares The upper part is all we need Conclusion The potential of a point charge above an in nite grounded conductor is given by Eq 39 for z 2 0 Notice the crucial role played by the uniqueness theorem in this argument without it no one would believe this solution since it was obtained for a completely different charge distribution But the uniqueness theorem certi es it If it satis es Poisson s equation in the region of interest and assumes the correct value at the boundaries then it must be right 32 THE METHOD OF IMAGES 123 322 Induced Surface Charge Now that we know the potential it is a straightforward matter to compute the surface charge 0 induced on the conductor According to Eq 249 605 where 8 V8n is the normal derivative of V at the surface In this case the normal direction is the z direction so 6 8V 039 0 az z0 From Eq 39 gz 1 qu d mzm 3Z 47160 x2 y2 z 1232 x2 y2 z 1232 SO x gt qd a v 271x2 y2 d232 As expected the induced charge is negative assuming q is positive and greatest at x 0 While we re at it let s compute the total induced charge szada This integral over the x y plane could be done in Cartesian coordinates with da 2 dx d y but it s a little easier to use polar coordinates r with r2 x2 y2 and da r dr dab Then 310 4d 00quot 2nltr2d232 and 2 7T 00 00 qd qd d d 31 Q f0 f0 277r2d232r r as Vr2d20 q Evidently the total charge induced on the plane is q as with bene t of hindsight you can perhaps convince yourself it had to be 323 Force and Energy The charge q is attracted toward the plane because of the negative induced charge Let s calculate the force of attraction Since the potential in the vicinity of q is the same as in the analog problem the one with q and q but no conductor so also is the eld and therefore the force 1 q2 A 124 CHAPTER 3 SPECIAL TECHNIQUES Beware It is easy to get carried away and assume that everything is the same in the two problems Energy however is not the same With the two point charges and no conductor Eq 242 gives 1 q2 313 4713960 2d But for a single charge and conducting plane the energy is half of this 1 2 w q 314 47160 4d Why half Think of the energy stored in the elds Eq 245 W 6 0 f E2 dt 2 In the rst case both the upper region z gt 0 and the lower region z lt 0 contribute and by symmetry they contribute equally But in the second case only the upper region contains a nonzero eld and hence the energy is half as great Of course one could also determine the energy by calculating the work required to bring q in from in nity The force required to oppose the electrical force in Eq 312 is 14713960 q24z22 so d 1 d qz W f F dl f dz 00 4713960 00 4Z2 1 42 d 1 42 4713960 4z 00 4713960 4d 39 As I m0ve q toward the conductor I do work only on q It is true that induced charge is moving in over the conductor but this costs me nothing since the whole conductor is at potential zero By contrast if I simultaneously bring in two point charges with no conductor I do work on both of them and the total is twice as great 324 Other Image Problems The method just described is not limited to a single point charge any stationary charge distribution near a grounded conducting plane can be treated in the same way by introducing its mirror image hence the name method of images Remember that the image charges have the opposite sign this is what guarantees that the xy plane will be at potential zero There are also some exotic problems that can be handled in similar fashion the nicest of these is the following Example 32 A point charge q is situated a distance a from the center of a grounded Conducting sphere of radius R Fig 312 Find the potential outside the sphere 32 THE METHOD OF IMAGES 125 Figure 312 Figure 313 Solution Examine the completely di erent con guration consisting of the point charge q together with another point charge R 1 q 315 placed a distance R2 b 316 a to the right of the center of the sphere Fig 313 No conductor now just the two point charges The potential of this con guration is 1 q q Vr 47160 lt4 4 317 where a and L are the distances from q and 1 respectively Now it happens see Prob 37 that this potential vanishes at all points on the sphere and therefore ts the boundary conditions for our original problem in the exterior region Conclusion Eq 317 is the potential of a point charge near a grounded conducting sphere Notice that b is less than R so the image charge 1 is safely inside the sphere you cannot put image charges in the region where you are calculating V that would change 0 and you d be solving Poisson s equation with the wrong source In particular the force of attraction between the charge and the sphere is 2 Fi2 a 318 47160 a b2 47160 a2 R22 This solution is delightfully simple but extraordinarily lucky There s as much art as science in the method of images for you must somehow think up the right aUXiliary problem to look at The rst person who solved the problem this way cannot have known in advance what image charge q to use or where to put it Presumably he she started with an arbitrary charge at an arbitrary point inside the sphere calculated the potential on the sphere and then discovered that with q and b just right the potential on the sphere vanishes But it is really a miracle that any choice does the job with a cube instead of a sphere for example no single charge anywhere inside would make the potential zero on the surface CHAPTER 3 SPECIAL TECHNIQUES Z 3d q d 2q Figure 314 Problem 36 Find the force on the charge q in Fig 314 The xy plane is a grounded conductor Problem 37 a Using the law of cosines show that Eq 317 can be written as follows 1 q q Vr9 319 47750 r2a2 2ra c0s6 R2raR2 2ra c056 where r and 6 are the usual spherical polar coordinates with the z axis along the line through q In this form it is obvious that V 0 on the sphere r R b Find the induced surface charge on the sphere as a function of 6 Integrate this to get the total induced charge What should it be C Calculate the energy of this con guration Problem 38 In Ex 32 we assumed that the conducting sphere was grounded V 0 But with the addition of a second image charge the same basic model will handle the case of a sphere at any potential V0 relative of course to in nity What charge should you use and where should you put it Find the force of attraction between a point charge q and a neutral conducting sphere Problem 39 A uniform line charge A is placed on an in nite straight wire a distance d above a grounded conducting plane Let s say the wire runs parallel to the xaxis and directly above it and the conducting plane is the x y plane a Find the potential in the region above the plane b Find the charge density 0 induced on the conducting plane Problem 310 Two semiin nite grounded conducting planes meet at right angles In the region between them there is a point charge q situated as shown in Fig 315 Set up the image con guration and calculate the potential in this region What charges do you need and where should they be located What is the force on q How much work did it take to bring q in from in nity Suppose the planes met at some angle other than 90 would you still be able to solve the problem by the method of images If not for what particular angles does the method work 33 SEPARATION OF VARIABLES 127 y 9 cy 21 x V0 V0 Figure 315 Figure 316 Problem 311 Two long straight coppe r pipes each of radius R are held a distance 2d apart One is at potential V0 the other at V0 Fig 316 Find the potential everywhere Suggestiom Exploit the result of Prob 247 33 Separation of Variables In this section we shall attack Laplace s equation directly using the method of separation of variables which is the physicist s favorite tool for solving partial differential equations The method is applicable in circumstances where the potential V or the charge density a is speci ed on the boundaries of some region and we are asked to nd the potential in the interior The basic strategy is very simple We look for solutions that are products of functions each of which depends on only one of the coordinates The algebraic details however can be formidable so I m going to develop the method through a sequence of examples We ll start with Cartesian coordinates and then do spherical coordinates I ll leave the cylindrical case for you to tackle on your own in Prob 323 331 Cartesian Coordinates Example 33 Two in nite grounded metal plates lie parallel to the xz plane one at y 0 the other at y a Fig 317 The left end at x 0 is closed off with an in nite strip insulated from the two plates and maintained at a speci c potential V0 y Find the potential inside this slot Solution The con guration is independent of z so this is really a twodimensional problem In mathematical terms we must solve Laplace s equation 32 V 32 V CHAPTER 3 SPECIAL TECHNIQUES Figure 317 subject to the boundary conditions i V0wheny0 ii V 0 when y a iii V V0y when x 0 iv V gt0asx gtoo 321 The latter although not explicitly stated in the problem is necessary on physical grounds as you get farther and farther away from the hot strip at x 0 the potential should drop to zero Since the potential is speci ed on all boundaries the answer is uniquely determined The rst step is to look for solutions in the form of products Vx y XxYy 322 On the face of it this is an absurd restriction the overwhelming majority of solutions to Laplace s equation do not have such a form For example Vx y 5x 6y satis es Eq 320 but you can t express it as the product of a function x times a function y Obviously we re only going to get a tiny subset of all possible solutions by this means and it would be a miracle if one of them happened to t the boundary conditions of our problem But hang on because the solutions we do get are very special and it turns out that by pasting them together we can construct the general solution Anyway putting Eq 322 into Eq 320 we obtain d2X dZY W XE 0 The next step is to separate the variables that is collect all the xdependence into one term and all the ydependence into another Typically this is accomplished by dividing through by V 1 dZX 1 sz r a r cry 2 Here the rst term depends only on x and the second only on y in other words we have an equation of the form 323 f x gv 0 324 33 SEPARATION OF VARIABLES 129 Now there s only one way this could possibly be true f and g must both be constant For what if f x changed as you vary x then if we held y xed and ddled with x the sum fx gx would change in violation of Eq 324 which says it s always zero That s a simple but somehow rather elusive argument don t accept it without due thought because the whole method rides on it It follows from Eq 323 then that 1 d2X 1 d2Y EE 2 C1 and C2 Wlth C1 C2 One of these constants is positive the other negative or perhaps both are zero In general one must investigate all the possibilities however in our particular problem we need C1 positive and C2 negative for reasons that will appear in a moment Thus dZX 2 sz 2 kX k Y 326 dx2 d y2 Notice what has happened A partial differential equation 320 has been converted into two ordinary differential equations 326 The advantage of this is obvious ordinary differential equations are a lot easier to solve Indeed Xx M BK Yy Csin ky Dcosky so that Vx y Aekx BekxC sin ky Dcosky 327 This is the appropriate separable solution to Laplace s equation it remains to impose the boundary conditions and see what they tell us about the constants To begin at the end condition iv requires that A equal zero4 Absorbing B into C and D we are left with Vx y ekxC sin ky Dcosky Condition i now demands that D equal zero so Vx y Cekx sin ky 328 Meanwhile ii yields sin ka 2 0 from which it follows that nrt k nl23 329 a At this point you can see why I chose C1 positive and C2 negative If X were sinusoidal we could never arrange for it to go to zero at in nity and if Y were exponential we could not make it vanish at both 0 and a Incidentally n 0 is no good for in that case the potential vanishes everywhere And we have already excluded negative n s That s as far as we can go using separable solutions and unless V0 yjust happens to have the form sinnny a for some integer n we simply can t t the nal boundary condition at x 0 But now comes the crucial step that redeems the method Separation of variables has given us an in nite set of solutions one for each n and whereas none of them by itser satis es 4I m assuming k is positive but this involves no loss of generality negative k gives the same solution 327 only with the constants shuf ed A lt9 B C gt C Occasionally but not in this example k 0 must also be included see Prob 347 130 CHAPTER 3 SPECIAL TECHNIQUES the nal boundary condition it is possible to combine them in a way that does Laplace s equation is linear it the sense that if V1 V2 V3 satisfy it so does any linear combination V 011 V1 012 V2 013 V3 where 011 012 are arbitrary constants For V2V011V2V1012V2V20a100120 Exploiting this fact we can patch together the separable solutions 328 to construct a much more general solution 00 Vx y Z cneema sinnrtva 330 nl This still satis es the rst three boundary conditions the question is can we by astute choice of the coef cients C t the last boundary condition 00 m y 2 Cquot sintnnyai V0y 331 nl Well you may recognize this sum it s a Fourier sine series And Dirichlet s theorem5 guaran tees that virtually any function V0 y it can even have a nite number of diScontinuities can be expanded in such a series But how do we actually determine the coef cients C n buried as they are in that in nite sum The device for accomplishing this is so lovely it deserves a name I call it Fourier s trick though it seems Euler had used essentially the same idea somewhat earlier Here s how it goes Multiply Eq 331 by sinn Try a where n is a positive integer and integrate from 0 to a 00 a a Z CnO sinnnya sinn nyady 2 V0y sinn rtya dy 332 nl 0 You can work out the integral on the left for yourself the answer is a 0 if n 75 n sinnnya sinn rtya dy a 333 O E if n n Thus all the terms in the series drop out save only the one where n n and the left side of Eq 332 reduces to a2C Conclusionquot6 2 a Cquot O V0y s1nn7tyady 334 That does it Eq 330 is the solution with coef cients given by Eq 334 As a concrete example suppose the strip at x 0 is a metal plate with constant potential V0 remember it s insulated from the grounded plates at y 0 and y a Then 2V 0 2 0 ifn is even V C 0 s1nnnya dy Ol cosnrt 335 a 0 mt 4V0 ifn is odd nn 5 Boas M Mathematical Methods in the Physical Sciences 2nd ed New York John Wiley 1983 6For aesthetic reasons I ve dropped the prime Eq 334 holds for n l 2 3 and it doesn t matter obviously what letter you use for the dummy index 33 SEPARATION OF VARIABLES 131 08 0 02 0 0 04 08 02 2 0 yl xa Figure 318 Evidently 1 VO 4E Z e n xa sinmtya N n1 3 5 n Figure 318 is a plot of this potential Fig 319 shows how the rst few terms in the Fourier series combine to make a better and better approximation to the constant V0 a is n 1 only b includes n up to 5 c is the sum of the rst 10 terms and d is the sum of the rst 100 terms V Vo 0 02 04 06 08 1 ya Figure 319 132 CHAPTER 3 SPECIAL TECHNIQUES Incidentally the in nite series in Eq 336 can be summed explicitly try your hand at it if you like the result is 2V sin 71 a Vx y 9 tan 1 337 71 sinhnxa In this form it is easy to check that Laplace s equation is obeyed and the four boundary conditions 321 are satis ed The success of this method hinged on two extraordinary properties of the separable solutions 328 completeness and orthogonality A set of functions fn y is said to be complete if any other function f y can be expressed as a linear combination of them fy Z Cnfno 338 711 The functions sinmTya are complete on the interval 0 5 y 5 a It was this fact guaranteed by Dirichlet s theorem that assured us Eq 331 could be satis ed given the proper choice of the coef cients C The proof of completeness for a particular set of functions is an extremely dif cult business and I m afraid physicists tend to assume it s true and leave the checking to others A set of functions is orthogonal if the integral of the product of any two different members of the set is zero 0 mom dy 0 forn 7A n 339 The sine functions are orthogonal Eq 333 this is the property on which Fourier s trick is based allowing us to kill off all terms but one in the in nite series and thereby solve for the coef cients C Proof of orthogonality is generally quite simple either by direct integration or by analysis of the differential equation from which the functions came Example 34 Two in nitely long grounded metal plates again at y 0 and y a are connected at x ib by metal strips maintained at a constant potential V0 as shown in Fig 320 a thin layer of insulation at each comer prevents them from shorting out Find the potential inside the resulting rectangular pipe Solution Once again the con guration is independent of z Our problem is to solve Laplace s equation subject to the boundary conditions i V0wheny0 ii V 0wheny 51 iii V 2 V0 when x b iv V V0 when x b 340 33 SEPARATION OF VARIABLES 133 Figure 320 The argument runs as before up to Eq 327 V05 y Aekx BekxC sin ky Dcos ky This time however we cannot set A 0 the region in question does not extend to x 00 so ekx is perfectly acceptable On the other hand the situation is symmetric with respect to x so V x y Vx y and it follows that A B Using ekx ekx 2cosh kx and absorbing 2A into C and D we have Vx y cosh kx C sin ky Dcos ky Boundary conditions i and ii require as before that D 0 and k mra so Vx y Ccoshm39txa sinm39tya 341 Because V x y is even in x it will automatically meet condition iv if it ts iii It remains therefore to construct the general linear combination OO Vx y E C coshm39txa sinmTya n21 and pick the coef cients C in such a way as to satisfy condition iii 00 Vb y 2 C coshm39tba sinm39tya V0 nl This is the same problem in Fourier analysis that we faced before I quote the result from Eq 335 0 ifn is even Cn coshm39tba 4V0 n71 if n is odd CHAPTER 3 SPECIAL TECHNIQUES Figure 321 Conclusion The potential in this case is given by 4V0 1 coshnnxa VX y quot2135 n coshm39tba sinnnya 342 This function is shown in Fig 321 Example 35 An in nitely long rectangular metal pipe sides a and b is grounded but one end at x 0 is maintained at a speci ed potential V0 y z as indicated in Fig 322 Find the potential inside the pipe Solution This is a genuinely threedimensional problem 2 0 343 Figure 322 3 3 SEPARATION OF VARIABLES 135 136 subject to the boundary conditions i V0wheny0 ii V 0 when y a iii V 0 when z 0 iv V 0 when z b 33944 V V gt 0 as x gt 00 vi V V0y z when x 0 As always we look for solutions that are products Vx y z XxYyZz 345 Putting this into Eq 343 and dividing by V we nd lde 1d2Y 1d220 dez idy2 Zdz2quot It follows that 1d2x 1d2Y IdZZ C C ithC C C 0 dez iny2 zzdzz 3W 12 3 Our previous experience Ex 33 suggests that C1 must be positive C2 and C3 negative Setting C2 k2 and C3 12 we have C1 k2 12 and hence dZX d2Y dZZ 2 k2 12x k2Y T 422 z 346 dx dyz 2 Once again separation of variables has turned a partial differential equation into ordinary differential equations The solutions are Aexk2l2x Be xszzx CHAPTER 3 SPECIAL TECHNIQUES by appr0priate choice of the coef cients Cnsm To determine these constants we multiply by sinn nya sinm nzb where n and m are arbitrary positive integers and integrate nlm1 0 b a sinmrya sinnerady sinm7rzb sinmnzbdz 0 a b 2 V0yz sinn nya sinm397rzbdy dz 0 0 Quoting Eq 333 the left side is ab4Cnsm so 4 a b Cmm E V0yzsinnnya sinmJIzbdydz 350 O 0 Equation 348 with the coef cients given by Eq 350 is the solution to our problem For instance if the end of the tube is a conductor at constant potential V0 4V0 a b Cnm 2 s1nn7rya dy s1nmnzb dz ab 0 0 0 if n or m is even 2 351 16V 7i ifn and m are odd IT nm In this case 16V0 00 l 2 e H quota2mb2xs1nnnya sinmnzb 352 nm Vx y z T It nml35 Notice that the successive terms decrease rapidly a reasonable approximation would be ob tained by keeping only the rst few Xx Yy C sinky Dcos ky 22 2 Esinlz Fcoslz Boundary condition v implies A 0 i gives D 0 and iii yields F 0 whereas ii and iv require that k 2 mt a andl mnb where n and m are positive integers Combining the remaining constants we are left with Vx y z Ce quota2mb2x sinmrya sinmnzb 347 This solution meets all the boundary conditions except vi It contains two unspeci ed integers n and m and the most general linear combination is a double sum 00 CO Vx y z Z Cmme V quota2mb2x sinmrya sinmnzb 348 n1ml We hope to t the remaining boundary condition o0 oo V0 y z Z 2 CM sinmrya sinmnzb V0y z 349 nl m1 Problem 312 Find the potential in the in nite slot of Ex 33 if the boundary at x 0 consists of two metal strips one from y 0 to y a 2 is held at a c0nstant potential V0 and the other from y a2 to y a is at potential V0 Problem 313 For the in nite slot Ex 33 determine the charge density 0y on the strip at x 0 assuming it is a conductor at constant potential V0 Problem 314 A rectangular pipe running parallel to the zaxis from 00 to 00 has three grounded metal sides at y 0 y a and x 0 The fourth side at x b is maintained at a speci ed potential V0y a Develop a general formula for the potential within the pipe b Find the potential explicitly for the case V0 y V0 a constant Problem 315 A cubical box sides of length a consists of ve metal plates which are welded together and grounded Fig 323 The top is made of a separate sheet of metal insulated from the others and held at a constant potential V0 Find the potential inside the box 33 SEPARATION OF VARIABLES 137 Figure 323 332 Spherical Coordinates In the examples considered so far Cartesian coordinates were clearly appropriate since the boundaries were planes For round objects spherical coordinates are more natural In the spherical system Laplace s equation reads 18ltr28V 1 3 WW 1 32v 0 353 s1n r2 Br Br r2 sin6 86 86 r2 sin2 6 8 2 39 39 I shall assume the problem has azimuthal symmetry so that V is independent of 7 in that case Eq 353 reduces to 3 rzav 1 a 681 0 35 ar ar sin686 51 E 39 394 As before we look for solutions that are products Vr 6 Rr 6 355 Putting this into Eq 354 and dividing by V 1 1 erR l d 39 6 19 0 356 s n R dr dr 9 sin 6 d6 1 d Since the rst term depends only on r and the second only on 6 it follows that each must be a constant 1d erR ll1 1 d 39 6 d9 ll 1 35 Sln R dr dr s1n6 d6 d6 7 Here 1 1 is just a fancy way of writing the separation constant you ll see in a minute why this is convenient 7The general case for ii dependent potentials is treated in all the graduate texts See for instance I D Jackson s Classical Electrodynamics 3rd ed Chapter 3 New York John Wiley 1999 138 CHAPTER 3 SPECIAL TECHNIQUES As always separation of variables has converted a partial differential equation 354 into ordinary differential equations 357 The radial equation d 2dR 1 358 dr r dr 1 R has the general solution B W Arl m 359 as you can easily check A and B are the two arbitrary constants to be expected in the solution of a secondorder differential equation But the angular equation d d9 ltsin6 gt ll 1 sin6 9 360 is not so simple The solutions are Legendre polynomials in the variable cos 6 6 134cos6 361 P1x is most conveniently de ned by the Rodrigues formula 1 d l 2 Plx 2 ll ltE x 1 362 The rst few Legendre polynomials are listed in Table 31 P0x 1 Pfx x P2x ltst 1gt2 Pm 5x3 3x2 134x 35x4 30x2 38 P5x 63x5 70x3 15x8 Table 31 Legendre Polynomials Notice that Plx is as the name suggests an lthorder polynomial in x it contains only even powers if I is even and odd powers if I is odd The factor in front 121 was chosen in order that 1311 1 363 The Rodrigues formula obviously works only for nonnegative integer values of l More over it provides us with only one solution But Eq 360 is secondorder and it should pos sess two independent solutions for every value of l It turns out that these other solutions 33 SEPARATION OF VARIABLES 139 blow up at 6 0 andor 6 7t and are therefore unacceptable on physical grounds8 For instance the second solution forl O is 96 111 tan 2 364 You might want to check for yourself that this Satis es Eq 360 In the case of azimuthal symmetry then the most general separable solution to Laplace s equation consistent with minimal physiCal requirements is B Vr 6 ArZ gtP1cos6 r There was no need to include an overall constant in Eq 361 because it can be absorbed into A and B at this stage As before separation of variables yields an in nite set of solutions one for each I The general solution is the linear combination of separable solutions B Vr a 2 AM Plcos6 365 0 The following examples illustrate the power of this important result Example 36 The potential V0 6 is speci ed on the surface of a hollow sphere of radius R Find the potential inside the sphere Solution In this case B 0 for all l otherwise the potential would blow up at the origin Thus 00 Vr 9 Z AlrlPlcos6 366 0 At r R this must match the speci ed function V0 6 OO VR 9 Z AlRlPcos6 V06 367 0 Can this equation be satis ed for an appropriate choice of coef cients A 1 Yes The Legendre polynomials like the sines constitute a complete set of functions on the interval 1 f x 5 l 8In rare cases where the z axis is for some reason inaccessible these other solutions may have to be considered 140 CHAPTER 3 SPECIAL TECHNIQUES 0 f 6 5 IT How do we determine the constants Again by Fourier s trick for the Legendre polynomials like the sines are orthogonal functions9 11P1xPlxdx 2 f0 Plcos 6Plcos 6 sin6d6 0 ifl 751 368 677 Thus multiplying Eq 367 by Pl cos 6 sin 6 and integrating we have 1 2 Aer m 0 V06Plcos6s1n6d6 or 21 1 7 l V06Pl cos 6 sin6 d6 369 2Rl 0 Equation 366 is the solution to our problem with the coef cients given by Eq 369 In can be dif cult to evaluate integrals of the form 369 analytically and in practice it is often easier to solve Eq 367 by eyeball 10 For instance suppose we are told that the potential on the Sphere is V06 k sin262 370 where k is a constant Using the halfangle formula we rewrite this as k k V06 51 cos6 P0cos6 P1cos6 Putting this into Eq 367 we read off immediately that A0 k2 A1 k 2R and all other A l s vanish Evidently l Vr 9 g rOP0cos6 P1cos6l g 1 cos6 371 Example 37 The potential V0 6 is again speci ed on the surface of a sphere of radius R but this time we are asked to nd the potential outside assuming there is no charge there Solution In this case it s the A l s that must be zero or else V would not go to zero at 00 so 00 B Vr 6 Z ml lPlcos6 372 i0 9M Boas Mathematical Methods in the Physical Sciences 2nd ed Section 127 New York John Wiley 1983 10This is certainly true whenever V0 6 can be expressed as a polynomial in cos 6 The degree of the polynomial tells us the highest 1 we require and the leading coef cient determines the corresponding A Subtracting off AIRl Pl cos 6 and repeating the process we systematically work our way down to A0 Notice that if V0 is an even function of cos 6 then only even terms will occur in the sum and likewise for Odd functions 33 SEPARATION OF VARIABLES 141 At the surface of the sphere we require that 00 B VR 6 Z i lmcose V06 i0 Multiplying by Pl cos 9 sin 9 and integrating exploiting again the orthogonality relation 368 we have By 2 71V 9 P RlH ZII1 0 0 lcos9s1n9d9 or 21 1 Bl TR1 f0 V09P1cos9sin6d9 373 Equation 372 with the coef cients given by Eq 373 is the solution to our problem Example 38 An uncharged metal sphere of radius R is placed in an otherwise uniform electric eld E 2 E02 The eld will push positive charge to the northern surface of the sphere leaving a negative charge on the southern surface Fig 324 This induced charge in turn distorts the eld in the neighborhood of the sphere Find the potential in the region outside the sphere Solution The sphere is an equipotential we may as well set it to zero Then by symmetry the entire xy plane is at potential zero This time however V does not go to zero at large 1 In fact far from the sphere the eld is E02 and hence V gt E0zC Figure 324 142 CHAPTER 3 SPECIAL TECHNIQUES Since V 0 in the equatorial plane the constant C must be zero Accordingly the boundary conditions for this problem are i V 0 whenr R ii V gt E0r cos9 for r gtgt R 33974 We must t these boundary conditions with a function of the form 365 The rst condition yields A R1 Bl 0 l Rl1 or B AR2 1 375 so 00 l RZHl Vr 9 2A r rl1 Plcos 0 For r gtgt R the second term in parentheses is negligible and therefore condition ii requires that 00 Z AlrlPl cos 9 E0r cos 9 0 Evidently only one term is present I 1 In fact since P1 cos 9 cos 9 we can read off immediately A1 E0 all other Al s zero Conclusion R3 Vr 9 E0 r 2 cos9 376 r The rst term E0r cos 9 is due to the external eld the contribution attributable to the induced charge is evidently R3 E0 2 cos 9 r If you want to know the induced charge density it can be calculated in the usual way av R3 09 60 60E0 1 2 cos9 360E0 c0s9 377 3r rR r3 rR As expected it is positive in the northern hemisphere 0 5 9 5 712 and negative in the southern 712 5 9 5 71 Example 39 A speci ed charge density 00 9 is glued over the surface of a spherical shell of radius R Find the resulting potential inside and outside the Sphere Solution You could of course do this by direct integration 1 a V ida 47160 0 33 SEPARATION OF VARIABLES 143 but separation of variables is often easier For the interior region we have 00 Vr 9 Z AlrlP1cos9 r 5 R 378 0 no 3 terms they blow up at the origin in the exterior region Vr 9 Z mP cosB r 3 R 379 0 no A l terms they don t go to zero at in nity These two functions must be joined together by the appropriate boundary conditions at the surface itself First the potential is continuous at r R Eq 234 00 00 B l 1 AR Pcos9 i Wmoose 380 0 It follows that the coef cients of like Legendre polynomials are equal 3 AR2 1 381 To prove that formally multiply both sides of Eq 380 by P1 cos 9 sin 9 and integrate from 0 to IT using the orthogonality relation 368 Second the radial derivative of V suffers a discontinuity at the surface Eq 236 Blout aVin Br Br B 00 1 Zl1RT r2Pcos9 ZIAZRl l Plcos9 a009 0 0 l 2 009 382 60 r Thus or using Eq 381 00 1 Z21A1Rl 1Pcos9 2 009 383 10 60 From here the coef cients can be determined using Fourier s trick 1 7r 2 009P1cos6 sin9d9 384 Al Equations 378 and 379 constitute the solution to our problem with the coef cients given by Eqs 381 and 384 For instance if 009 kcos9 kP1 cos 9 385 for some constant k then all the A s are zero except forl l and A k P 92 39 9d9 k COS SH 2 1 260 0 1 360 144 CHAPTER 3 SPECIAL TECHNIQUES The potential inside the sphere is therefore k Vr 9 rcos9 r 5 R 386 360 whereas outside the sphere kR3 1 Vr 9 cos9 r 3 R 387 360 r2 In particular if 00 9 is the induced charge on a metal sphere in an external eld E02 so that k 360E0 Eq 377 then the potential inside is EOr cos9 Eoz and the eld is E02 exactly right to cancel off the external eld as of course it should be Outside the sphere the potential due to this surface charge is R3 E0 2 COS 9 r consistent with our conclusion in Ex 38 Problem 316 Derive P3 x from the Rodrigues formula and check that P3 cos 9 satis es the angular equation 360 forl 3 Check that P3 and P1 are orthogonal by explicit integration Problem 317 a Suppose the potential is a constant V0 over the surface of the sphere Use the results of Ex 36 and Ex 37 to nd the potential inside and outside the sphere Of course you know the answers in advance this is just a consistency check on the method b Find the potential inside and outside a spherical shell that carries a uniform surface charge 00 using the results of Ex 39 Problem 318 The potential at the surface of a sphere radius R is given by V0 2 k cos 39 where k is a constant Find the potential inside and outside the sphere as well as the surface charge density a 6 on the sphere Assume there s no charge inside or outside the sphere Problem 319 Suppose the potential V0 9 at the surface of a sphere is speci ed and there is no charge inside or outside the Sphere Show that the charge density on the sphere is given by 00 60 2 09 ZQH 1 CPcos9 388 0 where C V09Plcos9sin9d9 389 O 33 SEPARATION OF VARIABLES 145 Problem 320 Find the potential outside a charged metal sphere charge Q radius R placed in an otherwise uniform electric eld E0 Explain clearly where you are setting the zero of potential Problem 321 In Prob 225 you found the potential on the axis of a uniformly charged disk Vr 0 i xrz R2 r 0 a Use this together with the fact that P1l l to evaluate the rst three terms in the expansion 372 for the potential of the disk at points off the axis assuming r gt R b Find the potential for r lt R by the same method using 366 Notes You must break the interior region up into two hemispheres above and below the disk Do not assume the coef cients A are the same in both hemispheres Problem 322 A spherical shell of radius R carries a uniform surface charge 00 on the north ern hemisphere and a uniform surface charge 00 on the southern hemisphere Find the potential inside and outside the sphere calculating the coef cients explicitly up to A6 and B6 Problem 323 Solve Laplace s equation by separation of variables in cylindrical coordinates assuming there is no dependence on z cylindrical symmetry Make sure you nd all solutions to the radial equation in particular your result must accommodate the case of an in nite line charge for which of course we already know the answer Problem 324 Find the potential outside an in nitely long metal pipe of radius R placed at right angles to an otherwise uniform electric eld E0 Find the surface charge induced on the pipe Use your result from Prob 323 Problem 325 Charge density 0 a sin 5 gt where a is a constant is glued over the surface of an in nite cylinder of radius R Fig 325 Find the potential inside and outside the cylinder Use your result from Prob 323 Figure 325 146 CHAPTER 3 SPECIAL TECHNIQUES 34 Multipole Expansion 341 Approximate Potentials at Large Distances If y0u are very far away from a localized charge distribution it looks like a point charge and the potential is to good approximation 1 47160Q r where Q is the total charge We have often used this as a check on formulas for V But what if Q is zero You might reply that the potential is then approximately zero and of course you re right in a sense indeed the potential at large r is pretty small even if Q is not zero But we re looking for something a bit more informative than that Example 310 A physical electric dipole consists of two equal and opposite charges lzq separated by a distance d Find the approximate potential at points far from the dipole Solution Let 7L be the distance from q and L the distance from q Fig 326 Then 1 q q Vr 47160 4 L and from the law of cosines d d2 4 r2 d22 rdcos6 r2lt12F 7 cos6 We re interested in the regime r gtgt d so the third term is negligible and the binomial expansion yields 12 l d i211212icos6 E li cosB 4i r r r 2r Thus 1 1 d cos 6 4 7L r2 and hence I qd COS 6 Vr E 2 390 47160 r Figure 326 34 M ULTIPOLE EXPANSION 147 11 o 0 4 Monopole Dipole Quadrupole Octopole V lr V lrz V1r3 V1r4 Figure 327 Evidently the potential of a dipole goes like 1 r2 at large r as we might have anticipated it falls off more rapidly than the potential of a point charge Incidentally if we put together a pair of equal and opposite dipoles to make a quadrupole the potential goes like lr3 for back to back quadrupoles an octopole it goes like 1r4 and so on Figure 327 summarizes this hierarchy for completeness I have included the electric monopole point charge whose potential of course goes like 1 r Example 310 pertained to a very special charge con guration I propose now to develop a systematic expansion for the potential of an arbitrary localized charge distribution in powers of 1 r Figure 328 de nes the appropriate variables the potential at r is given by l 1 Vr F60 prdr 391 Using the law of cosines 2 42 r2 r 2 2rr cos9 1 2ll L 2 L COSQ r r Ii rv1 E 392 r r E E gt lt 2cos6 r r For points well outside the charge distribution 6 is much less than 1 and this invites a binomial expansion 0139 where 1 1 1 1 3 5 1 12 1 23 L r 6 rlt 2686 166 393 Figure 328 148 CHAPTER 3 SPECIAL TECHNIQUES or in terms of r r and 9 lr t ll I 39I ra l MM l A 11 V A l N 0 O V a V gton A l V A l I N 0 O I a V N 2 11 C cose C 3 cosze 12 r r r r 3 3 l lt7 5cos 6 3cos62 In the last step I have collected together like powers of r r surprisingly their coef cients the terms in parentheses are Legendre polynomials The remarkable result11 is that 1 1 r quot Z Z gt Pncos9 394 n0 r where 6 is the angle between 1 and r Substituting this back into Eq 391 and noting that r is a constant as far as the integration is concerned I conclude that 1 1 Vr mgrm H r Pncos6 pr a t 395 or more explicitly 1 1 l 1 I l l Vl 0rdr r cos6prdr 47160 r r2 1 l2 3 2 1 7 r cos6 prdr 396 r 2 2 This is the desired result the multipole expansion of V in powers of l r The rst term n 0 is the monopole contribution it goes like 1 r the second n l is the dipole it goes like 1 r2 the third is quadrupole the fourth octopole and so on As it stands Eq 395 is exact but it is useful primarily as an approximation scheme the lowest nonzero term in the expansion provides the approximate potential at large r and the successive terms tell us how to improve the approximation if greater precision is required Incidentally this affords a second way of obtaining the Legendre polynomials the rst being Rodn39gues formula 1 L is called the generating function for Legendre polynomials 34 M ULTIPOLE EXPANSION 149 Problem 326 A sphere of radius R centered at the origin carries charge density R 0r v9 k 2R 2r s1n6 r where k is a constant and r 6 are the usual spherical coordinates Find the approximate potential for points on the z axis far from the sphere 342 The Monopole and Dipole Terms Ordinarily the multipole expansion is dominated at large r by the monopole term Q Vmon139 41160 7 397 where Q f 0 dr is the total charge of the con guration This is just what we expected for the approximate potential at large distances from the charge Incidentally for a point charge at the origin Vmon represents the exact potential everywhere not merely a rst approximation at large r in this case all the higher multipoles vanish If the total charge is zero the dominant term in the potential will be the dipole unless of course it also vanishes 1 1 Vdipr m r cose pr dt Since 6 is the angle between Iquot and r Fig 328 r cos6l f r and the dipole potential can be written more succinctly 1 1A Vdip mgrWW dri This integral which does not depend on r at all is called dipole moment of the distribution pE fr pr dr 398 and the dipole contribution to the potential simpli es to 399 150 CHAPTER 3 SPECIAL TECHNIQUES The dipole moment is determined by the geometry size shape and density of the charge distribution Equation 398 translates in the usual way Sect 214 for point line and surface charges Thus the dipole moment of a collection of point charges is p qir 3100 For the physical dipole equal and opposite charges tq pqr3 qr qr rL qd 3101 where dis the vector from the negative charge to the positive one Fig 329 Is this consistent with what we got for a physical dipole in Ex 310 Yes If you put Eq 3100 into Eq 399 you recover Eq 390 Notice however that this is only the approximate potential of the physical dipole evidently there are higher multipole contributions Of course as you go farther and farther away Vdip becomes a better and better approximation since the higher terms die off more rapidly with increasing r By the same token at a xed r the dipole approximation impr0ves as you shrink the separation d To construct a pure dipole whose potential is given exactly by Eq 399 y0u d have to let d approach zero Unfortunately you then lose the dipole term too unless you simultaneously arrange for q to go to in nity A physical dipole becomes a pure dipole then in the rather arti cial limit d gt 0 q gt 00 with the product qd p held xed When someone uses the word dipole you can t always tell whether they mean a physical dipole with nite separation between the charges or a pure point dipole If in doubt assume that d is small enough compared to r that you can safely apply Eq 399 Dipole moments are vectors and they add accordingly if you have two dipoles p1 and p2 the total dipole moment is p1 p2 For instance with four charges at the corners of a square as shown in Fig 330 the net dipole moment is zero You can see this by combining the charges in pairs vertically t T 0 or horizontally gt lt 0 or by adding up the four contributions individually using Eq 3100 This is a quadrupole as I indicated earlier and its potential is dominated by the quadrupole term in the multipole expansion 4 4 Figure 329 Figure 330 34 M ULTIPOLE EXPANSION 151 Figure 331 Problem 327 Four particles one of charge q one of charge 3g and two of charge 2q are placed as shown in Fig 331 each a distance a from the origin Find a simple approximate formula for the potential valid at points far from the origin Express your answer in spherical coordinates Problem 328 In Ex 39 we derived the exact potential for a spherical shell of radius R which carries a surface charge a k cos 6 a Calculate the dipole moment of this charge distribution b Find the approximate potential at points far from the sphere and compare the exact answer 387 What can you conclude about the higher multipoles Problem 329 For the dipole in Ex 310 expand l Wig to order d r3 and use this to determine the quadrupole and octopole terms in the potential 343 Origin of Coordinates in Multipole Expansions I mentioned earlier that a point charge at the origin constitutes a pure monopole If it is not at the origin it s no longer a pure monopole For instance the charge in Fig 332 has a dipole moment p qdy and a corresponding dipole term in its potential The monopole potential 1 4JT 0q r is not quite correct for this con guration rather the exact potential is l 4ne0q L The multipole expansion is remember a series in inverse powers of r the distance to the origin and when we expand lIi we get all powers not just the rst So moving the origin or what amounts to the same thing moving the charge can radically alter a multipole expansion The monopole moment Q does not change since the total charge is obviously independent of the coordinate system In Fig 332 the monopole term was unaffected when we moved q away from the origin it s just that it was no longer the whole story a dipole term and for that matter all higher poles appeared as well Ordinarily the dipole moment does change when you shift the origin but there is an important exception If the total charge is zero then the dipole moment is independent of 152 CHAPTER 3 SPECIAL TECHNIQUES lt d39c gtlt Figure 332 Figure 333 the choice of origin For suppose we displace the origin by an amount a Fig 333 The new dipole moment is then quot5 ll fiquotor d1rl r aor dr fr or drl apr dr p Qa II In particular if Q 0 then p p So if someone asks for the dipole moment in Fig 334a you can answer with con dence qd but if you re asked for the dipole moment in Fig 334b the appropriate response would be With respect to what originquot aa d 0 q q qaq a b Figure 334 Problem 330 Two point charges 3g and q are separated by a distance a For each of the arrangements in Fig 335 nd i the monopole moment ii the dipole moment and iii the approximate potential in spherical coordinates at large r include both the monopole and dipole contributions 34 M ULTIPOLE EXPANSION 153 z z z 3c a 361 q q a y a 3 3g y x x q x a b C Figure 335 344 The Electric Field of a Dipole So far we have worked only with potentials Now I would like to calculate the electric eld of a pure dipole If we choose coordinates so that p lies at the origin and points in the z direction Fig 336 then the potential at r 6 is Eq 399 I p p cos9 V 9 3102 dlpr 4n60r2 4ne0r2 To get the eld we take the negative gradient of V 8V 2p cos 6 Er 9 8r 471601quot3 1 8 V p sin 6 E0 a r 86 471601quot3 1 8 V E 0 r s1n 6 21 Thus Emma p 3 2cos9rsine 3103 47160quot Figure 336 154 CHAPTER 3 SPECIAL TECHNIQUES This formula makes explicit reference to a particular coordinate system spherical and assumes a particular orientation for p along 2 It can be recast in a coordinate free form analogous to the potential in Eq 399 see Prob 333 Notice that the dipole eld falls off as the inverse cube of r the monopole eld Q 4neor2f goes as the inverse square of course Quadrupole elds go like 1r4 oc topole like 1 r5 and so on This merely re ects the fact that monopole potentials fall off like 1 r dipole like 1 r2 quadrupole like 1r3 and so on the gradient introduces another factor of 1r Figure 337a shows the eld lines of a pure dipole Eq 3103 For comparison I have also sketched the eld lines for a physical dipole in Fig 337b Notice how similar the two pictures become if you blot out the central region up close however they are entirely different Only for points r gtgt d does Eq 3103 represent a valid approximation to the eld of a physical dipole As I mentioned earlier this regime can be reached either by going to large r or by squeezing the charges very close together12 Z Z a Field of a quotpurequot dipole a Field of a quotphysicalquot dipole Figure 337 Problem 331 A pure dipole p is situated at the origin pointing in the z direction a What is the force on a point charge q at a 0 0 Cartesian coordinates b What is the force on q at 0 0 a c How much work does it take to move q from a 0 0 to O 0 a 12Even in the limit there remains an in nitesimal region at the origin where the eld of a physical dipole points in the wrong direction as you can see by walking down the z axis in Fig 335b If you want to explore this subtle and important point work Prob 342 156 34 M ULTIPOLE EXPANSION 1 5 5 Figure 338 Problem 332 Three point charges are located as shown in Fig 338 each a distance a from the origin Find the approximate electric eld at points far from the origin Express your answer in spherical coordinates and include the two lowest orders in the multipole expansion Problem 333 Show that the electric eld of a pure dipole Eq 3103 can be written in the coordinate free form 1 l A Edipm m amor p 3104 More Problems on Chapter 3 Problem 334 A point Charge 1 of mass m is released from rest at a distance d from an in nite grounded conducting plane How long will it take for the Charge to hit the plane Answer ndq27reomd Problem 335 Two in nite parallel grounded conducting planes are held a distance a apart A point charge 1 is placed in the region between them a distance x from one plate Find the force on 11 Check that your answer is correct for the special cases a gt 00 and x 112 Obtaining the induced surface is not so easy See B G Dick Am J Phys 41 1289 1973 M Zahn Am J Phys 44 1132 1976 J Pleines and S Mahajan Am J Phys 45 868 1977 and Prob 344 below Problem 336 Two long straight wires carrying opposite uniform line Charges le are situated on either side of a long conducting cylinder Fig 339 The cylinder which carries no net charge has radius R and the wires are a distance a from the axis Find the potential at point r A 1 s2 a2 Zsa cos 1saR2 R2 Zsa cos 1 n 47160 s2 a2 2sa cos saR2 R2 2m cos 45 Answen39 Vs 1 Problem 337 A conducting sphere of radius a at potential V0 is surrounded by a thin con centric spherical shell of radius b over which someone has glued a surface charge 76 kcosd CHAPTER 3 SPECIAL TECHNIQUES A Figure 339 Figure 340 where k is a constant and 9 is the usual spherical coordinate a Find the potential in each region i r gt b and ii a lt r lt b b Find the induced surface charge 019 0n the conductor c What is the total Charge of this system Check that your answer is consistent with the behavior of V at large r aVOr b3 a3kcos93r260 r 2 b Answer Vr 9 aVOr r3 a3kcos93r260 r 5 b Problem 338 A Charge Q is distributed uniformly along the z axis from z a to z a Show that the electric potential at a point r is given by q 1 1 a 2 l a 4 V 9 l P r 47160 r 3 r 2COS9 5 r P4COSQ forr gt a Problem 339 A long cylindrical shell of radius R carries a uniform surface charge 00 on the upper half and an opp0site charge 00 on the lower half Fig 340 Find the electric potential inside and outside the cylinder Problem 340 A thin insulating rod running from z a to z 11 carries the indicated line charges In each case nd the leading term in the multipole expansion of the potential a A kc0s7rz2a b A k sin7rza c A kcos7rza where k is a constant Problem 341 Show that the average eld inside a sphere of radius R due to all the charge within the sphere is 1 p E 4m R3 3 105 where p is the total dipole moment There are several ways to prove this delightfully simple result Here s one method 3 4 M ULTIPOLE EXPANSION 157 a Show that the average eld due to a single charge 1 at point r inside the sphere is the same as the eld at r due to a uniformly charged sphere with 0 q7TR3 namely 1 l A 4 iznd 4760 jnR3 I where is the vector from r to dT b The latter can be found from Gauss s law see Prob 212 Express the answer in terms of the dipole moment of q C Use the superposition principle to generalize to an arbitrary charge distribution d While you re at it show that the average eld over the sphere due to all the charges outside is the same as the eld they produce at the center Problem 342 Using Eq 3103 calculate the average electric eld of a dipole over a spherical volume of radius R centered at the origin Do the angular intervals rst Note39 You must express 1quot and 5 in terms of 2 y and 2 see back cover before integrating If you don t understand why reread the discussion in Sect 141 Compare your answer with the general theorem Eq 3105 The discrepancy here is related to the fact that the eld of a dipole blows up at r 0 The angular integral is zero but the radial integral is in nite so we really don t know what to make of the answer To resolve this dilemma let s say that Eq 3103 applies outside a tiny sphere of radius e its contribution to E ave is then unambiguously zero and the whole answer has to come from the eld inside the e sphere b What must the eld inside the esphere be in order for the general theorem 3105 to hold Hint39 since 6 is arbitrarily small we re talking about something that is in nite at r 0 and whose integral over an in nitesimal volume is nite Answer p36083r Evidently the true eld of a dipole is E r 1 1 3 r r 1 531 3106 dlp 47160 r3 p p 360p You may well wonder how we missed the delta function term when we calculated the eld back in Sect 344 The answer is that the differentiation leading to Eq 3103 is perfectly valid except at r 0 but we should have known from our experience in Sect 151 that the point r 0 is problematic See C P Frahm Am J Phys 51 826 1983 or more recently R Estrada and R P Kanwal Am J Phys 63 278 1995 For further details and applications see D J Grif ths Am J Phys 50 698 1982 Problem 343 a Suppose a charge distribution p r produces a potential V1 r and some other charge dis tribution p2l produces a potential V2r The two situations may have nothing in common for all I care perhaps number 1 is a uniformly charged sphere and number 2 is a parallel plate capacitor Please understand that p1 and p2 are not present at the same time we are talking about two di erent problems one in which only 0 is present and another in which only p2 is present Prove Green s reciprocity theorem p1V2d12 ngldr all space all space CHAPTER 3 SPECIAL TECHNIQUES a b Figure 341 Hints Evaluate E E2 dr two ways rst writing E1 V V and using integrationby parts to transfer the derivative to E2 then writing E2 VV2 and transferring the derivative 0 E b Suppose now thatyou have two separated conductors Fig 341 If you charge up conductor a by amount Q leaving b uncharged the resulting potential of b is say Vab On the other hand if you put that same charge Q on conductor b leaving a uncharged the potential of a would be Vba Use Green s reciprocity theorem to show that Va Vim an astonishing result since we assumed nothing about the shapes or placement of the conductors Problem 344 Use Green s reciprocity theorem Prob 343 to solve the following two prob lems Hint39 for distribution 1 use the actual situation for distribution 2 remove 11 and set one of the conductors at potential V0 a Both plates of a parallelplate capacitor are grounded and a point charge 11 is placed between them at a distance x from plate 1 The plate separation is d Find the induced charge on each plate Answer Q1 2 qxd 1 Q2 2 qxd b Two concentric spherical conducting shells radii a and b are grounded and a point charge 1 is placed between them at radius r Find the induced charge on each sphere Problem 345 a Show that the quadrupole term in the multipole expansion can be written 3 1 1 unadr 4 7160 2r3 H on Qtj where Qij E 3rir r 26ijprl dr Here 1 ifi j 51739 0 ifi j is the Kronecker delta and Qij is the quadrupole moment of the charge distribution Notice the hierarchy ZA39PI 1 iththj 1 V V l V mon 47160 r dlp 47160 r2 quad 47160 r3 3 4 M ULTIPOLE EXPANSION 159 The monopole moment Q is a scalar the dipole moment p is a vector the quadrupole moment Qij is a secondrank tensor and so on b Find all nine components of Qij for the con guration in Fig 330 assume the square has side a and lies in the xy plane centered at the origin C Show that the quadrupole moment is independent of origin if the monopole and dipole moments both vanish This works all the way up the hierarchy the lowest nonzero multipole moment is always independent of origin d How would you de ne the octopole moment Express the octopole term in the multipole expansion in terms of the octopole moment Problem 346 In Ex 38 we determined the electric eld outside a spherical conductor radius R placed in a uniform external eld E0 Solve the problem now using the method of images and check that your answer agrees with Eq 376 Hint39 Use Ex 32 but put another charge q diametrically opposite 1 Let a gt 00 with 1471602qa2 2 E0 held constant Problem 347For the in nite rectangular pipe in Ex 34 suppose the potential on the bottom y 0 and the two sides x lb is zero but the potential on the top y a is a nonzero constant V0 Find the potential inside the pipe N0te This is a rotated version of Prob 3 14b but set it up as in Ex 34 using sinusoidal functions in y and hyperbolics in x It is an unusual case in which k 0 must be included Begin by nding the general solution to Eq 326 when k 0 For further discussion see S Hassani Am J Phys 59 470 1991 Answer VO 221 11 b g1 x sinn7Tya Problem 348 a A long metal pipe of square cross section side a is grounded on three sides while the fourth which is insulated from the rest is maintained at constant potential V0 Find the net charge per unit length on the side opposite to V0 Hint Use your answer to Prob 314 or Prob 347 b A long metal pipe of circular crosssection radius R is divided lengthwise into four equal sections three of them grounded and the fourth maintained at constant pbtential V0 Find the net charge per unit length on the section opposite to V0 Answer to both a and b A 60V0 In 2113 Problem 349 An ideal electric dipole is situated at the origin and points in the z direction as in Fig 336 An electric charge is released from rest at a point in the xy plane Show that it swings back and forth in a semicircular arc as though it were a pendulum supported at the origin This charming result is due to R S Jones Am J Phys 63 1042 1995 13These are special cases of the ThompsonLampard theorem see J D Jackson Am J Phys 67 107 1999 Chapter 4 Electric Fields in Matter L1 Polarization 411 Dielectrics In this chapter we shall study electric elds in matter Matter of course comes in many varieties solids liquids gases metals woods glasses and these substances do not all respond in the same way to electrostatic elds Nevertheless most everyday objects belong at least in good approximation to one of two large classes conductors and insulators or dielectrics We have already talked about conductors these are substances that contain an unlimited supply of charges that are free to move about through the material In practice what this ordinarily means is that many of the electrons one or two per atom in a typical metal are not associated with any particular nueleus but roam around at will In dielectrics by contrast all charges are attached to speci c atoms or molecules they re on a tight leash and all they can do is move a bit within the atom or molecule Such microscopic displacements are not as dramatic as the wholesale rearrangement of charge in a conductor but their cumulative effects account for the characteristic behavior of dielectric materials There are actually two principal mechanisms by which electric elds can distort the charge distribution of a dielectric atom or molecule stretching and rotating In the next two sections I ll discuss these processes 412 Induced Dipoles What happens to a neutral atom when it is placed in an electric eld E Your rst guess might well be Absolutely nothing since the atom is not charged the eld has no effect on it But that is incorrect Although the atom as a whole is electrically neutral there is a positively charged core the nucleus and a negatively charged electron cloud surrounding it These two regions of charge within the atom are in uenced by the eld the nucleus is pushed in the direction of the eld and the electrons the opposite way In principle if the eld is large enough it can pull the atom apart completely ionizing it the substance 160 4 1 POLARIZATION 1 6 1 then becomes a conductor With less extreme elds however an equilibrium is soon established for if the center of the electron cloud does not coincide with the nucleus these positive and negative charges attract one another and this holds the atoms together The two opposing forces E pulling the electrons and nucleus apart their mutual attraction drawing them together reach a balance leaving the atom polarized with plus charge shifted slightly one way and minus the other The atom now has a tiny dipole moment p which points in the same direction as E Typically this induced dipole moment is approximately proportional to the eld as long as the latter is not too strong p 2 dB 41 The constant of proportionality at is called atomic polarizability Its value depends on the detailed structure of the atom in question Table 41 lists some experimentally determined atomic polarizabilities H He Li Be C Ne Na Ar K Cs 0667 0205 243 560 176 0396 241 164 434 596 Table 41 Atomic Polarizabilities oz4760 in units of 1030 m3 Source Handbook of Chemistry and Physics 78th ed Boca Raton CRC Press Inc 1997 Example 41 A primitive model for an atom consists of a point nucleus 11 surrounded by a uniformly charged spherical cloud q of radius a Fig 41 Calculate the atomic polarizability of such an atom Solution In the presence of an external eld E the nucleus will be shifted slightly to the right and the electron cloud to the left as shown in Fig 42 Because the actual displacements 11 Figure 41 Figure 42 162 CHAPTER 4 ELECTRIC FIELDS IN MATTER involved are extremely small as you ll see in Prob 41 it is reasonable to assume that the electron cloud retains its spherical shape Say that equilibrium occurs when the nucleus is displaced a distance d from the centei of the sphere At that point the external eld pushing the nucleus to the right exactly balances the internal eld pulling it to the left E Ee where Ee is the eld produced by the electron cloud Now the eld at a distance d from the center of a uniformly charged sphere is 1 qd Ee 47160 a3 Prob 212 At equilibrium then 1 E t 47160 a3 or p qd 4neoa3E The atomic polarizability is therefore at 4neoa3 3601 42 where v is the volume of the atom Although this atomic model is extremely crude the result 42 is not too bad it s accurate to within a factor of four or so for many simple atoms For molecules the situation is not quite so simple because frequently they polarize more readily in some directions than others Carbon dioxide Fig 43 for instance has a polarizability of 45 X 10 40 C2mN when you apply the eld along the axis of the molecule but only 2 X 10 40 for elds perpendicular to this direction When the eld is at some angle to the axis you must resolve it into parallel and perpendicular components and multiply each by the pertinent polarizability pOtJELOtHEp In this case the induced dipole moment may not even be in the same direction as E And C02 is relatively simple as molecules go since the atoms at least arrange themselves in a straight line for a completely asymmetrical molecule Eq 41 is replaced by the most general linear relation between E and p quotU x 2 at E axy Ey aszz y or E anyy OtyZEZ 43 72 xx Ex azy Ey ZZ Ez 6 Figure 43 4 I POLARIZATION l 63 The set of nine constants or constitute the polarizability tensor for the molecule Their actual values depend on the orientation of the axes you chose though it is always possible to choose principal axes such that all the off diagonal terms axy Olzx etc vanish leaving just three nonzero polarizabilities or Olyy and an Problem 41 A hydrogen atom with the Bohr radius of half an angstrom is situated between two metal plates 1 mm apart which are connected to opposite terminals of a 500 V battery What fraction of the atomic radius does the separation distance d amount to roughly Estimate the voltage you would need with this apparatus to ionize the atom Use the value of oz in Table 41 Moral The displacements we re talking about are minute even on an atomic scale Problem 42 According to quantum mechanics the electron cloud for a hydrogen atom in the ground state has a charge density q pr 36 2 7TH where q is the charge of the electron and a is the Bohr radius Find the atomic polarizability of such an atom Hint First calculate the electric eld of the electron cloud E e r then expand the exponential assuming r ltlt a For a more sophisticated approach see W A Bowers Am J Phys 54 347 1986 Problem 43 According to Eq 41 the induced dipole moment of an atom is proportional to the external eld This is a rule of thumb not a fundamental law and it is easy to concoct exceptions in theory Suppose for example the charge density of the electron cloud were proportional to the distance from the center out to a radius R To what power of E would p be proportional in that case Find the condition on pr such that Eq 41 will hold in the weak eld limit Problem 44 A point charge q is situated a large distance r from a neutral atom of polarizability or Find the force of attraction between them 413 Alignment of Polar Molecules The neutral atom discussed in Sect 412 had no dipole moment to start with p was induced by the applied eld Some molecules have builtin permanent dipole moments In the water molecule for example the electrons tend to cluster around the oxygen atom Fig 44 and since the molecule is bent at 105 this leaves a negative charge at the vertex and a net positive charge at the opposite end The dipole moment of water is unusually large 61 X 10 30 Cm in fact this is what accounts for its effectiveness as a solvent What happens when such molecules called polar molecules are placed in an electric eld 164 CHAPTER 4 ELECTRIC FIELDS IN MATTER Figure 44 Figure 45 If the eld is uniform the force on the positive end F qE exactly cancels the force on the negative end F qE Fig 45 However there will be a torque N r X F r X F d2 X qE d2 X qE qd X E Thus a dipole p qd in a uniform eld E experiences a torque 44 Notice that N is in such a direction as to line p up parallel to E a polar molecule that is free to rotate will swing around until it points in the direction of the applied eld If the eld is nonuniform so that F does not exactly balance F there will be a net force on the dipole in addition to the torque Of course E must change rather abruptly for there to be signi cant variation in the space of one molecule so this is not ordinarily a major consideration in discussing the behavior of dielectrics Nevertheless the formula for the force on a dipole in a nonuniform eld is of some interest F F F qE E RAE where AE represents the difference between the eld at the plus end and the eld at the minus end Assuming the dipole is very short we may use Eq 135 to approximate the small change in Ex AEX E VEX d with corresponding formulas for E y and E z More compactly AE d VE 4 1 POLARIZATION 1 65 and thereforeI F p VE 45 For a perfect dipole of in nitesimal length Eq 44 gives the torque about the center of the dipole even in a nonuniform eld about any other point N p X E r X F Problem 45 In Fig 46 p1 and p2 are perfect dipoles a distance r apart What is the torque on p1 due to p2 What is the torque on p2 due to p In each case I want the torque on the dipole about its own center If it bothers you that the answers are not equal and opposite see Prob 429 Figure 46 Figure 47 Problem 46 A perfect dipole p is situated a distance z above an in nite grounded conducting plane Fig 47 The dipole makes an angle 6 with the perpendicular to the plane Find the torque on p If the dipole is free to rotate in what orientation will it come to rest Problem 47 Show that the energy of an ideal dipole p in an electric eld E is given by lt46gt Problem 48 Show that the interaction energy of two dipoles separated by a displacement r is l A A 3P1 39P2 3P1 rP2 rl 47 47160 r Hint use Prob 47 and Eq 3104 Problem 49 A dipole p is a distance r from a point charge q and oriented so that p makes an angle 6 with the vector r from q to p a What is the force on p b What is the force on q 1In the present context Eq 45 could be written more conveniently as F Vp E However it is safer to stick with p VE because we will be applying the formula to materials in which the dipole moment per unit volume is itself a function of position and this second expression would imply incorrectly that p too is to be differentiated 166 CHAPTER 4 ELECTRIC FIELDS IN MATTER 414 Polarization In the previous two sections we have considered the effect of an external electric eld on an individual atom or molecule We are now in a position to answer qualitatively the original question What happens to a piece of dielectric material when it is placed in an electric eld If the substance consists of neutral atoms or nonpolar molecules the eld will induce in each a tiny dipole moment pointing in the same direction as the eld2 If the material is made up of polar molecules each permanent dipole will experience a torque tending to line it up along the eld direction Random thermal motions compete with this process so the alignment is never complete especially at higher temperatures and disappears almost at once when the eld is rem0ved Notice that these two mechanisms produce the same basic result a lot of little dipoles pointing along the direction of the eld the material becomes polarized A convenient measure of this effect is P E dipole moment per unit volume which is called the polarization From now on we shall not worry much about how the polarization got there Actually the two mechanisms I described are not as clearcut as I tried to pretend Even in polar molecules there will be some polarization by displacement though generally it is a lot easier to rotate a molecule than to stretch it so the second mechanism dominates It s even possible in some materials to freeze in polarization so that it persists after the eld is rem0ved But let s forget for a moment about the cause of the polarization and study the eld that a chunk of polarized material itself produces Then in Sect 43 we ll put it all together the original eld which was responsible for P plus the new eld which is due to P 42 The Field of a Polarized Object 421 Bound Charges Suppose we have a piece of polarized material that is an object containing a lot of micro scopic dipoles lined up The dipole moment per unit volume P is given QuestiOn What is the eld produced by this object not the eld that may have caused the polarization but the eld the polarization itself causes Well we know what the eld of an individual dipole looks like so why not chop the material up into in nitesimal dipoles and integrate to get the total As usual it s easier to work with the potential For a single dipole p we have equation Eq 399 Vr 48 47T 0 t2 2In asymmetric molecules the induced dipole moment may not be parallel to the eld but if the molecules are randomly oriented the perpendicular contributions will average to zero Within a single crystal the orientations are certainly not random and we would have to treat this case separately 42 THE FIELD OF A POLARIZED OBJECT 167 Figure 48 where a is the vector from the dipole to the point at which we are evaluating the potential Fig 48 In the present context we have a dipole moment p Pdr in each volume element d 1quot so the total potential is 1 IZP That does it in principle But a little sleightof hand casts this integral into a much more illuminating form Observing that V122 a 2 where unlike Prob 113 the differentiation is with respect to the source coordinates r we have 1 l V PV dr 47T 0 a V Integrating by parts using product rule number 5 gives 1 P 1 l V fV dr VPd1 47T 0 a a V V or using the divergence theorem l 1 1 1 P da f V Pdr 410 47160 7L 47T 0 a S v The rst term looks like the potential of a surface charge abEP 411 168 CHAPTER 4 ELECTRIC FIELDS IN MATTER where n is the normal unit vector while the second term looks like the potential of a volume charge 0 E V P 412 With these de nitions Eq 410 becomes 1 1 Vr f3 da amp dr 413 47T 0 a 47160 a s v What this means is that the potential and hence also the eld of a polarized object is the same as that produced by a volume charge density 0 V P plus a surface charge density a P Instead of integrating the contributions of all the in nitesimal dipoles as in Eq 49 we just nd those bound charges and then calculate the elds they produce in the same way we calculate the eld of any other volume and surface charges for example using Gauss s law Example 42 Find the electric eld produced by a uniformly polarized sphere of radius R Solution We may as well choose the z axis to coincide with the direction of polarization Fig 49 The volume bound charge density 0 is zero since P is uniform but I P PcosO where 6 is the usual spherical coordinate What we want then is the eld produced by a charge density P cos 6 plastered over the surface of a sphere But we have already computed the potential of such a con guration in Ex 39 P rcos forr f R 360 Vr9 R3 cos forr 2 R 360 r2 Figure 49 42 THE FIELD OF A POLARIZED OBJECT 169 Since r cos 6 z the eld inside the sphere is uniform P l E VV i P f R 360 360 orr lt 414 This remarkable result will be very useful in what follows Outside the sphere the potential is identical to that of a perfect dipole at the origin 1 p A f gt R 47160 r2 orr 4 15 whose dipole moment is not surprisingly equal to the total dipole moment of the sphere p NR3P 416 The eld of the uniformly polarized sphere is shown in Fig 410 l D Figure 410 Problem 410 A sphere of radius R carries a polarization Pr kr where k is a c0nstant and r is the vector from the center a Calculate the bound charges a and pb b Find the eld inside and outside the sphere 170 CHAPTER 4 ELECTRIC FIELDS IN MATTER Problem 411 A short cylinder of radius a and length L carries a frozenin uniform polar ization P parallel to its axis Find the bound charge and sketch the electIic eld i for L gtgt 0 ii for L ltlt 1 and iii for L R a This device is known as a bar electret it is the electrical analog to a bar magnet In practice only very special materials barium titanate is the most familiar example will hold a permanent electric polarization That s why you can t buy electrets at the toy store Problem 412 Calculate the potential of a uniformly polarized sphere Ex 42 directly from Eq 49 422 Physical Interpretation of Bound Charges In the last section we found that the eld of a polarized object is identical to the eld that would be produced by a certain distribution of bound charges 0 and pb But this conclusion emerged in the course of abstract manipulations 0n the integral in Eq 49 and left us with no clue as to the physical meaning of these bound charges Indeed some authors give you the impression that bound charges are in some sense ctitious mere bookkeeping devices used to facilitate the calculation of elds Nothing could be farther from the truth 0 and 0 represent perfectly genuine accumulations of charge In this section I ll explain how polarization leads to such accumulations of charge The basic idea is very simple Suppose we have a long string of dipoles as shown in Fig 411 Along the line the head of one effectively cancels the tail of its neighbor but at the ends there are two charges left over plus at the right end and minus at the left It is as if we had peeled off an electron at one end and carried it all the way down to the other end though in fact no single electron made the whole trip a lot of tiny displacements add up to one large one We call the net charge at the ends bound charge to remind ourselves that it cannot be rem0ved in a dielectric every electron is attached to a speci c atom or molecule But apart from that bound charge is no different from any other kind HCHCHCHCHCH gt 7 Figure 411 To calculate the actual amount of bound charge resulting from a given polarization examine a tube of dielectric parallel to P The dipole moment of the tiny chunk shown in Fig 412 is PAd where A is the cross sectional area of the tube and d is the length of the chunk In terms of the charge q at the end this same dipole moment can be written qd The bound charge that piles up at the right end of the tube is therefore q 2 PA If the ends have been sliced off perpendicularly the surface charge density is 0 1 p A 4 2 THE FIELD OF A POLARIZED OBJECT 171 Figure 412 Figure 413 For an oblique cut Fig 413 the charge is still the same but A Aend cos 6 so Pcos6P Ob 1 Aend The effect of the polarization then is to paint a bound charge 0 P Over the surface of the material This is exactly what we found by more rigorous means in Sect 421 But now we know where the bound charge comes from If the polarization is nonuniform we get accumulations of bound charge within the material as well as on the surface A glance at Fig 414 suggests that a diverging P results in a pileup of negative charge Indeed the net bound charge f pb d1 in a given volume is equal and opposite to the amount that has been pushed out through the surface The latter by the same reasoning we used before is P per unit area so fpbdrz fPda VPdr V S V Since this is true for any volume we have Pb V P con rming again the more rigorous conclusion of Sect 42 l Figure 4 l 4 172 CHAPTER 4 ELECTRIC FIELDS IN MATTER Example 43 There is another way of analyzing the uniformly polarized sphere Ex 42 which nicely illustrates the idea of a bound charge What we have really is two spheres of charge a positive sphere and a negative sphere Without polarization the two are superimposed and cancel completely But when the material is uniformly polarized all the plus charges move slightly upward the z direction and all the minus charges move slightly downward Fig 415 The two spheres no longer overlap perfectly at the top there s a cap of leftover positive charge and at the bottom a cap of negative charge This leftover charge is precisely the bound surface charge 01 Figure 415 In Prob 218 you calculated the eld in the region of overlap between two uniformly charged spheres the answer was l qd 47160 where q is the total charge of the positive sphere d is the vector from the negative center the positive center and R is the radius of the sphere We can express this in terms of the polarization of the sphere p qd gnR3P as l E P 360 Meanwhile for points outside it is as though all the charge on each sphere were concentrated at the respective center We have then a dipole with potential 1 v p 47160 r2 Remember that d is some small fraction of an atomic radius Fig 415 is grossly exaggerated These answers agree of course with the results of EX 42 42 THE FIELD OF A POLARIZED OBJECT 173 Problem 413 A very long cylinder of radius a carries a uniform polarization P perpendicular to its axis Find the electric eld inside the cylinder Show that the eld outside the cylinder can be expressed in the form a2 Er 2P P 26032 Careful I said uniform not radial Problem 414 When you polarize a neutral dielectric charge moves a bit but the total remains zero This fact should be re ected in the bound charges 0 and p1 Prove from Eqs 41 l and 412 that the total bound charge vanishes 423 The Field Inside 2 Dielectric I have been sloppy about the distinction between pure dipoles and physical dipoles In developing the theory of bound charges I assumed we were working with the pure kind indeed I started with Eq 48 the formula for the potential of a pure dipole And yet an actual polarized dielectric consists of physical dipoles albeit extremely tiny ones What is more I presumed to represent discrete molecular dipoles by a continuous density function P How can I justify this method Outside the dielectric there is no real problem here we are far away from the molecules It is many times greater than the separation distance between plus and minus charges so the dipole potential dominates overwhelmingly and the detailed graininess of the source is blurred by distance Inside the dielectric however we can hardly pretend to be far from all the dipoles and the procedure I used in Sect 421 is open to serious challenge In fact when you stop to think about it the electric eld inside matter must be fantas tically complicated on the microscopic level If you happen to be very near an electron the eld is gigantic whereas a short distance away it may be small or point in a totally different direction Moreover an instant later as the atoms move about the eld will have altered entirely This true microscopic eld would be utterly impossible to calculate nor would it be of much interest if you could Just as for macroscopic purposes we regard water as a continuous uid ignoring its molecular structure so also we can ignore the microscopic bumps and wrinkles in the electric eld inside matter and concentrate on the macroscopic eld This is de ned as the average eld Over regions large enough to contain many thousands of atoms so that the uninteresting microscopic uctuations are smoothed over and yet small enough to ensure that we do not wash out any signi cant large scale variations in the eld In practice this means we must average over regions much smaller than the dimensions of the object itself Ordinarily the macroscopic eld is what people mean when they speak of the eld inside matter3 3In case the introduction of the macroscopic eld sounds suspicious to you let me point out that you do exactly the same averaging whenever you speak of the density of a material 174 CHAPTER 4 ELECTRIC FIELDS IN MATTER Figure 416 It remains to show that the macroscopic eld is what we actually obtain when we use the methods of Sect 421 The argument is subtle so hang on Suppose I want to calculate the macroscopic eld at some point r Within a dielectric Fig 416 I know I must average the true microscopic eld Over an appropriate volume so let me draw a small sphere about r of radius say a thousand times the size of a molecule The macroscopic eld at r then consists of two parts the average eld over the sphere due to all charges outside plus the average due to all charges inside E Eout Ein Now you pr0ved in Prob 34ld that the average eld Over a sphere produced b charges outside is equal to the eld they produce at the center so Eout is the eld at r due to the dipoles exterior to the sphere These are far enough away that we can safely use Eq 49 1 3 P r V0 dr 417 a outside The dipoles inside the sphere are too close to treat in this fashion But fortunately all we need is their average eld and that according to Eq 3105 is 1 P E m 47160 R37 regardless of the details of the charge distribution within the sphere The only relevant quantity is the total dipole moment p gnR3 P l E in 360 418 Now by assumption the sphere is small enough that P does not vary signi cantly over its volume so the term left out of the integral in Eq 417 corresponds to the eld at the center of a uniformly polarized sphere to wit 1 360P Eq 414 But this is precisel what Em Eq 418 puts back in The macroscopic eld then is given by the potential 1 AP Vr fquot 2 dr 419 47160 a 43 THE ELECTRIC DISPLACEMENT 175 where the integral runs over the entire volume of the dielectric This is of course what we used in Sect 421 without realizing it we were correctly calculating the averaged macroscopic eld for points inside the dielectric You may have to reread the last couple of paragraphs for the argument to sink in Notice that it all revolves around the curious fact that the average eld over any sphere due to the charge inside is the same as the eld at the center of a uniformly polarized sphere with the same total dipole moment This means that no matter how crazy the actual microscopic charge con guration we can replace it by a nice smooth distribution of perfect dipoles if all we want is the macroscopic average eld Incidentally while the argument ostensibly relies on the spherical shape I chose to average over the macroscopic eld is certainly independent of the geometry of the averaging region and this is re ected in the nal answer Eq 419 Presumably one could reproduce the same argument for a cube or an ellipsoid or whatever the calculation might be more dif cult but the conclusion would be the same 43 The Electric Displacement 431 Gauss s Law in the Presence of Dielectrics In Sect 42 we found that the effect of polarization is to produce accumulations of bound charge 0 V P within the dielectric and ab P ii on the surface The eld due to polarization of the medium is just the eld of this bound charge We are now ready to put it all together the eld attributable to bound charge plus the eld due to everything else which for want of a better term we call free charge The free charge might consist of electrons on a conductor or ions embedded in the dielectric material or whatever any charge in other words that is not a result of polarization Within the dielectric then the total charge density can be written 0 Pb pf 420 and Gauss s law reads EOV39Ep PbPf VPpf where E is now the total eld not just that portion generated by polarization It is convenient to combine the two divergence terms V 60E P pf The expression in parentheses designated by the letter D 1 E 60E P 421 is known as the electric displacement In terms of D Gauss s law reads 22gt 176 CHAPTER 4 ELECTRIC FIELDS IN MATTER or in integral form yin da Qfm 423 where Qfenc denotes the total free charge enclosed in the volume This is a particularly useful way to express Gauss s law in the context of dielectrics because it makes reference only to free charges and free charge is the stuff we control Bound charge comes along for the ride when we put the free charge in place a certain polarization automatically ensues by the mechanisms of Sect 41 and this polarization produces the bound charge In a typical problem therefore we know pf but we do not initially know Ob Eq 423 lets us go right to work with the information at hand In particular whenever the requisite symmetry is present we can immediately calculate D by the standard Gauss s law methods Example 44 A long straight wire carrying uniform line charge A is surrounded by rubber insulation out to a radius a Fig 417 Find the electric displacement Figure 417 Solution Drawing a cylindrical Gaussian surface of radius s and length L and applying Eq 423 we nd D271sL 2 AL Therefore D 424 271s Notice that this formula holds both within the insulation and outside it In the latter region P 0 so 1 A A E s for s gt a 60 27160 Inside the rubber the electric eld cannot be determined since we do not know P It may have appeared to you that I left out the surface bound charge ab in deriving Eq 422 and in a sense that is true We cannot apply Gauss s law precisely at the surface of a dielectric for here 0 blows up taking the divergence of E with it But everywhere else the logic is sound and in fact if we picture the edge of the dielectric as having some nite thickness within which the polarization tapers off to zero probably a more realistic model 43 THE ELECTRIC DISPLACEMENT 177 than an abrupt cutoff anyway then there is no surface bound charge 0 varies rapidly but smoothly within this skin and Gauss s law can be safely applied everywhere At any rate the integral form Eq 423 is free from this defect Problem 415 A thick spherical shell inner radius a outer radius b is made of dielectric material with a frozenin polarization where k is a constant and r is the distance from the center Fig 418 There is no free charge in the problem Find the electric eld in all three regions by two different methods a Locate all the bound Charge and use Gauss s law Eq 2 13 to calculate the eld it produces b Use Eq 423 to nd D and then get E from Eq 421 Notice that the second method is much faster and avoids any explicit reference to the bound charges Problem 416 Suppose the eld inside a large piece of dielectric is E0 so that the electric displacement is D0 60E0 P a Now a small spherical cavity Fig 419a is hollowed out of the material Find the eld at the center of the cavity in terms of E0 and P Also nd the displacement at the center of the cavity in terms of D0 and P b Do the same for a long needleshaped cavity running parallel to P Fig 41 c Do the same for a thin wafershaped cavity perpendicular to P Fig 419c Assume the cavities are small enough that P E0 and D0 are essentially uniform Hint Carving out a cavity is the same as superimposing an object of the same shape but opposite polarization a Sphere b Needle c Wafer Figure 418 Figure 419 178 CHAPTER 4 ELECTRIC FIELDS IN MATTER 432 A Deceptive Parallel Equation 422 looks just like Gauss s law only the total charge density p is replaced by the free charge density pf and D is substituted for 60E For this reason you may be tempted to conclude that D is just like E apart from the factor 60 except that its source is o instead of p To solve problems involving dielectrics you just forget all about the bound charge calculate the eld as you ordinarily would only call the answer D instead of This reasoning is seductive but the conclusion is false in particular there is no Coulomb39s law for D 1 91 D0 75 EEpfrd7 The parallel between E and D is more subtle than that For the divergence alone is insuf cient to determine a vector eld you need to know the curl as well One tends to forget this in the case of electrostatic elds because the curl of E is always zero But the curl of D is not always zero VXD60VXEVXPVXP 4251 and there is no reason in general to suppose that the curl of P vanishes Sometimes it does as in Ex 44 and Prob 415 but more often it does not The bar electret of Prob 411 is a case in point here there is no free charge anywhere so if you really believe that the onl source of D is pf you will be forced to conclude that D O everywhere and hence that E 160P inside and E 0 outside the electret which is obviously wrong I leave it for you to nd the place where V X P 75 0 in this problem Because V X D are 0 moreover D cannot be expressed as the gradient of a scalar there is no potential for D Advice When you are asked to compute the electric displacement rst look for sym metry If the problem exhibits spherical cylindrical or plane symmetry then you can get D directly from Eq 423 by the usual Gauss s law methods Evidently in such cases V X P is automatically zero but since symmetry alone dictates the answer you re not really obliged to worry about the curl If the requisite symmetry is absent you ll have to think of another approach and in particular you must not assume that D is determined exclusively by the free charge 433 Boundary Conditions The electrostatic boundary conditions of Sect 235 can be recast in terms of D Equation 423 tells us the discontinuity in the component perpendicular to an interface J J Dabove Dbelow af 43926 while Eq 425 gives the discontinuity in parallel components ll ll ll lbelow P above Pbelow39 Dll above 427 44 LINEAR DIELECTRICS 179 In the presence of dielectrics these are sometimes more useful than the corresponding boundary conditions on E Eqs 231 and 223 l L J Eabove Ebelow ga 43928 and En below Ell above 0 429 You might try applying them for example to Probs 416 and 417 Problem 417 For the bar electret of Prob 411 make three careful sketches one of P one of E and one of D Assume L is about 2a Hint E lines terminate on charges D lines terminate on free charges 44 Linear Dielectrics 441 Susceptibility Permittivity Dielectric Constant In Sects 42 and 43 we did not commit ourselves as to the cause of P we dealt only with the e ects of polarization From the qualitative discussion of Sect 41 though we know that the polarization of a dielectric ordinarily results from an electric eld which lines up the atomic or molecular dipoles For many substances in fact the polarization is proportional to the eld provided E is not too strong P Goer 430 The constant of proportionality Xe is called the electric susceptibility of the medium a factor of 60 has been extracted to make Xe dimensionless The value of Xe depends on the microscopic structure of the substance in question and also on external conditions such as temperature I shall call materials that obey Eq 430 linear dielectrics4 Note that E in Eq 430 is the total eld it may be due in part to free charges and in part to the polarization itself If for instance we put a piece of dielectric into an external eld E0 we cannot compute P directly from Eq 430 the external eld will polarize the material and this polarization will produce its own eld which then contributes to the total eld and this in turn modi es the polarization which Breaking out of this in nite regress is not always easy You ll see some examples in a moment The simplest approach is to begin with the displacement at least in those cases where D can be deduced directly from the free charge distribution 4In modern optical applications especially nonlinear materials have become increasingly important For these there is a second term in the formula for P as a function of E typically a cubic one In general Eq 430 can be regarded as the rst nonzero term in the Taylor expansion of P in powers of E 180 CHAPTER 4 ELECTRIC FIELDS IN MATTER In linear media we have D 60E P 60E EOXeE 6011 XeE 431 so D is also proportional to E 1 2 6E 432 where e E 601 Xe 433 This new constant 6 is called the permittivity of the material In vacuum where there is no matter to polarize the susceptibility is zero and the permittivity is 60 That s why 60 is called the permittivity of free space I dislike the term for it suggests that the vacuum is just a special kind of linear dielectric in which the permittivity happens to have the value 885 X 10 12 C 2Nm2 If you remove a factor of 60 the remaining dimensionless quantity 6 a 1 X 434 is called the relative permittivity or dielectric constant of the material Dielectric con stants for some conrrnon substances are listed in Table 42 Of course the permittivity and the dielectric constant do not convey any information that was not already available in the susceptibility nor is there anything essentially new in Eq 432 the physics of linear dielectrics is all contained in Eq 4305 Material Dielectric Constant Material Dielectric Constant Vacuum 1 Benzene 228 Helium 1000065 Diamond 5 7 Neon 100013 Salt 59 Hydrogen 100025 Silicon l 18 Argon 1 00052 Methanol 33 0 Air dry 100054 Water 80 1 Nitrogen 100055 Ice 30 C 99 Water vapor 100 C 100587 KTaNbO3 0 C 34000 Table 42 Dielectric Constants unless otherwise speci ed values given are for 1 atm 20 C Source Handbook of Chemistry and Physics 78th ed Boca Raton CRC Press Inc 1997 5As long as we are engaged in this orgy of unnecessary terminology and notation I might as well mention that formulas for D in terms of E Eq 432 in the case of linear dielectrics are called constitutive relations 44 LINEAR DIELECTRICS 1 81 Figure 420 Example 45 A metal sphere of radius a carries a charge Q Fig 420 It is surrounded out to radius b by linear dielectric material of permittivity 6 Find the potential at the center relative to in nity Solution To compute V we need to know E to nd E we might rst try to locate the bound charge we could get the bound charge from P but we can t calculate P unless we already know E Eq 430 We seem to be in a bind What we do know is the free charge Q and fortunately the arrangement is spherically symmetric so let s begin by calculating D using Eq 423 Q 4ttr2 Inside the metal sphere of course E P D 0 Once we know D it is a trivial matter to obtain E using Eq 432 i39 for all points r gt a Q A mr fOI39Cl lt r lt b E if forr gt b 47teor2 The potential at the center is therefore 0 b a 0 L L 00 E dl foo 4ne0r2 dr A 47102 dr Q 0 dr Q 1 1 1 471 60bea 617 As it turns out it was not necessary for us to compute the polarization or the bound charge explicitly though this can easily be done V EOXle P e E oxe 47t6r2 9 182 CHAPTER 4 ELECTRIC FIELDS IN MATTER in the dielectric and hence pb V i P 2 07 while Q 6 OXe at the outer surface A 4716b2 7 P n e Lag at the inner surface 4716a Notice that the surface bound charge at a is negative in points outward with respect to the dielectric which is i39 at b but i39 at a This is natural since the Charge on the metal sphere attracts its Opposite in all the dielectric molecules It is this layer of negative charge that reduces the eld within the dielectric from 1 47160Qr2f39 to 14tteQr2i39 In this respect a dielectric is rather like an imperfect conductor on a conducting shell the induced surface charge would be such as to cancel the eld of Q completely in the region a lt r lt b the dielectric does the best it can but the cancellation is only partial You might suppose that linear dielectrics would escape the defect in the parallel between E and D Since P and D are now proportional to E does it not follow that their curls like E s must vanish Unfortunately it does not for the line integral of P around a closed path that straddles the boundary between one type of material and another need not be zero even though the integral of E around the same loop must be The reason is that the proportionality factor Goxe is different on the two sides For instance at the interface between a polarized dielectric and the vacuum Fig 421 P is zero on one side but not on the other Around this loop 55 P dl 75 O and hence by Stokes theorem the curl of P cannot vanish everywhere within the loop in fact it is in nite at the boundary P 0 Vacuum Dielectric P 0 Figure 421 Of course if the space is entirely lled with a homogeneous6 linear dielectric then this objection is void in this rather special circumstance Vszf and VXDO so D can be found from the free charge just as though the dielectric were not there D EOEvac where EVle is the eld the same free charge distribution would produce in the absence of any dielectric According to Eqs 432 and 434 therefore 1 1 E D Evac 435 6 er 6A homogeneous medium is one whose properties in this case the susceptibility do not vary with position 44 LINEAR DIELECTRICS 183 Conclusion When all space is lled with a homogeneous linear dielectric the eld every where is simply reduced by a factor of one over the dielectric constant Actually it is not necessary for the dielectric to ll all space in regions where the eld is zero anyway it can hardly matter whether the dielectric is present or not since there s no polarization in any event For example if a free charge q is embedded in a large dielectric the eld it produces is E 4 f 436 that s 6 not 60 and the force it exerts on nearby charges is reduced accordingly But it s not that there is anything wrong with Coulomb s law rather the polarization of the medium partially shields the charge by surrounding it with bound charge of the opposite sign Fig 4227 qo Figure 422 Example 46 A parallelplate capacitor Fig 423 is lled with insulating material of dielectric constant er What effect does this have on its capacitance Solution Since the eld is con ned to the space between the plates the dielectric will reduce E and hence also the potential difference V by a factor 16 Accordingly the capacitance C Q V is increased by a factor of the dielectric constant C Er Cvac This is in fact a common way to beef up a capacitor 7In quantum electrodynarniCS the vacuum itself can be polarized and this means that the effective or renor malized charge of the electron as you might measure it in the laboratory is not its true bare value and in fact depends slightly on how far away you are 184 CHAPTER 4 ELECTRIC FIELDS IN MATTER lt Dielectric Figure 423 By the way a crystal is generally easier to polarize in some directions than in others8 and in this case Eq 430 is replaced by the general linear relation Px 50Xe Ex XexyEy Xeszz Py 60Xeyx Ex Xeyy Ey 139 XeyzEz a 438 Pz 500er Ex 139 XezyEy 1 XeZZ Ez just as Eq 41 was superseded by Eq 43 for asymmetrical molecules The nine coef cients Xe Xe constitute the susceptibility tensor Problem 418 The space between the plates of a parallelplate capacitor Fig 424 is lled with two slabs of linear dielectric material Each slab has thickness a so the total distance between the plates is 2a Slab 1 has a dielectric constant of 2 and slab 2 has a dielectric constant of 15 The free charge density on the top plate is a and on the bottom plate a a Find the electric displacement D in each slab b Find the electric eld E in each slab c Find the polarization P in each slab d Find the potential difference between the plates e Find the location and amount of all bound charge f Now that you know all the charge free and bound recalculate the eld in each slab and con rm your answer to b 8A medium is said to be isotropic if its properties such as susceptibility are the same in all directions Thus Eq 430 is the special case of Eq 438 that holds for isotropic media Physicists tend to be sloppy with their language and unless otherwise indicated the term linear dielectricquot certainly means isotropic linear dielectricquot and probably means homogeneous isotropic linear dielectric 44 LINEAR DIELECTRICS 185 6 lt Slab l lt Slab 2 Figure 424 Problem 419 Suppose you have enough linear dielectric material of dielectric constant Er to half ll a parallelplate capacitor Fig 425 By what fraction is the capacitance increased when you distribute the material as in Fig 425a How about Fig 425b For a given potential difference V between the plates nd E D and P in each region and the free and bound charge on all surfaces for both cases Figure 425 Problem 420 A sphere of linear dielectric material has embedded in it a uniform free charge density 0 Find the potential at the center of the sphere relative to in nity if its radius is R and its dielectric constant is Er Problem 421 A certain coaxial cable consists of a copper wire radius a surrounded by a concentric copper tube of inner radius c Fig 426 The space between is partially lled from b out to c with material of dielectric constant 6 as shown Find the capacitance per unit length of this cable 186 CHAPTER 4 ELECTRIC FIELDS IN MATTER Figure 426 442 Boundary Value Problems with Linear Dielectrics In a homogeneous linear dielectric the bound charge density p1 is proportional to the free charge density 0 f9 Xe Xe pb VP V606D 1Xegtpf 439 In particular unless free charge is actually embedded in the material 0 0 and any net charge must reside at the surface Within such a dielectric then the potential obeys Laplace s equation and all the machinery of Chapter 3 carries over It is convenient however to rewrite the boundary conditions in a way that makes reference only to the free charge Equation 426 says Gabove Eli 30 Ebelow Eli 610w Ufa 440 or in terms of the potential 8 V 8 V 6above Barlow Ebelow Of 441 whereas the potential itself is of course continuous Eq 234 Vabove Vbelow 442 Example 47 A sphere of homogeneous linear dielectric material is placed in an otherwise uniform electric eld E0 Fig 427 Find the electric eld inside the sphere 9This does not apply to the surface charge 0 because Xe is not independent of position obviously at the boundary 44 LINEAR DIELECTRICS 187 1 Figure 427 F Solution This is reminiscent of Ex 38 in which an uncharged conducting sphere was introduced into a uniform eld In that case the eld of the induced charge completely canceled E0 within the sphere in a dielectric the cancellation from the bound charge is only partial Our problem is to solve Laplace s equation for Vinr 6 when r 5 R and Voutr 6 when r 2 R subject to the boundary conditions i Vin Vout atVZR 3V 3V ii ea m 603 2 atrR 443 r iii Vout gt E0rcos6 forr gtgt R The second of these follows from Eq 441 since there is no free charge at the surface Inside the sphere Eq 365 says 00 Vin 9 2 A1 r1 P1cos 6 444 0 outside the sphere in view of iii we have V0utra 9 E0r cos6 E H1 P1cos6 445 739 0 188 CHAPTER 4 ELECTRIC FIELDS IN MATTER Boundary condition 1 requires that 2A R1 Plcos6 E0Rcos6 Z WPlcos6 0 0 so10 AIR forl 94 1 R1 446 Bl A1R EOR Meanwhile condition ii yields l1B Er ZIA1R11Plcos 6 2 E0 cos6 Z 1P1cos6 0 12 0 R so I 1 3 ErlAlRl 1 Rl 2 f01 l7 l 447 281 ErA E It follows that AlBl0 forla l 3 448 5r 1 3 A1 mE0 B1 6r2R Evidently 3E0 3E0 e Vmr 6 r2rcos 6r2z and hence the eld inside the sphere is surprisingly uniform E 3 E 4 49 Q 2 0 Example 48 Suppose the entire region below the plane z 0 in Fig 428 is lled with uniform linear dielectric material of susceptibility Xe Calculate the force on a point charge q situated a distance d above the origin Solution The surface bound charge on the xy plane is of opposite sign to q so the force will be attractive In view of Eq 439 there is no volume bound charge Let us rst calculate 01 using Eqs 411 and 430 ab P39 Pz 50X2Ezs 10Remember P1 cos 6 cos 6 and the coef cients must be equal for each I as you could prove by multiplying by Pl cos 6 sin 6 integrating from 0 to 71 and invoking the orthogonality of the Legendre polynomials Eq 3 68 44 LINEAR DIELECTRICS 189 Figure 428 where Ez is the zcomponent of the total eld just inside the dielectric at z 0 This eld is due in part to q and in part to the bound charge itself From Coulomb s law the former contribution is 1 q 1 qd c 9 4m r2 d2 09 47160 r2 d232 where r jx2 y2 is the distance from the origin The z component of the eld of the bound charge meanwhile is Gb260 see footnote 6 p 89 Thus 44 0b 0 Z 60 47160 r2 d232 Q which we can solve for 01 1 Xe qd 450 0 221 Xe2gtr2d232 Apart from the factor Xe Xe 2 this is exactly the same as the induced charge on an in nite conducting plane under similar circumstances Eq 31011 Evidently the total bound charge is Xe 451 Qb Xe2gtq We could of course obtain the eld of a by direct integration 1 2 E abda 47160 2 11For some purposes a conductor can be regarded as the limiting case of a linear dielectric with Xe v 00 This is often a useful checkvtry applying it to Exs 45 46 and 47 190 CHAPTER 4 ELECTRIC FIELDS IN MATTER But as in the case of the conducting plane there is a nicer solution by the method of images Indeed if we replace the dielectric by a single point charge qb at the image position 0 0 d we have 1 q qb V 452 47160 x2y2z d2 x272zd2 in the region z gt 0 Meanwhile a charge q qb at 0 0 0 yields the potential 1 H qb 453 47 0 xxz lyzz a 2 for the region z lt 0 Taken together Eqs 452 and 453 constitute a function which satis es Poisson s equation with a point charge q at 00 d which goes to zero at in nity which is continuous at the boundary z 0 and whose normal derivative exhibits the discontinuity appropriate to a surface charge 0 at z 0 6 EV gt 1 lt Xe gt qd 0 Bz z0 21 Xe2 x2y2d23239 Accordingly this is the correct potential for our problem In particular the force on q is 1 1 2 F i Xe q i 454 47160 2d2 47160 Xe 2 4d2 3V z0 BZ I do not claim to have provided a compelling motivation for Eqs 452 and 453 like all image solutions this one owes its justi cation to the fact that it works it solves Poisson39s equation and it meets the boundary conditions Still discovering an image solution is not entirely a matter of guesswork There are at least two rules of the game 1 You must never put an image charge into the region where you re computing the potential Thus Eq 452 gives the potential for z gt 0 but this image charge qb is at z d when we turn to the region z lt 0 Eq 453 the image charge q qb is at z d 2 The image charges must add up to the correct total in each region That s how I knew to use qb to account for the charge in the region z 5 0 and q qb to cover the region z 2 0 Problem 422 A very long cylinder of linear dielectric material is placed in an otherwise uniform electric eld E0 Find the resulting eld within the cylinder The radius is a the susceptibility Xe and the axis is perpendicular to E0 Problem 423 Find the eld inside a sphere of linear dielectric material in an otherwise uniform electric eld E0 Ex 47 by the following method of successive approximations First pretend the eld inside is just E0 and use Eq 430 to write down the resulting polarization P0 This polarization generates a eld of its own E1 Ex 42 which in turn modi es the polarization by an amount P1 which further changes the eld by an amount E2 and so on The resulting eld is E0 E E2 Sum the series and compare your answer with Eq 449 Problem 424 An uncharged conducting sphere of radius a is coated with a thick insulating shell dielectric constant Er out to radius b This object is now placed in an otherwise uniform electric eld E0 Find the electric eld in the insulator 44 LINEAR DIELECTRICS 191 Problem 425 Suppose the region above the xy plane in Ex 48 is also lled with linear dielectric but of a different susceptibility x Find the potential everywhere 443 Energy in Dielectric Systems It takes work to charge up a capacitor Eq 255 W C V2 If the capacitor is lled with linear dielectric its capacitance exceeds the vacuum value by a factor of the dielectric constant C Er Cvac as we found in Ex 46 Evidently the work necessary to charge a dielectric lled capacitor is increased by the same factor The reason is pretty clear you have to pump on more free charge to achieve a given potential because part of the eld is canceled off by the bound charges In Chapter 2 I derived a general formula for the energy stored in any electrostatic system Eq 245 W 630 Ezdr 455 The case of the dielectric lled capacitor suggests that this should be changed to 1 W 6 06rE2dr DEdr 2 2 in the presence of linear dielectrics To prove it suppose the dielectric material is xed in position and we bring in the free charge a bit at a time As p f is increased by an amount Apf the polarization will change and with it the bound charge distribution but we re interested only in the work done on the incremental free charge AW ApfVdT 456 Since V D pf Apf V AD where AD is the resulting change in D so AW 2 V ADVdr Now V ADV V ADV AD VV and hence integrating by parts AWVADVdrADEdr 192 CHAPTER 4 ELECTRIC FIELDS IN MATTER The divergence theorem turns the rst term into a surface integral which vanishes if we integrate over all of space Therefore the work done is equal to AW 2 AD Edr 457 So far this applies to any material Now if the medium is a linear dielectric then D 2 6E so mn E A6E2 6AE E AD E for in nitesimal increments Thus AWADEdrgt The total work done then as we build the free charge up from zero to the nal con guration is 1 W EDEarr 458 as anticipated12 It may puzzle you that Eq 455 which we derived quite generally in Chapter 2 does not seem to apply in the presence of dielectrics where it is replaced by Eq 458 The point is not that one or the other of these equations is wrong but rather that they speak to somewhat different questions The distinction is subtle so let s go right back to the beginning What do we mean by the energy of a system Answer It is the work required to assemble the system Very well but when dielectrics are involved there are two quite different ways one might construe this process 1 We bring in all the charges free and bound one by one with tweezers and glue each one down in its proper nal location If this is what you mean by assemble the system the Eq 455 is your formula for the energy stored Notice however that this will not include the work involved in stretching and twisting the dielectric molecules if we picture the positive and negative charges as held together by tiny springs it does not include the spring energy kx2 associated with polarizing each molecule13 2 With the unpolarized dielectric in place We bring in the free charges one by one allowing the dielectric to respond as it sees t If this is what you mean by assemble the systemquot and ordinarily it is since free charge is what We actually push around then Eq 458 is the formula you want In this case the spring energy is included albeit indirectly because the force you must apply to the free charge depends on the disposition of the bound charge as you move the free charge you are automatically stretching those springs To put it another 12In case you are wondering why I did not do this more simply by the method of Sect 243 starting with W f pf V d1 the reason is that this formula is untrue in general Study the derivation of Eq 242 and you will see that it applies only to the total charge For linear dielectrics it happens to hold for the free charge alone but this is scarcely obvious a priori and in fact is most easily con rmed by working backward from Eq 458 13The spring itself may be electrical in nature but it is still not included in Eq 455 if E is taken to be the macroscopic eld 44 LINEAR DIELECTRICS 193 way in method 2 the total energy of the System consists of three parts the electrostatic energy of the free charge the electrostatic energy of the bound charge and the spring energy Wtot Wfree l Wbound l Wspring The last two are equal and opposite in procedure 2 the bound charges are always in equilibrium and hence the net work done on them is zero thus method 2 in calculating Wfree actually delivers Wtot whereas method 1 by calculating Wfree i39WbOund leaves out Wspring Incidentally it is sometimes alleged that Eq 458 represents the energy even for nonlinear dielectrics but this is false To proceed beyond Eq 457 one must assume linearity In fact for dissipative systems the whole notion of stored energy loses its meaning because the work done depends not only on the nal con guration but on how it got there If the molec ular springs are allowed to have some friction for instance then Wspring can be made as large as you like by assembling the charges in such a way that the spring is obliged to expand and contract many times before reaching its nal state In particular you get nonsensical results if you try to apply Eq 458 to electrets with frozenin polarization see Prob 427 Problem 426 A spherical conductor of radius a carries a charge Q Fig 429 It is surrounded by linear dielectric material of susceptibility Xe out to radius b Find the energy of this con guration Eq 458 Figure 429 Problem 427 Calculate W using both Eq 455 and Eq 458 for a sphere of radius R with frozenin uniform polarization P Ex 42 Comment on the discrepancy Which if either is the true energy of the system 444 Forces on Dielectrics Just as a conductor is attracted into an electric eld Eq 251 so too is a dielectric and for essentially the same reason the bound charge tends to accumulate near the free charge of the opposite sign But the calculation of forces on dielectrics can be surprisingly tricky 194 CHAPTER 4 ELECTRIC FIELDS IN MATTER Dielectric Figure 430 Consider for example the case of a slab of linear dielectric material partially inserted between the plates of a parallelplate capacitor Fig 430 We have always pretended that the eld is uniform inside a parallelplate capacitor and zero outside If this were literally true there would be no net force on the dielectric at all since the eld everywhere would be perpendicular to the plates However there is in reality a fringing eld around the edges which for most purposes can be ignored but in this case is responsible for the whole effect Indeed the eld could not terminate abruptly at the edge of the capacitor for if it did the line integral of E around the closed loop shown in Fig 431 would not be zero It is this nonuniform fringing eld that pulls the dielectric into the capacitor Fringing elds are notoriously dif cult to calculate luckily we can avoid this altogether by the following ingenious method Let W be the energy of the system it depends of course on the amount of overlap If I pull the dielectric out an in nitesimal distance dx the energy is changed by an amount equal to the work done d W Fme dx 459 44 LINEAR DIELECTRICS 195 ytt Figure 431 where Fme is the force I must exert to counteract the electrical force F on the dielectric Fme F Thus the electrical force on the slab is F d W 460 dx Now the energy stored in the capacitor is W lcvz 461 and the capacitance in this case is C erz M 462 where l is the length of the plates Fig 430 Let s assume that the total charge on the plates Q C V is held constant as the dielectric moves In terms of Q 2 W lQ 463 2 C SO 2 lQdClV2dC 464 dx 2 C2 dx 2 dx But dC eoxew dx d i and hence w F 60 v2 465 2d 196 CHAPTER 4 ELECTRIC FIELDS IN MATTER The minus sign indicates that the force is in the negative x direction the dielectric is pulled into the capacitor It is a common error to use Eq 461 with V constant rather than Eq 463 with Q constant in computing the force One then obtains 1 dC F V2 2 dx which is off by a sign It is of course possible to maintain the capacitor at a xed potential by connecting it up to a battery But in that case the battery also does work as the dielectric moves instead of Eq 459 we now have dW Fme dx VdQ 466 where V d Q is the work done by the battery It follows that 1 2d C 2 dC l V2dC dW E F dx dx 2 dx dx 2 dx 7 467 the same as before Eq 464 with the correct sign Please understand the force on the dielectric cannot possibly depend on whether you plan to hold Q constant or V constant it is determined entirely by the distribution of charge free and bound It s simpler to calculate the force assuming constant Q because then you don t have to worry about work done by the battery but if you insist it can be done correctly either way Notice that we were able to determine the force without knowing anything about the fringing elds that are ultimately responsible for it Of course it s built into the whole structure of electrostatics that V x E 0 and hence that the fringing elds must be present we re not really getting something for nothing here just cleverly exploiting the internal consistency of the theory The energy stored in the fringing elds themselves which was not accounted for in this derivation stays constant as the slab moves what does change is the energy well inside the capacitor where the eld is nice and uniform Problem 428 Two long coaxial cylindrical metal tubes inner radius a outer radius b stand vertically in a tank of dielectric oil susceptibility Xe mass density p The inner one is maintained at potential V and the outer one is grounded Fig 432 To what height h does the oil rise in the space between the tubes 4 4 LINEAR D IEL ECTR ICS 197 198 CHAPTER 4 ELECTRIC FIELDS IN MATTER Problem 431 A dielectric cube of side a centered at the origin carries a frozen in polar ization P kr Where k is a constant Find all the bound charges and check that they add up to zero Problem 432 A point charge q is imbedded at the center of a sphere of linear dielectric material with susceptibility Xe and radius R Find the electric eld the polarization and the bound charge densities pi and 0 What is the total bound charge on the surface Where is the compensating negative bound charge located Problem 433 At the interface between one linear dielectric and another the electric eld lines bend see Fig 434 Show that tan 92 tan 9 6261 468 assuming there is no free charge at the boundary Comment Eq 468 is reminiscent of Snell s law in optics Would a convex lens of dielectric material tend to focus or defocus the electric eld Figure 432 More Problems on Chapter 4 Problem 429 a For the con guration in Prob 45 calculate the force on p2 due to p1 and the force on p1 due to p2 Are the answers consistent with Newton s third law b Find the total torque on p2 with respect to the center of p1 and compare it with the torque on p1 about that same point Hint combine your answer to a with the result of Prob 45 Problem 430 An electric dipole p pointing in the y direction is placed midway between two large conducting plates as shown in Fig 433 Each plate makes a small angle 6 with respect to the x axis and they are maintained at potentials EV What is the direction of the net force on p There s nothing to calculate here but do explain your answer qualitatively Figure 434 2 Problem 434 A point dipole p is imbedded at the center of a sphere of linear dielectric material with radius R and dielectric constant 6 Find the electric potential inside and outside the sphere a 3 1 6 3 Answer pcos 1 2r er r f R pcos r 2 R 47192 R3 er 2 47160r2 er 2 Problem 435 Prove the following uniqueness theorem A volume V contains a speci ed free charge distribution and various pieces of linear dielectric material with the susceptibility of each one given If the potential is speci ed on the boundaries 8 of V V 0 at in nity would be suitable then the potential throughout V is uniquely determined Hint integrate Figure 433 V V3133 W V 44 LINEAR DIELECTRICS 199 Figure 435 Problem 436 A conducting sphere at potential V0 is half embedded in linear dielectric material of susceptibility Xe which occupies the region z lt 0 Fig 435 Claim the potential everywhere is exactly the same as it would have been in the absence of the dielectric Check this claim as follows a Write down the formula for the proposed potential Vr in terms of V0 R and r Use it to determine the eld the polarization the bound charge and the free charge distribution on the sphere b Show that the total charge con guration would indeed produce the potential Vr c Appeal to the uniqueness theorem in Prob 435 to complete the argument d Could you solve the con gurations in Fig 436 with the same potential If not explain why Figure 436 Problem 437 According to Eq 45 the force on a single dipole is p VE so the net force on a dielectric object is F P VEexdr 469 Here Eext is the eld of everything except the dielectric You might assume that it wouldn t matter if you used the total eld after all the dielectric can t exert a force on itself However because the eld of the dielectric is discontinuous at the location of any bound surface charge the derivative introduces a spurious delta function and you must either add a compensating surface term or better stick with Eext which suffers no such discontinuity Use Eq 469 to determine the force on a tiny sphere of radius R composed of linear dielectric material of susceptibility Xe which is situated a distance s from a ne wire carrying a uniform line charge A 200 CHAPTER 4 ELECTRIC FIELDS IN MATTER Problem 438 In a linear dielectric the polarization is proportional to the eld P 60 XeE If the material consists of atoms or nonpolar molecules the induced dipole moment of each one is likewise proportional to the eld p otE Question What is the relation between the atomic polarizability oz and the susceptibility Xe Since P the dipole moment per unit volume is p the dipole moment per atom times N the number of atoms per unit volume P Np N 01E one s rst inclination is to say that N a Xe 470 60 And in fact this is not far off if the density is low But closer inspection reveals a subtle problem for the eld E in Eq 430 is the total macroscopic eld in the medium whereas the eld in Eq 41 is due to everything except the particular atom under consideration polarizability was de ned for an isolated atom subject to a speci ed external eld call this eld E6156 Imagine that the space allotted to each atom is a sphere of radius R and show that N E 1 1 E6156 471 360 Use this to conclude that Not60 X8 l Not360 or 3 l 60 Gr 472 a N 5r 2 Equation 472 is known as the ClausiusMossotti formula or in its application to optics the LorentzLorenz equation Problem 439 Check the ClausiusMossotti relation Eq 472 for the gases listed in Table 4 l Dielectric constants are given in Table 42 The densities here are so small that Eqs 470 and 472 are indistinguishable For experimental data that con rm the ClausiusMossotti correction term see for instance the rst edition of Purcell s Electricity and Magnetism Problem 92814 Problem 440 The ClausiusMossotti equation Prob 438 tells you how to calculate the susceptibility of a nonpolar substance in terms of the atomic polarizability a The Langevin equation tells you how to calculate the susceptibility of a polar substance in terms of the permanent molecular dipole moment p Here s how it goes a The energy of a dipole in an external eld E is u p E Eq 46 it ranges from pE to pE depending on the orientation Statistical mechanics says that for a material in equilibrium at absolute temperature T the probability of a given molecule having energy u is proportional to the Boltzmann factor exp u k T The average energy of the dipoles is therefore Lie du folHm ltugt 14E M Purcell Electricity and Magnetism Berkeley Physics Course Vol 2 New York McGraw Hill 1963 44 LINEAR DIELECTRICS 201 where the integrals run from pE to pE Use this to show that the polarization of a substance containing N molecules per unit volume is P NpcothpEkT kTpE 473 That s the Langevin formula Sketch P N p as a function of pEkT b Notice that for large eldslow temperatures virtually all the molecules are lined up and the material is nonlinear Ordinarily however kT is much greater than pE Show that in this r gime the material is linear and calculate its susceptibility in terms of N p T and k Compute the susceptibility of water at 20 C and compare the experimental value in Table 42 The dipole moment of water is 61 X 10 30 Cm This is rather far off because we have again neglected the distinction between E and E6156 The agreement is better in lowdensity gases for which the difference between E and E61se is negligible Try it for water vapor at 100 and 1 atm Chapter 5 Magnetostatics 51 The Lorentz Force Law 511 Magnetic Fields Remember the basic problem of classical electrodynamics We have a collection of charges q q2 q3 the source charges and we want to calculate the force they exert on some other charge Q the test charge See Fig 51 According to the principle of superposition it is suf cient to nd the force of a single source charge the total is then the vector sum of all the individual forces Up to now we have con ned our attention to the simplest case electrostatics in which the source charge is at rest though the test charge need not be The time has come to consider the forces between charges in motion 39ql Q 11239 C o 13 Source charges Test charge Figure 51 To give you some sense of what is in store imagine that I set up the following demon stration Two wires hang from the ceiling a few centimeters apart when I turn on a current so that it passes up one wire and back down the other the wires jump apart they evidentl repel one another Fig 52a How do you explain this Well you might suppose that the battery or whatever drives the current is actually charging up the wire and that the force is simply due to the electrical repulsion of like charges But this explanation is in correct I could hold up a test charge near these wires and there would be no force on it 202 51 THE LORENTZ FORCE LAW 203 a Currents in opposite b Currents in same directions repel directions attract Figure 52 for the wires are in fact electrically neutral It s true that electrons are owing down the line that s what a current is but there are just as many stationary plus charges as moving minus charges on any given segment Moreover I could hook up my demonstration so as to make the current ow up both wires Fig 52b in this case they are found to attract Whatever force accounts for the attraction of parallel currents and the repulsion of antiparallel ones is not electrostatic in nature It is our rst encounter with a magnetic force Whereas a stationary charge produces only an electric eld E in the space around it a moving charge generates in addition a magnetic eld B In fact magnetic elds are a lot easier to detect in practice all you need is a Boy Scout compass How these devices work is irrelevant at the moment it is enough to know that the needle points in the direction of the local magnetic eld Ordinarily this means north in response to the earth s magnetic eld but in the laboratory where typical elds may be hundreds of times stronger than that the compass indicates the direction of whatever magnetic eld is present Now if you hold up a tiny compass in the vicinity of a currentcarrying wire you quickly discover a very peculiar thing The eld does not point toward the wire nor away from it but rather it circles around the wire In fact if you grab the wire with your right 204 CHAPTER 5 MAGNETOSTATICS Current Wire 1 Wire 2 Figure 53 Figure 54 hand thumb in the direction of the current your ngers curl around in the direction of the magnetic eld Fig 53 How can such a eld lead to a force of attraction on a nearby parallel current At the second wire the magnetic eld points into the page Fig 54 the velocity of the charges is upward and yet the resulting force is to the left It s going to take a strange law to account for these directions I ll introduce this law in the next section Later on in Sect 52 we ll return to what is logically the prior question How do you calculate the magnetic eld of the rst wire 512 Magnetic Forces It may have occurred to you that the combination of directions in Fig 54 is just right for a cross product In fact the magnetic force in a charge Q moving with velocity v in a magnetic field B is1 Fmg Qv X B 51 This is known as the Lorentz fOI Ce law In the presence of both electric and magnetic elds the net force on Q would be F QE v x 3 52 I do not pretend to have derived Eq 51 of course it is a fundamental axiom of the theor whose justi cation is to be found in experiments such as the one I described in Sect 5 l 1 Our main job from now on is to calculate the magnetic eld B and for that matter the electric eld E as well for the rules are more complicated when the source charges are in motion But before we proceed it is worthwhile to take a closer look at the Lorentz force law itself it is a peculiar law and it leads to some truly bizarre particle trajectories 1Since F and v are vectors B is actually a pseudovector 51 THE LORENTZFORCE LAW 205 206 CHAPTER 5 MAGNETOSTATICS Z Example 51 I E Cyclotron motion The archetypical motion of a charged particle in a magnetic eld is circular with the magnetic force providing the centripetal acceleration In Fig 55 a uniform magnetic eld points into 0 b the page if the charge Q moves counterclockwise with speed v around a circle of radius R a C y the magnetic force 51 points inward and has a xed magnitude QvB just right to sustain A uniform circular motion 2 v QvB m or p QBR 53 where m is the particle s mass and p 2 mt is its momentum Equation 53 is known as the cyclotron formula because it describes the motion of a particle in a cyclotron the rst of the modem particle accelerators It also suggests a simple experimental technique for nding the momentum of a particle send it through a region of known magnetic eld and measure the radius of its circular trajectory This is in fact the standard means for determining the momenta of elementary particles Figure 57 to the right The faster it goes the stronger Fmag becomes eventually it curves the particle back around towards the y axis At this point the charge is moving against the electrical force so it begins to slow down the magnetic force then decreases and the electrical force takes over bringing the charge to rest at point a in Fig 57 There the entire process commences Inc1dentally I assumed that the charge moves in a plane perpendicular to B If It starts out anew carrying the particle over to point b and SO on with some additional speed 11quot parallel to B this component of the motion is unaffected by the magnetic eld and the particle moves in a helix Fig 56 The radius is still given by Eq 53 but the velocity in question is now the component perpendicular to B UL Now let s do it quantitatively There being no force in the xdirection the position of the particle at any time t can be described by the vector 0 311 zt the velocity is therefore V 0 y39 2 where dots indicate time derivatives Thus I y 2 va O y 2 23239 3512 B O O and hence applying Newton s second law F QEvgtltB QE2Bz39y Byi mamyy22 B U Or treating the y and 2 components separately Q32 2 QE Q8 2 For convenience let Figure 55 Figure 56 w E g 54 m This is the cyclotron frequency at which the particle would revolve in the absence of an electric eld Then the equations of motion take the form Example 52 E ywz zw y 55 Cycloid Motion B A more exotic trajectory occurs if we include a uniform electric eld at right angles to the Their general solution2 is magnetic one Suppose for instance that B points in the xdirection and E in the z direction ya C1 COS w C2 sin w EBt C3 56 as shown in Fig 57 A particle at rest 18 released from the or1g1n what path W111 it follow Z0 C2 COS wt C1 sin w C4 Solution Let s think it through qualitatively rst Initially the particle is at rest so the i p magnetic force is zero and the electric eld accelerates the charge in the zdirection AS it 2As coupled differential equations they are easdy solved by differentiating the rst and usmg the second In picks up speed a magnetic force develops which according to Eq 5 l pulls the charge around ehmmate 3 51 THE LOREN TZ FORCE LAW 207 But the particle started from rest 0 20 O at the origin y0 z0 0 these four conditions determine the constants C1 C2 C3 and C4 yt w Bwt sin wt zt 2 0 coswt 57 In this form the ansWer is not terribly enlightening but if we let R E i 5 8 L03 39 and eliminate the sines and cosines by exploiting the trigonometric identity sin2 wt cos2 cat 1 we nd that y sz2 z R2 R2 59 This is the formula for a circle of radius R whose center 0 th R travels in the ydirection at a constant speed E The particle moves as though it were a spot on the rim of a wheel rolling down the y axis at speed 1 The curve generated in this way is called a cycloid Notice that the overall motion is not in the direction of E as you might suppose but perpendicular to it One feature of the magnetic force law Eq 51 warrants special attention Magnetic forces do no work l For if Q moves an amount d1 v dt the work done is dWmldngmagdlzQvgtltBvdt0 511 This follows because v x B is perpendicular to v so v x B v 0 Magnetic forces may alter the direction in which a particle moves but they cannot speed it up or slow it down The fact that magnetic forces do no work is an elementary and direct consequence of the Lorentz force law but there are many situations in which it appears so manifestly false that one s con denCe is bound to waver When a magnetic crane lifts the carcass of a junked Car for instance something is obviously doing work and it seems perverse to deny that the magnetic force is responsible Well perverse or not deny it we must and it can be a very subtle matter to gure out what agency does deserve the credit in such circumstances I ll show you several examples as we go along Problem 51 A particle of charge q enters a region of uniform magnetic eld B pointing into the page The eld de ects the particle a distance d ab0ve the original line of ight as shown in Fig 58 Is the charge positive or negative In terms of a d B and q nd the momentum of the particle Problem 52 Find and sketch the trajectory of the particle in Ex 52 if it starts at the origin with velocity 1 V0 EBW 17 WW EZBW C V0 IEBM i 208 CHAPTER 5 MAGNETOSTATICS Field region Figure 58 Problem 53 In 1897 I I Thomson discovered the electron by measuring the chargeto mass ratio of cathode rays actually streams of electrons with charge q and mass m as follows a First he passed the beam through uniform crossed electric and magnetic elds E and B mutually perpendicular and both of them perpendicular to the beam and adjusted the electric eld until he got zero de ection What then was the speed of the particles in terms of E and B b Then he turned off the electric eld and measured the radius of curvature R of the beam as de ected by the magnetic eld alone In terms of E B and R what is the chargeto mass ratio q m of the particles 513 Currents The current in a wire is the charge per unit time passing a given point By de nition negative charges moving to the left count the same as positive ones to the right This conveniently re ects the physical fact that almost all phenomena involving moving charges depend on the product of charge and velocity if you change the sign of q and v you get the same answer so it doesn t really matter which you have The Lorentz force la is a case in point the Hall effect Prob 539 is a notorious exception In practice it is ordinarily the negatively charged electrons that do the moving in the direction opposite the electric current To avoid the petty complications this entails I shall often pretend it39s the positive charges that move as in fact everyone assumed they did for a century or so after Benjamin Franklin established his unfortunate convention3 Current is measured in coulombsper second or amperes A lAlCs 512 A line charge A traveling down a wire at speed 1 Fig 59 constitutes a current I A1 513 because a segment of length vAt carrying charge AvAt passes point P in a time interval At Current is actually a vector I 2 Av 514 3If we called the electron plus and the proton minus the problem would never arise In the Context of Franklin39s experiments with cat s fur and glass rods the choice was completely arbitrary 51 THE LORENTZ FORCE LAW 209 vAt 7 A P Figure 59 since the path of the ow is dictated by the shape of the wire most people don t bother to display the vectorial character of I explicitly but when it comes to surface and volume currents we cannot afford to be so casual and for the sake of notational consistency it is a good idea to acknowledge this right from the start A neutral wire of course contains as many stationary positive charges as mobile negative ones The former do not contribute to the current the charge density A in Eq 513 refers only to the moving charges In the unusual situation where both types move I Av Av The magnetic force on a segment of current carrying wire is evidently Fmag v x B dq V X BM dl fa x B dl 515 Inasmuch as I and d1 both point in the same direction we can just as well write this as Typically the current is constant in magnitude along the wire and in that case I comes outside the integral Fmg 1 d1 x B 517 Example 53 A rectangular loop of wire supporting a mass m hangs vertically with one end in a uniform magnetic eld B which points into the page in the shaded region of Fig 510 For what current I in the loop would the magnetic force upward exactly balance the gravitational force downward Solution First of all the current must circulate clockwise in order for I X B in the horizontal segment to point upward The force is Fmag I Ba where a is the width of the loop The magnetic forces on the two vertical segments cancel For Fmag to balance the weight mg we must therefore have mg Ba 39 The weight just hangs there suspended in midair I 518 210 CHAPTER 5 MAGNETOSTATICS Figure 510 What happens if we now increase the current Then the upward magnetic force exceeds the downward force of gravity and the loop rises lifting the weight Somebody s doing work and it sure looks as though the magnetic force is responsible Indeed one is tempted to write Wmg Fmagh IBah 519 where h is the distance the loop rises But we know that magnetic forces never do work What s going on here Well when the loop starts to rise the charges in the wire are no longer moving ho zontally their velocity now acquires an upward component u the speed of the loop Fig 5 1 l in addition to the horizontal component to associated with the current I Aw The magnetic force which is always perpendicular to the velocity no longer points straight up but tilts back It is perpendicular to the net displacement of the charge which is in the direction of v and therefore it does no work on q It does have a vertical component qu indeed the net vertical force on all the charge la in the upper segment of the loop is Fvert AawB IBa 5201 as before but now it also has a horizontal component tu which opposes the ow of current Whoever is in charge of maintaining that current therefore must now push those charges along against the backward component of the magnetic force tu Figure 511 51 THE LORENTZ FORCE LAW 211 Figure 512 The total horizontal force on the t0p segment is evidently FhoriZ AauB 521 In a time d t the charges move a horizontal distance w dt so the work done by this agency presumably a battery or a generator is Wbattery AaB uwdt IBah which is precisely what we naively attributed to the magnetic force in Eq 519 Was work done in this process Absolutely Who did it The battery What then was the role of the magnetic force Well it redirected the horizontal force of the battery into the vertical motion of the 100p and the weight It may help to consider a mechanical analogy Imagine you re pushing a trunk up a frictionless ramp by pushing on it horizontally with a m0p Fig 512 The normal force N does no work because it is perpendicular to the displacement But it does have a vertical component which in fact is what lifts the trunk and a backward horizontal component which you have to overcome by pushing on the mop Who is doing the work here You are obviously and yet your force which is purely horizontal is not at least not directly what lifts the box The normal force plays the same passive but crucial role as the magnetic force in Ex 53 while doing no work itself it redirects the efforts of the active agent you or the battery as the case may be from horizontal to vertical When charge ows over a surface we describe it by the surface current density K de ned as follows Consider a ribbon of in nitesimal width dlL running parallel to the ow Fig 513 If the current in this ribbon is all the surface current density is all K E 522 dlL In words K is the current per unit width perpendicularto ow In particular if the mobile surface charge density is o and its velocity is v then K 2 av 523 In general K will vary from point to point over the surface re ecting variations in 0 andor v The magnetic force on the surface current is Fmag v X Bo da 2 K X B da 524 212 CHAPTER 5 MAGNETOSTATICS Figure 513 Caveat Just as E suffers a discontinuity at a surface charge so B is discontinuous at a surface current In Eq 524 you must be careful to use the average eld just as we did in Sect 253 When the ow of charge is distributed throughout a threedimensional region we de scribe it by the volume current density J de ned aS follows Consider a tube of in nitesimal cross section dai running parallel to the ow Fig 514 If the Current in this tube is all the volume current density is J E 525 In words J is the current per unit areaperpendicularto ow If the mobile volume charge density is p and the velocity is v then J pv 526 The magnetic force on a volume current is therefore Fmag v X Bpd739 O X Bd739 527 Flow Figure 514 51 THE LORENTZ FORCE LAW 213 Figure 515 Figure 516 Example 54 a A current I is uniformly distributed over a wire of circular cross section with radius a Fig 515 Find the volume current density J Solution The areaperpendicular to ow is naQ so This was trivial because the current density was uniform b Suppose the current density in the wire is proportional to the distance from the axis J ks for some constant k Find the total current in the wire Solution Because J varies with s we must integrate Eq 525 The current in the shaded patch Fig 516 is JdaL and daL sds dqb So 2739rka3 a 1 kssdsd 2nk szds 3 0 According to Eq 525 the current crossing a surface 8 can be written as I JdaL2Jda 528 S S The dot product serves neatly to pick out the appropriate component of da In particular the total charge per unit time leaving a volume V is yiJdaVJdr S V 214 CHAPTER 5 MAGNETOSTATICS Because charge is conserved whatever ows out through the surface must come at the expense of that remaining inside i 3 fvVJdr dtvpdr fvltatgtdt The minus sign re ects the fact that an outward ow decreases the charge left in V Since this applies to any volume we conclude that 3 V J 529 at This is the precise mathematical statement of local charge conservation it is called the continuity equation For future reference let me summarize the dictionary We have implicitly developed for translating equations into the forms appropriate to point line surfaCe and volume currents Z CIiViN IdIN 1 line 5 urface Kda Jdr 530 volume This correspondence which is analogous to q A d I a da p d r for the various charge distributions generates Eqs 515 524 and 527 from the original Lorentz force law 51 Problem 54 Suppose that the magnetic eld in some region has the form B kz x where k is a constant Find the force on a square loop side a lying in the yz plane and centered at the Origin if it carries a current I owing counterclockwise when you look dOWn the x axis Problem 55 A current I ows down a wire of radius a a If it is uniformly distributed over the surface what is the surface current density K 7 b If it is distributed in such a way that the volume current density is inversely pr0portional to the distance from the axis what is J Problem 56 a A phonograph record carries a uniform density of static electricity 0 If it rotates at angular velocity u what is the surface current density K at a distance r from the center b A uniformly charged solid sphere of radius R and total charge Q is centered at the origin and spinning at a constant angular velocity u about the z axis Find the current density J at any point r 6 15 within the sphere Problem 57 For a con guration of charges and currents con ned within a volume V sho that Jdt dpdt V where p is the total dipole moment Hint evaluate fv V xJdr 52 THE BIOTSAVART LAW 215 52 The BiotSavart Law 521 Steady Currents Stationary charges produce electric elds that are constant in time hence the term electro statics4 Steady currents produce magnetic elds that are constant in time the theory of steady currents is called magnetostatics Stationary charges gt constant electric elds electrostatics Steady currents 39 gt constant magnetic elds magnetostatics By steady current I mean a continuous ow that has been going on forever with out change and without charge piling up anywhere Some people call them stationary currents to my ear that s a contradiction in terms Of course there s no such thing in practice as a truly steady current any more than there is a truly stationary charge In this sense both electrostatics and magnetostatics describe arti cial worlds that exist only in textbooks However they represent suitable approximations as long as the actual uctua tions are reasonably slow in fact for most purposes magnetostatics applies very well to household currents which alternate 60 times a second Notice that a moving point charge cannot possibly constitute a steady current If it s here one instant it s gone the next This may seem like a minor thing to you but it s a major headache for me I developed each topic in electrostatics by starting out with the simple case of a point charge at rest then I generaliged to an arbitrary charge distribution by invoking the superposition principle This approachiis not open to us in magnetostatics because a moving point charge does not produce a static eld in the rst place We are forced to deal with extended current distributions right from the start and as a result the arguments are bound to be more cumbersome When a steady current ows in a wire its magnitude I must be the same all along the line otherwise charge would be piling up somewhere and it wouldn t be a steady current By the same token Zip8t 0 in magnetostatics and hence the continuity equation 529 becomes V 39J 0 531 522 The Magnetic Field of a Steady Current The magnetic eld of a steady line current is given by the BiotSavart law uo I X fl uo d X I B dl I 532 r 4n t2 39 4n t2 39 4Actually it is not necessary that the charges be stationary but only that the charge density at each point be constant For example the sphere in Prob 56b produces an electrostatic eld 1 471 60Q r2i39 even though it is rotating because p does not depend on t 216 CHAPTER 5 MAGNETOS TATICS dl Figure 517 The integration is along the current path in the direction of the ow a39l is an element of length along the wire and It as always is the vector from the source to the point r Fig 517 The constant uo is called the permeability of free space5 M0 471 x 10 7 NAZ 533 These units are such that B itself comes out in newtons per ampere meter as required by the Lorentz force law or teslas T6 1 T 1 NA m 534 As the starting point for magnetostatics the Biot Savart law plays a role analogous to Coulomb s law in electrostatics Indeed the 142 dependence is common to both laws Example 55 1Find the magnetic eld a distance s from a long straight wire carrying a steady current I EFig 518 Solution In the diagram dl x 12 points out of the page and has the magnitude dl sinot dl cosQ AlSO l s tanQ so s md6 andsoCoseso 1 cos26 42 s2 5This is an exact number not an empirical constant It serves via Eq 537 to de ne the ampere and the ampere in turn de nes the coulomb 6For some reason in this one case the cgs unit the gauss is more commonly used than the SI unit 1 tesla 104 gauss The earth s magnetic eld is about half a gauss a fairly strong laboratory magnetic eld is say 10000 gauss 52 THE BI OT SAVART LAW 217 P Wire segment Figure 518 Figure 519 Thus 62 2 B M 01 COS 6 cos6d6 4H 91 s2 cos2 6 1 92 1 2 Eq cos6 d6 L Q sin 62 s1n 61 535 471 s 91 47m Equation 535 gives the eld of any straight segment of wire in terms of the initial and nal angles 61 and 62 Fig 519 Of course a nite segment by itself could never support a steady current where would the charge go when it got to the end but it might be a piece of some closed circuit and Eq 535 would then represent its contribution to the total eld In the case of an in nite wire 61 7r2 and 62 7r2 so we obtain I B 32 536 27m Notice that the eld is inversely proportional to the distance from the wire just like the electric eld of an in nite line charge In the region below the wire B points into the page and in general it circles around the wire in accordance with the righthand rule stated earlier Fig 53 it As an application let s nd the force of attraction between two long parallel wires a distance if d apart carrying currents 11 and 12 Fig 520 The eld at 2 due to l is 2 E91 3 27rd and it points into the page The Lorentz force law in the form appropriate to line currents Eq 517 predicts a force directed towards l of magnitude I F212 The total force not surprisingly is in nite but the force per unit length is 1112 537 27r d 21 8 CHAPTER 5 MAGNETOS TATI CS 1 2 Figure 520 If the currents are antiparallel one up one down the force is repulsive consistent again with the qualitative observations in Sect 511 Example 56 Find the magnetic eld a distance z above the center of a circular loop of radius R which carries a steady current I Fig 521 Figure 521 Solution The eld dB attributable to the segment dl points as shown As we integrate dl around the loop dB sweeps out a cone The horizontal components cancel and the vertical components combine to give 0 dl B z I cos 6 4 f L2 Notice that dl and i are perpendicular in this case the factor of cos 0 projects out the vertical component Now cos 6 and a2 are constants and f dl is simply the Circumference 2n R so no c050 Ruol R2 4yr 2 7 2 R2z23239 Bz 538 52 THE BI OT SAVART LAW 219 For surface and volume currents the Biot Savart law becomes K A A BO Mda and Br Mdr 539 471 42 471 42 respectively You might be tempted to write down the corresponding formula for a moving point charge using the dictionary 530 0 W W 471 2 Br 540 but this is simply wrong7 As I mentioned earlier a point charge does not constitute a steady current and the BiotSavart law which only holds for steady currents does not correctly determine its eld Incidentally the superposition principle applies to magnetic elds just as it does to electric elds If you have a collection of source currents the net eld is the vector sum of the elds due to each of them taken separately Problem 58 a Find the magnetic eld at the center of a square loop which carries a steady current I Let R be the distance from center to side Fig 522 b Find the eld at the center of a regular n sided polygon carrying a steady current I Again let R be the distance from the center to any side c Check that your formula reduces to the eld at the center of a circular loop in the limit n gt 00 Problem 59 Find the magnetic eld at point P for each of the steady current con gurations shown in Fig 523 7 I a b Figure 522 Figure 523 71 say this loud and clear to emphasize the point of principle actually Eq 540 is approximately right for nonrelativistic charges u ltlt c under conditions where retardation can be neglected see Ex 104 220 CHAPTER 5 MAGNETOSTATICS Figure 524 Problem 510 a Find the force on a square loop placed as shown in Fig 524a near an in nite straight wire Both the loop and the wire carry a steady current I b Find the force on the triangular loop in Fig 524b Figure 525 Problem 511 Find the magnetic eld at point P on the axis of a tightly wound solenoid helical coil consisting of It turns per unit length wrapped around a cylindrical tube of radius a and carrying current I Fig 525 Express your answer in terms of 61 and 62 it s easiest that way Consider the turns to be essentially circular and use the result of Ex 56 What is the eld on the axis of an in nite solenoid in nite in both directions Figure 526 Problem 512 Suppose you have two in nite straight line charges A a distance d apart moving along at a constant speed 1 Fig 526 How great would u have to be in order for the magnetic attraction to balance the electrical repulsion Work out the actual number Is this a reasonable sort of speed8 8If you ve studied special relativity you may be tempted to look for complexities in this problem that are not really there A and v are both measured in the laboratory frame and this is ordinary electrostatics see footnote 4 340 73 MAX WELL S EQUATIONS 339 C Using the result in a solve these equations for 1t and 12t Assume 11 has no DC component 1 Show that the output voltage Vout 12 R divided by the input voltage Vin is equal to the turns ratio Vout Vin N2N1 e Calculate the input power Pin 2 Vin 1 and the output power Pout Voutlz and show that their averages over a full cycle are equal Problem 755 Suppose J r is constant in time but 0 r t is not conditions that might prevail for instance during the charging of a capacitor a Show that the charge density at any particular point is a linear function of time MIquot I p02 0 x301 0t where pa 0 is the time derivative of 0 at t 0 This is not an electrostatic or magnetostatic con guration21 nevertheless rather surprisingly both Coulomb s law in the form of Eq 28 and the Biot Savart law Eq 539 hold as you can con rm by showing that they satisfy Maxwell s equations In particular b Show that 0 J0quot X 2 B r d 471 f a2 I obeys Ampere s law with Maxwell s displacement current term Problem 756 The magnetic eld of an in nite straight wire carrying a steady current I can be obtained from the displacement current term in the AmpereMaxwell law as follows Picture the current as consisting of a uniform line charge A moving along the z axis at speed 1 so that 2 Av with a tiny gap of length 6 which reaches the origin at time t 0 In the next instant up to t 61 there is no real current passing through a circular Amperian loop in the xy plane but there is a displacement current due to the missing charge in the gap a Use Coulomb s law to calculate the 2 component of the electric eld for points in the xy plane a distance s from the ori gin due to a segment of wire with uniform density A extending from z vt e to Q vt b Determine the ux of this electric eld through a circle of radius a in the x y plane c Find the displacement current through this circle Show that Id is equal to I in the limit as the gap width 6 goes to zero For a slightly different approach to the same problem see w K Terry Am J Phys 50 742 1982 Problem 757 The magnetic eld outside a long straight wire carrying a steady current 1 is of course I A B 271 s 2 Some authors would regard this as magnetostalic since B is independent of t For them the Biot Savart law is a general rule of ma netostatics but V J 0 and V x B 1 J a l on under the additional assumption that g 0 PP y y CHAPTER 7 ELECTRODYNAMICS Figure 755 The electric eld inside the wire is uniform where 0 is the resistivity and a is the radius see Exs 71 and 73 Question What is the electric eld outside the wire This is a famous problem rst analyzed by Sommerfeld and known in its most recent incarnation as Merzbacher s puzzle 3 22 The answer depends on how you complete the circuit Suppose the current returns along a perfectly conducting grounded coaxial cylinder of radius 17 Fig 755 In the region a lt s lt b the potential Vs z satis es Laplace s equation with the boundary conditions i Vaz 1pz ii Vbz0 71112 Unfortunately this does not suf ce to determine the answer we still need to specify boundary conditions at the two ends In the literature it is customary to sweep this ambiguity under the rug by simply asserting in so many words that Vs z is proportional to z Vs z z f s On this assumption a Determine Vs z b Find Es z c Calculate the surface charge density 0z on the wire Answer V Izpna2lnsb lnab This is apeculiar result since E5 and 02 are not independent of z as one would certainly expect for a truly in nite wire Problem 758 A certain transmission line is constructed from two thin metal ribbons of width w a very small distance h ltlt w apart The current travels down one strip and back along the other In each case it spreads out uniformly over the surface of the ribbon a Find the capacitance per unit length C b Find the inductance per unit length L c What is the product LC numerically L and C will of course vary from one kind of transmission line to another but their product is a universal constant check for example the cable in Ex 713 provided the space between the conductors is a vacuum In the theory of transmission lines this product is related to the speed with which a pulse propagates down the line 1 1v C o is constant In such a formulation Maxwell s displacement term can in this very special case be derived from the Biot Savart law by the method of part b See D F Bartlett Am J Phys 58 1168 1990 D J Grif ths and M A Heald Am J Phys 59 111 1991 22A Sommerfeld Electrodynamics p 125 New York Academic Press 1952 E Merzbacher Am J Phys 48 104 1980 further references in M A Heald Am J Phys 52 522 1984 a 5 3 R for in nitesimal dt Use the method of Sect 713 to rewrite the second integral as dtf B gtltvdl 73 and invoke Stokes theorem to conclude that dd 3B V dt 19at gtlt V X B da Together with the result in a this proves the theorem Figure 756 342 CHAPTER 7 ELECTRODYNAMICS Problem 760 a Show that Maxwell s equations with magnetic charge Eq 743 are invariant under the duality transformation 13 2 E0080 cBSin01 CB CB cos or E Sing 768 cqe cqe cos 0 qm 31 05 qr n qm cosoz qu Sin 0 where c E l 0 MO and a is an arbitrary rotation angle in EB space Charge and current densities transform in the same way as qe and qm This means in particular that if you know the elds produced by a con guration of electric charge you can immediately using a 90 write down the elds produced by the corresponding arrangement of magnetic charge b Show that the force law Prob 735 1 FqeEVgtltBqmB 2vgtltE 769 c is also invariant under the duality transformation Intermission All of our cards are now on the table and in a sense my job is done In the rst seven chapters we assembled electrodynamics piece by piece and now with Maxwell s equations in their nal form the theory is complete There are no more laws to be learned no further generalizations to be considered and with perhaps one exception no lurking inconsistencies to be resolved If yours is a onesemester course this would be a reasonable place to stop But in another sense we have just arrived at the starting point We are at last in possession of a full deck and we know the rules of the game it s time to deal This is the fun part in which one comes to appreciate the extraordinary power and richness of electrodynamics In a full year course there should be plenty of time to cover the remaining chapters and perhaps to supplement them with a unit on plasma physics say or AC circuit theory or even a little General Relativity But if you have room only for one topic I d recommend Chapter 9 on Electromagnetic Waves you ll probably want to skim Chapter 8 as preparation This is the segue to Optics and is historically the most important application of Maxwell s theory Chapter 8 Conservation Laws 81 Charge and Energy 811 The Continuity Equation In this chapter we study conservation of energy momentum and angular momentum in electrodynamics But I want to begin by reviewing the conservation of charge because it is the paradigm for all conservation laws What precisely does conservation of charge tell us That the total charge in the universe is constant Well sure that s global conservation of charge but local conservation of charge is a much stronger statement If the total charge in some volume changes then exactly that amount of charge must have passed in or out through the surface The tiger can t simply rematerialize outside the cage if it got from inside to outside it must have found a hole in the fence Formally the charge in a volume V is Qt V prr 0dr 81 and the current owing out through the boundary 8 is f5 J da so local conservation of charge says d Q W LJda 82 Using Eq 81 to rewrite the left side and invoking the divergence theorem on the right we have 3 dr V Jdr 83 12 3t 12 and since this is true for any volume it follows that 8x V at J 84 345 346 CHAPTER 8 CONSERVATION LAWS This is of course the continuity equation the precise mathematical statement of local conservation of charge As I indicated earlier it can be derived from Maxwell s equations conservation of charge is not an independent assumption but a consequence of the laws of electrodynamics The purpose of this chapter is to construct the corresponding equations for conservation of energy and conservation of momentum In the process and perhaps more important we will learn how to express the energy density and the momentum density the analogs t0 0 as well as the energy current and the momentum current analogous to J 812 Poynting s Theorem In Chapter 2 we found that the work necessary to assemble a static charge distribution against the Coulomb repulsion of like charges is Eq 245 We 6 0E2dr 2 where E is the resulting electric eld Likewise the work required to get currents going against the back emf is Eq 734 1 2 Wm B d1 220 where B is the resulting magnetic eld This suggests that the total energy stored in elec tromagnetic elds is l l Uem 60E2 B2gt d1 85 2 MO I propose to derive Eq 85 more generally now in the context of the energy conservation law for electrodynamics Suppose we have some charge and current con guration which at time 1 produces elds E and B In the next instant d t the charges move around a bit Question How much work d W is done by the electromagnetic forces acting on these charges in the interval d1 According to the Lorentz force law the work done on a charge 61 is FdlqEvavdtqEvdt Now q pair and ov J so the rate at which work is done on all the charges in a volume V is dW E J d 8 6 7quot dt V Evidently E J is the work done per unit time per unit volume which is to say the power delivered per unit volume We can express this quantity in terms of the elds alone using the AmpereMaxwell law to eliminate J 1 8E EJ EVxB 60E M0 3 81 CHARGE AND ENERGY 347 From product rule 6 VEXBBV XE EV XB Invoking Faraday s law V x E 8B8t it follows that E39VXB B39 VEXB Meanwhile 8B 18B2 d E BE 1 8E2 87 an 8t 28 8t 28 SO 18 2 l 2 l EJ 60E B VEgtltB 88 28t no M0 Putting this into Eq 86 and applying the divergence theorem to the second term we have dW d 1 1 1 E2 32 d lE Bd 89 dt tit12260 Mo 1 M0 8 X a where S is the surface bounding V This is Poynting s theorem it is the work energy theorem of electrodynamics The rst integral on the right is the total energy stored in the elds Uem Eq 85 The second term evidently represents the rate at which energy is carried out of V across its boundary surface by the electromagnetic elds Poynting s theorem says then that the work done on the charges by the electromagnetic force is equal to the decrease in energy stored in the eld less the energy that owed out through the surface The energy per unit time per unit area transported by the elds is called the Poynting vector S E E x B 810 Speci cally S da is the energy per unit time crossing the in nitesimal surface da the energy ux if you like so S is the energy ux density1 We will see many applications of the Poynting vector in Chapters 9 and ll but for the moment I am mainly interested in using it to express Poynting s theorem more compactly dW dU em yisda 811 dt dt 5 1If you re very fastidious you ll notice a small gap in the logic here We know from Eq 89 that f S da is the total power passing through a closed surface but this does not prove that f S da is the power passing through any open surface there could be an extra term that integrates to zero over all closed surfaces This is however the obvious and natural interpretation as always the precise location of energy is not really determined in electrodynamics see Sect 244 348 CHAPTER 8 CONSERVATION LAWS Of course the work W done on the charges will increase their mechanical energy kinetic potential or whatever If we let umech denote the mechanical energy density so that dW d I E Vumech d7 812 and use u6m for the energy density of the elds 1 2 1 2 uem 60E B 813 2 0 then d mech emd7 2 S39d3 VSd139 dt v s v and hence 3 EWmeCh l em V 8J4 This is the differential version of Poynting s theorem Compare it with the continuity equation expressing conservation of charge Eq 84 8x I V at J the charge density is replaced by the energy density mechanical plus electromagnetic and the current density is replaced by the Poynting vector The latter represents the ow of energy in exactly the same way that J describes the ow of charge2 Example 81 When current ows down a wire work is done which shows up as Joule heating of the wire Eq 77 Though there are certainly easier ways to do it the energy per unit time delivered to the wire can be calculated using the Poynting vector A5suming it s uniform the electric eld parallel to the wire is V E L where V is the potential difference between the ends and L is the length of the wire Fig 81 The magnetic eld is circumferential at the surface radius a it has the value 01 B 27m Accordingly the magnitude of the Poynting vector is 1Vu01 VI pg L 27m 27taL 2In the presence of linear media one is typically interested only in the Work done on free charges and currents see Sect 443 In that case the appropriate energy density is E D B H and the Poynting vector becomes E X H See J D Jackson Classical Electrodynamics 3rd ed Sect 67 New York John Wiley 1999 82 MOMENTUM 349 L Figure 81 and it points radially inward The energy per unit time passing in through the surface of the wire is therefore Sda S27taL 2 VI which is exactly what we concluded on much more direct grounds in Sect 711 Problem 81 Calculate the power energy per unit time transported down the cables of Ex 713 and Prob 7 5 8 assuming the two conductors are held at potential difference V and carry current 1 down one and back up the other Problem 82 Consider the charging capacitor in Prob 731 a Find the electric and magnetic elds in the gap as functions of the distance s from the axis and the time t Assume the charge is zero at t 0 b Find the energy density uem and the Poynting vector S in the gap Note especially the direction of S Check that Eq 814 is satis ed c Determine the total energy in the gap as a function of time Calculate the total power owing into the gap by integrating the Poynting vector over the appropriate surface Check that the power input is equal to the rate of increase of energy in the gap Eq 89 in this case W 0 because there is no charge in the gap If you re worried about the fringing elds do it for a volume of radius 7 lt a well inside the gap 82 Momentum 821 Newton s Third Law in Electrodynamics Imagine a point charge q traveling in along the x axis at a constant speed 11 Because it is moving its electric eld is not given by Coulomb s law nevertheless E still points radially outward from the instantaneous position of the charge Fig 82a as we ll see in Chapter 10 Since moreover a moving point charge does not constitute a steady current its magnetic 350 CHAPTER 8 CONSERVATION LAWS E B v z m a x KiddI x a b Figure 82 eld is not given by the Biot Savart law Nevertheless it s a fact that B still circles around the axis in a manner suggested by the right hand rule Fig 82b again the proof will come in Chapter 10 Now suppose this charge encounters an identical one proceeding in at the same speed along the y axis Of course the electromagnetic force between them would tend to drive them off the axes but let s assume that they re mounted on tracks or something so they re forced to maintain the same direction and the same speed Fig 83 The electric force between them is repulsive but how about the magnetic force Well the magnetic eld of 11 points into the page at the position of 2 so the magnetic force on 12 is toward the right whereas the magnetic eld of q is out of the page at the position of ql and the magnetic force on q is upward The electromagnetic force of ql on 612 is equal but not opposite to the force of 2 on 611 in violation of Newton s third law In electrostatics and magnetostatics the third law holds but in electrodynamics it does not Fequoty B 72 Fm v Fm V2 41 Figure 83 82 MOMENTUM 35 1 Well that s an interesting curiosity but then how often does one actually use the third law in practice Answer All the time For the proof of conservation of momentum rests on the cancellation of internal forces which follows from the third law When you tamper with the third law you are placing conservation of momentum in jeopardy and there is no principle in physics more sacred than that Momentum conservation is rescued in electrodynamjcs by the realization that the elds themselves carry momentum This is not so surprising when you consider that we have already attributed energy to the elds In the case of the two point charges in Fig 83 whatever momentum is lost to the particles is gained by the elds Only when the eld momentum is added to the mechanical momentum of the charges is momentum conservation restored You ll see how this works out quantitatively in the following sections 822 Maxwell s Stress Tensor Let s calculate the total electromagnetic force on the charges in volume V FZEVXBpdTf0EJXBdT 815 V V The force per unit volume is evidently fpEJxB 816 As before I propose to write this in terms of elds alone eliminating p and J by using Maxwell s equations i and iv 1 E f60VEE V xB eoa XB M0 3 3 8E 8B E B B E 3t X th X3tgt and Faraday s law says 3B V XE 3 S0 BE 8 Thus f 60V EE E x V x E 1 B x V x B 60E x B 817 0 352 CHAPTER 8 CONSERVATION LAWS Just to make things look more symmetrical let s throw in a term V BB since V B 0 this costs us nothing Meanwhile product rule 4 says VE2 203 VE 2E x v x E E x V x E VE2 EVE and the same goes for B Therefore f 601V EE E VE luv BB B VB 0 818 l l 8 V E2 B2 E B 2 60 M0 60 at X Uglyl But it can be simpli ed by introducing the Maxwell stress tensor 1 2 1 1 2 The indices i and j refer to the coordinates x y and z so the stress tensor has a total of nine components Txx Tyy sz Tyx and so on The Kronecker delta 81 is 1 if the indices are the same 6 5 5Z1 l and zero otherwise SW 6 SW 0 Thus 1 1 2 2 2 2 2 2 TH 260w Ey Ez 2was 15y BZ l Txy 60ExEy BxBys MO and so on Because it carries two indices where a vector has only one Ti is sometimes written with a double arrow One can form the dot product of T with a vector 3 2 Z aiTij 820 ixyz the resulting object which has one remaining index is itself a vector In particular the divergence of T has as its j th component 1 1 1 VBBBV B VB2 MO I J J 2 J Thus the force per unit volume Eq 818 can be written in the much simpler form as fV 60 L0 at 821 where S is the Poynting vector Eq 810 82 MOMENTUM 353 The total force on the charges in V Eq 815 is evidently d F 2 7 da Woo f Sdr 822 5 dt v I used the divergence theorem to convert the rst term to a surface integral In the static case or more generally whenever f S d r is independent of time the second term drops out and the electromagnetic force on the charge con guration can be expressed entirely in terms of the stress tensor at the boundary Physically T is the force per unit area or stress acting on the surface More precisely Ti is the force per unit area in the i th direction acting on an element of surface oriented in the jth direction diagonal elements Txx Tyy Tu represent pressures and off diagonal elements Txy T XZ etc are shears Example 82 Determine the net force on the northern hemisphere of a uniformly charged solid sphere of radius R and charge Q the same as Prob 243 only this time we ll use the Maxwell stress tensor and Eq 822 Disk Figure 84 Solution The boundary surface consists of two parts a hemispherical bowl at radius R and a circular disk at 6 n2 Fig 84 For the bowl da R2 sin6d6 d f and 1 Q A r 47160 R2 In Cartesian components f sin6cos xsin6sin cos i Q 2 T 6 E E 6 s1n0cos0cos zx 0 z x 0lt4NEOR2 P Q 2 T 6 E E 6 sin6cosesin zy 0 1 v 04HEOR2 Z5 60 E2 E2 2 60 Q 2 2 26 823 TZZ3 Z x E7 4n60R2 cos 6 s1n 354 CHAPTER 8 CONSERVATION LAWS The net force is obviously in the zdirection so it suf ces to calculate 4f 60 Q 2 daZ sz daX TZ v day TZZ daZ 4n60R s1n6cos6d6 d The force on the bowl is therefore 2 7r2 2 E0 Q 1 Q F 2 2 bowl 2 47mm nO s1n6 c0s6d6 4H60 8R2 824 Meanwhile for the equatorial disk da r dr d 2 825 and since we are now inside the sphere 1 Q 1 Q A A 47T 0 F r m Rj cosq x s1n y Thus 2 E0 2 2 2 60 Q 2 T E E E ZZ z x 2 r 7 andhence 2 Tda 7 6 0 h Q 3d d 2 4nEOR3 r r 4539 The force on the disk is therefore 2 R 2 60 Q r 1 Q F 27f r d 2 1 2 4neOR3 0 r 47T 0 16R2 8 6 Combining Eqs 824 and 826 I conclude that the net force on the northern hemisphere is 1 392 47r60 827 Incidentally in applying Eq 822 any volume that encloses all of the charge in question and no other charge will do the job For example in the present case we could use the whole region 2 gt 0 In that case the boundary surface consists of the entire x y plane plus a hemisphere at r 00 but E 0 out there anyway so it contributes nothing In place of the bowl we now have the outer portion of the plane r gt R Here 60 Q 2 l Tzz T 2 47160 r Eq 823 with 6 n2 and R gt r and dais given by Eq 825 so 2 T daz6 0 Q idrdqs 2 47TEO r3 and the contribution from the plane for r gt R is 2 oo 2 E l l 22 Q 271 dr amp 2 27r 0 R r3 47r60 8R2 the same as for the bowl Eq 824 82 MOMENTUM 355 I hope you didn t get too bogged down in the details of Ex 82 If so take a moment to appreciate what happened We were calculating the force on a solid object but instead of doing a volume integral as you might expect Eq 822 allowed us to set it up as a surface integral somehow the stress tensor sniffs out what is going on inside Problem 83 Calculate the force of magnetic attraction between the northern and southern hemispheres of a uniformly charged spinning spherical shell with radius R angular velocity to and surface charge density a This is the same as Prob 542 but this time use the Maxwell stress tensor and Eq 822 Problem 84 a Consider two equal point charges q separated by a distance 2a Construct the plane equidistant from the two charges By integrating Maxwell s stress tensor over this plane determine the force of one charge on the other b Do the same for charges that are opposite in sign 823 Conservation of Momentum According to Newton s second law the force on an object is equal to the rate of change of its momentum F dpmech dt Equation 822 can therefore be written in the form dpmech d f i Sdr Toa 828 dt 60 dt V s where pmech is the total mechanical momentum of the particles contained in the volume V This expression is similar in structure to Poynting s theorem Eq 89 and it invites an anal ogous interpretation The rst integral represents momentum stored in the electromagnetic elds themselves Pem MoeoVSdr 829 while the second integral is the momentum per unit time owing in through the surface Equation 828 is the general statement of conservation of momentum in electrodynamics Any increase in the total momentum mechanical plus electromagnetic is equal to the momentum brought in by the elds If V is all of space then no momentum ows in or out and pmech pem is constant As in the case of conservation of charge and conservation of energy conservation of momentum can be given a differential formulation Let pmech be the density of mechanical momentum and pen the density of momentum in the elds 83m 356 CHAPTER 8 CONSERVATION LAWS Then Eq 828 in differential form says a 560mb seem V T 831 Evidently T is the momentum ux density playing the role of J current density in the continuity equation or S energy ux density in Poynting s theorem Speci cally Ti is the momentum in the i direction crossing a surface oriented in the j direction per unit area per unit time Notice that the Poynting vector has appeared in two quite different roles S itself is the energy per unit area per unit time transported by the electromagnetic elds while uoeos is the momentum per unit volume stored in those elds Similarly T plays a dual role T itself is the electromagnetic stress force per unit area acting on a surface and T describes the ow of momentum the momentum current density transported by the elds Example 83 A long coaxial cable of length l consists of an inner conductor radius a and an outer condUCtor radius 3 It is connected to a battery at one end and a resistor at the other Fig 85 The inner conductor carries a uniform charge per unit length A and a steady current I to the right the outer conductor has the opposite charge and current What is the electromagnetic momentum stored in the elds Solution The elds are I A A I A E s B go 27160 s 271 s The Poynting vector is therefore A1 A S z 47260s2 Evidently energy is owing down the line from the battery to the resistor In fact the power transported is A b 1 A P S da 2 2739rs ds lnha 2 IV 471260 a s2 27550 Figure 85 82 MOMENTUM 357 as it should be But that s not what we re interested in right now The momentum in the elds is b pem 050 f 5 dr 13 20 127 ds 3 lnba 2 This is an astonishing result The cable is not moving and the elds are static and yet we are asked to believe that there is momentum in the system If something tells you this cannot be the whole story you have sound intuitions In fact if the center of mass of a localized system is at rest its total momentum must be zero In this case it turns out that there is hidden mechanical momentum associated with the ow of current and this exactly cancels the momentum in the elds But locating the hidden momentum is not easy and it is actually a relativistic effect so I shall save it for Chapter 12 Ex 1212 Suppose now that we turn up the resistance so the current decreases The changing magnetic eld will induce an electric eld Eq 719 Mod A E l K 271611 n i2 This eld exerts a force on le 0 d A 0 d A MOM d A FAl ln K tl lnb K z lnb 2 271m Jr 2 271m 271 dz U The total momentum imparted to the cable as the current drops from I to 0 is therefore Ml A Pmech det Mo lnwa Z 271 which is precisely the momentum originally stored in the elds The cable will not recoil however because an equal and opposite impulse is delivered by the simultaneous disappearance of the hidden momentum Problem 85 Consider an in nite parallel plate capacitor with the lower plate at z d 2 carrying the charge density 0 and the upper plate at z 61 2 carrying the charge density 0 a Determine all nine elements of the stress tensor in the region between the plates Display your answer as a 3 X 3 matrix Txx Txy TXZ Tyx Tyy Tyz sz sz Tzz b Use Eq 822 to determine the force per unit area on the top plate Compare Eq 251 c What is the momentum per unit area per unit time crossing the xy plane or any other plane parallel to that one between the plates d At the plates this momentum is absorbed and the plates recoil unless there is some nonelectrical force holding them in position Find the recoil force per unit area on the top 358 CHAPTER 8 CONSERVATION LAWS Figure 86 plate and compare your answer to b Note39 This is not an additional force but rather an alternative Way of calculating the same force in b we got it from the force law and in d we did it by conservation of momentum Problem 86 A charged parallel plate capacitor with uniform electric eld E E 2 is placed in a uniform magnetic eld B B 2 as shown in Fig 863 a Find the electromagnetic momentum in the space between the plates b Now a resistive wire is connected between the plates along the z axis so that the capacitor slowly discharges The current through the wire will experience a magnetic force what is the total impulse delivered to the system during the discharge c Instead of turning off the electric eld as in b suppose we slowly reduce the magnetic eld This will induce a Faraday electric eld which in turn exerts a force on the plates Show that the total impulse is again equal to the momentum originally stored in the elds 824 Angular Momentum By now the electromagnetic elds which started out as mediators of forces between charges have taken on a life of their own They carry energy Eq 813 1 1 uem 60192 B2 832 2 M0 and momentum Eq 830 pm M0608 60E X B 833 and for that matter angular momentum em r X pem 60 r X E X B 834 3See F S Johnson B L Cragin and R R Hodges Am J Phys 62 33 1994 82 MOMENTUM 359 Even perfectly static elds can harbor momentum and angular momentum as long as E X B is nonzero and it is only when these eld contributions are included that the classical conservation laws hold Example 84 Imagine a very long solenoid with radius R n turns per unit length and current I Coaxial with the solenoid are two long cylindrical shells of length l one inside the solenoid at radius a carries a charge Q uniformly distributed over its surface the other outside the solenoid at radius 13 carries charge Q see Fig 87 l is supposed to be much greater than b When the current in the solenoid is gradually reduced the cylinders begin to rotate as we found in Ex 78 Question Where does the angular momentum come from4 Solution It was initially stored in the elds Before the current was switched off there was an electric eld Q l altsltb E 271501 s Figure 87 4This is a variation on the Feynman disk paradox R P Feynman R B Leighton and M Sands The Feynman Lectures vol 2 pp 17 5 Reading Mass Addison Wesley 1964 suggested by F L Boos Jr Am J Phys 52 756 1984 A similar model was proposed earlier by R H Romer Am J Phys 34 772 1966 For further references see T C E Ma Am J Phys 54 949 1986 360 CHAPTER 8 CONSERVATION LAWS in the region between the cylinders and a magnetic eld B u0n12s lt R inside the solenoid The momentum density Eq 833 was therefore M0le A 2711s pem in the region a lt s lt R The angular momentum density was 1amp0le 2 2711 lemrgtltpem which is constant as it turns out to get the total angular momentum in the elds we simply multiply by the volume 739rR2 12 1 Lem u0n1QR2 a2 2 835 When the current is turned off the changing magnetic eld induces a circumferential electric eld given by Faraday s law 1 d1 R2 R Zuondt s b Sgt 1 d1 4 R n 39 2M0 dis a A lt E Thus the torque on the outer cylinder is l d I N r X E R2A b Q 2 0quot Q d 2 and it picks up an angular momentum d1 l 611 2A d 2MOnIQR z 1 0 Lb u0nQR22 2 I Similarly the torque on the inner cylinder is l d I N n 2 2 a 2 0 Qa dt and its angular momentum increase is l 2 A La EMonIQa Z So it all works out Lem 2 L Lb The angular momentum lost by the elds is precisely equal to the angular momentum gained by the cylinders and the total angular momentum elds plus matter is conserved 82 MOMENTUM 361 Incidentally the angular case is in some respects cleaner than the linear analog Ex 83 because there is no hidden angular momentum to compensate for the angular momentum in the elds and the cylinders really do rotate when the magnetic eld is turned off If a localized system is not moving its total linear momentum has to be zero5 but there is no corresponding theorem for angular momentum and in Prob 812 you will see a beautiful example in which nothing at all is moving not even currents and yet the angular momentum is nonzero Problem 87 In Ex 84 suppose that instead of turning off the magnetic eld by reducing I we turn off the electric eld by connecting a weakly6 conducting radial spoke between the cylinders We ll have to cut a slot in the solenoid so the cylinders can still rotate freely From the magnetic force on the current in the spoke determine the total angular momentum delivered to the cylinders as they discharge they are now rigidly connected so they rotate together Compare the initial angular momentum stored in the elds Eq 835 Notice that the mechanism by which angular momentum is transferred from the elds to the cylinders is entirely different in the two cases in Ex 84 it was Faraday s law but here it is the Lorentz force law Problem 887 Imagine an iron sphere of radius R that carries a charge Q and a uniform magnetization M M i The sphere is initially at rest a COmpute the angular momentum stored in the electromagnetic elds b Suppose the sphere is gradually and uniformly demagnetized perhaps by heating it up past the Curie point Use F araday s law to determine the induced electric eld nd the torque this eld exerts on the sphere and calculate the total angular momentum imparted to the sphere in the course of the demagnetization c Suppose instead of demagnetizing the sphere we discharge it by connecting a grounding wire to the north pole Assume the current ows over the surface in such a way that the charge density remains uniform Use the Lorentz force law to determine the torque on the sphere and calculate the total angular momentum imparted to the sphere in the course of the discharge The magnetic eld is discontinuous at the surface does this matter Answers 6qu 9R2 More Problems on Chapter 8 Problem 898 A very long solenoid of radius a with n turns per unit length carries a current I S Coaxial With the solenoid at radius 3 gtgt a is a circular ring of wire with resistance R When the current in the solenoid is gradually decreased a current I is induced in the ring 362 CHAPTER 8 CONSERVATION LAWS a Calculate 1 in terms of dISdt b The power I2 R delivered to the ring must have come from the solenoid Con rm this by calculating the Poynting vector just outside the solenoid the electric eld is due to the changing ux in the solenoid the magnetic eld is due to the current in the ring Integrate over the entire surface of the solenoid and check that you recover the correct total power Problem 8109 A sphere of radius R carries a uniform polarization P and a uniform magneti zation M not necessarily in the same direction Find the electromagnetic momentum of this con guration Answers 49na0R3M X P Problem 81110 Picture the electron as a uniformly charged spherical shell with charge e and radius R Spinning at angular velocity to a Calculate the total energy contained in the electromagnetic elds b Calculate the total angular momentum contained in the elds c According to the Einstein formula E mcz the energy in the elds should contribute to the mass of the electron Lorentz and others speculated that the entire mass of the electron might be accounted for in this way Uem mecz Suppose moreover that the electron s spin angular momentum is entirely attributable to the electromagnetic elds Lem 112 On these two assumptions determine the radius and angular velocity of the electron What is their product wR Does this classical model make sense Problem 81211 Suppose you had an electric charge qe and a magnetic monopole qm The eld of the electric charge is 1 qe A 47TEO a2 of course and the eld of the magnetic monopole is a 471 a2 Find the total angular momentum stored in the elds if the two charges are separated by a distance d Answer39 MO4nqeqm2 Problem 813 Paul DeYoung of Hope College points out that because the cylinders in Ex 84 are left rotating at angular velocities wa and tab say there is actually a residual magnetic eld and hence angular momentum in the elds even after the current in the solenoid has been extinguished If the cylinders are heavy this correction will be negligible but it is interesting to do the problem without making that assumption 9For an interesting discussion and references see R H Romer Am J Phys 63 777 1995 10566 J Higbie Am J Phys 56 378 1988 11This system is known as Thomson s dipole See 1 Adawi Am J Phys 44 762 1976 and Phys Rev D31 5S Coleman and J H van Vleck Phys Rev 171 1370 1968 6In Ex 84 We turned the current off slowly to keep things quasistatic here we reduce the electric eld slowly to keep the displacement current negligible 7This version of the Feynman disk paradox was proposed by N L Sharma Am J Phys 56 420 1988 similar models were analyzed by E M Pugh and G E Pugh Am J Phys 35 153 1967 and by R H Romer Am J Phys 35 445 1967 8For extensive discussion see M A Heald Am J Phys 56 540 1988 3301 1985 and K R Brownstein Am J Phys 57420 1989 for discussion and references 12Note that this result is independent of the separation distance d 1 it points from qe toward qm In quantum mechanics angular momentum comes in half integer multiples of 71 so this result suggests that if magnetic monopoles exist electric and magnetic charge must be quantized uoqeqm 4n nit2 for n 1 2 3 an idea rst proposed by Dirac in 1931 If even one monopole exists somewhere in the universe this would explain why electric charge comes in discrete units 82 MOMENTUM 363 a Calculate in terms of cod and rub the nal angular momentum in the elds b As the cylinders begin to rotate their changing magnetic eld induces an extra azimuthal electric eld which in turn will make an additional contribution to the torques Find the re sulting extra angular momentum and compare it to your result in a Answer39 0 szb 72 2 a 4711 Problem 81413 A point charge q is a distance a gt R from the axis of an in nite solenoid radius R n turns per unit length current 1 Find the linear momentum and the angular momentum in the elds Put q on the x axis With the solenoid along z treat the solenoid as a nonconductor so you don t need to worry about induced charges on its surface Answer Pem M04n1R220 Lem 01 Problem 81514 a Carry through the argument in Sect 812 starting with Eq 86 but using J f in place of J Show that the Poynting vector becomes S E X H and the rate of change of the energy density in the elds is auem EH a a a For linear media show that l b In the same spirit reproduce the argument in Sect 822 starting with Eq 815 with p f and J f in place of p and J Don t bother to construct the Maxwell stress tensor but do show that the momentum density is p D X B 13See F S Johnson B L Cragin and R R Hodges Am J Phys 62 33 1994 for a discussion of this and related problems 14This problem was suggested by David Thouless of the University of Washington Refer to Sect 443 for the meaning of energy in this context 91 Chapter 9 Electromagnetic Waves Waves in One Dimension 911 The Wave Equation What is a wave Idon t think can give you an entirely satisfactory answer the concept is intrinsically somewhat vague but here s a start A wave is a disturbance of a continuous medium that propagates with a xed shape at constant velocity Immediately I must add quali ers In the presence of absorption the wave will diminish in size as it moves if the medium is dispersive different frequencies travel at different speeds in two or three dimensions as the wave spreads out its amplitude will decrease and of course standing waves don t propagate at all But these are re nements let s start with the simple case xed shape constant speed Fig 91 How would you represent such an object mathematically In the gure I have drawn the wave at two different times once att 0 and again at some later time t each point on the wave form simply shifts to the right by an amount vt where v is the velocity Maybe the wave is generated by shaking one end of a taut string f z t represents the displacement of the string at the point z at time t Given the initial shape of the string gz E f z 0 z 0 z t 2 gt 1 z m Figure 91 364 91 WAVES IN ONE DIMENSION 365 what is the subsequent form f z t Evidently the displacement at point z at the later time t is the same as the displacement a distance vt to the left ie at z vt back at time t 0 fzatfzvt70gzvt That statement captures mathematically the essence of wave motion It tells us that the function f z t which might have depended on z and t in any old way in fact depends on them only in the very special combination z vt when that is true the function f z t represents a wave of xed shape traveling in the z direction at speed 1 For example if A and b are constants with the appropriate units A ZU 2 f z I Ae b t f2z As1nbz vt f3zt m all represent waves with different shapes of course but f4z t Aebbzz and f5z t A sinbz cosbvt3 do not Why does a stretched string support wave motion Actually it follows from Newton s second law Imagine a very long string under tension T If it is displaced from equilibrium the net transverse force on the segment between z and z AZ Fig 92 is AF 2 Tsin6 Tsin6 where 6 is the angle the string makes with the z direction at point z Az and 6 is the corresponding angle at point z Provided that the distortion of the string is not too great these angles are small the gure is exaggerated obviously and we can replace the sine by the tangent 3 AF 2 Ttan6 tan6 T l zAz az H2 Bz 32f Z z zAz Z Figure 92 366 CHAPTER 9 ELECTROMAGNETIC WAVES If the mass per unit length is u Newton s second law says 32 f and therefore 32f e32f az2 T 32 Evidently small disturbances on the string satisfy 2 I 32 3f f 92 3Z2 v2 31 2 where U which as we ll soon see represents the speed of propagation is T v 93 u Equation 92 is known as the classical wave equation because it admits as solutions all functions of the form fZ t gz vt 94 that is all functions that depend on the variables z and t in the special combination u E Z vt and we have just learned that such functions represent waves propagating in the z direction with speed 1 For Eq 94 means 3fdg3udg 3fdg3uvdg azduazdu 3tdu3t du and 32f a dg d2g3u d2g 3Z2 Bz du 39 du2 Bz duz 32f a dg dzg Bu 2ng 2 1 32 var du Uduz 3t duz SO 2 2 2 dg 3f L3f qed d L12 322 v2 at2 Note that gu can be any differentiable function whatever If the disturbance propagates without changing its shape then it satis es the wave equation But functions of the form gz vt are not the only solutions The wave equation involves the square of 1 so we can generate another class of solutions by simply changing the sign of the velocity fz t hz vt 95 This of course represents a wave propagating in the negative z direction and it is certainly reasonable on physical grounds that such solutions would be allowed What is perhaps 91 WAVES IN ONE DIMENSION 367 surprising is that the most general solution to the wave equation is the sum of a wave to the right and a wave to the left fzt gZ vt hzvt 96 Notice that the wave equation is linear The sum of any two solutions is itself a solution Every solution to the wave equation can be expressed in this form Like the simple harmonic oscillator equation the wave equation is ubiquitous in physics If something is vibrating the oscillator equation is almost certainly responsible at least for small amplitudes and if something is waving whether the context is mechanics or acoustics optics or oceanography the wave equation perhaps with some decoration is bound to be involved Problem 91 By explicit differentiation check that the functions f1 f2 and f3 in the text satisfy the wave equation Show that f4 and f5 do not Problem 92 Show that the standing wave f z t A sinkz coskvt satis es the wave equation and express it as the sum of a wave traveling to the left and a wave traveling to the right Eq 96 912 Sinusoidal Waves i Terminology Of all possible wave forms the sinusoidal one fz t Acoskz vt 8 97 is for good reason the most familiar Figure 93 shows this function at time t 0 A is the amplitude of the wave it is positive and represents the maximum displacement from equilibrium The argument of the cosine is called the phase and 6 is the phase constant obviously you can add any integer multiple of 271 to 8 without changing f z t ordinarily one uses a value in the range 0 f 8 lt 271 Notice that at z vt 8 k the phase is zero let s call this the central maximum If 8 0 the central maximum passes the origin at time t 0 more generally 8 k is the distance by which the central maximum and Central maximum f Z 0 Figure 93 368 CHAPTER 9 ELECTROMAGNETIC WAVES therefore the entire wave is delayed Finally k is the wave number it is related to the wavelength A by the equation 271 k for when z advances by 271 k the cosine executes one complete cycle As time passes the entire wave train proceeds to the right at speed 1 At any xed point z the string vibrates up and down undergoing one full cycle in a period A 98 T 99 kv The frequency 17 number of oscillations per unit time is uzlzk v3 910 T 271 A For our purposes a more convenient unit is the angular frequency a so called because in the analogous case of uniform circular motion it represents the number of radians swept out per unit time w271vkv 911 Ordinarily it s nicer to write sinusoidal waves Eq 97 in terms of a rather than 1 fz t Acoskz at6 912 A sinusoidal oscillation of wave number k and angular frequency a traveling to the left would be written fzt Acoskzat 6 913 The sign of the phase constant is chosen for consistency with our previous convention that 6 k shall represent the distance by which the wave is delayed since the wave is now moving to the left a delay means a shift to the right Att 2 0 the wave looks like Fig 94 Because the cosine is an even function we can just as well write Eq 913 thus fz t A cos kz cot 6 914 Comparison with Eq 912 reveals that in effect we could simply switch the sign of k to produce a wave with the same amplitude phase constant frequency and wavelength traveling in the opposite direction f z 0 Central Figure 94 9 WAVES IN ONE DIMENSION 369 ii Complex notation In view of Euler s formula 6 ei cos0 i sin 6 915 the sinusoidal wave Eq 912 can be written f z t ReAeiquot Z 5 916 where Re denotes the real part of the complex number S This invites us to introduce the complex wave function f z t E Ae U Z M 917 with the complex amplitude A E Aeits absorbing the phase constant The actual wave function is the real part of f fz t Refz t 918 If you know f it is a simple matter to nd f the advantage of the complex notation is that exponentials are much easier to manipulate than sines and cosines Example 91 Suppose you want to combine two sinusoidal waves f3 f1 f2 Rafi Reltf2gt Retfi f2 Ref3 with f3 fl 2 You simply add the corresponding complex wave functions and then take the real part In particular if they have the same frequency and wave number f3 ileum wt berth wt A3eikz wt where i 8 A3 A1A2 or 1436183 A1e151 A2612 919 evidently you just add the complex amplitudes The combined wave still has the same frequency and wavelength f3z t A3 COS kz wt 83 and you can easily gure out A3 and 83 from Eq 919 Prob 93 Try doing this without using the complex notation you will nd yourself looking up trig identities and slogglng through nasty algebra iii Linear combinations of sinusoidal waves Although the sinusoidal function 917 is a very special wave form the fact is that any wave can be expressed as a linear combination of sinusoidal ones 00 fz t AkeiltkZ w gt dk 920 Here a is a function of k Eq 911 and I have allowed k to run through negative values in order to include waves going in both directions1 1This does not mean that A and a are negative wavelength and frequency are always positive If we allow negative wave numbers then Eqs 98 and 911 should really be written A 271 k and w kv 370 CHAPTER 9 ELECTROMAGNETIC WAVES The formula for Age in terms of the initial conditions fz 0 and fz 0 can be obtained from the theory of Fourier transforms see Prob 932 but the details are not relevant to my purpose here The point is that any wave can be written as a linear combination of sinusoidal waves and therefore if you know how sinusoidal waves behave you know in principle how any wave behaves So from now on we shall con ne our attention to sinusoidal waves Problem 93 Use Eq 919 to determine A3 and 83 in terms of A1 A2 81 and 82 Problem 94 Obtain Eq 920 directly from the wave equation by separation of variables 913 Boundary Conditions Re ection and Transmission So far I have assumed the string is in nitely long or at any rate long enough that we don t need to worry about what happens to a wave when it reaches the end As a matter of fact what happens depends a lot on how the string is attached at the end that is on the speci c boundary conditions to which the wave is subject Suppose for instance that the string is simply tied onto a second string The tension T is the same for both but the mass per unit length a presumably is not and hence the wave velocities v1 and 122 are different remember 12 W Let s say for convenience that the knot occurs at z 0 The incident wave f1z t Aequotltklz wtgt z lt 0 921 coming in from the left gives rise to a re ected wave fRz t AR ik1Zw z lt 0 922 traveling back along string 1 hence the minus sign in front of k1 in addition to a trans mitted wave fTz r ATe39U ZZ w z gt 0 923 Which continues on to the right in string 2 The incident wave f1z t is a sinusoidal oscillation that extends in principle all the way back to z 2 oo and has been doing so for all of history The same goes for fR and fT except that the latter of course extends to z 00 All parts of the system are oscillating at the same frequency a a frequency determined by the person at z 2 00 who is shaking the string in the rst place Since the wave velocities are different in the two strings however the wavelengths and wave numbers are also different Q 9 Ag k1 122 Of course this situation is pretty arti cial what s more with incident and re ected waves of in nite extent traveling on the same piece of string it s going to be hard for a spectator to 91 WAVES IN ONE DIMENSION 371 tell them apart You might therefore prefer to consider an incident wave of nite extent say the pulse shown in Fig 95 You can work out the details for yourself if you like Prob 95 The trouble with this approach is that no nite pulse is truly sinusoidal The waves in Fig 95 may look like sine functions but they re not they re little pieces of sines joined onto an entirely di erent function namely zero Like any other waves they can be built up as linear combinations of true sinusoidal functions Eq 920 but only by putting together a whole range of frequencies and wavelengths If you want a single incident frequency as we shall in the electromagnetic case you must let your waves extend to in nity In practice if you use a very long pulse with many oscillations it will be close to the ideal of a single frequency f a Incident pulse b Re ected and transmitted pulses Figure 95 For a sinusoidal incident wave then the net disturbance of the string is Ale lk Z w AReik Zw for z lt 0 m t g 925 ATe39U ZZ w for z gt 0 At the join 2 0 the displacement just slightly to the left z 0 must equal the displacement slightly to the right 2 0 or else there would be a break between the two strings Mathematically f z t is continuous at z 0 f 0 t f 0 r 926 If the knot itself is of negligible mass then the derivative of f must also be continuous 3 3 f f 927 Bz 0 32 0 Otherwise there would be a net force on the knot and therefore an in nite acceleration Fig 96 These boundary conditions apply directly to the real wave function f z t But since the imaginary part of f differs from the real part only in the replacement of cosine by sine Eq 915 it follows that the complex wave function f z t obeys the same rules 6f 6f a Z 928 3z 0 0 372 CHAPTER 9 ELECTROMAGNETIC WAVES T T Knot T Knot z z a Discontinuous slope force on knot at Continuous slope no force on knot Figure 96 When applied to Eq 925 these boundary conditions determine the outgoing amplitudes A R and AT in terms of the incoming one A 1 fl AR AT kid AR k2z lr from which it follows that kl kz 2kl A A A A 2 R k1k2 I T k1k2 I 9 9 Or in terms of the velocities Eq 924 122 121 2122 AR A A A 930 ltvZvl 1 T v2Ul 1 The real amplitudes and phases then are related by AReiaR A1651 ATeiST 214651 931 122 121 v2 v1 If the second string is lighter than the rst M2 lt u so that 122 gt 121 all three waves have the same phase angle 6R 2 6T 2 61 and the outgoing amplitudes are 122 121 2122 R ltvZvl I T ltvZvl I If the second string is heavier than the rst 122 lt 121 the re ected wave is out of phase by 180 6R 71 2 6T 2 61 In other words since cos k1z cot 61 71 cos k1z cot 61 the re ected wave is upside down The amplitudes in this case are 2 AR A1 and AT lt 2 gt141 933 v2v1 v2v1 9 WAVES IN ONE DIMENSION 373 In particular if the second string is in nitely massive or what amounts to the same thing if the rst string is simply nailed down at the end then AR A1 and AT 0 Naturally in this case there is no transmitted wave all of it re ects back Problem 95 Suppose you send an incident wave of speci ed shape g1 z 121 t down string number 1 It gives rise to a re ected wave h R z vlt and a transmitted wave gTz vzt By imposing the boundary conditions 926 and 927 nd 11R and gT Problem 96 a Formulate an appropriate boundary condition to replace Eq 927 for the case of two strings under tension T joined by a knot of mass m b Find the amplitude and phase of the re ected and transmitted waves for the case where the knot has a mass m and the Second string is massless Problem 97 Suppose string 2 is embedded in a viscous medium such as molasses which imposes a drag force that is proportional to its transverse speed af Adeg EAZ a Derive the modi ed wave equation describing the motion of the string b Solve this equation assuming the string oscillates at the incident frequency a That is look for solutions of the form f z t e39w Fz c Show that the waves are attenuated that is their amplitude decreases with increasing 2 Find the characteristic penetration distance at which the amplitude is l e of its original value in terms of y T M and a d If a wave of amplitude A I phase 81 0 and frequency a is incident from the left string 1 nd the re ected wave s amplitude and phase 914 Polarization The waves that travel down a string when you shake it are called transverse because the displacement is perpendicular to the direction of propagation If the string is reasonably elastic it is also possible to stimulate compression waves by giving the string little tugs Compression waves are hard to see on a string but if you try it with a slinky they re quite noticeable Fig 97 These waves are called longitudinal because the displacement from equilibrium is along the direction of propagation Sound waves which are nothing but compression waves in air are longitudinal electromagnetic waves as we shall see are transverse 374 CHAPTER 9 ELECTROMAGNETIC WAVES 39U WHO 0 Hillbilly 0 0 HHNWJJMH 0 O HHHWM Figure 97 Now there are of course two dimensions perpendicular to any given line of propagation Accordingly transverse waves occur in two independent states of polarization you can shake the string upanddown vertical polarization Fig 98a he I Aei kz w 2 934 or leftandright horizontal polarization Fig 98b Em I flew a i 935 or along any other direction in the xy plane Fig 98c iz t Aei zrw n 936 The polarization vector 13 de nes the plane of vibration2 Because the waves are transverse fl is perpendicular to the direction of propagation 2 0 937 In terms of the polarization angle 6 cos fr sin6y 938 Thus the wave pictured in Fig 98c can be considered a superposition of two waves one horizontally polarized the other vertically iz z A cos mathwt a A sin 9eiltkZw gt 9 939 Problem 98 Equation 936 describes the most general linearly polarized wave on a string Linear or plane polarization so called because the displacement is parallel to a xed vector results from the combination of horizontally and vertically polarized waves of the same phase Eq 939 If the two components are of equal amplitude but out of phase by 90 say 8 0 8 90 the result is a circularly polarized wave In that case a At a xed point z show that the string moves in a circle about the z axis Does it go clockwise or counterclockwise as you look down the axis toward the origin How would you construct a wave circling the other way In Optics the clockwise case is called right circular polarization and the counterclockwise left circular polarization b Sketch the string at time t 0 c How would you shake the string in order to produce a circularly polarized wave 2Notice that you can always switch the Sign of f1 provided you simultaneously advance the phase constant by 180 since both operations change the sign of the wave 92 ELECTROMAGNETIC WAVES IN VACUUM 375 X X 1 1 b Horizontal polarization c Polarization vector Figure 98 92 Electromagnetic Waves in Vacuum 921 The Wave Equation for E and B In regions of space where there is no charge or current Maxwell s equations read 8B O VE0 iii VxE 940 8E 1i V B 0 1v V x B M0605 They constitute a set of coupled rstorder partial differential equations for E and B They can be decoupled by applying the curl to iii and iv 2 8B VxVxE VVE VEVX E a V B 32E 6 8t X Mooat2 2 8E VxVxB VVB VBVX M0605 82B 8 MOEOEV X E M060W 376 CHAPTER 9 ELECTROMAGNETIC WAVES OrsinceVE0andVB0 32E 82B V2E e V2B e 941 L M0 0 at Mo 0 atz We now have separate equations for E and B but they are of second order that s the price you pay for decoupling them In vacuum then each Cartesian component of E and B satis es the threedimensional wave equation V2f This is the same as Eq 92 except that 82 f azz is replaced by its natural generaliza tion V2 f So Maxwell s equations imply that empty space supports the propagatiOn of electromagnetic waves traveling at a speed 1 V EOMO which happens to be precisely the velocity of light c The implication is astounding Perhaps light is an electromagnetic wave3 Of course this conclusion does not surprise anyone today but imagine what a revelation it was in Maxwell s time Remember how 60 and 0 came into the theory in the rst place they were 00nstants in Coulomb s law and the Biot Savart law respectively You measure them in experiments involving charged pith balls batteries and wires experiments having nothing whatever to do with light And yet according to Maxwell s theory you can calculate c from these two numbers Notice the crucial role played by Maxwell s contribution to Ampere s law 1106an it without it the wave equation would not emerge and there would be no electromagnetic theory of light 300 x 108 ms 942 v 922 Monochromatic Plane Waves For reasons discussed in Sect 912 we may con ne our attention to sinusoidal waves of frequency a Since different frequencies in the visible range correspond to different colors such waves are called monochromatic Table 91 Suppose moreover that the waves are traveling in the z direction and have no x or y dependence these are called plane waves4 because the elds are uniform over every plane perpendicular to the direction of propagation Fig 99 We are interested then in elds of the form EZ EoelkZ wt oeikzwt 3As Maxwell himself put it We can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena See Ivan Tolstoy James Clerk Maxwell A Biography Chicago University of Chicago Press 1983 4For a discussion of spherical waves at this level see J R Reitz F J Milford and R W Christy Foundations of Electromagnetic Theory 3rd ed Sect 17 5 Reading MA Addison Wesley 1979 Or work Prob 933 Of course over small enough regions any wave is essentially plane as long as the Wavelength is much less than the radius of the curvature of the wave front 92 ELECTROMAGNETIC WAVES IN VACUUIVI 377 The Electromagnetic Spectrum Frequency Hz Type Wavelength m 1022 10 13 1021 gamma rays 10 12 1020 10 11 1019 10 10 1018 x rays 10 9 1017 lo8 1016 ultraviolet 10 7 1015 visible 106 1014 infrared 10 5 1013 lo 4 1012 lo3 1011 102 1010 microwave 10391 109 l 108 TV FM 10 107 lo2 106 AM 103 105 lo4 104 RF 105 103 106 The Visible Range Frequency Hz Color Wavelength In 10 x 1015 near ultraviolet 30 x 10 7 75 x 1014 shortest visible blue 40 x 10 7 65 x 1014 blue 46 x 107 56 x 1014 green 54 x 10 7 51 x 1014 yellow 59 x 10 7 49 x 1014 orange 61 x 10 7 39 x 1014 longest visible red 76 x 107 30 x 1014 near infrared 10 x 10 6 Table 91 where E0 and B0 are the complex amplitudes the physical elds of course are the real parts ofE and B Now the wave equations for E and B Eq 941 were derived from Maxwell s equations However whereas every solution to Maxwell s equations in empty space must obey the wave equati0n the converse is not true Maxwell s equations impose extra constraints on 378 CHAPTER 9 ELECTROMAGNETIC WAVES Figure 99 E0 and B0 In particular since V E 0 and V B 0 it follows5 that mf4ampm0 ampMgt That is electromagnetic waves are transverse the electric and magnetic elds are per pendicular to the direction of propagation Moreover Faraday s law V x E 8B8t implies a relation between the electric and magnetic amplitudes to wit Mnwnuhnwampn a or more compactly k B0 X E0 946 Evidently E and B are in phase and mutually perpendicular their real amplitudes are related by k 1 BO 2 E0 E0 a C The fourth of Maxwell s equations V x B M0608E8t does not yield an independent condition it simply reproduces Eq 945 Example 92 If E points in the x direction then B points in the y direction Eq 946 A t 1 s A Ea t Eoetkz wtX Bz7 t ZEoelkz wty7 or taking the real part 1 Ez t 2 E0 coskz cot 81 Bz t 2E0 003kz cot Dijl 948 5 Because the real part of E differs from the imaginary part only in the replacement of sine by cosine if the former obeys Maxwell s equations so does the latter and hence E as well 92 ELECTROMAGNETIC WAVES IN VACUUM 379 Figure 910 This is the paradigm for a monochromatic plane wave see Fig 910 The wave as a whole is said to be polarized in the x direction by convention we use the direction of E to specify the polarization of an electromagnetic wave There is nothing special about the z direction of course we can easily generalize to monochromatic plane waves traveling in an arbitrary direction The notation is facilitated by the introduction of the propagation or wave vector k pointing in the direction of propagation whose magnitude is the wave number k The scalar product kr is the appropriate generalization of kz Fig 911 so rm t roamwt ii 949 1 k wt A 1 quot Br t E0el r k x n k x E c c where n is the polarization vector Because E is transverse 1 0 950 Figure 911 380 CHAPTER 9 ELECTROMAGNETIC WAVES The transversality of B follows automatically from Eq 949 The actual real electric and magnetic elds in a monochromatic plane wave with propagation vector k and polarization are Er t Eocos kr wt 5 951 Br t 2 1E0 cos k r wt 61A x 952 c Problem 99 Write down the real electric and magnetic elds for a monochromatic plane wave of amplitude E0 frequency a and phase angle zero that is a traveling in the negative x direction and polarized in the z direction b traveling in the direction from the origin to the point 1 I 1 with polarization parallel to the x z plane In each case sketch the wave and give the explicit Cartesian components of k and f1 923 Energy and Momentum in Electromagnetic Waves According to Eq 813 the energy per unit volume stored in electromagnetic elds is 1 2 1 2 u 60E B 2 0 In the case of a monochromatic plane wave Eq 948 l B2 2E2 M060E2 954 c so the electric and magnetic contributions are equal 2 2 2 u 60E 60E0 cos kz wt 6 955 As the wave travels it carries this energy along with it The energy ux density energy per unit area per unit time transported by the elds is given by the Poynting vector Eq 810 S iE X B 956 0 For monochromatic plane waves propagating in the z direction S ceoEg cos2 kz wt 8 2 cu 2 957 Notice that S is the energy density u times the velocity of the waves c z as it should be For in a time At a length c At passes through area A Fig 912 carrying with it an energy uAc At The energy per unit time per unit area transported by the wave is therefore uc 92 ELECTROMAGNETIC WAVES IN VACUUM 381 Figure 912 Electromagnetic elds not only carry energy they also carry momentum In fact we found in Eq 830 that the momentum density stored in the elds is 1 p 2 ZS 958 C For monochromatic plane waves then 1 2 2 A 1 A p 60E0 cos kz wt 8 z 2 Eu 2 959 C In the case of light the wavelength is so short 5 x 10 7 m and the period so brief 10 15 s that any macroscopic measurement will encompass many cycles Typically therefore we re not interested in the uctuating cosinesquared term in the energy and momentum densities all we want is the average value Now the average of cosinesquared 6 over a complete cycle is so 1 u 560155 960 1 2 A S EceoEO z 961 1 2 A p 2 ceoE0 z 962 I use brackets to denote the time average over a complete cycle or many cycles if you prefer The average power per unit area transported by an electromagnetic wave is called the intensity 1 I E 5 EceOEg 963 6There is a cute trick for doing this in your head sin2 6 cos2 6 1 and over a complete cycle the average of sin2 6 is equal to the average of cos2 6 so sinz cosz 12 More formally 1 T 0052kz 2mT 3 dt 212 0 382 CHAPTER 9 ELECTROMAGNETIC WAVES When light falls on a perfect absorber it delivers its momentum to the surface In a time At the momentum transfer is Fig 912 Ap pAc At so the radiation pressure average force per unit area is I C On a perfect re ector the pressure is twice as great because the momentum switches direction instead of simply being absorbed We can account for this pressure qualitatively as follows The electric eld Eq 948 drives charges in the x direction and the magnetic eld then exerts on them a force q Ix B in the z direction The net force on all the charges in the surface produces the pressure Problem 910 The intensity of sunlight hitting the earth is about 1300 Wmz If sunlight strikes a perfect absorber what pressure does it exert How about a perfect re ector What fraction of atmOSpheric pressure does this amount to Problem 911 In the complex notation there is a clever device for nding the time average of a product Suppose fr I Acos k r 01 8 and gr t B cos k t r a Sb Show that f g l 2Re 1757 where the star denotes complex conjugation Note that this only works if the two waves have the same k and a but they need not have the same amplitude or phase For example I 1 I t u ReeOE E B o B and S ReE X B 4 0 2H0 Problem 912 Find all elements of the Maxwell stress tensor for a monochromatic plane wave traveling in the z direction and linearl polarized in the x direction Eq 948 Does your answer make sense Remember that represents the momentum ux density How is the momentum ux density related to the energy density in this case 93 Electromagnetic Waves in Matter 931 Propagation in Linear Media Inside matter but in regions where there is no free charge or free current Maxwell s equa tions become 8B 1 VD0 111 VXE2 5 965 8D 11 VB0 1v VXH If the medium is linear l D 2 6E H B 966 93 ELECTROMAGNETIC WAVES IN MATTER 383 and homogeneous so 6 and u do not vary from point to point Maxwell s equations reduce to 3B 1 VE0 111 V sz E 967 8E ii VB0 iv VXBueE which remarkably differ from the vacuum analogs Eqs 940 only in the replacement of 7 M060 by Me Evidently electromagnetic waves propagate through a linear homogeneous medium at a speed 1 c 9 68 v MM n where n E 6 969 EOMO is the index of refraction of the material For most materials it is very close to no so n E er 970 where e is the dielectric constant Eq 434 Since 6 is almost always greater than 1 light travels more slowly through matter a fact that is well known from optics All of our previous results carry over with the simple transcription 60 a e no gt M and hence c gt U see Prob 815 The energy density is8 l 1 u 6E2 492 971 2 M and the Poynting vector is l S E x B 972 n For monochromatic plane waves the frequency and wave number are related by a 2 kn Eq 911 the amplitude of B is l v times the amplitude of E Eq 947 and the intensity IS 1 I EevEg 973 7This observation is mathematically pretty trivial but the physical implications are astonishing As the wave passes through the elds busily polarize and magnetize all the molecules and the resulting oscillating dipoles create their own electric and magnetic elds These combine with the original elds in such a way as to create a single wave with the same frequency but a different speed This extraordinary conspiracy is responsible for the phenomenon of transparency It is a distinctly nontrivial consequence of the linearity of the medium For further discussion see M B James and D J Grif ths Am J Phys 60 309 1992 8Refer to Sect 443 for the precise meaning of energy density in the context of linear media 384 CHAPTER 9 ELECTROMAGNETIC WAVES The interesting question is this What happens when a wave passes from one transparent medium into another air to water say or glass to plastic As in the case of waves on a string we expect to get a re ected wave and a transmitted wave The details depend on the exact nature of the electrodynamic boundary conditions which we derived in Chapter 7 Eq 764 r elEli ezEzi iii E39 E 1 1 974 11 13i 32 iv 31 B39239 M1 2 These equations relate the electric and magnetic elds just to the left and just to the right of the interface between two linear media In the following sections we use them to deduce the laws governing re ection and refraction of electromagnetic waves 932 Re ection and Transmission at Normal Incidence Suppose the xy plane forms the boundary between two linear media A plane wave of frequency a traveling in the z direction and polarized in the x direction approaches the interface from the left Fig 913 E1 Z t Eole er w i 1 975 BIZ t EolelkZ wt v1 It gives rise to a re ected wave IER Z t EOReikrz wt 2 976 1 BRZ t vlE0Rel krz wt y X E D ET 1 1 B BT ER Z J BR v Interface y Figure 913 93 ELECTROMAGNETIC WAVES IN MATTER 385 which travels back to the left in medium 1 and a transmitted wave ETZ t EOTeiUQz wt g 977 ik2z wt A 1 BTZat E0Te Ya U2 which continues on the the right in medium 2 Note the minus sign in ER as required by Eq 949 or if you prefer by the fact that the Poynting vector aims in the d1rectlon of propagation N 0 At z 0 the combined elds on the left E1 E R and B 1 B R must JOlll the elds on the right ET and ET in accordance with the boundary conditions 974 In this case there are no components perpendicular to the surface so i and ii are trivial However 111 requires that 50 150 Eor 978 while iv says 1 1 1 ii190 EOR lt E0T 979 1 U1 UI M2 v2 or E0 EoR 3E0p 980 where E mm M1n2 981 MM Mznl Equations 978 and 980 are easily solved for the outgoing amplitudes in terms of the incident amplitude 2 1 3 2 2 gt 2 E E E0 982 E0 17 0 0T 19 These results are strikingly similar to the ones for waves on a string Indeed if the permittivities M are close to their values in vacuum as remember they are for most media then 8 111112 and we have v2 v1 2112 E E E0 983 EOR v2v1 0 0T UziUr I which are identical to Eqs 930 In that case as before the re ected wave is in phase right side up if 112 gt U and out of phase upside down if 172 lt 111 the real amplitudes are related by 21 2 E0 EOT lt E01 984 172 171 v2v1 E 0R U2 U1 or in terms of the indices of refraction 2 E0 E0Tlt quot1 E0 985 n1 2 I 11 12 E0 R n1n2 386 CHAPTER 9 ELECTROMAGNETIC WAVES What fraction of the incident energy is re ected and what fraction is transmitted According to Eq 973 the intensity average power per unit area is I 1 E2 2 EU 2 0 If again LL 2 M2 no then the ratio of the re ected intensity to the incident intensity is I E 2 2 R R quot1 quot2 986 1 EO n1n2 whereas the ratio of the transmitted intensity to the incident intensity is 2 1 E 4 T E T lt 0T quotW 987 I 61111 E0 quot1 nz2 R is called the re ection coef cient and T the transmission coef cient they measure the fraction of the incident energy that is re ected and transmitted respectively Notice that R T l 988 as conservation of energy of course requires For instance when light passes from air n1 1 into glass n2 15 R 004 and T 096 Not surprisingly most of the light is transmitted Problem 913 Calculate the exact re ection and transmission coef cients without assuming HI 2 0 Con rm that R T 1 Problem 914 In writing Eqs 976 and 977 l tacitly assumed that the re ected and transmitted waves have the same polarization as the incident wave alon g the x direction Prove that this must be so Hint Let the polarization vectors of the transmitted and re ected waves be T cos6T 2 sinQT y R cos6R 2 sin 6R y and prove from the boundary conditions that 6T 2 6R 2 0 933 Re ection and Transmission at Oblique Incidence In the last section I treated re ection and transmission at normal incidence that is when the incoming wave hits the interface headon We now turn to the more general case of oblique incidence in which the incoming wave meets the boundary at an arbitrary angle 9 Fig 914 Of course normal incidence is really just a special case of oblique incidence with 9 0 but I wanted to treat it separately as a kind of warmup because the algebra is now going to get a little heavy 93 ELECTROMAGNETIC WAVES IN MATTER 387 kR kT Z Plane of Incidence 9 k1 Figure 914 Suppose then that a monochromatic plane wave EmnnEmwh W unng ampxEn 03 approaches from the left giving rise to a re ected wave ER EWJWRFWL ERQJEampRXE 99m and a transmitted wave 1 A c 17mm EoTequotquot Bra 0 v kT x Er 991 2 All three waves have the same frequency w that is determined once and for all at the source the ashlight or whatever that produces the incident beam The three wave numbers are related by Eq 911 U2 n1 kip va kTv2 a 01 k CR 2 U lkT EkT The combined elds in medium 1 E ER and f ER must now bejoined to the elds ET and ET in medium 2 using the boundary conditions 974 These all share the generic structure eikr wt eikRr wt eukrr wr at Z 0 993 I ll ll in the parentheses in a moment for now the important thing to notice is that the x y and t dependence is con ned to the exponents Because the boundary conditions must hold at all points on the plane and for all times these exponential factors must be equal Otherwise a slight change in x say would destroy the equality see Prob 915 Of 388 CHAPTER 9 ELECTROMAGNETIC WAVES course the time factors are already equal in fact you could regard this as an independent con rmation that the transmitted and re ected frequencies must match the incident one As for the spatial terms evidently klorszrszr whenz0 994 or more explicitly xklx Ykly xkRx YkRy xkTx YkTyv 995 for all x and all y But Eq 995 can only hold if the components are separately equal for if x 0 we get kny kRy kTya while y 0 gives 11 kRx kTx 997 We may as well orient our axes so that k 1 lies in the x z plane ie k1 y 0 according to Eq 996 so too will kR and kT Conclusion First Law The incident re ected and transmitted wave vectors form a plane called the plane of incidence which also includes the normal to the surface here the z axis Meanwhile Eq 997 implies that 1 81119 2 CR sin HR 2 7 sin 9T 998 where 91 is the angle of incidence OR is the angle of re ection and HT is the angle of transmission more commonly known as the angle of refraction all of them measured with respect to the normal Fig 914 In view of Eq 992 then Second Law The angle of incidence is equal to the angle of re ection 91 2 1 999 This is the law of re ection As for the transmitted angle Third Law l9 S T wm s1n 9 n2 This is the law of refraction or Snell s law 93 ELECTROMAGNETIC WAVES IN MATTER 389 These are the three fundamental laws of geometrical optics It is remarkable how little actual electrodynamics went into them we have yet to invoke any specific boundary conditions all we used was their generic form Eq 993 Therefore any other waves water waves for instance or sound waves can be expected to obey the same optical laws when they pass from one medium into another Now that we have taken care of the exponential factors they cancel given Eq 994 the boundary conditions 974 become 0 61E0 Eogh 62E0Tz wgtdmnwzdet 9101 E01 EORxy EOTxy 1 1 iv B B BO m 01 0Rxy M r xy where 130 1 vk X E0 in each case The last two represent pairs of equations one for the xcomponent and one for the ycomponent SuppOse that the polarization of the incident wave is parallel to the plane of incidence the x z plane in Fig 915 it follows see Prob 914 that the re ected and transmitted waves are also polarized in this plane I shall leave it for you to analyze the case of polarization perpendicular to the plane of incidence see Prob 916 Then i reads 61 E0 sine EOR sin 0R 2 EOT sin 6T 9102 ii adds nothing 0 0 since the magnetic elds have no z components iii becomes E0 cos 9 EOR cos 19R 2 EDT cos19T 9103 BR x k R ER ET kT 6R 9T ET 2 E 91 quot1 lt9 BI Figure 915 390 CHAPTER 9 ELECTROMAGNETIC WAVES and iv says 1 1 E0 EOR E0T 9l04 M1111 2112 Given the laws of re ection and refraction Eqs 9102 and 9104 both reduce to EU EOR 1350 9105 where as before u n SEMIM12 9106 2112 mm and Eq 9103 says E01 EOR OlEOT Where 6 a 2 C05 T 9108 cos 9 Solving Eqs 9105 and 9107 for the re ected and transmitted amplitudes we obtain E r s 2 5 9m9 oR a 0 0T lta 0 These are known as Fresnel s equations for the case of polarization in the plane of inci dence There are two other Fresnel equations giving the re ected and transmitted ampli tudes when the polarization is perpendicular to the plane of incidence see Prob 916 Notice that the transmitted wave is always in phase with the incident one the re ected wave is either in phase right side up if a gt 8 or 180 out of phase upside down if a lt 39 The amplitudes of the transmitted and re ected waves depend on the angle of incidence because a is a function of 191 1 sinzer 1 n1n2sin912 a cos19 cos 9 9110 In the case of normal incidence 191 0 a l and we recover Eq 982 At grazing incidence 19 90 or diverges and the wave is totally re ected a fact that is painfully familiar to anyone who has driven at night on a wet road Interestingly there is an in termediate angle 63 called Brewster s angle at which the re ected wave is completely extinguished10 According to Eq 9109 this occurs when a 8 or 1 2 quotln2232 9There is an unavoidable ambiguity in the phase of the re ected wave since as I mentioned in footnote 2 changing the sign of the polarization vector is equivalent to a 1800 phase shift The convention I adopted in Fig 915 with E R positive upward is consistent with some but not all of the standard optics texts 10Because waves polarized perpendicular to the plane of incidence exhibit no corresponding quenching of the re ected component an arbitrary beam incident at Brewster s angle yields a re ected beam that is totally polarized parallel to the interface That s why Polaroid glasses with the transmission axis vertical help to reduce glare off a horizontal surface sin263 9111 93 ELECTROMAGNETIC WAVES IN MATTER 391 10 r 08 r 02 0396 E01 04 7 6 02 r B T l l i l l l i l l i I i I J l 00 wise 80 02 V 04 E0 7 Figure 916 For the typical case it 2 112 so 8 E n2n1SiIl2 03 2 821 82 and hence tan 193 g quot 2 9112 quot1 Figure 916 shows a plot of the transmitted and re ected amplitudes as functions of 91 for light incident on glass n2 15 from air n1 1 On the graph a negative number indicates that the wave is 180 out of phase with the incident beam the amplitude itself is the absolute value The power per unit area striking the interface is S 2 Thus the incident intensity is 1 I 561111193 cos61 9113 while the re ected and transmitted intensities are 1 1 2 IR 2 EEIUIEgR cos 19R and IT EezngOT cos 0T 9114 The cosines are there because I am talking about the average power per unit area of interface and the interface is at an angle to the wave front The re ection and transmiss1on coef cients for waves polarized parallel to the plane of incidence are 2 2 RE RZ 3 9115 I Eo 13 I v E 2c0519 2 2 TET 22lt 0r T a 9116 I 6101 E0 cosi9 a 392 CHAPTER 9 ELECTROMAGNETIC WAVES i 08 06 7 r 04 A E R E 02 r 00 l i l i i 1 0 10 20 30 40 50 60 70 80 90 Figure 917 They are plotted as functions of the angle of incidence in Fig 917 for the air glass inter face R is the fraction of the incident energy that is re ected naturally it goes to zero at Brewster s angle T is the fraction transmitted it goes to 1 at 193 Note that R T 1 as required by conservation of energy the energy per unit time reaching a particular patch of area on the surface is equal to the energy per unit time leaving the patch Problem 915 Suppose Aei Be 2 Cei for some nonzero constants A B C a b c and forallx Prove thata b candA B C Problem 916 Analyze the case of polarization perpendicular to the plane of incidence ie electric elds in the y direction in Fig 915 Impose the boundary conditions 9101 and obtain the Fresnel equations for EOR and EDT Sketch EOR E0 and EDT E01 as functions of 61 for the case n2n1 15 Note that for this the re ected wave is always 180 out of phase Show that there is no Brewster s angle for any n1 and n2 EOR is never zero unless of course n1 2 n2 and m M2 in which case the two media are optically indistinguishable Con rm that your Fresnel equations reduce to the proper forms at normal incidence Compute the re ection and transmission coef cients and check that they add up to 1 Problem 917 The index of refraction of diamond is 242 Construct the graph analogous to Fig 916 for the airdiamond interface Assume 11 2 no In particular calculate a the amplitudes at normal incidence b Brewster s angle and c the crossover angle at which the re ected and transmitted amplitudes are equal 94 Absorption and Dispersion 941 Electromagnetic Waves in Conductors In Sect 93 I stipulated that the free charge density pf and the free current density J f are zero and everything that followed was predicated on that assumption Such a restriction 94 ABSORPTION AND DISPERSION 393 is perfectly reasonable when you re talking about wave propagation through a vacuum or through insulating materials such as glass or pure water But in the case of conductors we do not independently control the ow of charge and in general J f is certainly not zero In fact according to Ohm s law the free current density in a conductor is proportional to the electric eld Jf aE 9117 With this Maxwell s equations for linear media assume the form 39 V E 1 quot V E BB 1 111 x e pf at 9118 8E 11 VB0 1v VXBZMOEM 57 Now the continuity equation for free charge apf V J 9119 f at together with Ohm s law and Gauss s law i gives Bpf a V E at 0 6 pf for a homogeneous linear medium from which it follows that p f t equot p f 0 9120 Thus any initial free charge density pf 0 dissipates in a characteristic time 1 E 60 This re ects the familiar fact that if you put some free charge on a conductor it will ow out to the edges The time constant 1 affords a measure of how good a conductor is For a perfect conductor a 00 and 1 0 for a good conductor 1 is much less than the other relevant times in the problem in oscillatory systems that means 1 ltlt lw for a poor conductor 1 is greater than the characteristic times in the problem 1 gtgt 1 a11 At present we re not interested in this transient behavior we ll wait for any accumulated free charge to disappear From then on pf 0 and we have 8B 1 VE0 111 V XE a t 9121 8E ii VB0 iv VXBM a tMOE 11N Ashby Am J Phys 43 553 1975 points out that for good conductors r is absurdly short 10 19 s for copper whereas the time between collisions is rc 10 14 s The problem is that Ohm s law itself breaks down on time scales shorter than TC actually the time it takes free charge to dissipate in a good conductor is of order 15 not I Moreover H C Ohanian Am J Phys 51 1020 1983 shows that it takes even longer for the elds and currents to equilibrate But none of this is relevant to our present purpose the free charge density in a conductor does eventually dissipate and exactly how long the process takes is beside the point 394 CHAPTER 9 ELECTROMAGNETIC WAVES These differ from the corresponding equations for nonconducting media 967 only in the addition of the last term in iv Applying the curl to iii and iv as before we obtain modi ed wave equations for E and B V2Eue IZEMU VZBM68827123Mag 9122 These equations still admit planewave solutions Ez t Eoeidzr 13z t BoeiUEZ w 9123 but this time the wave number I is complex 122 2 ea i1ww 9124 as you can easily check by plugging Eq 9123 into Eq 9122 Taking the square root k k ic 9125 12 12 en a 2 en a 2 E E l 126 k a 216w1 K a 216w 9 The imaginary part of 12 results in an attenuation of the wave decreasing amplitude with increasing z Ez r E0equotzeikz Bz t Roe 2quotkZ w 9127 The distance it takes to reduce the amplitude by a factor of l e about a third is called the skin depth d 2 9128 7 Rlv d it is a measure of how far the wave penetrates into the conductor Meanwhile the real part of 12 determines the wavelength the propagation speed and the index of refraction in the usual way 271 a ck A v n k a The attenuated plane waves Eq 9127 satisfy the modi ed wave equation 9122 for any E0 and E0 But Maxwell s equations 9121 impose further constraints which serve to determine the relative amplitudes phases and polarizations of E and B As before i and ii rule out any 2 components the elds are transverse We may as well orient our axes so that E is polarized along the x direction 9129 Ez t E0eKzeikz a 9130 94 ABSORPTION AND DISPERSION 395 Then iii gives I Bz t Eoequotze kzwt 9 9131 a Equation iv says the same thing Once again the electric and magnetic elds are mutually perpendicular Like any complex number k can be expressed in terms of its modulus and phase 12 Keiq 9132 where 2 K E lk xk2lt2 w feL 1 1 9133 em and q E tan 1Kk 9134 According to Eq 9130 and 9131 the complex amplitudes E0 Eoei sE and 130 Boei sB are related by K 30658 6 E0435 9135 Evidently the electric and magnetic fields are no longer in phase in fact 58 5E E 9136 the magnetic eld lags behind the electric eld Meanwhile the real amplitudes of E and B are related by 30 K 039 2 1 E0 w Jew w 9137 The real electric and magnetic elds are nally Ez t Eoe cos kz wt 6E 12 9138 Bz t Bee K2 cos kz wt 85 4 9 These elds are shown in Fig 918 Problem 918 a Suppose you imbedded some free charge in a piece of glass About how long would it take for the charge to ow to the surface b Silver is an excellent conductor but it s expensive Suppose you were designing a mi crowave experiment to operate at a frequency of 1010 Hz How thick would you make the silver coatings c Find the wavelength and propagation speed in copper for radio waves at 1 MHz Compare the corresponding values in air or vacuum 396 CHAPTER 9 ELECTROMAGNETIC WAVES Figure 918 Problem 919 a Show that the skin depth in a poor conductor 0 ltlt we is 2 74 e M independent of frequency Find the skin depth in meters for pure water b Show that the skin depth in a good conductor a gtgt we is AZn where t is the wavelength in the conductor Find the skin depth in nanometers for a typical metal a k 107 S2 m1 in the visible range a 10158 assuming e 56 60 and u R M0 Why are metals opaque c Show that in a good conductor the magnetic eld lags the electric eld by 45 and nd the ratio of their amplitudes For a numerical example use the typical metal in part b Problem 920 a Calculate the time averaged energy density of an electromagnetic plane wave in a conduct ing medium Eq 9138 Show that the magnetic contribution always dominates Answer kZZMwZE8e 2KZ b Show that the intensity is k2twE8e2quotz 942 Re ection at a Conducting Surface The boundary conditions we used to analyze re ection and refraction at an interface between two dielectrics do not hold in the presence of free charges and currents Instead we have the more general relations 763 i 61EL 62E 0 m 131 E1 0 9139 1 1 ii 3 132i 0 iv EB EB39Z39 K x n where If not to be confused with conductivity is the free surface charge K f the free surface current and not to be confused with the polarization of the wave is a unit 94 ABSORPTION AND DISPERSION 397 vector perpendicular to the surface pointing from medium 2 into medium 1 For ohmic conductors J f 2 0E there can be no free surface current since this would require an in nite electric eld at the boundary Suppose now that the xy plane forms the boundary between a nonconducting linear medium 1 and a conductor 2 A monochromatic plane wave traveling in the z direction and polarized in the x direction approaches from the left as in Fig 913 1 A E1z t E0e39ltk1Z wtgt 2 B1z z E0e klz y 9140 U1 This incident wave gives rise to a re ected wave 1 A EN 0 EORe lt k gt 2 Rm 0 Eoe k quot gt y 9141 U1 propagating back to the left in medium 1 and a transmitted wave quot 102 wt A I22 102 z wt A ETZI E0Te zz X BTZI E0Te 2 339 9142 0 which is attenuated as it penetrates into the conductor At z 0 the combined wave in medium 1 must join the wave in medium 2 pursuant to the boundary conditions 9139 Since E i 0 on both sides boundary condition i yields of 0 Since BL 0 ii is automatically satis ed Meanwhile iii gives E01 EUR EOT and iv with K f 0 says 1 I E01 E01 2E01 0 9144 mm M260 or E0 EOR EoT 9145 where M v N 0 E 19 9146 M260 It follows that 1 3 t 2 E0 E0 E0 E0 9147 W These results are formally identical to the ones that apply at the boundary between nonconductors Eq 982 but the resemblance is deceptive since 5 is now a complex number For a perfect conductor a 00 k2 00 Eq 9126 so 13 is in nite and EOR 420 E0 0 9148 398 CHAPTER 9 ELECTROMAGNETIC WAVES In this case the wave is totally re ected with a 180 phase shift That s why excellent conductors make good mirrors In practice you paint a thin coating of silver onto the back of a pane of glass the glass has nothing to do with the re ection it s just there to support the silver and to keep it from tarnishing Since the skin depth in silver at optical frequencies is on the order of 100 A you don t need a very thick layer Problem 921 Calculate the re ection coef cient for light at an airto silver interface u 2 11061 60 a 6 x 1079 m1 at optical frequencies a 4 x 1015s 943 The Frequency Dependence of Permittivity In the preceding sections we have seen that the propagation of electromagnetic waves through matter is governed by three properties of the material which we took to be constants the permittivity e the permeability u and the conductivity 0 Actually each of these parameters depends to some extent on the frequency of the waves you are considering Indeed if the permittivity were truly constant then the index of refraction in a transparent medium n 2 would also be constant But it is well known from optics that n is a function of wavelength Fig 919 shows the graph for a typical glass A prism or a raindrop bends blue light more sharply than red and spreads white light out into a rainbow of colors This phenomenon is called dispersion By extension whenever the speed of a wave depends on its frequency the supporting medium is called dispersive12 1480 C 2 E 2 1470 21 O gtlt G 2 1460 1450 4000 5000 6000 7000 Angstroms Wavelength 1 in air if V Figure 919 12Conductors incidentally are dispersive see Eqs 9126 and 9129 94 ABSORPTION AND DISPERSION 399 Because waves of different frequency travel at different speeds in a dispersive medium a wave form that incorporates a range of frequencies will change shape as it propagates A sharply peaked wave typically attens out and whereas each sinusoidal component travels at the ordinary wave or phase velocity 14 U k 9 9 the packet as a whole the envelope travels at the socalled group velocity13 do 9150 vs dk You can demonstrate this by dropping a rock into the nearest pond and watching the waves that form While the disturbance as a whole spreads out in a circle moving at speed vg the ripples that go to make it up will be seen to travel twice as fast 1 2vg in this case They appear at the back end of the packet growing as they move forward to the center then shrinking again and fading away at the front Fig 920 We shall not concern ourselves with these matters I ll stick to monochromatic waves for which the problem does not arise But I should just mention that the energy carried by a wave packet in a dispersive medium ordinarily travels at the group velocity not the phase velocity Don t be too alarmed therefore if in some circumstances 1 comes out greater than c14 Figure 920 My purpose in this section is to account for the frequency dependence of e in nonconduc tors using a simpli ed model for the behavior of electrons in dielectrics Like all classical models of atomic scale phenomena it is at best an approximation to the truth nevertheless it does yield qualitatively satisfactory results and it provides a plausible mechanism for dispersion in transparent media The electrons in a nonconductor are bound to speci c molecules The actual binding forces can be quite complicated but we shall picture each electron as attached to the end of an imaginary spring with force constant ksp ng Fig 921 Fbinding kspringx mw2X 9151 13See A P French Vibrations and Waves p 230 New York W W Norton amp Co 1971 or F S Crawford Jr Waves Sect 62 New York McGraw Hill 1968 14Even the group velocity can exceed c in special cases see P C Peters Am J Phys 56 129 1988 Incidentally if two different Speeds of light are not enough to satisfy you check out S C Bloch Am J Phys 45 538 1977 in which no fewer than eight distinct velocities are identi ed 400 CHAPTER 9 ELECTROMAGNETIC WAVES Electron x W U ksprmg l W Z Figure 921 where x is displacement from equilibrium m is the electron s mass and coo is the natural oscillation frequency kspnng m If this strikes you as an implausible model look back at Ex 41 where we were led to a force of precisely this form As a matter of fact practically any binding force can be approximated this way for suf ciently small displacements from equilibrium as you can see by expanding the potential energy in a Taylor series about the equilibrium point 1 Ux U0 xU 0 ExZU KO The rst term is a constant with no dynamical signi cance you can always adjust the zero of potential energy so that U0 0 The second term automatically vanishes since d U dx F and by the nature of an equilibrium the force at that point is zero The third term is precisely the potential energy of a spring with force constant ksp ng d2 U a x2 0 the second derivative is positive for a point of stable equilibrium As long as the displacements are small the higher terms in the series can be neglected Geometrically all I am saying is that virtually any function can be t near a minimum by a suitable parabola Meanwhile there will presumably be some damping force on the electron Fdamping my 9152 Again I have chosen the simplest possible form the damping must be opposite in direction to the velocity and making it proportional to the velocity is the easiest way to accomplish this The cause of the damping does not concern us here among other things an oscillating charge radiates and the radiation siphons off energy We will calculate this radiation damping in Chapter 11 In the presence of an electromagnetic wave of frequency a polarized in the x direction Fig 921 the electron is subject to a driving force Fdriving qE qu 9039 9153 where q is the charge of the electron and E0 is the amplitude of the wave at the point where the electron is situated Since we re only interested in one point I have reset the clock so that the maximum E occurs there at t 0 Putting all this into Newton s second law gives d 2x 771W Ftot Fbinding Fdamping Fdriving 94 ABSORPTIONAND DISPERSION 401 0139 d2x dx 2 E t 9154 m m quot10 X COS a dtz y d t 0 q 0 Our model then describes the electron as a damped harmonic oscillator driven at frequency a I assume that the much more massive nuclei remain at rest Equation 9154 is easier to handle if we regard it as the real part of a complex equation d d 2 q Wyd tw0xEoe W 9155 In the steady state the system oscillates at the driving frequency 20 x0ei 9156 Inserting this into Eq 9155 we obtain 20 Z qQLEO 9157 coo a lya The dipole moment is the real part of 2 pz qm ail 9158 0 The imaginary term in the denominator means that p is out of phase with E lagging behind by an angle tan1yaa3 02 that is very small when u ltlt mg and rises to 7 when a gtgt coo In general differently situated electrons within a given molecule experience different natural frequencies and damping coef cients Let s say there are f j electrons with frequency wj and damping yj in each molecule If there are N molecules per unit volume the polarization P is given by15 the real part of N2 P q E 9159 m J wj a zyja Now I de ned the electric susceptibility as the proportionality constant between P and E speci cally P so XeE In the present case P is not proportional to E this is not strictly speaking a linear medium because of the difference in phase However the complex polarization P is proportional to the complex eld E and this suggests that we introduce a complex susceptibility Ze P 60er 9160 15 This applies directly to the case of a dilute gas for denser materials the theory is modi ed slightly in accordance with the Clausius Mossotti equation Prob 438 By the way don t confuse the polarization of a medium P with the polarization of a wave same word but two completely unrelated meanings 402 CHAPTER 9 ELECTROMAGNETIC WAVES All of the manipulations we went through before carry over on the understanding that the physical polarization is the real part of P just as the physical eld is the real part of E In particular the proportionality between i and E is the complex permittivity E 601 g and the complex dielectric constant in this model is N 2 g 1 412 f1 9161 meo j wj wz zij Ordinarily the imaginary term is negligible however when a is very close to one of the resonant frequencies wj it plays an important role as we shall see In a dispersive medium the wave equation for a given frequency reads 321 v21 037 9162 it admits plane wave solutions as before Ez t EoeiUEZ wt 9163 with the complex wave number k E MEMO a 9164 Writing 1 in terms of its real and imaginary parts k kiK 9165 Eq 9163 becomes Ez t E0e e39kz 9166 Evidently the wave is attenuated this is hardly surprising since the damping absorbs energy Because the intensity is proportional to E 2 and hence to e72 the quantity 05 E 2K 9167 is called the absorption coef cient Meanwhile the wave velocity is w k and the index of refraction is n 9168 a Ihave deliberately used notation reminiscent of sect 941 However in the present case k and K have nothing to do with conductivity rather they are determined by the parameters of our damped harmonic oscillator For gases the second term in Eq 9161 is small and we can approximate the square root Eq 9164 by the rst term in the binomial expansion 15 21s Then N92 fj 1 9169 2m60 wz 1 ij W1 11 le 112 le 94 ABSORPTION AND DISPERSION 403 8 Dj Figure 922 ck qu 190 w2 n E 1 9170 and 2 2 N a 2K q w Z z Q g H 9171 meoc j wj w yja In Fig 922 I have plotted the index of refraction and the absorption coef cient in the Vicinity of one of the resonances Most of the time the index of refraction rises gradually with increasing frequency consistent with our experience from optics Fig 919 However in the immediate neighborhood of a resonance the index of refraction drops sharply Because this behavior is atypical it is called anomalous dispersion Notice that the region of anomalous dispersion col lt a lt 602 in the gure coincides with the region of maximum absorption in fact the material may be practically opaque in this frequency range The reason is that we are now driving the electrons at their favorite frequency the amplitude of their oscillation is relatively large and a correspondingly large amount of energy is dissipated by the damping mechanism In Fig 922 n runs below 1 above the resonance suggesting that the wave speed exceeds c As I mentioned earlier this is no cause for alarm since energy does not travel at the wave velocity but rather at the group velocity see Prob 925 Moreover the graph does not include the contributions of other terms in the sum which add a relatively constant background that in some cases keeps n gt 1 on both sides of the resonance 404 CHAPTER 9 ELECTROMAGNETIC WAVES If you agree to stay away from the resonances the damping can be ignored and the formula for the index of refraction simpli es qu fj 1 n 2m 0w w2 9172 For most substances the natural frequencies wj are scattered all over the spectrum in a rather chaotic fashion But for transparent materials the nearest signi cant resonances typically lie in the ultraviolet so that a lt wj In that case 1 l l 1 02 N l 1w2 zj 2 2 2 co co wj co co co and Eq 9172 takes the form M j 2 M f l 9173 n Z wz a 2meo C04 J J j J Or in terms of the wavelength in vacuum A 2n cw B nl A 1F 9174 This is known as Cauchy s formula the constant A is called the coef cient of refraction and B is called the coef cient of dispersion Cauchy s equation applies reasonably well to most gases in the optical region What I have described in this section is certainly not the complete story of dispersion in nonconducting media Nevertheless it does indicate how the damped harmonic motion of electrons can account for the frequency dependence of the index of refraction and it explains why n is ordinarily a slowly increasing function of a with occasional anomalous regions where it precipitously drops Problem 922 a Shallow water is nondispersive the waves travel at a speed that is proportional to the square root of the depth In deep water however the waves can t feel all the way down to the bottom they behave as though the depth were proportional to A Actually the distinction between shallow and deep itself depends on the wavelength If the depth is less than A the water is shallow if it is substantially greater than A the water is deep Show that the Wave velocity of deep water waves is twice the group Velocity b In quantum mechanics a free particle of mass m traveling in the x direction is described by the wave function IJx t AeipxEth where p is the momentum and E p2 2m is the kinetic energy Calculate the group velocity and the wave velocity Which one corresponds to the classical speed of the particle Note that the wave velocity is half the group velocity 95 GUIDED WAVES 405 Problem 923 If you take the model in Ex 41 at face value what natural frequency do you get Put in the actual numbers Where in the electromagnetic spectrum does this lie assuming the radius of the atom is 05 A Find the coef cients of refraction and dispersion and compare them with those for hydrogen at 0 C and atmospheric pressure A 136 X 10 4 B 77 X10 15m2 Problem 924 Find the width of the anomalous dispersion region for the case of a single resonance at frequency 00 Assume 3 ltlt 00 Show that the index of refraction assumes its maximum and minimum values at points where the absorption coef cient is at halfmaximum Problem 925 Assuming negligible damping yj 0 calculate the group velocity vg dwdk of the waves described by Eqs 9166 and 9169 Show that vg lt 6 even when U gt c Guided Waves 951 Wave Guides So far we have dealt with plane waves of in nite extent now we consider electromagnetic waves con ned to the interior of a hollow pipe or wave guide Fig 923 We ll assume the wave guide is a perfect conductor so that E 0 and B 0 inside the material itself and hence the boundary conditions at the inner wall are16 i E 0 9175 ii 13L 0 Figure 923 16See Eq 9139 and Prob 742 In a perfect conductor E 2 0 and hence by Faraday s law 3331 0 assuming the magnetic eld started out zero then it will remain so 406 CHAPTER 9 ELECTROMAGNETIC WAVES Free charges and currents will be induced on the surface in such a way as to enforce these constraints We are interested in monochromatic waves that propagate down the tube so E and B have the generic form i EOE y z t 2 E006 yeikz at 9176 ii OC y z t og yeikZ wt For the cases of interest k is real so I shall dispense with the tilde The electric and magnetic elds must of course satisfy Maxwell s equations in the interior of the wave guide 3B iV39E0 111VgtltE 9177 x 11 IV 02 at The problem then is to nd functions E0 and 30 such that the elds 9176 obey the differential equations 9177 subject to boundary conditions 9175 As we shall soon see con ned waves are not in general transverse in order to t the boundary conditions we shall have to include longitudinal components E Z and 15217 E0ExxEyyEzi Bosz ByyBzi 9178 where each of the components is a function of x and y Putting this into Maxwell s equations iii and iv we obtain Prob 926a 8E 8Er BB 83 la 1 8x leZa 1V 8x c2 Z9 3E BB iw u a zkEy szx v szy C2E 9179 3E BB iw 111 zkEx 8 szy v1 szx 8 c2Ey 17To avoid cumbersome notation I shall leave the subscript 0 and the tilde off the individual components 95 GUIDED WAVES 407 Equations ii iii v and vi can be solved for Ex Ey Bx and By i E i kwa BZ x wc2 k2 8x 3y l 8EZ 881 11 El wc2 k2 3y 0 3x 9180 iii B i ka BZ aEz x wc2 k2 3x c2 3y i 33 a 8E 39 B k1 Z w y wc2 k2 3y c2 3x It suf ces then to determine the longitudinal components EZ and BZ if we knew those we could quickly calculate all the others just by differentiating Inserting Eq 9180 into the remaining Maxwell equations Prob 926b yields uncoupled equations for E Z and Bz 32 32 i 2 cuc2 k2 E 0 3y 817 9181 32 32 2 2 11 d l Uc kle039 If EZ 0 we call these TE transverse electric waves if BZ 0 they are called TM transverse magnetic waves if both EZ 0 and BZ 0 we call them TEM waves18 It turns out that TEM waves cannot occur in a hollow wave guide Proof If E1 0 Gauss s law Eq 9177i says aEx 3Ey 0 8x 8y and if 81 0 Faraday s law Eq 9177iii says BEX 0 8x 8y Indeed the vector E0 in Eq 9178 has zero divergence and zero curl It can therefore be written as the gradient of a scalar potential that satis es Laplace s equation But the boundary condition on E Eq 9175 requires that the surface be an equipotential and since Laplace s equation admits no local maxima or minima Sect 314 this means that the potential is constant throughout and hence the electric eld is zero no wave at all qed 181n the case of TEM waves including the uncon ned plane waves of Sect 92 k wc Eqs 9180 are indeterminate and you have to go back to Eqs 9179 408 CHAPTER 9 ELECTROMAGNETIC WAVES Notice that this argument applies only to a completely empty pipe if you run a separate conductor down the middle the potential at its surface need not be the same as on the outer wall and hence a nontrivial potential is possible We ll see an example of this in Sect 95 3 Problem 926 a Derive Eqs 9179 and from these obtain Eqs 9180 b Put Eq 9180 into Maxwell s equations i and ii to obtain Eq 9181 Check that you get the same results using 1 and iv of Eq 9179 952 TE Waves in a Rectangular Wave Guide Suppose we have a wave guide of rectangular shape Fig 924 with height a and width b and we are interested in the propagation of TE waves The problem is to solve Eq 91811i subject to the boundary condition 9175ii We ll do it by separation of variables Let 310530 XxYy sothat 2 2 d X d Y 2 2 Divide by X Y and note that the x and ydependent terms must be constant 1 de 1 dZY 1 EW ki 11 Yd yz k 9182 with k k3 wc2 k2 0 9183 V unmmmmun i I llllllllllllllllll l 1L V Figure 924 95 GUIDED WAVES 409 The general solution to Eq 9182i is Xx A sin kxx B cos kxx But the boundary conditions require that Bx and hence also Eq 9180iii dXdx vanishes at x 0 and x a So A 0 and kx m71a m012 9184 The same goes for Y with ky mrb 110 12 9185 and we conclude that BZ Bo cos mmca cos mTyb 9186 This solution is called the TEmn mode The rst index is conventionally associated with the larger dimension so we assume a z b By the way at least one of the indices must be nonzero see Prob 927 The wave number k is obtained by putting Eqs 9184 and 9185 into Eq 9183 k Awoz Muma2 nbgt21 9187 w lt c71 ma2nb2 Ewmn 9188 the wave number is imaginary and instead of a traveling wave we have exponentially attenuated elds Eq 9176 For this reason comquot is called the cutoff frequency for the mode in question The lowest cutoff frequency for a given wave guide occurs for the mode TE10 com 2 CJTa 9189 Frequencies less than this will not propagate at all The wave number can be written more simply in terms of the cutoff frequency k 21 602 602 9190 c The wave velocity is 8 lt1 9191 v k 1 wmnw2 which is greater than c However see Prob 929 the energy carried by the wave travels at the group velocity Eq 9150 1 2 192 vg dkd c 1 wmnw lt c 9 410 CHAPTER 9 ELECTROMAGNETIC WAVES Wave fronts Figure 925 There s another way to visualize the propagation of an electromagnetic wave in a rect angular pipe and it serves to illuminate many of these results Consider an ordinary plane wave traveling at an angle 9 to the z axis and re ecting perfectly off each conducting surface Fig 925 In the x and y directions the multiply re ected waves interfere to form standing wave patterns of wavelength Ax 2a m and M 2b n hence wave num ber kx 2712 2 nm a and ky nnb respectively Meanwhile in the z direction there remains a traveling wave with wave number kZ k The propagation vector for the original plane wave is therefore and the frequency is w ck c k2 739r2ma2 nb2 ck2 wmn2 Only certain angles will lead to one of the allowed standing wave patterns cos9 1 wmnw2 The plane wave travels at speed c but because it is going at an angle 9 to the z axis its net velocity down the wave guide is vg ccos9 c ll wmnw2 The wave velocity on the other hand is the speed of the wave fronts A say in Fig 925 down the pipe Like the intersection of a line of breakers with the beach they can move much faster than the waves themselves in fact C C v 0 59 1 wmnw2gt39 95 GUIDED WAVES 411 Problem 927 Show that the mode TEOO cannot occur in a rectangular wave guide Hint In this case wc k so Eqs 9180 are indeterminate and you must go back to 9179 Show that BZ is a constant and hence applying Faraday s law in integral form to a cross section that BZ 0 so this would be a TEM mode Problem 928 Consider a rectangular wave guide with dimensions 228 cm X 101 cm What TE modes will propagate in this wave guide if the driving frequency is 170 X 1010 Hz Suppose you wanted to excite only one TE mode what range of frequencies could you use What are the corresponding wavelengths in open space Problem 929 Con rm that the energy in the TEM mode travels at the group velocity Hints Find the time averaged Poynting vector S and the energy density u use Prob 911 if you wish Integrate over the cross section of the wave guide to get the energy per unit time and per unit length carried by the wave and take their ratio Problem 930 Work out the theory of TM modes for a rectangular wave guide In particular nd the longitudinal electric eld the cutoff frequencies and the wave and group velocities Find the ratio of the lowest TM cutoff frequency to the lowest TE cutoff frequency for a given wave guide Cautionz What is the lowest TM mode 953 The Coaxial Transmission Line In Sect 951 I showed that a hollow wave guide cannot support TEM waves But a coaxial transmission line consisting of a long straight wire of radius a surrounded by a cylindrical conducting sheath of radius 1 Fig 926 does admit modes with E Z 0 and BZ 0 In this case Maxwell s equations in the form 9179 yield k wc 9193 so the waves travel at speed c and are nondispersive cBy Ex and ch Ey 9194 so E and B are mutually perpendicular and together with V E 0 V B 0 8Exg u 3 8Ex0 8x 8y 8x 8y 9195 81 8By 88x 0 3x 3y 8x 3y 39 Figure 926 412 CHAPTER 9 ELECTROMAGNETIC WAVES These are precisely the equations of electrostatics and magnetostatics for empty space in two dimensions the solution with cylindrical symmetry can be borrowed directly from the case of an in nite line charge and an in nite straight current respectively A A A A Eos s Bos 9196 S CS for some constant A Substituting these into Eq 9176 and taking the real part A cos kz wt s 1 E3 z t 9197 A cos kz wt A 5 CS Bs d z t Problem 931 a Show directly that Eqs 9197 satisfy Maxwell s equations 9177 and the boundary con ditions 9175 b Find the charge density Az t and the current I Z t on the inner conductor More Problems on Chapter 9 Problem 932 The inversion theorem for Fourier transforms states that 00 1 00 ltzgt dgtke kzdk ltgt dgtk mag CZ dz 919s 00 277 00 Use this to determine Ak in Eq 920 in terms of fz 0 and fz 0 Answer 127f3 oo1fz 0 iwfz one Z dz Problem 933 Suppose sin Er 6 I t A 6 cos kr wt lkr sin kr 01 with i C r This is incidentally the simplest possible spherical wave For notational convenience let kr wt 5 u in your calculations a Show that E obeys all four of Maxwell s equations in vacuum and nd the associated magnetic eld b Calculate the Poynting vector Average S over a full cycle to get the intensity vector I Does it point in the expected direction Does it fall off like r Z as it should c Integrate I da over a spherical surface to determine the total power radiated Answer 47rA23MOC 95 GUIDED WAVES 413 Problem 934 Light of angular frequency 0 passes from medium 1 through a slab thickness d of medium 2 and into medium 3 for instance from water through glass into air as in Fig 927 Show that the transmission coef cient for normal incidence is giVen by 2 2 2 2 1 n n n n d T1 quot1n32123m2ltquot2w 9199 4n1n3 quotg c Hint39 To the left there is an incident wave and a re ected wave to the right there is a transmitted wave inside the slab there is a wave going to the right and a wave going to the left Express each of these in terms of its complex amplitude and relate the amplitudes by imposing suitable boundary conditions at the two interfaces All three media are linear and homogeneous assume m 2 M3 0 x 11 Water Glass Air z Figure 927 Problem 935 A microwave antenna radiating at 10 GHz is to be protected from the environment by a plastic shield of dielectric constant 25 What is the minimum thickness of this shielding that will allow perfect transmission assuming normal incidence Hint39 use Eq 9199 Problem 936 Light from an aquarium Fig 927 goes from water n 31 through a plane of glass n into air n 1 Assuming it s a monochromatic plane wave and that it strikes the glass at normal incidence nd the minimum and maximum transmission coef cients Eq 9199 You can see the sh clearly how well can it see you Problem 937 According to Snell s law when light passes from an optically dense medium into a less dense one n1 gt n the propagation vector k bends away from the normal Fig 928 In particular if the light is incident at the critical angle 9C 2 sin 1n2n1 9200 414 CHAPTER 9 ELECTROMAGNETIC WAVES kT 6 Z 1 1 Figure 928 on which light pipes and ber optics are based But the elds are not zero in medium 2 what we get is a socalled evanescent wave which is rapidly attenuated and transports no energy into medium 21 A quick way to construct the evanescent wave is simply to quote the results of Sect 933 with kT wnzc and kT kT sin 6T 2 cos 6T 2 the only change is that n srn GT 2 1 sine n2 is now greater than 1 and c0s67 1 sin2 9T ivsinz 6T 1 is imaginary Obviously QT can no longer be interpreted as an angle a Show that ETr r Bore Kzelkxwt 9201 l K 2 n1 sine2 n and k E w quotl sine 9202 C C This is a wave propagating in the x direction parallel to the interface and attenuated in the z direction where b Noting that 0 Eq 9108 is now imaginary use Eq 9109 to calculate the re ection coef cient for polarization parallel to the plane of incidence Notice that you get 100 re ection which is better than at a conducting surface see for example Prob 921 c Do the same for polarization perpendicular to the plane of incidence use the results of Prob 916 l9The evanescent elds can be detected by placing a second interface a short distance to the right of the rst in a close analog to quantum mechanical tunneling the wave crosses the gap and reassembles to the right See F o then 67 90 and the transmitted ray Just grazes the surface If 6 exceeds 6c there is no Albiol S Navas and M V Andres Am39 1 Phys 61 165 1993 refracted ray at all only a re ected one this is the phenomenon of total internal re ection 95 GUIDED WAVES 415 d In the case of polarization perpendicular to the plane of incidence show that the real evanescent elds are Er t Eoe39CZ coskx wt 9 9203 E0 Br t e quotZ K sinkx wt 9K kcoskx wt 2 a e Check that the elds in d satisfy all of Maxwell s equations 967 i For the elds in 1 construct the Poynting vector and show that on average no energy is transmitted in the z direction Problem 938 Consider the resonant cavity produced by closing off the two ends of a rect angular wave guide at z 0 and at z d making a perfectly conducting empty box Show that the resonant frequencies for both TE and TM modes are given by col ova1M2 ma2 nb2 9204 for integers l m and n Find the associated electric and magnetic elds 101 Chapter 10 Potentials and Fields The Potential Formulation 1011 Scalar and Vector Potentials In this chapter we ask how the sources p and J generate electric and magnetic elds in other words we seek the general solution to Maxwell s equations 1 BB 1 V39Ez p 111 VXE 60 at 101 8E 11 V B 0 1v V x B qu 060 3 Given pr t and Jr I what are the elds Er t andBr t In the static case Coulomb s law and the BiotSavart law provide the answer What we re looking for then is the generalization of those laws to timedependent con gurations This is not an easy problem and it pays to begin by representing the elds in terms of potentials In electrostatics V x E 0 allowed us to write E as the gradient of a scalar potential E VV In electrodynamics this is no longer possible because the curl of E is nonzero But B remains divergenceless so we can still write as in magnetostatics Putting this into Faraday s law iii yields 3 V E V A x at x 8A VX E 0 at 01 416 10 THE POTENTIAL FORMULATION 417 Here is a quantity unlike E alone whose curl does vanish it can therefore be written as the gradient of a scalar 8A E VV at In terms of V and A then 8A E VV 103 at This reduces to the old form of course when A is constant The potential representation Eqs 102 and 103 automatically ful lls the two homoge neous Maxwell equations ii and iii How about Gauss s law i and the AmpereMaxwell law iv Putting Eq 103 into i we nd that 8 1 V2V VA 0 104 at 60 this replaces Poisson s equation to which it reduces in the static case Putting Eqs 102 and 103 into iv yields 8V 82A V X V X A MN MOEOV MOEO z at at or using the vector identity V x V x A VV A VZA and rearranging the terms a bit 32A av VZA M060 V V 39A 0603 MOJ 105 Equations 104 and 105 contain all the information in Maxwell s equations Example 10 Find the charge and current distributions that would give rise to the potentials k Ziwt Ix2 z for x lt ct v 0 A C 0 for x gt ct where k is a constant and c 1 mono Solution First we ll determine the electric and magnetic elds using Eqs 102 and 103 8A k A E 2 E 39uct x 2 k 8 A k A B V x A J i cz x2y i m xy 4c 8x 2c 418 CHAPTER 10 POTENTIALS AND FIELDS Ez By Hokt 2 Ct Ct Ct C x fHOkt Ck x T 2 Figure 101 plus for x gt 0 minus forx lt 0 These are for x lt ct when xf gt ct E B 0 Fig101 Calculating every derivative in sight I nd VE0 VB0 VXEquotT k VXB M2 0ki c 8E kc A 3B k A A O z Zi y at 2 8t 2 As you can easily check Maxwell s equations are all satis ed with p and J both zero Notice however that B has a discontinuity at x 0 and this signals the presence of a surface current K in the yz plane boundary condition iv in Eq 763 gives ktyKgtltx and hence K kt Evidently we have here a uniform surface current owing in the z direction over the plane x 0 which starts up at t 0 and increases in proportion to t Notice that the news travels out in both directions at the Speed of light for points x gt ct the message that current is now owing has not yet arrived so the elds are zero Problem 101 Show that the differential equations for V and A Eqs 104 and 105 can be written in the more symmetrical form 8L 60 106 EIZA VL qu where 2 2 2 3 8V Dzv e and LEV39A 0 0812 M0 0 at 10 THE POTENTIAL FORMULATION 419 Figure 102 Problem 102 For the con guration in Ex 10 I consider a rectangular box of length 1 width w and height h situated a distance d above the yz plane Fig 102 a Find the energy in the box at time t dc and at 12 d hc b Find the Poynting vector and determine the energy per unit time owing into the b0x during the interval 11 lt t lt t2 c Integrate the result in b from t1 to 2 and con rm that the increase in energy part a equals the net in ux 1012 Gauge Transformations Equations 104 and 105 are ugly and you might be inclined at this stage to abandon the potential formulation altogether However we have succeeded in reducing six problems nding E and B three components each down to four V one component and A three more Moreover Eqs 102 and 103 do not uniquely de ne the potentials we are free to impose extra conditions on V and A as long as nothing happens to E and B Let s work out precisely what this gauge freedom entails Suppose we have two sets of potentials V A and V A which correspond to the same electric and magnetic elds By how much can they differ Write A Aa and V V Since the two A s give the same B their curls must be equal and hence V x a 0 We can therefore write a as the gradient of some scalar aVl 420 CHAPTER 10 POTENTIALS AND FIELDS The two potentials also give the same E so 80 V 0 39B8t 8A wimp The term in parentheses is therefore independent of position it could however depend on time call it kt 01 8quot k0 at 39 Actually we might as well absorb kt into A de ning a new 1 by adding ktdt to the old one This Will not affect the gradient of A it just adds kt to EBABr It follows that AAVt 107 V al at39 Conclusion For any old scalar function A we can with impunity add V1 to A provided we simultaneously subtract 8 at from V None of this will affect the physical quantities E and B Such changes in V and A are called gauge transformations They can be exploited to adjust the divergence of A with a View to simplifying the ugly equations 104 and 105 In magnetostatics it was best to choose V A 0 Eq 561 in electrodynamics the situation is not so clear cut and the most convenient gauge depends to some extent on the problem at hand There are many famous gauges in the literature I ll show you the two most popular ones Problem 103 Find the elds and the charge and current distributions corresponding to l t q r Vr z 0 Ar t 4n 0 r2 Problem 104 Suppose V 0 and A A0 sin kx wt 9 where A0 a and k are constants Find E and B and check that they satisfy Maxwell s equations in vacuum What condition must you impose on a and k Problem 105 Use the gauge function A 14neoqtr to transform the potentials in Prob 103 and comment on the result 101 THE POTENTIAL FORMULATION 421 1013 Coulomb Gauge and Lorentzquot Gauge The Coulomb Gauge As in magnetostatics we pick VA0 108 With this Eq 104 becomes 1 VZV 0 109 60 This is Poisson s equation and we already know how to solve it setting V 0 at in nity 1 r t Vr t f M dr 1010 47160 a Don t be fooled though unlike electrostatics V by itself doesn t tell you E you have to know A as well Eq 103 There is apeculiar thing about the scalar potential in the Coulomb gauge it is determined by the distribution of charge right now If I move an electron in my laboratory the potential V on the moon immediately records this change That sounds particularly odd in the light of special relativity which allows no message to travel faster than the speed of light The point is that V by itself is not a physically measurable quantity all the man in the moon can measure is E and that involves A as well Somehow it is built into the vector potential in the Coulomb gauge that whereas V instantaneously re ects all changes in p the combination V V BA it does not E will change only after suf cient time has elapsed for the news to arrive1 The advantage of the Coulomb gauge is that the scalar potential is particularly simple to calculate the disadvantage apart from the acausal appearance of V is that A is particularly dz cult to calculate The differential equation for A 105 in the Coulomb gauge reads 82A av VZA 11060 u0J MOEOV 1011 The Lorentz gauge In the Lorentz gauge we pick 8V This is designed to eliminate the middle term in Eq 105 in the language of Prob 101 it sets L 0 With this 82A VZA 1013 060 a qu Meanwhile the differential equation for V 104 becomes 82V 1 V2V uoeo p 1014 3t2 60 There is some question whether this should be attibuted to H A Lorentz or to L V Lorenz see J Van Blade IEEE Antennas and Propagation Magazine 332 69 1991 But all the standard textbooks include the t and to avoid possible confusion I shall adhere to that practice lSee O L Brill and B Goodman Am J Phys 35 832 1967 422 CHAPTER 10 POTENTIALS AND FIELDS The virtue of the Lorentz gauge is that it treats V and A on an equal footing the same differential operator 82 v2 060872 E 132 1015 called the d Alembertian occurs in both equations 2 1 l D V p 60 1016 11 DZA 01 This democratic treatment of V and A is particularly nice in the context of special relativity where the d Alembertian is the natural generalization of the Laplacian and Eqs 1016 can be regarded as fourdimensional versions of Poisson s equation In this same spirit the wave equation for propagation speed c E12 f 0 might be regarded as the fourdimensional version of Laplace s equation In the Lorentz gauge V and A satisfy the inhomogeneous wave equation with a source term in place of zero on the right From now on I shall use the Lorentz gauge exclusively and the whole of electrodynamics reduces to the problem of solving the inhomogeneous wave equation for speci ed sources That s my project for the next section Problem 106 Which of the potentials in Ex 101 Prob 103 and Prob 104 are in the Coulomb gauge Which are in the Lorentz gauge Notice that these gauges are not mutually exclusive Problem 107 In Chapter 5 1 showed that it is always possible to pick a vector potential whose divergence is zero Coulomb gauge Show that it is always possible to choose V A u0608 V at as required for the Lorentz gauge assuming you know how to solve equations of the form 1016 Is it always possible to pick V 0 How about A 0 Continuous Distributions 1021 Retarded Potentials 1n the static case Eqs 1016 reduce to four copies of Poisson s equation 2 1 2 V V p V A u0J 60 with the familiar solutions 1 VI f drh Ar Z EJiidr 1017 47160 2 102 CONTINUOUS DIS TRIB UTIONS 423 Figure 103 where L as always is the distance from the sourCe point r to the eld point r Fig 103 Now electromagnetic news travels at the speed of light In the nonstatic case therefore it s not the status of the source right now that matters but rather its condition at some earlier time t called the retarded time when the message left Since this message must travel a distance a the delay is ac 1 Ex 1018 C The natural generalization of Eq 1017 for nonstatic sources is therefore 1 Gt 2 Vr t dr Ar t Mdrk 1019 47160 a 471 2 Here pr t is the charge density that prevailed at point r at the retarded time t Because the integrands are evaluated at the retarded time these are called retarded potentials I speak of the retarded time but of course the most distant parts of the charge distribution have earlier retarded times than nearby ones It s just like the night sky The light we see now left each star at the retarded time correSponding to that star s distance from the earth Note that the retarded potentials reduce properly to 1017 in the static case for which p and J are independent of time Well that all sounds reasonable and surprisingly simple But are we sure it s right I didn t actually derive these formulas for V and A all I did was invoke a heuristic argument electromagnetic news travels at the speed of light to make them seem plausible To prove them I must show that they satisfy the inhomogeneous wave equation 1016 and meet the Lorentz condition 1012 In case you think I m being fussy let me warn you that if you apply the same argument to the elds you ll get entirely the wrong answer 1 0rIA u Jr9IXamp Ert 7e 4mm Tradr Brt 7E 25 dt 424 CHAPTER 10 POTENTIALS AND FIELDS as you would expect if the same logic worked for Coulomb s law and the Biot Savart law Let s stop and check then that the retarded scalar potential satis es Eq 1016 essentially the same argument would serve for the vector potential2 I shall leave it for you Prob 108 to check that the retarded potentials obey the Lorentz condition In calculating the Laplacian of Vr t the crucial point to notice is that the integrand in Eq 1019 depends on r in two places explicitly in the denominator a r r l and implicitly through tr t 2L c in the numerator Thus 1 1 1 vv v AMIGO393 ma pV 4dr 1020 and 1 Vp thr 0VL 1021 the dot denotes differentiation with respect to time3 Now Va 2 and Vla 22Lz Prob 113 so 1 p I2 I2 V 4n60 c a p42dt 1022 Taking the divergence VZV If li W39 39V 2 2 v 2 4160 c i P H v W W But 1 1 Vp 0VL p4 C C as in Eq 1021 and Prob 162 whereas v A 4 3 72 7154 Eq 1100 So 1 1 p 1 32v 1 V2V 4 53 47160 c2 2 71p 4 dr c2 12 600 t con rming that the retarded potential 1019 satis es the inhomogeneous wave equation 1016 qed 2I ll give you the straightforward but cumbersome proof for a clever indirect argument see M A Heald and J B Marion Classical Electromagnetic Radiation 3d ed Sect 81 Orlando FL Saunders 1995 3Note that 33 88t since 1 z Lc andL is independent oft 102 CONTINUOUS DISTRIBUTIONS 425 Incidentally this proof applies equally well to the advanced potentials 1 Vart p r t dr Aart J r t dr 1023 47160 a 471 a in which the charge and the current densities are evaluated at the advanced time ta 21 2 1024 A few signs are changed but the nal result is unaffected Although the advanced potentials are entirely consistent with Maxwell s equations they Violate the mOst sacred tenet in all of physics the principle of causality They suggest that the potentials now depend on what the charge and the current distribution will be at some time in the future the effect in other words precedes the cause Although the advanced potentials are of some theoretical interest they have no direct physical signi cance4 Example 102 An in nite straight wire carries the current 0 fort 5 0 2 10 for t gt 0 That is a constant current 10 is turned on abruptly at t 0 Find the resulting electric and magnetic elds Solution The wire is presumably electrically neutral so the scalar potential is zero Let the wire lie along the z axis Fig 104 the retarded vector potential at point P is It Ast i d 00 For I lt sc the news has not yet reached P and the potential is zero For t gt sc only the segment lzl s t 602 s2 1025 contributes outside this range tr is negative so I tr 2 0 thus 2 2 As r 0102 2 f a s dZ 47T 0 s2 Z2 u2 s2 M0101nct C292 32 A 1 2n 1 A MUM 32Z2Z 2H s O 4Because the d Alembenian involves t2 as Opposed to t the theory itself is timereversal invariant and does not distinguish past from future Time asymmetry is introduced when we select the retarded potentials in preference to the advanced ones re ecting the not unreasonable belief that electromagnetic in uences propagate forward not backward in time 426 CHAPTER 10 POTENTIALS AND FIELDS I dz 4 P Figure 104 The electric eld is 3A I 0 OC Est 2 3t 27txct2 32 and the magnetic eld is 8A A I t A BstVxA Z C 3s 27m 12 s2 Notice that as t gt 00 we recover the static case E 0 B polo27m Problem 108 Con rm that the retarded potentials satisfy the Lorentz gauge condition Hints First show that I 1 l I I l v v n w J v where V denotes derivatives with respect to r and V denotes derivatives with respect to 1 Next noting that J I t Lc depends on r both explicitly and through ft whereas it depends on r only through ft con rm that 1 1 V J JV V J p ZJV a Use this to calculate the divergence of A Eq 1019 Problem 109 a Suppose the wire in Ex 102 carries a linearly increasing current I t kt for t gt 0 Find the electric and magnetic elds generated b Do the same for the case of a sudden burst of current 11 q05t 102 CONTINUOUS DISTRIBUTIONS 427 I 39 n Figure 105 Problem 1010 A piece of wire bent into a loop as shown in Fig 105 carries a current that increases linearly with time I t kt Calculate the retarded vector potential A at the center Find the electric eld at the center Why does this neutral wire produce an electric eld Why can t you determine the magnetic eld from this expression for A 1022 Je menko s Equations Given the retarded potentials 1 t r t Vr t Mdr Ar t J r2dr 1026 47160 a 471 a it is in principle a straightforward matter to determine the elds 8A E VV a BVgtltA 1027 t But the details are not entirely trivial because as I mentioned earlier the integrands depend on 1 both explicitly through a r r in the denominator and implicitly through the retarded time t t ac in the argument of the numerator I already calculated the gradient of V Eq 1022 the time derivative of A is easy 5 3 Edi 1028 at 471 ft Putting them together and using c2 l LEI Er t WM 22 pr t amp mam dr 1029 42 ea c a This is the timedepende nt generalization of Coulomb s law to which it reduces in the static case where the second and third terms drop out and the rst term loses its dependence on tr 428 CHAPTER 10 POTENTIALS AND FIELDS As for B the curl of A contains two terms VxA wab JxVQHdT 471 Now BJZ BJy V X Jx 3y az and BJZ j at 1 84 By Zay c Zay so 1 84 84 1 V X Dx kg 0 Z J X V x But Va 2 Prob 113 so 1 A V x J J x a 1030 C Meanwhile V1t amp42 again Prob 113 and hence Br t 5 72 x Ildt 1031 42 ca This is the time dependent generalization of the Biot Savart law to which it reduces in the static case Equations 1029 and 1031 are the causal solutions to Maxwell s equations For some reason they do not seem to have been published until quite recently the earliest explicit statement of which I am aware was by Oleg Je menko in 19665 In practice Je menko s equations are of limited utility since it is typically easier to calculate the retarded potentials and differentiate them rather than going directly to the elds Nevertheless they provide a satisfying sense of closure to the theory They also help to clarify an observation I made in the previous section To get to the retarded potentials all you do is replace t by t in the electrostatic and magnetostatic formulas but in the case of the elds not only is time replaced by retarded time but completely new terms involving derivatives of p and J appear And they provide surprisingly strong support for the quasistatic approximation see Prob 1012 5O D Je menko Electricity and Magnetism Sect 157 New York Appleton CenturyCrofts 1996 Closely related expressions appear in W K H Panofsky and M Phillips Classical Electricity and Magnetism Sect 143 Reading MA AddisonWesley 1962 See K T McDonald Am J Phys 65 1074 1997 for illuminating commentary and references I 03 POINT CHARGES 429 Problem 1011 Suppose Jr is constant in time so Prob 755 pr t pr 0 r 0t Show that Ert l M3411 47re0 02 that is Coulomb s law holds with the charge density evaluated at the nonretarded time Problem 1012 Suppose the current density changes slowly enough that we can to good approximation ignore all higher derivatives in the Taylor expansion 10 M tr 010 for clarity I suppress the rdependence which is not at issue Show that a fortuitous cancel lation in Eq 1031 yields A 1031 X Ldro 42 That is the BiotSavart law holds with J evaluated at the nonretarded time This means that the quasistatic approximation is actually much better than we had any right to expect the two errors involved neglecting retardation and dropping the second term in Eq 1031 cancel one another to rst order Br t 7 103 Point Charges 1031 Li nardWiechert Potentials My next project is to calculate the retarded potentials Vr t and Ar t of a point charge q that is moving on a speci ed trajectory wt 2 position of q at time t 1032 The retarded time is determined implicitly by the equation r Wtr 6t tr 1033 for the left side is the distance the news must travel and t tr is the time it takes to make the trip Fig 106 I shall call wtr the retarded position of the charge L is the vector from the retarded p0sition to the eld point r 4 r wtr 1034 It is important to note that at most one point on the trajectory is in communication with r at any particular time t For suppose there were two such points with retarded times t1 and t2 ft Cl t1 and 42 Cl 1 2 430 CHAPTER 10 POTENTIALS AND FIELDS Retarded position Particle trajectory Present q position Figure 106 Then a 42 ct2 t1 so the average velocity of the particle in the direction of 1 would have to be c and that s not counting whatever velocity the charge might have in other directions Since no charged particle can travel at the speed of light it follows that only one retarded point contributes to the potentials at any given moment6 Now a naive reading of the formula 1 r t Vr t ELI dr 1035 47160 4 might suggest to you that the retarded potential of a point charge is simply I q 47160 a the same as in the static case only with the understanding that a is the distance to the retarded position of the charge But this is wrong for a very subtle reason It is true that for a point source the denominator 4 comes outside the integral7 but what remains fpu maz 1036 is not equal to the charge of the particle To calculate the total charge of a con guration you must integrate 0 over the entire distribution at one instant of time but here the retardation tr t tc obliges us to evaluate p at di erent times for different parts of the con guration If the source is moving this will give a distorted picture of the total charge You might 6For the same reason an observer at r sees the particle in only one place at a time By contrast it is possible to hear an object in two places at once Consider a bear who growls at you and then runs toward you at the speed 01 sound and growls again you hear both growls at the same time coming from two different locations but there only one bear 7There is however an implicit change in its functional dependence Before the integration 4 r r l is 1 function of r and r after the integration which xes r wtr a 2 Ir wtr is like tr a function of r and z 103 POINT CHARGES 431 think that this problem would disappear for point charges but it doesn t In Maxwell s electrodynamics formulated as it is in terms of charge and current densities a point charge must be regarded as the limit of an extended charge when the size goes to zero And fOr an extended particle no matter how small the retardation in Eq 1036 throws in a factor 1 2 vc1 where v is the velocity of the charge at the retarded time pr tdt39 f 1037 l ILVc Proof This is a purely geometrical effect and it may help to tell the story in a less abstract context You will not have noticed it for obvious reasons but the fact is that a train coming towards you looks a little longer than it really is because the light you receive from the caboose left earlier than the light you receive simultaneously from the engine and at that earlier time the train was farther away Fig 107 In the interval it takes light from the caboose to travel the extra distance L the train itself moves a distance L L L L L L or L c v 1 vc So approaching trains appear longer by a factor 1 vc By contrast a train going away from you looks shorter8 by a factor 1 vc1 In general if the train s velocity makes an angle 6 with your line of sight9 the extra distance light from the caboose mustcover is L cos 6 Fig 108 In the time L cos Qc then the train moves a distance L L L cosl9 L L L 2 Or L c v l UCOsQc 9 fquot quot1 a LR 7r LC7l 39 Ll Figure 107 8Please note that this has nothing whatever to do with special relativity or Lorentz contractionnL is the length of the moving train and its rest length is not at issue The argument is somewhat reminiscent of the Doppler effect 9I assume the train is far enough away or more to the point short enough so that rays from the caboose and engine can be considered parallel 432 CHAPTER 1 0 POTENTIALS AND FIELDS 2 To observer Figure 108 Notice that this effect does not distort the dimensions perpendicular to the motion the height and width of the train Never mind that the light from the far side is delayed in reaching you relative to light from the near side since there s no motion in that direction they ll still look the same distance apart The apparent volume I of the train then is related to the actual volume I by r 4 1038 1 a vc where 2 is a unit vector from the train to the observer In case the connection between moving trains and retarded potentials escapes you the point is this Whenever you do an integral of the type 1037 in which the integrand is evaluated at the retarded time the effective volume is modi ed by the factor in Eq 1038 just as the apparent volume of the train was and for the same reason Because this correction factor makes no reference to the size of the particle it is every bit as signi cant for a point charge as for an extended charge qed It follows then that C V t r 471 0iC V 1039 where v is the velocity of the charge at the retarded time and L is the vector from the retarded position to the eld point 1 Meanwhile since the current density of a rigid object is pv Eq 526 we also have AW 2 of WW I XWl W 471 4 4H 4 103 POINT CHARGES 433 E9 ch V kn t 471 r a V C Vr t 1040 Equations 1039 and 1040 are the famous Li nardWiechert potentials for a moving point charge 10 Example 103 Find the potentials of a point charge moving with constant velocity Solution For convenience let s say the particle passes through the origin at time t 0 so that wt Vt We rst compute the retarded time using Eq 1033 lr Vtrl Ct tr or squaring r2 2r W Uzi2 c202 2n t3 Solving for tr by the quadratic formula I nd that c2t r V t c2t r V2 c2 v2r2 c2t2 tr C2 v2 1041 To x the sign consider the limit v 0 r trtt C In this case the charge is at rest at the origin and the retarded time should be t r c evidently we want the minus sign Now from Eqs 1033 and 1034 rVtr a ct tr and Cttr so 2 A t 41 4Vc ct t1 Ywcrt z H Lz c Ct tr c c nch r v c2 vzm CZI r V2 c2 v2r2 c2t2 10There are many ways to obtain the Li nardWiechert potentials I have tried to emphasize the geometrical origin of the factor 1 amp vc 1 for illuminating commentary see W K H Panofsky and M Phillips Classical Electricity and Magnetism 2d ed pp 3423 Reading MA AddisonWesley 1962 A more rigorous derivation is provided by J R Reitz F J Milford and R W Christy Foundations of Electromagnetic Theory 3d ed Sect 211 Reading MA AddisonWesley 1979 or M A Heald and J B Marion Classical Electromagnetic Radiation 3d ed Sect 83 Orlando FL Saunders 1995 434 CHAPTER 1 0 POTENTIALS AND FIELDS I used Eq 1041 with the minus sign in the last step Therefore 1 qc V 1042 r t 47r 0 62t r V2 C2 v2r2 C212 and Eq 1040 0 WV A 1043 r 0 4n C2t r V2 C2 v2r2 C22 Problem 1013 A particle of charge q moves in a circle of radius a at constant angular velocity u Assume that the circle lies in the x y plane centered at the origin and at time t 0 the charge is at a 0 on the positive x axis Find the Lienard Wiechert potentials for points on the z axis Problem 1014 Show that the scalar potential of a point charge moving with constant velocity Eq 1042 can be written equivalently as Va 1 1 4 1044 4 Rl 12 sin2 Qc2 where R E r Vt is the vector from the present I position of the particle to the eld point r and 0 is the angle between R and v Fig 109 Evidently for nonrelativistic velocities v2 ltlt c2 V t 1 1 r 471601 R 0 q v Figure109 Problem 1015 I showed that at most one point on the particle trajectory communicates with r at any given time In some cases there may be no such point an observer at r would not see the particle in the colorful language of General Relativity it is beyond the horizon As an example consider a particle in hyperbolic motion along the x axis wt le ct2fr oo lt t lt 00 1045 103 POINT CHARGES 435 In Special Relativity this is the trajectory of a particle subject to a constant force F 2 me2 b Sketch the graph of w versus t At four or ve representative points on the curve draw the trajectory of a light signal emitted by the particle at that point both in the plus x direction and in the minus x direction What region on your graph corresponds to points and times x t from which the particle cannot be seen At what time does someone at point x rst see the particle Prior to this the potential at x is evidently zero Is it possible for a particle once seen to disappear from view Problem 1016 Determine the Li nard Wiechert potentials for a charge in hyperbolic motion Eq 1045 Assume the point r is on the x axis and to the right of the charge 1032 The Fields of a Moving Point Charge We are now in a position to calculate the electric and magnetic elds of a point charge in arbitrary motion using the Li nard Wiechert potentials11 c v Vr t q Ar t Vr I 1046 4713960 ac a v c2 and the equations for E and B 8A E VV BVgtltA at The differentiation is tricky however because a r wtr and v v39vt 1047 are both evaluated at the retarded time and tr de ned implicitly by the equation r wt ct tr 1048 is itselfa function of 1 and t12 So hang on the next two pages are rough going but the answer is worth the effort Let s begin with the gradient of V 6 VV 4713960 ac a v2 VZC ILv 1049 You can get the elds directly from Je menko s equations but it s not easy See for example M A Heald and J B Marion Classical Electromagnetic Radiation 3d ed Sect 84 Orlando FL Saunders 1995 12The following calculation is done by the most direct brute force method For a more clever and ef cient approach see I D Jackson Classical Electrodynamics 3d ed Sect 141 New York John Wiley 1999 436 CHAPTER I 0 POTENTIALS AND FIELDS Sincei ct tr Va thr 1050 As for the second term product rule 4 gives VOLv 4 Vv v A Vaa x V x v v x V x a 1051 Evaluating these terms one at a time aV aaaaaa t O v Xax yay Zaz vr L dv Btr dv Btr dv 3t quotdt 3x ydt 8y Zdtr 8z 2 34 Vtr 1052 where a E v is the acceleration of the particle at the retarded time Now v VIL v V v Vw 1053 and v Vr v3 vyi v23 xiyyzi 3x 8y dz vxxvyyvziv 1054 while v A Vw vv Vtr same reasoning as Eq 1052 Moving on to the third term in Eq 1051 3U 8U 8U 8U 3U 81 V zy A xz A yX A XV By azxlt3z 3x y 3x 8y 2 dvz 8t dvy at A dvx at dvz at dvy at dvx at A X y z dtr 3y dt Bz dt Bz dt 3x dt 3x dt 3y a x Vtr 1055 Finally aner wa 1056 but V x r 0 while by the same argument as Eq 1055 V x w v x Vtr 1057 103 POINT CHARGES 437 Putting all this back into Eq 1051 and using the BACCAB rule to reduce the triple cross products V4v a4VtrV vvVt 4x ax Vtrvx vx Vtr v 4a v2vn 1058 Collecting Eqs 1050 and 1058 together we have qc 1 2 2 42160 40 4402 VH5 U 4 aWtr 1059 To complete the calculation we need to know Vtr This can be found by taking the gradient of the de ning equation 1048 which we have already done in Eq 1050 and expanding out V4 1 th V4 Vx4 mVMw 1 4 V44 x V x 4 1060 But 4 V4 4 v4 Vtr same idea as Eq 1053 while from Eq 1056 and 1057 Vx4vatr Thus 1 l th 4 v4 Vtr 4 x v x Vtr 4 4vVt and hence 4 Vt 1061 40 4 v Incorporating this resu1t into Eq 1059 I conclude that 1 qc V 2 2 47160 M a W3 4c 4 vv c 0 4 a4 1062 A similar calculation which I sha111eave for you Prob 1017 yields 8A 1 qc 3t 47160 46 4 v3 w a v v wC 4 Ec2 02 4 av 1063 438 CHAPTER 10 POTENTIALS AND FIELDS Combining these results and introducing the vector ch4 v 1064 I nd Er t f Eomf upucz v2u 4 x u x a 1065 Meanwhile V XA iZV X VV i2VV XV vgtlt VV c c We have already calculated V X v Eq 1055 and V V Eq 1062 Putting these together VgtltA l q 4xc2 v2v4av4ua c 4760 11 43 The quantity in brackets is strikingly similar to the one in Eq 1065 which can be written using the BAC CAB rule as c2 122u 4 au 4 ua the main difference is that we have v s instead of u s in the rst two terms In fact since it s all crossed into 4 anyway we can with impunity change these V s into u s the extra term proportional to 2 disappears in the cross product It follows that 1 Br t 4 x Er t 1066 C Evidently the magnetic eld of a point charge is always perpendicular to the electric eld and to the vector from the retarded paint The rst term in E the one involving c2 v2u falls off as the inverse square of the distance from the particle If the velocity and acceleration are both zero this term alone survives and reduces to the old electrostatic result 1 471604 m 4 N For this reason the rst term in E is sometimes called the generalized Coulomb eld Because it does not depend on the acceleration it is also known as the velocity eld The second term the one involving 4 x u x a falls off as the inverse rst power of 4 and is therefore dominant at large distances As we shall see in Chapter 11 it is this term that is responsible for electromagnetic radiation accordingly it is called the radiation eld or since it is proportional to a the acceleration eld The same terminology applies to the magnetic eld Back in Chapter 2 I commented that if we could only write down the formula for the force one charge exerts on another we would be done with electrodynamics in principle That together with the superposition principle would tell us the force exerted on a test 103 POINT CHARGES 439 charge Q by any con guration whatsoever Well here we are Eqs 1065 and 1066 give us the elds and the Lorentz force law determines the resulting force IQ 4 2 2 47160 4103 c v uagtltuxa V A 2 2 F x axc v uagtlt uxa 1067 where V is the velocity of Q and 4 u v and a are all evaluated at the retarded time The entire theory of classical electrodynamics is contained in that equation but you see why I preferred to start out with Coulomb s law Example 104 Calculate the electric and magnetic elds of a point charge moving with constant velocity Solution Putting a 0 in Eq 1065 q CZ 02 47T60 Lu3 In this case using w Vt 4n ca LV cr vtr ct trv cr Vt In Ex 103 we found that ac dv4uc2t rv2c2 vZr2 c2t2 In Prob 1014 you showed that this radical could be written as Rc 1 02 sin2 0c2 REr vt where is the vector from the present location of the particle to r and 6 is the angle between R and v Fig 109 Thus q 1 vzc2 32 4n60 1 02 sin2 6c2 R I Er t 1068 N Notice that E points along the line from the present position of the particle This is an extraordinary coincidence since the message came from the retarded position Because of the sin2 9 in the denominator the eld of a fastmoving charge is attened out like a pancake in the direction perpendicular to the motion Fig 1010 In the forward and backward directions E is reduced by a factor 1 11262 relative to the eld of a charge at rest in the perpendicular direction it is enhanced by a factor 1 1 vzcz 440 CHAPTER I 0 POTENTIALS AND FIELDS E lllt Figure 1010 As for B we have r vtr r vtt trv A V a L L L C and therefore 1069 1i 1 B ILgtltE 2VXE c c Lines of B circle around the charge as shown in Fig 1011 Figure 1011 The elds of a point charge moving at constant velocity Eqs 1068 and 1069 were rst obtained by Oliver Heaviside in 188813 When u ltlt c2 they reduce to 113 Brt 5 0 q vx 10701 Em t 47T60 R2 71 E The rst is essentially Coulomb s law and the latter is the Biot Savart law for a point chargequot I warned you about in Chapter 5 Eq 540 13 For history and references see 0 J Je menko Am J Phys 62 79 1994 I 03 POINT CHARGES 4 41 442 Problem 1017 Derive Eq 1063 First show that arr ac a t 1071 Problem 1018 Suppose a point charge 1 is constrained to move along the x axis Show that the elds at points on the axis to the right of the charge are given by 1 E q 61 4 60 L2 c v What are the elds on the axis to the left of the charge Problem 1019 a Use Eq 1068 to calculate the electric eld a distance d from an in nite straight wire carrying a uniform line charge A moving at a constant speed 1 down the wire b Use Eq 1069 to nd the magnetic eld of this wire Problem 1020 For the con guration in Prob 1013 nd the electric and magnetic elds at the center From your formula for B determine the magnetic eld at the center of a circular CHAPTER 10 POTENTIALS AND FIELDS a What is the force F2 on 12 due to ql at time t b What total impulse 12 3 00 Fjdt is delivered by 12 t0 ql c What is the force F1 on ql due to 12 at time t d What total impulse 11 3 00 Fldt is delivered to ql by qz Hint It nlight help to review Prob 1015 before doing this integral Answer 12 11 q1q2460bc Problem 1025 A particle of charge 1 is traveling at constant speed 1 along the x axis Calculate the total power passing through the plane x a at the moment the particle itself is at the origin Answer 12v 327T6002 Problem 102614 A particle of charge ql is at rest at the origin A second particle of charge qz moves along the z axis at constant velocity v a Find the force F12t of ql on qz at time t when qz is at z vt b Find the force F21 t of qz on 11 at time t Does Newton s third law hold in this case cCalculate the linear momentum pt in the electromagnetic elds at time t Don t bother with any terms that are constant in time since you won t need them in part d Answer upturn47 2 d Show that the sum of the forces is equal to minus the rate of change of the momentum in the elds and interpret this result physically loop carrying a steady current I and compare your answer with the result of Ex 56 More Problems on Chapter 10 Problem 1021 Suppose you take a plastic ring of radius a and glue charge on it so that the line charge density is Aol sin62 Then you spin the loop about its axis at an angular velocity a Find the exact scalar and vector potentials at the center of the ring Answer A Mokowa3n sinwt ac ii c0swt ac 9 Problem 1022 Figure 235 summarizes the laws of electrostatics in a triangle diagram relating the source 0 the eld E and the potential V Figure 548 does the same for magnetostatics where the source is J the eld is B and the potential is A Construct the analogous diagram for electrodynamics with sources 0 and J constrained by the continuity equation elds E and B and potentials V and A constrained by the Lorentz gauge condition Do not include formulas for V and A in terms of E and B Problem 1023 Check that the potentials of a point charge moving at constant velocity Eqs 1042 and 1043 satisfy the Lorentz gauge condition Eq 1012 Problem 1024 One particle of charge ql is held at rest at the on gin Another particle of charge 12 approaches along the x axis in hyperbolic motion X0 b2 602 it reaches the closest point 19 at time t 0 and then returns out to in nity 14See J J G Scanio Am J Phys 43 258 1975 111 Chapter 11 Radiation Dipole Radiation 1111 What is Radiation In Chapter 9 we discussed the propagation of plane electromagnetic waves through various media but I did not tell you how the waves got started in the rst place Like all electro magnetic elds their source is some arrangement of electric charge But a charge at rest does not generate electromagnetic waves nor does a steady current It takes accelerating charges and changing currents as we shall see My purpose in this chapter is to show you how such con gurations produce electromagnetic waves that is how they radiate Once established electromagnetic waves in vacuum propagate out to in nity carrying energy with them the signature of radiation is this irreversible ow of energy away from the source Throughout this chapter I shall assume the source is localized1 near the origin Imagine a gigantic spherical shell out at radius r Fig 111 the total power passing out through this surface is the integral of the Poynting vector PrfsdaifExBda 111 0 The power radiated is the limit of this quantity as r goes to in nity Prad E lim Pr 112 race This is the energy per unit time that is transported out to in nity and never comes back Now the area of the Sphere is 4nr2 so for radiation to occur the Poynting vector must decrease at large r no faster than 1 r2 if it went like lr3 for example then Pr would go like 1 r and Prad would be zero According to Coulomb s law electrostatic elds fall off like 1 r2 or even faster if the total charge is zero and the BiotSavart law says 1For nonlocalized sources such as in nite planes wires or solenoids the whole concept of radiation must be reformulated see Prob 1124 444 CHAPTER II RADIATION Figure 111 that magnetostatic elds go like 1r2 or faster which means that S 1r4 for static con gurations So static sources do not radiate But Je menko s equations 1029 and 1031 indicate that timedependent elds include terms involving 039 and j that go like 1 r it is these terms that are responsible for electromagnetic radiation The study of radiation then involves picking out the parts of E and B that go like 1 r at large distances from the source constructing from them the 1r2 term in S integrating over a large spherical2 surface and taking the limit as r gt 00 I ll carry through this procedure rst for oscillating electric and magnetic dipoles then in Sect 112 we ll consider the more dif cult case of radiation from an accelerating point charge 1112 Electric Dipole Radiation Picture two tiny metal spheres separated by a distance d and connected by a ne wire Fig 112 at time t the charge on the upper sphere is qt and the charge on the lower sphere is qt Suppose that we drive the charge back and forth through the wire from one end to the other at an angular frequency a W qo coswt 113 The result is an oscillating electric dipole3 P0 100 coswt 1 114 where p0 E qod is the the maximum value of the dipole moment th doesn t have to be a sphere of course but this makes the calculations a lot easier 3It might occur to you that a more natural model would consist of equal and opposite charges mounted on a spring say so that q is constant while d oscillates instead of the other way around Such a model would lead to the same result but there is a subtle problem in calculating the retarded potentials of a moving point charge which I would prefer to save for Sect 112 I 11 DIPOLE RADIATION 445 Figure 112 The retarded potential Eq 1019 is 1 go cosat C 10 Cosh C 115 Vr t 4 W60 4 4 where by the law of cosines ti r2 q rd cos6 d22 ll6 Now to make this physical dipole into a perfect dipole we want the separation distance to be extremely small approximation 1 d ltlt r 117 Of course if d is zero we get no potential at all what we want is an expansion carried to rst order in d Thus d i rltlq cos6gt 118 Zr It follows that 1 1 d lt1l cos6gt 119 4i r Zr and cod cosat itic 2 cos coo rc 1 5 cos 6 c cod cud cosat rc cos c050 2F s1nat rcs1n cos6 2c 2c In the perfect dipole limit we have further approximation 2 d ltlt 3 1110 Cl 446 CHAPTER I 1 RADIATION Since waves of frequency a have a wavelength A 27rca this amounts to the requirement d ltlt t Under these conditions cosat iC E cosat rc 1 63 cos6 sinat rc 1111 Putting Eqs 119 and 1 111 into Eq 115 we obtain the potential of an oscillating perfect dipole p0 cos0 Vr0t sinwt rc cosot rc 1112 47T 0r In the static limit a gt 0 the second term reproduces the old formula for the potential of a stationary dipole Eq 399 p0 cos 6 47r 0r2 39 This is not however the term that concerns us now we are interested in the elds that survive at large distances from the source in the socalled radiation zone4 approximation 3 r gtgt 2 1113 61 Or in terms of the wavelength r gtgt A In this region the potential reduces to P060 47T 0C Vr a r 0 56 sinat rc 1114 r Meanwhile the vector potential is determined by the current owing in the wire at m if 2 q0asinati 1115 Referring to Fig 113 AW of WC 1116 47f d2 4 Because the integration itself introduces a factor of d we can to rst order replace the integrand by its value at the center Ar6t HTpowsinat rc2 1117 r Notice that whereas I implicitly used approximations l and 2 in keeping only the rst order in d Eq 1117 is not subject to approximation 3 4Note that approximations 2 and 3 subsume approximation 1 all together we have d ltlt A ltlt r 111 DIPOLE RADIATION 447 z q dz r 6 y 1 Figure 113 From the potentials it is a straightforward matter to compute the elds 3 V A 1 3 V A V V r 6 3r r 66 sin 6 P0 C056 lt i sinat rc g cosat rcgt f r2 sinat rc6 r2 pow2 cos 6 A 47r 0c2 lt r cosat rc r ll I dropped the rst and last terms in accordance with approximation 3 Likewise 3A 2 A 9 i cosat rccos6 f sin6 0 at 47rr and therefore 3A 2 6 A E vv Op0w 3 cosat rc6 1118 6t 47r r Meanwhile 1 a 3A A v A A X r Bra 9 86 u0p0a cu sin r sin6 cosat rc 0 sinat rc 47rr c The second term is again eliminated by approximation 3 so 2 BV XA M cosat rc 1119 47w r 448 CHAPTER I I RADIATION Equations 1 118 and 1119 represent monochromatic waves of frequency a traveling in the radial direction at the speed of light E and B are in phase mutually perpendicular and transverse the ratio of their amplitudes is EOBO c All of which is precisely what we expect for electromagnetic waves in free space These are actually spherical waves not plane waves and their amplitude decreases like 1 r as they progress But for large r they are approximately p1ane over small regions just as the surface of the earth is reasonably at locally The energy radiated by an oscillating electric dipole is determined by the Poynting vector 2 2 s in x B amp pow S n6coswr rc f 1120 110 c 4n r The intensity is obtained by averaging in time over a complete cycle 2 4 2 110 pow Sin 6 A S r 112 lt 327T2C r2 Notice that there is no radiation along the axis of the dipole here sin 6 0 the intensity pro le5 takes the form of a donut with its maximum in the equatorial plane Fig 114 The total power radiated is found by integrating S over a sphere of radius r H0P8 04 1122 127w 2 4 2 a 6 71c r Figure 1 14 5The radial coordinate in Fig 114 represents the magnitude of S at xed r as a function of 6 and q I 11 DIPOLE RADIATION 449 It is independent of the radius of the sphere as one would expect from conservation of energy with approximation 3 we were anticipating the limit r gt oo Example 111 The sharp frequency dependence of the power formula is what accounts for the blueness of the sky Sunlight passing through the atmosphere stimulates atoms to oscillate as tiny dipoles The incident solar radiation covers a broad range of frequencies white light but the energy absorbed and reradiated by the atmospheric dipoles is stronger at the higher frequencies because of the 04 in Eq 1 122 It is more intense in the blue then than in the red It is this reradiated light that you see when you look up in the sky unless of course you re staring directly at the sun Because electromagnetic waves are transverse the dipoles oscillate in a plane orthogonal to the sun s rays In the celestial arc perpendicular to these rays where the blueness is most pronounced the dipoles oscillating along the line of sight send no radiation to the observer because of the sin2 6 in equation Eq 1121 light received at this angle is therefore polarized perpendicular to the sun s rays Fig 1 15 Sun s rays gt This dipole does not radiate to the observer Figure 115 The redness of sunset is the other side of the same coin Sunlight coming in at a tangent to the earth s surface must pass through a much longer stretch of atmosphere than sunlight coming from overhead Fig 116 Accordingly much of the blue has been removed by scattering and what s left is red Problem 111 Check that the retarded potentials of an oscillating dipole Eqs 1 l 12 and 1 ll7 satisfy the Lorentz gauge condition Do not use approximation 3 Problem 112Equation 1 114 can be expressed in coordinate free form by writing p0 cos 6 2 p0 139 Do so and likewise for Eqs lll7 1118 l119 and 1121 CHAPTER II RADIATION Atmosphere thickness grossly exaggerated Sun s rays Figure 116 Problem 113 Find the radiation resistance of the wire joining the two ends of the dipole This is the resistance that would give the same average power loss to heat as the oscillating dipole in fact puts out in the form of radiation Show that R 790 d A2 2 where A is the wavelength of the radiation For the wires in an ordinary radio say d 5 cm should you worry about the radiative contribution to the total resistance 1 Figure 117 Problem 114 A rotating electric dipole can be thought of as the superposition of two oscillating dipoles one along the x axis and the other along the y axis Fig 117 with the latter out of phase by 90 p p0cosat x sinat 9 Using the principle of superposition and Eqs 1118 and 1119 perhaps in the form suggested by Prob 112 nd the elds of a rotating dipole Also nd the Poynting vector and the intensity of the radiation Sketch the intensity pro le as a function of the polar angle 6 and calculate the total power radiated DOes the answer seem reaSOnable Note that power being quadratic in the elds does not satisfy the superposition principle In this instance however it seems to Can you account for this 111 DIPOLE RADIATION 451 Figure 1 18 1113 Magnetic Dipole Radiation Suppose now that we have a wire loop of radius 1 Fig l 18 around which we drive an alternating current 1 t 10 cosat 1123 This is a model for an oscillating magnetic dipole mt yrb21t 2 mocosat 2 1124 where mo 2 7112210 1125 is the maximum value of the magnetic dipole moment The loop is uncharged so the scalar potential is zero The retarded vector potential is Art gWar 1126 For a point 139 directly above the x axis Fig 118 A must aim in the y direction since the x components from symmetrically placed points on either side of the x axis will cancel Thus 2n olob M cos d ll27 o M Mr I 4 cos d serves to pick out the ycomponent of dl By the law of cosines 4 r2b2 2rbcos1p 452 CHAPTER I 1 RADIATION where w is the angle between the vectors 139 and b r rsin6xrcos62 b bcos xbsin So rb cos w r b rb sin6 cos d and therefore r2b2 2rbsin6cos 1128 For a perfect dipole we want the loop to be extremely small approximation 1 b ltlt r 1129 To rst order in b then a g r 1 gsin0cos so 7 1 sin6cos 1 1130 r and 60 cosat LC cos at rc s1n6 cos c cob cub cosat rccos 7 s1n6 cosd s1nat rc sin lt sinO cos W c As before we also assume the size of the dipole is small compared to the wavelength radiated approximation 2 b ltlt 5 1131 a In that case N cub cosat 4c cosat rc 7 s1n6 cos d sinat rc 1132 Inserting Eqs 1 130 and 1132 into Eq 1127 and dropping the secondorder term I b 27139 Ar t E 7337 cosat rc O 1 b sin6 cos d cosat rc gsinat rcgt cos d The rst term integrates to zero 27r cos d 0 O 1 11 DIPOLE RADIATION 453 The second term involves the integral of cosine squared 271 cos211J drjl 7r 0 Putting this in and noting that in general A points in the iidirection I conclude that the vector potential of an oscillating perfect magnetic dipole is Ar 6 t Homo Sine cosat rc 6C2sinat rc 1133 47r r In the static limit a 0 we recover the familiar formula for the potential of a magnetic dipole Eq 585 uo m0 Sin 6 A Ar E r2 In the radiation zone C approximation 3 r gtgt a 1134 the rst term in A is negligible so 39 6 A Ar6 t Om0w lt31 gtsinat rc 1135 47m r From A we obtain the elds at large r 2 6 A E a A m cosat rc 1136 at 47rc r and 2 39 6 B V x A M 5 cosat rc6 1137 47rc2 r I used approximation 3 in calculating B These elds are in phase mutually perpendicular and transverse to the direction of propagation r and the ratio of their amplitudes is EOBO c all of which is as expected for electromagnetic waves They are in fact remarkably similar in structure to the elds of an oscillating electric dipole Eqs 1118 and 1119 only this time it is B that points in the 6 direction and E in the 1 direction whereas for electric dipoles it s the other way around The energy ux for magnetic dipole radiation is 2 gtcosat rc i 1138 1 2 in6 sExBampm0a s 0 C 47rc the intensity is 2 4 2 Homoa sin 6 A S r2 r 3939 454 CHAPTER 11 RADIATION and the total radiated power is P Momgw lt 127rc3 39 Once again the intensity pro le has the shape of a donut Fig 1 14 and the power radiated goes like w4 There is however one important difference between electric and magnetic dipole radiation For con gurations with comparable dimensions the power radiated elec trically is enormously greater Comparing Eqs 1122 and 1140 Pmagnetic 1 41 Pelectric POC 1 140 where remember me 7119210 and p0 2 god The amplitude of the current in the electrical case was 10 11061 Eq 1 115 Setting d 2 71b for the sake of comparison I get 2 m 1142 P electric C But cobc is precisely the quantity we assumed was very small approximation 2 and here it appears squared Ordinarily then one should expect electric dipole radiation to dominate Only when the system is carefully contrived to exclude any electric contribution as in the case just treated will the magnetic dipole radiation reveal itself Problem 115 Calculate the electric and magnetic elds of an oscillating magnetic dipole without using approximation 3 Do they look familiar Compare Prob 933 Find the Poynting vector and show that the intensity of the radiation is exactly the same as we got usng approximation 3 Problem 116 Find the radiation resistance Prob 1 13 for the oscillating magnetic dipole in Fig 1 18 Express your answer in terms of A and b and compare the radiation resistance of the electric dipole Answer 3 X 105 bA4 2 Problem 117 Use the duality transformation of Prob 760 together with the elds of an oscillating electric dipole Eqs 1 1 18 and l 119 to determine the elds that would be produced by an oscillating Gilbert magnetic dipole composed of equal and opposite magnetic charges instead of an electric current loop Compare Eqs 1 136 and l 137 and comment on the result 1114 Radiation from an Arbitrary Source In the previous sections we studied the radiation produced by two speci c systems os cillating electric dipoles and oscillating magnetic dipoles Now I want to apply the same procedures to a con guration of charge and current that is entirely arbitrary except that it is localized within some nite volume near the origin Fig 1 19 The retarded scalar potential is 1 Vr t mMdt 1143 111 DIPOLE RADIATION 455 Figure 119 where 4ir2r 2 2rr 1144 As before we shall assume that the eld point r is far away in comparison to the dimensions of the source approximation 1 z r ltlt r 1 145 Actually r is a variable of integration approximation 1 means that the maximum value of r as it ranges over the source is much less than r On this assumption r rltl r2gt 1146 r lgllt1r39 1147 4L r r SO and r r r pl l LC0 l t c Expanding p as a Taylor series in t about the retarded time at the origin IOEt 1148 we have pr t ic 2 pr to pr to r 39Cr gt 1149 where the dot signi es differentiation with respect to time The next terms in the series wouldbe 1 er 2 1m rr 3 2p 6 32p 6 456 CHAPTER 11 RADIATION We can afford to drop them provided c c c approxrmatlon 2 r ltlt libpl Iiiblur l pll For an oscillating system each of these ratios is ca and we recover the old approximation 2 In the general case it s more dif cult to interpret Eq 1150 but as a procedural matter approximations 1 and 2 amount to keeping only the rstorder terms in r Putting Eqs 1147 and 1149 into the formula for V Eq 1143 and again discarding the secondorder term 1quot A d pr to dr 39l pl todr E c E r pr to dr 1150 Vr t g 47160r The rst integral is simply the total charge Q at time to Because charge is conserved however Q is actually independent of time The other two integra1s represent the electric dipole moment at time to Thus Q 1quotPt0fquotl3to 1151 l V iz r 4neor r2 rc In the static case the rst two terms are the monopole and dipole contributions to the multipole expansion for V the third term of course would not be present Meanwhile the vector potential is r t L c Ar t 51 My 1152 471 L As you ll see in a moment to rst order in r it suf ces to replace L by r in the integrand Art mc t0dr 1153 471r According to Prob 57 the integral of J is the time derivative of the dipole moment so 1101 Ar t g 4 r 1 154 Now you see why it was unnecessary to carry the approximation of L beyond the zeroth order L E r p is already rst order in r and any re nements would be corrections of second order Next we must calculate the elds Once again we are interested in the radiation zone that is in the elds that survive at large distances from the source so we keep only those terms that go like 1 r approximation 3 discard 1 r2 terms in E and B 1155 For instance the Coulomb eld 1 2 f 47160 r2 I I 1 DIPOLE RADIATION 457 coining from the rst term in Eq 1 151 does not contribute to the electromagnetic radiation In fact the radiation comes entirely from those terms in which we differentiate the argument to From Eq 1148 it follows that 1 1 Vto Vr r C c and hence 1 A 39 A A vvsv it 1 r P00 Vt0 1 r PtoIA 4760 re 47160 rc 4n60C2 r Similarly A 0 M0 0 A V A V X 4 rl X 1300 4 rlV10 X Pto 4nrcr x pt0 whi1e N PUo 8t 471 r So N 0 A A 0 A A E t r 4mm Pr P 4mlrgtltrgtltp 1156 where p is evaluated at time to t r c and 1157 In particular if we use spherical polar coordinates with the z axis in the direction of ptothen Ema g M01700 SIM 6 471 r 1158 quott 39 6 A 306 Mopo dquot 47w r The Poynting vector is 1 no sin26 S g E B at 2 A M0 X 16712cllol r2 1 1159 and the total radiated power is 110132 P E Sd f a 6716 1160 Notice that E and B are mutually perpendicular transverse to the direction of propagation 1quot and in the ratio E B c as always for radiation elds 458 CHAPTER 11 RADIATION Example 112 a In the case of an oscillating electric dipole pa poooswr pm w2p0coswt and we recover the results of Sect 1112 b For a single point charge q the dipole moment is W qdt where dis the position of q with respect to the origin Accordingly 111 6131 where a is the acceleration of the charge In this case the power radiated Eq 1 160 is 061212 6m P 1161 This is the famous Larmor formula I ll derive it again by rather different means in the next section Notice that the power radiated by a point charge is proportional to the square of its acceleration What I have done in this section amounts to a multipole expansion of the retarded potentials carried to the lowest order in r that is capable of producing electromagnetic radiation elds that go like 1 r This turns out to be the electric dipole term Because charge is conserved an electric monopole does not radiate if charge were not conserved the rst term in Eq 1151 would read 1 Q00 47160 r Vmono and we would get a monopole eld proportional to 1 r 1 Q00 4JTEOC r Emono You might think that a charged sphere whose radius oscillates in and out would radiate but it doesn t the eld outside according to Gauss s law is exactly Q 4yreor2i39 regardless of the uctuations in size In the acoustical analog by the way monopoles do radiate witness the croak of a bullfrog If the electric dipole moment should happen to vanish or at any rate if its second time derivative is zero then there is no electric dipole radiation and one must look to the next term the one of second order in r As it happens this term can be separated into two parts one of which is related to the magnetic dipole moment of the source the other to its electric quadrupole moment The former is a generalization of the magnetic dipole radiation we considered in Sect 1113 If the magnetic dipole and electric quadrupole contributions vanish the r 3 term must be considered This yields magnetic quadrupole and electric octopole radiation and so it goes 111 DIPOLE RADIATION 459 Problem 118 Apply Eqs 1 159 and 1160 to the rotating dipole of Prob 114 Explain any apparent discrepancies with your previous answer Problem 119 An insulating circular ring radius b lies in the x y plane centered at the origin It carries a linear charge density A A0 sin where A0 is constant and is the usual azimuthal angle The ring is now set spinning at a constant angular velocity u about the z axis Calculate the power radiated Problem 1110 An electron is released from rest and falls under the in uence of gravity In the rst centimeter what fraction of the potential energy lost is radiated away Figure 1110 Problem 1111 As a model for electric quadrupole radiation consider two oppositely oriented oscillating electric dipoles separated by a distance d as shown in Fig 1110 Use the results of Sect 1112 for the potentials of each dipole but note that they are not located at the origin Keeping only the terms of rst order in d a Find the scalar and vector potentials b Find the electric and magnetic elds c Find the Poynting vector and the power radiated Sketch the intensity pro le as a function of 0 Problem 1112 A current I t ows around the circular ring in Fig 118 Derive the general formula for the power radiated analogous to Eq 1160 expressing your answer in terms of the magnetic dipole moment m t of the loop Answers P 01112 67rc3 112 460 CHAPTER 11 RADIATION Point Charges 1121 Power Radiated by a Point Charge In Chapter 10 we derived the elds of a point charge q in arbitrary motion Eqs 1065 and 1066 Er t hf eo cz v2u 4 x u x a 1162 where u cli v and Br t 2 12 X Er t 1163 The rst term in Eq 1162 is called the velocity eld and the second one with the triple crossproduct is called the acceleration eld The Poynting vector is SiltExBiExaxEiE2amp aEE 1164 0 M00 00 However not all of this energy ux constitutes radiation some of it is just eld energy carried along by the particle as it moves The radiated energy is the stuff that in effect detaches itself from the charge and propagates off to in nity It s like ies breeding on a garbage truck Some of them hover around the truck as it makes its rounds others y away and never come back To calculate the total power radiated by the particle at time tr we draw a huge sphere of radius a Fig 1111 centered at the position of the particle at time 1 wait the appropriate interval 2 1 3 1165 C for the radiation to reach the sphere and at that moment integrate the Poynting vector over the surface6 I have used the notation t because in fact this is the retarded time for all points on the sphere at time t Now the area of the sphere is proportional to 2 so any term in S that goes like 1 L2 will yield a nite answer but terms like 1 L3 or 1 L4 will contribute nothing in the limit a gt 00 For this reason only the acceleration elds represent true radiation hence their other name radiation elds Erad a X 11 X a 1166 1 0 47160 a u3 6Note the subtle change in strategy here In Sect 111 we worked from a xed point the origin but here it is more appropriate to use the moving location of the charge The implications of this change in perspective will become clearer in a moment 1 12 POINT CHARGES 461 Figure 1111 The velocity elds carry energy to be sure and as the charge moves this energy is dragged along but it s not radiation It s like the ies that stay with the garbage truck Now Erad is perpendicular to 2 so the second term in Eq 1164 vanishes 1 s d E2 22 1167 Ta Moc rad If the charge is instantaneously at rest at time tr then u 02 and q A A A A Bradm x4xaaaa a 1168 Inthatcase 1 M04 2 2 A 2 A 061202 sin26 A SIM Mm M 14 76 2 quot 3969 where 6 is the angle between Ii and a No power is radiated in the forward or backward direction rather it is emitted in a donut about the direction of instantaneous acceleration Fig 1112 Figure 1112 462 CHAPTER 11 RADIATION The total power radiated is evidently 2 2 2 5m 6 P Sraddaz oqa ezsin6d6d 16712c 22 Or 2 2 P M 1170 671C This again is the Larmor formula which we obtained earlier by another route Eq 1 161 Although I derived them on the assumption that v 0 Eqs 1169 and 1170 actually hold to good approximation as long as v ltlt c An exact treatment of the case v 7s 0 is more dif cult7 both for the obvious reason that Erad is more complicated and also for the more subtle reason that Srad the rate at which energy passes through the sphere is not the same as the rate at which energy left the particle Suppose someone is ring a stream of bullets out the window of a moving car Fig1113 The rate N at which the bullets strike a stationary target is not the same as the rate N g at which they left the gun because of the motion of the car In fact you can easily check that N g l vcN if the car is moving towards the target and a N 1 Vgt N c for arbitrary directions here v is the velocity of the car c is that of the bullets relative t0 the ground and Ii is a unit vector from car to target In our case if dWdt is the rate at which energy passes through the sphere at radius L then the rate at which energy left the charge was 1171 dW dWdt mu dW dtr 8181 ac W39 Figure 1113 7In the context of special relativity the condition 1 0 simply represents an astute choice of reference system with no essential loss of generality If you can decide how P transforms you can deduce the general Li nard result from the v 0 Larmor formula see Prob 1269 112 POINT CHARGES 463 I used Eq 1071 to express arr8 But mu licv 1 LC C x which is precisely the ratio of N g to N it s a purely geometrical factor the same as in the Doppler effect The power radiated by the particle into a patch of area a2 sin 6 d6 dd 2 a2 d S2 on the Sphere is therefore given by dP augt1 2 2 42 12xuxa2 1172 L A d9 ac 110C rad 1671260 au5 where d S2 sin 6 d6 dd is the solid angle into which this power is radiated Integrating over 6 and to get the total power radiated is no picnic and for once I shall simply quote the answer 2 1 173 where y E 1 1 v2c2 This is Li nard s generalization of the Larmor formula to which it reduces when u ltlt c The factor y6 means that the radiated power increases enormously as the particle velocity approaches the speed of light vxa 2 6 Pz oq 3 a2 67w C Example 113 Suppose v and a are instantaneously collinear at time tr as for example in straight line motion Find the angular distribution of the radiation Eq 1172 and the total power emitted Solution In this case u x a 2 camp x a so dP q2c2 22 x 2 x a2 E 167T2 0 c ziv5 Now 3x xaiaamp a soampxampxa2a2 ampa2 In particular if we let the z axis point along v then dP 110an2 sin26 d9 167T2C l cos65 1174 where f E vc This is consistent of course with Eq 1169 in the case 1 0 However for very large 1 3 8 1 the donut of radiation Fig 1112 is stretched out and pushed forward by the factor 1 f3 cos 65 as indicated in Fig 1114 Although there is still no radiation in precisely the forward direction most of it is concentrated within an increasingly narrow cone about the forward direction see Prob 1115 464 CHAPTER 11 RADIATION Figure 1114 The total power emitted is found by integrating Eq 1174 over all angles dP u0q2a2 sin26 P dSZ dSZ l67r2c 1 cos65 sin 6 d6 dgb The integral is 2n the 6 integral is simpli ed by the substitution x E cos 6 PM0512a2f1 I xz 8m 1 l x5 39 Inte ration b parts ields l 2 3 and I conclude that g y y 3 Moqzazy6 67rc P 1175 This result is consistent with the Lienard formula Eq 1 173 for the case of collinear v and a Notice that the angular distribution of the radiation is the same whether the particle is accel erating or deCelerating it only depends on the square of a and is concentrated in the forward direction with respect to the velocity in either case When a high speed electron hits a metal target it rapidly decelerates giving off what is called bremsstrahlung or braking radiation What I have described in this example is essentially the classical theory of bremsstrahlung Problem 1113 a Suppose an electron decelerated at a constant rate a from some initial velocity 120 down to zero What fraction of its initial kinetic energy is lost to radiation The rest is absorbed by whatever mechanism keeps the acceleration constant Assume 120 ltlt c so that the Larmor formula can be used b To get a sense of the numbers involved suppose the initial velocity is thermal around 105 ms and the distance the electron goes is 30 A What can you conclude about radiation losses for the electrons in an ordinary conductor Problem 1114 In Bohr s theory of hydrogen the electron in its ground state was supposed to travel in a circle of radius 5 x 10 11m held in orbit by the Coulomb attraction of the proton I 12 POINT CHARGES 465 According to classical electrodynamics this electron should radiate and hence spiral in to the nucleus Show that u ltlt c for most of the trip so you can use the Larmor formula and calculate the lifespan of Bohr s atom Assume each revolution is essentially circular Problem 1115 Find the angle Qmax at which the maximum radiation is emitted in Ex 113 see Fig 1114 Show that for ultrarelativistic speeds 11 close to c Hmax quotV l 62 What is the intensity of the radiation in this maximal direction in the ultrarelativistic case in proportion to the same quantity for a particle instantaneously at rest Give your answer in terms of y Figure 1115 Figure 1116 Problem 1116 In Ex 1 13 we assumed the velocity and acceleration were instantaneously at least collinear Carry out the same analysis for the case where they are perpendicular Choose your axes so that v lies along the z axis and a along the x axis Fig l 115 so that v v i a a fr and i sin 0 cos ab sin0 sin M cos 6 2 Check that P is consistent with the Li nard formula Answeri dP aqua2 1 6cos62 1 62 sin2 6cos2 P a0q2a2y4 E 167r2c 1 6cos65 6H6 For relativistic velocities 6 quota 1 the radiation is again sharply peaked in the forward direction Fig 1 116 The most important application of these formulas is to circular motion in this case the radiation is called synchrotron radiation For a relativistic electron the radiation sweeps around like a locomotive s headlight as the particle moves 1122 Radiation Reaction According to the laws of classical electrodynamics an accelerating charge radiates This radiation carries off energy which must come at the expense of the particle s kinetic energy Under the in uence of a given force therefore a charged particle accelerates less than a neutral one of the same mass The radiation evidently exerts a force Frad back on the charge a recoil force rather like that of a bullet on a gun In this section we ll derive the 466 CHAPTER I I RADIATION radiation reaction force from conservation of energy Then in the next section I ll show you the actual mechanism responsible and derive the reaction force again in the context of a simple model For a nonrelativistic particle u ltlt c the total power radiated is given by the Larmor formula Eq 1170 Moqza2 671c 39 Conservation of energy suggests that this is also the rate at which the particle loses energy under the in uence of the radiation reaction force Frad P 1176 uoqza2 1177 671c Frad39V I say suggests advisedly because this equation is actually wrong For we calculated the radiated power by integrating the Poynting vector over a sphere of in nite radius in this calculation the velocity elds played no part since they fall off too rapidly as a function of L to make any contribution But the velocity elds do carry energy they just don t transport it out to in nity As the particle accelerates and decelerates energy is exchanged between it and the velocity elds at the same time as energy is irretrievably radiated away by the acceleration elds Equation 1177 accounts only for the latter but if we want to know the recoil force exerted by the elds on the charge we need to consider the total power lost at any instant not just the portion that eventually escapes in the form of radiation The term radiation reaction is a misnomer We should really call it the eld reaction In fact we ll soon see that Frad is determined by the time derivative of the acceleration and can be nonzero even when the acceleration itself is instantaneously zero so that the particle is not radiating The energy lost by the particle in any given time interval then must equal the energy carried away by the radiatiOn plus whatever extra energy has been pumped into the velocity elds8 HOWever if we agree to consider only intervals over which the system returns to its initial state then the energy in the velocity elds is the same at both ends and the only net loss is in the form of radiation Thus Eq 1177 while incorrect instantaneously is valid on the average 1 2 2 2 Fradvdzz g q f azdt 1178 I 710 I with the stipulation that the state of the system is identical at 2 1 and t2 In the case of periodic motion for instance we must integrate over an integral number of full cycles9 Now the SActually while the total eld is the sum of velocity and acceleration elds E 2 EU Ea the energy is proportional to E 2 2 E3 2E1 Ea E3 and contains three terms energy stored in the velocity elds alone E 3 energy radiated away E3 and a cross term Ev Ea For the sake of simplicity I m referring to the combination E3 ZEU Ea as quotenergy stored in the velocity elds These terms go like 1M4 and lL z respectively so neither one contributes to the radiation For nonperiodic motion the condition that the energy in the velocity elds be the same at t1 and t2 is more dif cult to achieve It is not enough that the instantaneous velocities and accelerations be equal since the elds farther out depend on v and a at earlier times In principle then u and a and all higher derivatives must be identical at t1 and t2 In practice since the velocity elds fall off rapidly with a it is suf cient that v and a be the same over a brief interval prior to t1 and t2 112 POINT CHARGES 467 right side of Eq 1178 can be integrated by parts 2 2 2 2 Md V t1 1 dt dt dt The boundary term drops out since the velocities and accelerations are identical at t1 and t2 so Eq 1178 can be written equivalently as 1 2 2 Fwd 2 a vdt 0 1179 1 WC Equation 1179 will certainly be satis ed if 2 2 2 dZV f 1 dt 11 1180 This is the AbrahamLorentz formula for the radiation reaction force Of course Eq 1179 doesn t prove Eq 1180 It tells you nothing Whatever about the component of Frad perpendicular to v and it only tells you the time average of the parallel component the average moreover over very special time intervals As we ll see in the next section there are other reasons for believing in the AbrahamLorentz formula but for now the best that can be said is that it represents the simplest form the radiation reaction force could take consistent with conservation of energy The Abraham Lorentz formula has disturbing implications which are not entirely un derstood nearly a century after the law was rst proposed For suppose a particle is subject to no external forces then Newton s second law says Em Mwama 671 from which it follows that at a0e T 1181 where 2 r E 0 1 1182 6717116 In the case of the electron r 6 X 10 24s The acceleration spontaneously increases exponentially with time This absurd conclusion can be avoided if we insist that a0 0 but it turns out that the systematic exclusion of such runaway solutions has an even more unpleasant consequence If you do apply an external force the particle starts to respond before the force acts See Prob 1119 This acausal preacceleration jumps the gun by only a short time I nevertheless it is to my mind philosophically repugnant that the theory should countenance it at all 10 10These dif culties persist in the relativistic version of the Abraham Lorentz equation which can be derived by starting with Li nard s formula instead of Larrnor s see Prob 1270 Perhaps they are telling us that there can be no such thing as a point charge in Classical electrodynamics or maybe they presage the onset of quantum mechanics For guides to the literature see Philip Pearle s chapter in D Teplitz ed Electromagnetism Paths to Research New York Plenum 1982 and F Rohrlich Am J Phys 65 1051 1997 468 CHAPTER 11 RADIATION Example 114 Calculate the radiation damping of a charged particle attached to a spring of natural frequency coo driven at frequency w Solution The equation of motion is quotUquot Fspring Frad Fdriving mng quotquot3quot Fdriving With the system oscillating at frequency w x1 x0 coswt 8 so X CU2X Therefore mic39 myft mng Fd ving 1183 and the damping factor 7 is given by y wzr 1184 When I wrote Fdamping ymv back in Chap 9 Eq 9152 I assumed for simplicity that the damping was proportional to the velocity We now know that radiation damping at least is proportional to ii But it hardly matters for sinusoidal oscillations any even number of derivatives of 11 would do since they re all proportional to v Problem 1117 a A particle of charge q moves in a circle of radius R at a constant speed 11 To sustain the motion you must of course provide a centripetal force mv2 R what additional force Fe must you exert in order to counteract the radiation reaction It s easiest to express the answer in terms of the instantaneous velocity v What power Pe does this extra force deliver Compare P6 with the power radiated use the Larrnor formula b Repeat part a for a particle in simple harmonic motion with amplitude A and angular frequency w wt A coswt 2 Explain the discrepancy 0 Consider the case of a particle in free fall constant acceleration g What is the radiation reaction force What is the power radiated Comment on these results Problem 1118 a Assuming implausibly that y is entirely attributable to radiation damping Eq 1184 show that for optical dispersion the damping is small y ltlt coo Assume that the releVant resonances lie in or near the optical frequency range b Using your results from Prob 924 estimate the width of the anomalous dispersion region for the model in Prob 923 1 12 POINT CHARGES 469 Problem 1119 With the inclusion of the radiation reaction force Eq I 180 Newton s second law for a charged particle becomes F a ra m where F is the external force acting on the particle a In contrast to the case of an uncharged particle a F m acceleration like position and velocity must now be a continuous function of time even if the force changes abruptly Physically the radiation reaction damps out any rapid change in a Prove that a is continuous at any time t by integrating the equation of motion above from t e to t e and taking the limit 6 gt 0 b A particle is subjected to a constant force F beginning at time t 0 and lasting until time T Find the most general solution a t to the equation of motion in each of the three periods it lt0ii0lttlt Tiiit gt T c Impose the continuity condition a at t 0 and I T Show that you can either eliminate the runaway in region iii or avoid preacceleration in region i but not both d If you choose to eliminate the runaway what is the acceleration as a function of time in each interval How about the velocity The latter must of course be continuous at t 0 and t T Assume the particle was originally at rest v oo 0 e Plot a t and vt both for an uncharged particle and for a nonrunaway charged particle subject to this force 1123 The Physical Basis of the Radiation Reaction In the last section I derived the AbrahamLorentz formula for the radiation reaction using conservation of energy I made no attempt to identify the actual mechanism responsible for this force except to point out that it must be a recoil effect of the particle s own elds acting back on the charge Unfortunately the elds of a point charge blow up right at the particle so it s hard to see how one can calculate the force they exert11 Let s avoid this problem by considering an extended charge distribution for which the eld is nite everywhere at the end we ll take the limit as the size of the charge goes to zero In general the electromagnetic force of one part A on another part B is not equal and opposite to the force of B on A Fig 1117 If the distribution is divided up into in nitesimal chunks and the imbalances are added up for all such pairs the result is a net force of the charge on itself It is this selfforce resulting from the breakdown of Newton s third law within the structure of the particle that accounts for the radiation reaction Lorentz originally calculated the electromagnetic selfforce using a spherical charge distribution which seems reasonable but makes the mathematics rather cumbersome12 Because I am only trying to elucidate the mechanism involved I shall use a less realistic model a dumbbell in which the total charge q is divided into two halves separated by 11It can be done by a suitable averaging of the eld but it s not easy See T H Boyer Am J Phys 40 1843 1972 and references cited there 12See J D Jackson Classical Electrodynamics 3rd ed Sect 163 New York John Wiley 1999 470 CHAPTER 11 RADIATION Retarded Present position xt position xt Figure 1117 Figure 1118 a xed distance d Fig 1118 This is the simplest possible arrangement of the charge that permits the essential mechanism imbalance of internal electromagnetic forces to function Never mind that it s an unlikely model for an elementary particle in the point limit d gt 0 any model must yield the AbrahamLorentz formula to the extent that conservation of energy alone dictates that answer Let s assume the dumbbell moves in the x direction and is instantaneously at rest at the retarded time The electric eld at 1 due to 2 is 2 01 L c2ILau ILua 1185 47160 Lll3 Eq 1065 where ucil and 4l d u 1186 sothat LllCL ILala and a12d2 1187 Actually we re only interested in the x component of E1 since the y components will cancel when we add the forces on the two ends for the same reason we don t need to worry about the magnetic forces N ow cl ux 1188 L and hence q ch adz E 1189 1quot 8716062 12 d232 By symmetry E2 E 1x so the net force on the dumbbell is 2 2 2 q q lc ad A F E E 1190 112 POINT CHARGES 471 So far everything is exact The idea now is to expand in powers of d when the size of the particle goes to zero all positive powers will disappear Using Taylor s theorem x1 xtr5ctrl tr x1r1 tr2 i gitUr tr3 i 39 7 we have 1 2 1 3 lxt xtrEaT 6aT 1191 where T E t tr for short Now T is determined by the retarded time condition CT2 12 d2 1192 so 2 2 aT aT2 2 a2 3 4 dCT l cT 1 2 6 cT T T C C C This equation tells us d in terms of T we need to solve it for T as a function of d There s a systematic procedure for doing this known as reversion of series13 but we can get the rst couple of terms more informally as follows Ignoring all higher powers of T dgcT TEE c using this as an approximation for the cubic term 2 3 2 3 a d d a d d T gt TE C 8cc3 c 8c5 and so on Evidently l a2 T d 5d3 d4 1193 c 8c Returning to Eq 1191 we construct the power series for l in terms of d 1id2id3d4 1194 2c2 6C3 Putting this into Eq 1190 I conclude that F 12 a d d 1195 x Se 4260 4c2d 1203 Here a and a are evaluated at the retarded time tr but it s easy to rewrite the result in terms of the present time t aUr 00 illl Ir 00 5IIT 00 ilti 13See for example the CRC Standard Mathematical Tables Cleveland CRC Press 472 CHAPTER 1 I RADIATION and it follows that 2 Fself q a M d 2 1196 47160 4C2d 3c3 The rst term on the right is proportional to the acceleration of the charge if we pull it over to the other side of Newton s second law it simply adds to the dumbbell s mass In effect the total inertia of the charged dumbbell is q2 47160 4dc2 1197 m2m0 where m0 is the mass of either end alone In the context of special relativity it is not surprising that the electrical repulsion of the charges should enhance the mass of the dumbbell FOr the potential energy of this con guration in the static case is 1 122 118 41160 d 9 and according to Einstein s formula E mcz this energy contributes to the inertia of the object14 The second term in Eq 1 196 is the radiation reaction rad 1211c 39 1199 It alone apart from the mass correction survives in the point dumbbell limit d gt 0 Unfortunately it differs from the AbrahamLorentz formula by a factor of 2 But then this is only the selfforce associated with the interaction between 1 and 2 hence the superscript int There remains the force of each end on itself When the latter is included see Prob 1120 the result is uoq2d 11100 6nc Frad 7 reproducing the Abraham Lorentz formula exactly Conclusion The radiation reaction is due to the force of the charge on itself or more elaborately the net force exerted by the elds generated by different parts of the charge distribution acting on one another 14The fact that the numbers work out perfectly is a lucky feature of this con guration If you do the same calculation for the dumbbell in longitudinal motion the mass correction is only half of what it should be there s a 2 instead of a 4 in Eq 1197 and for a sphere it s off by a factor of 34 This notorious paradox has been the subject of much debate over the years See D J Grif ths and R E Owen Am J Phys 51 1120 1983 15Of course the limit d gt 0 has an embarrassing effect on the mass term In a sense it doesn t matter since only the total mass m is observable maybe m0 somehow has a compensating negative in nity so that m comes out nite This awkward problem persists in quantum electrodynamics where it is swept under the mg in a process known as mass renormalization I 12 POINT CHARGES 473 Problem 1120 Deduce Eq 11100 from Eq 1199 as follows a Use the AbrahamLorentz formula to determine the radiation reaction on each end of the dumbbell add this to the interaction term Eq 1199 b Method a has the defect that it uses the AbrahamLorentz formula the very thing that we were trying to derive To avoid this smear out the charge along a strip of length L oriented perpendicular to the motion the charge density then is A q L nd the cumulative interaction force for all pairs of segments using Eq 1199 with the correspondence q2 gt A d yl at one end and q 2 gt A d yz at the other Make sure you don t count the same pair tw1ce More Problems on Chapter 11 Problem 1121 A particle of mass m and charge q is attached to a spring with force constant k hanging from the ceiling Fig 1 119 Its equilibrium position is a distance h above the oor It is pulled down a distance d below equilibrium and released at time t 0 a Under the usual assumptions d ltlt A ltlt h calculate the intensity of the radiation hitting the oor as a function of the distance R from the point directly below q Note The intensity here is the average power per unit area of oor At what R is the radiation most intense Neglect the radiative damping of the oscillator Answerz quZdZw4R2h327r2cR2 h252 b As a check on your formula assume the oor is of in nite extent and calculate the average energy per unit time striking the entire oor Is it what you d expect c Because it is losing energy in the form of radiation the amplitude of the oscillation will gradually decrease After what time I has the amplitude been reduced to d e Assume the fraction of the total energy lost in one cycle is very small Problem 1122 A radio tower rises to height h above at horizontal ground At the top is a magnetic dipole antenna of radius b with its axis vertical FM station KRUD broadcasts from this antenna at angular frequency w with a total radiated power P that s averaged of course over a full cycle Neighbors have complained about problems they attribute to excessive Figure 1119 474 CHAPTER 11 RADIATION radiation from the tower interference with their stereo systems mechanical garage doors opening and closing mysteriously and a variety of suspicious medical problems But the city engineer who measured the radiation level at the base of the tower found it to be well below the accepted standard You have been hired by the Neighborhood Association to assess the engineer s report a In terms of the variables given not all of which may be relevant of course nd the formula for the intensity of the radiation at ground level a distance R from the base of the tower You may assume that a ltlt cw ltlt h Note we are interested only in the magnitude of the radiation not in its direction when measurements are taken the detector will be aimed directly at the antenna b How far from the base of the tower should the engineer have made the measurement What is the formula for the intensity at this location c KRUD s actual power output is 35 kilowatts its frequency is 90 MHZ the antenna s radius is 6 cm and the height of the tower is 200 m The city s radioemission limit is 200 microwattscmz Is KRUD in compliance Problem 1123 As you know the magnetic north pole of the earth does not coincide with the geographic north pole in fact it s off by about 11 Relative to the xed axis of rotation therefore the magnetic dipole moment vector of the earth is changing with time and the earth must be giving off magnetic dipole radiation a Find the formula for the total power radiated in terms of the following parameters 111 the angle between the geographic and magnetic north poles M the magnitude of the earth39s magnetic dipole moment and w the angular velocity of rotation of the earth Hint refer to Prob 114 or Prob 1112 b Using the fact that the earth s magnetic eld is about half a gauss at the equator estimate the magnetic dipole moment M of the earth c Find the power radiated Answers 4 x 105 W d Pulsars are thought to be rotating neutron stars with a typical radius of 10 km a rotational period of 10 3s and a surface magnetic eld of 108 T What sort of radiated power would you expect from such a star See J P Ostriker and J E Gunn Astrophys J 157 1395 1969 Answers 2 X 1036 W Problem 1124 Suppose the electrically neutral y z plane carries a timedependent but uni form surface current K t i a Find the electric and magnetic elds at a height x above the plane if i a constant current is turned on at r 0 Kt 476 112 POINT CHARGES 475 b Show that the retarded vector potential can be written in the form 00 Axt ampi Kt u du 2 0 C and from this determine E and B c Show that the total power radiated per unit area of surface is w 2 Explain what you mean by radiation in this case given that the source is not localized For discussion and related problems see B R Holstein Am J Phys 63 217 1995 T A Abbott and D J Grif ths Am J Phys 53 1203 1985 Ian Problem 1125 When a charged particle approaches or leaves a conducting surface radiation is emitted associated with the changing electric dipole moment of the charge and its image If the particle has mass m and charge q nd the total radiated power as a function of its height z above the plane Answer uocq247t36m2z Problem 1126 Use the duality transformation Prob 760 to construct the electric and mag netic elds of a magnetic monopole qm in arbitrary motion and nd the Larmor formula for the power radiated For related applications see J A Heras Am J Phys 63 242 1995 Problem 1127 Assuming you exclude the runaway solution in Prob 1 119 calculate a the work done by the external force b the nal kinetic energy assume the initial kinetic energy was zero c the total energy radiated Check that energy is conserved in this process16 3 Problem 1128 a Repeat Prob 1119 but this time let the external force be a Dirac delta function F t 2 k6 t for some constant k17 Note that the acceleration is now discontinuous at t 0 though the velocity must still be continuous use the method of Prob 1 119 a to show that Aa kmr In this problem there are only two intervals to consider i t lt 0 and ii I gt 0 b As in Prob 1 127 check that energy is conserved in this process Problem 1129 A charged particle traveling in from 00 along the x axis encounters a rectangular potential energy barrier U0 Ux0 1f0ltxltL otherwise Show that because of the radiation reaction it is possible for the particle to tunnel through the barrier that is even if the incident kinetic energy is less than U0 the particle can pass CHAPTER 11 RADIATION through See F Denef et al Phys Rev E 56 3624 1997 Hint Your task is to solve the equation F a ta m subject to the force FCC U01 505 5x L Refer to Probs 1119 and 1 128 but notice that this time the force is a speci ed function of x not t There are three regions to consider i x lt 0 ii 0 lt x lt L iii x gt L Find the general solution for at vt and xt in each region exclude the runaway in region iii and impose the appropriate boundary conditions at x 0 and x L Show that the nal velocity of is related to the time T spent traversing the barrier by the equation U Lva 0 re TrlT t mvf and the initial velocity atx 2 00 is U0 1 1 vi v f mvf 1 U02 e Tr 1 mvf To simplify these results since all we re looking for is a speci c example suppose the nal kinetic energy is half the barrier height Show that in this case 11 f 1 L U f I In particular if you choose L UfT4 then Ul39 43vf the initial kinetic energy is 89 U0 and the particle makes it through even though it didn t have suf cient energy to get over the barrier Problem 1130 a Find the radiation reaction force on a particle moving with arbitrary velocity in a straight line by reconstructing the argument in Sect 1123 without assuming vtr 0 Answer39 uoq2y467rcxd 3y2a2vCz b Show that this result is consistent in the sense of Eq 1 178 with the power radiated by such a particle Eq 1175 Problem 1131 a Does aparticle in hyperbolic motion Eq 1045 radiate Use the exact formula Eq 1 175 to calculate the power radiated b Does a particle in hyperbolic motion experience aradiation reaction Use the exact formula Prob 1130 to determine the reaction force Comment39 These famous questions carry important implications for the principle of equiv alence See T Fulton and F Rohrlich Annals of Physics 9 499 1960 J Cohn Am J Phys 46 225 1978 Chapter 8 of R Peierls Surprises in Theoretical Physics Princeton Princeton University Press 1979 and the article by P Pearle in Electromagnetism Paths to Research ed D Teplitz New York Plenum Press 1982 l6Problems l 127 and 1128 were suggested by G L Pollack 17This example was rst analyzed by P A M Dirac Proc Roy Soc A167 148 1938 121 Chapter 12 Electrodynamics and Relativity The Special Theory of Relativity 1211 Einstein s Postulates Classical mechanics obeys the principle of relativity the same laws apply in any inertial reference frame By inertial I mean that the system is at rest or moving with constant velocityI Imagine for example that you have loaded a billiard table onto a railroad car and the train is going at constant speed down a smooth straight track The game would proceed exactly the same as it w0uld if the train were parked in the stati0n you don t have to correct your shots for the fact that the train is moving indeed if you pulled all the curtains you would have no way of knowing whether the train was moving or not Notice by contrast that you would know it immediately if the train sped up or slowed dOWn or turned a corner or went over a bump the billiard balls would roll in weird curved trajectories and you yourself would feel a lurch The laws of mechanics then are certainly not the same in accelerating reference frames In its application to classical mechanics the principle of relativity is hardly new it was stated clearly by Galileo Question does it also apply to the laws of electrodynamics At rst glance the answer would seem to be no After all a charge in motion produces a magnetic eld whereas a charge at rest does not A charge carried along by the train would generate a magnetic eld but someone on the train applying the laws of electrodynamics 1This raises an awkward problem If the laws of physics hold just as well in a uniformly moving frame then we have no way of identifying the rest frame in the rst place and hence no way of checking that some other frame is moving at constant velocity To avoid this trap we de ne an inertial frame formally as one in which Newton 39s rst law holds If you want to know whether you re in an inertial frame throw some rocks around if they travel in straight lines at constant speed you ve got yourself an inertial frame and any frame moving at constant velocity with respect to you will be another inertial frame see Prob l21 477 478 CHAPTER 12 ELEC TRODYNAMI C S AND RELATI VI TY Wire loop V 39O Figure 121 in that system would predict no magnetic eld In fact many of the equations of elec trodynamiCs starting with the Lorentz force law make explicit reference to the velocity of the charge It certainly appears therefore that electromagnetic theory presupposes the existence of a unique stationary reference frame with respect to which all velocities are to be measured And yet there is an extraordinary coincidence that gives us pause Suppose we mount a wire loop on a freight car and have the train pass between the poles of a giant magnet Fig 121 As the loop rides through the magnetic eld a motional emf is established acc0rding to the ux rule Eq 713 5 9 dt This emf remember is due to the magnetic f0rce on charges in the wire loop which are moving along with the train On the other hand if someone on the train naively applied the laws of electrodynamics in that system what would the prediction be No magnetic force because the loop is at rest But as the magnet ies by the magnetic eld in the freight car will change and a changing magnetic eld induces an electric eld by Faraday s law The resulting electric force would generate an emf in the loop given by Eq 714 dd dt39 Because Faraday s law and the ux rule predict exactly the same emf people On the train will get the right answer even though their physical interpretation of the process is completely wrong Or is it Einstein could not believe this was a mere coincidence he took it rather as a clue that electromagnetic phenomena like mechanical ones obey the principle of relativity In his view the analysis by the observer on the train is just as valid as that of the observer on the ground If their interpretations differ one calling the process electric the other magnetic so be it their actual predictions are in agreement Here s what he wrote on the rst page of his 1905 paper introducing the special theory of relativity

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