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# Electric & Magnetic Fields PH 316

RHIT

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This 8 page Class Notes was uploaded by Jalon Willms on Monday October 19, 2015. The Class Notes belongs to PH 316 at Rose-Hulman Institute of Technology taught by Staff in Fall. Since its upload, it has received 7 views. For similar materials see /class/225116/ph-316-rose-hulman-institute-of-technology in Physics 2 at Rose-Hulman Institute of Technology.

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Date Created: 10/19/15

PH 316 MJM September 1 2006 WorkKE gradient of PE curl ofE Yesterday we talked about the gradient of a scalar function f and how I claim df E Vfodr In xyz df dX if x dy ifBy dz fBz Since dX XA dx yA dy zA dz we can conclude that VfxyZ XA if6x yA f y zA faz In cylindrical coordinates df dp 6f6p d9 if69 dz Bf z An in nitesimal displacement in cylindrical is dr pA dp 9A pde zA dz This means that the gradient in cylindrical has what components Now we show that network done on a particle is the change in its kinetic energy deet dK This is the famous workkinetic energy theorem deet deX Fydy dez Fodr where Fx Fy and FZ refer to the components of net force on a particle of mass m Fx dpxdt ddtmvx For constant m we can write Fx m dvxdt chain rule m dXdt dVxdX dXdt vx so we have Fx m vx dedX ddXmsz2 Then deX dmvx22 Finally for the total work dW we have dW dex Fydy dez dmvx22 mvyzZ vaZz dmv22 dK When we have only a conservative force acting the work done by this force is the change in the kinetic energy but because energy is conserved U K constant it is the negative change in the potential energy dWcons dK dU The change in a scalar function f may always be written as df grad fodr dWcons dK dU Making use of dWcons Fcons 0 dr and dU grad U 0 11 we conclude from dWcons dU that ans grad U VU This says a conservative force is the negative gradient of a potential energy U For example when U mgy the negative gradient gives Fy mg When U 12 sz the negative gradient gives F kX 2 If we consider the force F qE acting on a particle in the presence of some electrical potential energy U61 then we have qE grad U61 Dividing by q gives E grad Uelq But the electrical potential energy per unit charge is the electric potential39 V So gradVVV This relation lets us connect electric eld and electric potential E always points away from regions of positive voltage E points downhill direction of decresing V and is as mentioned before the direction a positive charge would travel if free to move We can calculate E if we are given V by the following means Ex 6V6X Ey 6V6y and EZ 6V62 On the other hand if we are given E we can nd differences in V and different points by Fcons odl dK dUel or q Eodl dUe1 dqV gt End dV Then vb va aIb Eodl If we make a round trip and return to our starting point Va Va 0 lclosedpmh End The circulation39 of a vector V is de ned as lclosed path Vodl and the 39curl39 of a vector has its components de ned as the ratio of circulationarea as area gt0 Since for electrostatic E elds the circulation is always zero the curl must always vanish Brief summary for static charges E VV circulation of E 0 curl E 0 This fails when we get to changing magnetic elds and induced E elds from BB t curl E 7i 0 Related items Stokes39 theorem states that I Edl I curl E 0 da for any closed path around area da There is a vector identity saying that curl grad f E 0 for any scalar function f PH 317 Why does light go slower in glass From Orear s Physics January 17 2001 Overview We will send a light beam in the x direction think of it starting along the x axis Ein E0y exp imt ikx See sketch on next page When this strikes a thin layer of material Ax thick in the yz plane it will set up a sheet current Ky which will in turn create a magnetic eld ABz on the side of the sheet where the wave is going This in turn creates an induced electric eld AEy which is 900 in phase behind E The total outgoing wave is the sum of En AEy This wave lags the original wave by a phase I AEyEin This phase delay is equivalent to a wave of amplitude Ein travelling the distance Ax at a speed V At speed c the time would be t0 Axc while at speed V the time would be tn Axv The phase difference between the two waves would be I n tn to n Ax 1v1c n cv 1 Axc n n1 Axc When we equate these phase differences we nd the index of refraction of the material Gory details The electrons in the material are bound to atoms in some springlike way Imagine a thin sheet of width Ax is struck at x0 by the incoming plane wave The electrons hit by the invading electric eld feel a total force Fuel qun Fspring may and since Fspring ky and k moan2 quy expint moan2 y mn2 y and y quy expint mn2 02 The current density Jy p vy p ion y imp quy expint mn2 432 We may imagine a sheet Ax thick of this currentarea and we get a sort of sheet current Ky Jy Ax i03pr quy expint mn2 432 On either side of a sheet current in the ydirection we get a magnetic eld BZ From curl B qu we nd that BZ on opposite sides of the sheet current has magnitude 12 uoKy On the side the wave is going remember it came in from the x direction B2 is negative Bz uoKy 2 From curl E 6B6t we nd BEyBx BExay in BZ all go like expint ikx Eyvexua 1D Bz The outgoing eld is then Eom E0y Eyyextm E0y cu0 2 i03pr quy expint mn2 02 Thus the outgoing wave is suppressing the expintikx Eout Eoy Eyextra Eoy l AE where AE c2 uo n pr quy mn2 432 Letting N be the number per unit volume of oscillating electrons we have p eN and q e This gives AE E0y cuoZ 03 Ne2 Ax mnoz 402 The atomic resonant frequencies 030 are generally a lot higher than visible frequencies on so AB is a positive quantity and we have a 900 phase lag between E0y and AE From a simple phasor diagram we observe that the tiny AE added to E0y produced a tiny phase delay I AEE0y E0y rotates ccw in the complex plane Im Re E0y r 3 E out AE Now we bring into play the arguments in the overview to obtain an expression for the index of refraction n 1 Nez 2s0 mam 432 This assumes only one resonant frequency and all N electronsvolume oscillating It is the little brother of Griffiths 9170 for the index of refraction on p 403 incoming E0y along x axis sheet current in the yz plane Magnetic eld BiotSavart etc PH 316 MJM 101605 The BiotSavart BS law says B is created from currents according to cm Ho41m 1 d x rr3 where 1 goes from d to point p where B is to be calculated In class we found the magnetic eld at the center of a current loop to be Bcemer HUI2R We will show see Grif ths p 218 that on the aXis of a current loop BZ 12 HQ IRzRzzz3z Note that this reduces to Bcemer when zgt 0 What about 2 increasing till zgtgtR Then we are very far from the loop on the axis The result is B m no 211 I nR2z3 no 411 2 I nR2z3 2 km 1 nR2z3 where Ampere claimed that little current loops acted like magnets his model of atomic magnetism was a bunch of magnetic dipoles each a little current loop each having a magnetic moment m I a magnetic moment m of a tiny cunrent loop of area a The area vector a is J to the area of the loop in accordance with the righthand rule In Ampere39s terms the B eld on the aXis of a current loop is B nyar zA 2 km I az3 2 km mz3 This should correspond to the electric eld of an electric dipole p with km substituted fo k 4T780 and m for p o E elemmipole kers 3 pCI 1 p r2 3104 p 155 When p is parallel to 1 r points along the aXis of the dipole this reduces to It appears likely that the magnetic eld of a dipole m is given by direct analogy with p by Bdipole kmr5 3 mr r m r2 Note that the interchange between km and k6 involves replacing 180 by uo Now to show two results or at least their main points in Griffiths diV B 0 follows from the BS law curl B uo J where J is the current density current per unit area Ampere39s law I Bdl Ha I enclosed from curl B uo J Ampere s law at a point First we generalize BS to 2D and 3D currents dB km 1 d x rr3 gt B km I Jr x HgI H39P d Then using product rule 6 from the inside cover diV B VOB km Ir r39I r r39l3 0VXJ d39 I Jr x Vor r39I r r39I3 d39 V does not act on J p 223 VxJ 0 so VIB km IJr 0 V x r r39I r r39l3 d39 This vanishes because curl r r39I r r39l3 0 To check this take the Xcomponent of the curl and verify that it is identically zero So we have showed that diV B 0 follows from the generalized BS law V x B km Vx IJr XV r r39I r r39l3 d39 From product rule 8 from the inside cover keeping in mind V doesn39t act on J V x B km I Jr V o r r39I mm d I Jr 0V r r39yI r r39I3 d1 Griffiths disposes of the 2nd term by a series of manipulations including V39oJ 0 for steady currents The first term will reduce to V x B 411km J because as Griffiths argued on p 50 and we shall argue from poisson s equation in a moment V0 r r39I r r39l3 diV grad lIr r39l V2 lIr r39l 4115r r39 We know that the electric potential V satisfies poisson39s equation V2 V ps0 411ke p and that the integral for V starts out as a sum over kqiri and comes out as Vr ke I d39c pr Ir r39l So we write compute VZV and deltafunction property gives us poisson s equation V2 Vr ke I d pr39 V21Ir r39I ke I d pr39 4115r r39 4nke p1 From V x B 411km J and stokes theorem I curl F 0 da lc Fodl you should be able to obtain Ic Bod HO 1enclosed AMPERE S LAW Vector potential A Since diV B 0 and in general diV curl F 0 we can imagine B to be generated by a vector potential B V x A The vector potential A will of course depend on the currents J which create B A also has the freedom to have the gradient of any scalar added to it because it won39t change B B V x AJ Vf V x AJ curl grad f and curl grad f is identically zero for any scalar function According to Griffiths we can use this freedom to make VIA be zero something we will use right now qu VxBVxVxAVVoAV2A So we know that V2llr r39l 4115r r39 and that V2Ar uo Jr This leads us to say A FLO41 I Jr lr r39l d1 because when V2 acts on A it goes inside the integral doesn39t bother with Jr and produces a 5 function when acting on llr r39l giVing us back V2 A 4 J Now we can write down some electrical and magnetic correspondences dV ke dqr gt V ke lpd clrr l ke 1411780 km uo4n dB km 1 d x rr3 gt A km IJ d39clrr l E VV B V X A VoEpso VxBu0J V2 V ps0 VZA 4 J Edipoles kc r5 31 1 P 39 Pr2 l Bdipoles km r5 31 r0m mr2 pqd mIa

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