Electromagnetic Fields ECE 340
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This 18 page Class Notes was uploaded by Dr. Reina Hane on Monday October 19, 2015. The Class Notes belongs to ECE 340 at Rose-Hulman Institute of Technology taught by Edward Wheeler in Fall. Since its upload, it has received 25 views. For similar materials see /class/225084/ece-340-rose-hulman-institute-of-technology in Electrical Engineering & Computer Science at Rose-Hulman Institute of Technology.
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Date Created: 10/19/15
Dynamic Fields Charge conservation Charge neutrality and relaxation time Faraday s law Transformer and motional emf Inductance Ideal transformer Maxwell s equations and the displacement current term Wave propagation Frequency domain description of TEM waves Electromagnetics in the world and beyond 21 28 32 Charge Conservation it Current density J is a vector valued current per area in SI units Am2 Performing a surface integral with J gives the total current passing through an area I j JDds surface 27r a For example the current in the az direction through the surface below is l J J J Dds 0p0 x If integration done about a closed surface the result is the net current leaving the volume enclosed by the closed surface Assuming that charge is neither created nor destroyed within the volume the net current out of the volume must equal the time rate of change of charge stored within the volume Here ds is assumed to be directed out of the volume dQ I0 t Jgds d surface Charge conservation is a generalization of KCL where it is assumed that nodes cannot store charge Point form of charge conservation it Like any vector the divergence theorem holds for current density V gJ dv J gds VJJJS 596 Using charge conservation Divergence holds for current density which is electric charge flux From the definition of charge and current density dQ J ds out 9 Surface dQ d 0 dt Elllfdv r W Epdv 9st stationary vulume lume Where p is volume charge density This result together with the divergence theorem ngJdv Jgds volume Surface gives the point form of charge conservation Consider the physical meaning of the divergence of J The divergence of the current density must be the net current out per unit volume The net current out per volume is the time rate of change of the net charge per unit volume leaving a point Example spherical shell it The current density in a spherical shell J 100 r ar Amz Given rims 01 m and t 1 cm answer the following questions by using surface integrals Check your answer using volume integrals i Is the charge within the spherical shell increasing or decreasing with time ii At what rate is the charge increasing or decreasing t thickness Forthis problem the surface integrals will be of the form lom t Q J gds The volume integrals will be of the form om Ct2 V gJ dv volume using surface integrals dQ 2n 2 2 2 j j1ooo11ago11 smeded a j j1ooo1ago1 sIn6d6dpa dt 6 0 e0 o 51d 100 0113 471 100 013 4739 E 2 O416Cs dt check using volume integrals To use lom Ct2 V gJ dv the divergence of J is needed volume 1 0 2 3 V gJ FED 100 r 300 Nm 7t 27 011 3 3 dQ j300r2sinedrd de3oomj4n dt eo zp0 ro1 3 E O416Cs dt The charge in the shell is decreasing at a rate of 0416 Cs That the charge is decreasing is reasonable since both J and the surface area increases as r increases More current is leaving the outer surface of the spherical shell where r is greater than is entering the inner surface Charge neutrality and relaxation time it If a net charge were introduced in a material does the material return to a neutral state or does the material remain charged The answer lies in the material s conductivity In a good conductor charge moves rapidly and a localized net charge is rapidly eliminated On the other hand if a net charge were introduced in a good dielectric perhaps by ion implantation the material would maintain a net charge for an extended period From charge conservation and Gauss law d VgJ Vchd VgDVgsEp Combining the two equations to eliminate E s d it pp0 gt ptp e8 p dt Where i 239 is the relaxation timequot the characteristic time required for material to return to p charge neutrality For copper o 58 108 Sm s 885410 12Fmr 15 1019 s For undoped silicon o 44 104 Sm s 1O451O3911Fmr 24 107 s For glass 0 103912 Sm s 885410 11Fmr 89 s Forfused quartz o 103917 Sm s 33651039 Fm r 3400 s 39 days Faraday s law it In the early 180039s Michael Faraday discovered that when a loop of wire encircles a timevarying magnetic field a voltage is induced about the loop of wire In the loop of wire shown below a voltage will be induced if a timevarying magnetic field passes through the area defined by the loop d ldt l How is V related to the changing flux and how does one predict its polarity Faraday found the relation to be V dt In terms of the electric field and magnetic flux density vectors NE gdl i B gds loop surface d and d5 are related by the righthand rule With the fingers of your right hand in the direction of dl yourthumb will be in the direction of ds The induced voltage is equal to the path integral of the electric field V NE gdl loop Consider for the moment that the wire is a perfect conductor and begin the path integral at the negative sign of V in the direction indicated below Putting the fingers of your right hand in the direction of dl results in the thumb of your right hand being directed upward which is the direction of ds lfthe path is taken within the PEG the path integral in the conductor will yield nothing and the induced voltage is V J39E gdl What is required for the induced voltage to be positive As can be seen from the diagram ds is upward this is determined by the RHR from dl so that B gds in the relation below is negative d v a jngds surface In order for V is be positive the time derivative must be positive So with B directed as above with ds and B in opposite directions V will be positive E B is increasing with time Another arrangement would also result in V being positive If B were directed upward rather than downward and were decreasing rather than increasing with time lfthese conditions were met V would again be positive B and d5 would be in the same direction so that B gds would be positive The only way to end with a positive result would be for the time derivative ddt to introduce a negative sign B must be decreasing with time in order for V to be positive From this discussion one can see that for V gt 0 two possibilities exist The first is that B is directed downward and is increasing with time d E B gds loop surface dd v Edl dt gt N9 The other possibility is that B be directed upward and is decreasing with time I BdBdt gt 0 BdBldt lt 0 dl dl Lenz s law it We are now in a position to appreciate the significance of the negative sign in Faraday s law dBIdt gt 0 referred to as Lenz s law Connect a resistance to one of the loops above One can see that for W0 R VgtO the direction of current flow will be in the same direction as dl m amp Now eliminate the resistor and consider the wire to be a nonperfect conductor The direction of current flow would be unchanged Consider this If the rate that B were increasing Q would rise iwould increase The magnetic field from iwould be directed opposite to the flux which i B dBd1 gt 0 caused it to increase There s a negative feedback mechanism at work Consider what would happen without Lenz s law In this case an increase in dBdt would result in a increase in a current whose flux would add to the flux which caused it thereby further increasing the current and so on This would violate the conservation of energy It is not physical When Faraday began his experiments it was not known what quantity was induced in a loop linked by a changing magnetic field Was the induced quantity a current or a potential To distinguish between the two possibilities Faraday used several rings of the same size and thickness but made from different materials and measured the current He found that the current was less in those rings with lower conductivity and higher in those rings made from a material with higher conductivity Upon further investigation Faraday found the current was proportional to the conductivity With this experimental evidence he proposed that the induced quantity in a loop was an electromotive force a voltage and that the current present in a shorted loop is simply Ohm s law at work Faraday39s law transformer and motional emf it Starting with the integral form of Faraday s law d4 d V dl B ds If the surface over which the surface integral is taken is stationary in space the only possible time variation of J B gds Surface is in the explicit time variation of B In this case the total time derivative outside the surface integral can be replaced by a partial derivative with respect to time operating on B V g gds transformer emf Surface This electromotive force is termed transformer emf since the coils of transformers are typically stationary Motional emf Consider a conductor moving in a magnetic field The charges within the conductor experience the magnetic component of the Lorentz force The force perunit charge is an electric field granted it is not an electrostatic field but it is an electric field and places a force on an electric charge Fqva E Eva q motional The induced motional emf is NE gdl NV x B gdl For a combination of motion and timevariation in the B field these components add to give the total expression for the induced voltage v ggds 3vade dl and d5 related by RHR Surface Example motional and transformer emf it A 100 T coil moves in a magnetic field produced by a current it The motion of the coil is in the x0 plane with a velocity v i Given an expression for B in the x 0 plane assume air ii find integras for vt in the diagram shown below i For a current on the zaxis in the az direction the resulting magnetic flux density is u i 7a 27rp D In this expression p is the distance from the v vy ay vz I current m is the permeability and a is determined 10mm from the right hand rule l Pm V93iquotKOP 6 vl Adapting this relation to the problem at hand 2 is w the distance from the current uo is the permeability of air and placing your thumb along x it reveals aX to be the direction of the field in the x0 plane 0 i 2712 aX ii vT gdu NV x Bng path surface at zhvzl Y1WVyl 1 di V 7a d dza 1T 2772 dt x 9 y x 2 le yy vyl zhvzl I i yvyl i O O vyay vzazx 2 ax gdz a2 J vyay vzazx 2 ax gdy ay zzvzl 772 yywvyl 772 zzhvzl z wzt ywvyl vava u a dza vava u a da y y z 2 X2 x g z y y z 2 X2 x g y y zzhvzl 772 yyvyl 772 zzvzl V1T MInEMJ d39 transformer emf 277 21 vzt a u i In M from first part of path integral 277 21 vzt u wv Al from second art 27721hvzt p ii In M from third part 0 I In LU ii In m 277 21 vzt 277 21 h vzt 277 21 vzt ampi from fourth part 277z1 vzt v1oov1T Understanding is deepened when the problem is viewed in a qualitative light For example given that the current is increasing with time should we expect the transformer emf to be positive or negative If the current is positive should we expect the motional emf to be positive or negative How do the four components of the path integral contribute to the overall result for motional emf X Transformer emf From the direction of ds and B it may be seen that the transformer emf should be negative if the field is increasing with time Motional emf Along path 1 the first part of the path integral from the negative sign of vt to the positive sign the motional emf term V x B will tend to push a positive charge toward the negative of vt This contributes negatively to vt so that in the sum its contribution should appear as negative As one can see above it does Similarly the motional term along path 2 should appear negative in the sum along path 3 positive and along path 4 positive Contributions from 1 and 3 will be equal they re the same distance from the current and so see the same field and opposite and will completely cancel Contributions from 2 and 4 will cancel but not completely Path 4 will dominate since it is closer to the current and sees a stronger field Example transformer and motional EMF it Transformer EMF Determine whether voltages v1 and v2 across the stationary loops are positive negative or zero for i cat 0 ii cat 112 and iii cat 3112 cosmtA I v1 V2 i for cat O cosoot is positive but its derivative is zero 2cos cot 2 V w m 7tl2 37rl2 This and the direction of B and ds provide the information needed to determine whether vt is positive negative or zero coswtA X X V dS B 391 v o o 2 ds B The determination centers on the evaluation of the integral stationary loops 6B 7 ds l a 9 surface From the diagrams above it can be seen that B and ds are in the same direction so that the dot product of B and ds is positive To cancel the negative sign and obtain a positive voltage ddt will need to introduce another negative sign B decreasing with time Here B is not changing with time Its derivative is zero B is at a maximum In this case therefore V 0 ii Here B is decreasing with time Following the reasoning given above V gt 0 iii Here B is increasing with time Following the reasoning given above V lt 0 Motional EMF Considerthe loops to be moving Assume the loop on the right is moving straight up parallel to the current and that the loop at the left is moving left away from the current Determine whether the motional EMF for v1 and v2 are positive negative or zero for iv cat O V cat 112 and Vi cat n Example inclass exercise Work with neighbors to develop a reasonable physicsbased circuit model of the system shown stIdtgt0 AE v N Faraday s law transformer and motional emf terms 1 pQEJE gdl dtsuLLB gds For a function of time and space the total time derivative can be split into two operators the first a partial derivative of time for explicit time dependence and the other linked to movement and spatial changes E Using this approach on the differential relation in Faraday s law E gdl B gds B gds v gVB gds 3 at Taking this result with the vector identity below v gVB Vxva B gVv BV gv vV gB which reduces to v gVB Vxva vV gB for constant v Using these results in the integral form of Faraday s law we have NE gdl pugs gds SUJ6VXBXV vV gB gds path NE gd gds Vxva gds since VgBO path surface surface NEgdl ggds VXVXB gds B xVVxB path surface surface NE gdl gds me gdl using Stoke39s theorem path surface at path Inductance In electrical circuits inductance is defined by the element relation V L2 dt ln magnetostatics selfinductance was defined as the ratio of flux linkage 7 to the current i L see magnetostatic notes for more on flux linkage I Let s trace the path from magnetostatics to the element relation 1 From either Biot Savart or from Ampere the magnetic field intensity is proportional to the current producing it H 0C i 2 For a constant permeability the magnetic flux density and therefore the flux passing through a loop are also proportional to the current producing the magnetic field oci 3 Taking a time derivative of the above equation and using Faraday s law we obtain the fact that the voltage induced in a current carrying loop is proportional to the time derivative of the current V 0C 3 the negative in Faraday s law is accounted for by labeling V and i with PSC 4 The proportionality constant is the inductance V Li dt Flux fluxlinkage and the selfinductance of an Nturn coil A new term flux linkage can help in discussions of inductance The flux linkage of a coil is the sum of the flux linking passing through the coil s loops As can be seen from Biot Savart or from Ampere s law the magnetic field Hr is determined by the current distribution lfthe permeability is known the magnetic flux density can be determined and so also the flux linking a single loop For a coil consisting of a single loop the flux linkage is equal to the flux linking the single loop 7 4 i 1T B d W V rrrnere are N turns eaen cammg a current L than the Arnbenan path Wbum enebse a tuta br Nr rnanner uxthatvwaps ran hP nPn P Pd P nmmm nf the WWE ean assurne thatthe turn are duse m bne anuther and any Eakage ux x r rn nurnber br bubs and the ux hnNng eaen bub Furthe Nrtum qu the tuta ux due m the current r rs v mm th Meter Nd neg ectmg Eakage ux The ux k hnkage fur the Nrtum ED 5 therefme N mrns A Numb N N 7 We se mndumance uf a can rs de ned astne ratm uf ux hnkage m current L 1 r Furthe Nrtum qu the ybnage mduced aerbss eaen bub s by Faraday s wavy vrwp T negabye m Faraday avv rs accuunted fur by abehng v and r PSC V Mm d Mr WP t at Fur 2 WM nnn the re umance br the ux path r Mn Vm sz N2 W93 dt dt 7 N2 m 7 m M a Mr by the urrent uvvmg m the uH rtsem ean ertner be de ned as v N Liurasii i mm a War r Example Ampere Faraday inductance and the magnetic circuit approximation it If a magnetic core has sufficient permeability it is possible x that the flux in the surrounding materials having a much lower permeability may be neglected This is assumed in this problem 2 9 MW thickness t This implies that the magnetic flux is assumed to be entirely N confined to the magnetic core Apparently the magnetic flux is mainly due to the orientation of internal magnetic dipoles within the core That is thinking of the magnetic flux as B uH uo H u0 ur 1H uo H M it is assumed here that the magnetization term dominates the flux This being true it can therefore be argued that the symmetry of the flux density vector must be that of the material So that from symmetry B Bp a H Hp a 27r By Ampere J Haw gpd a Ni 70 H a 3 BMa 277p a 277p a b t Nz I Nyt b a dz d a ln I 11 n w p w 2 a V N Lenz s law accounted for by labeling V and i P80 2 2 V N t mtg 2 L N t mtg 27139 a dt 27139 a What would the result have been if N 1 4 tln2 i and vT tln2 3 27 a 27 a dt V 1T zut b 2 L n In eneralLNL didt i 27 a g 9 The above method of finding the inductance is effective for those FEW cases in which Ampere s law can be utilized effectively for those few cases where simple symmetry exists For similar cases where there is a highpermeability core additional assumptions can be made leading to the magnetic circuit approximation 1 The magnetic field H is parallel to the path meaning the Amperian path 2 The magnetic circuit approximation assumes the flux density B is uniform across the core s crosssectional area For a narrow washer shaped core with the average radius much largerthan the width of the core the magnetic flux density would be approximately uniform across its crosssectional area recall from earlierthat this is the basis of the magnetic circuit approximation n r quot a 3994 Y thickness t Rewriting the previous result for A pav i I N utln2 i LNI tln 2 uNl tln p 277 a 2 7139 A 2 7139 A paV 1 2 paV Ifthe washer thickness is much smaller than the average radius we have 1A2 A A A mi In1p31 4 LDN ln1 7 L 1 72 pav av A In 1Anglne av A pav pav I LNltiz m AtN A 2 pav 2 pav e A flux path39s area 6 path39s length By Faraday39s law each turn will have an induced voltage VW 00p 1N HA N E 3 dt l l dt yA di llolal N Vper loop N 7 a
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