Griffiths電動(dòng)力學(xué)習(xí)題解答 Chapter 5 Magnetostatics.pdf

Chapter5 Magnetostatics Problem5 1 Sincev x B points upward and that is also the direction of the force q must beIpositive ITofindR in termsof a and d use the pythagoreantheorem a2 d2 R d 2 a2 R2 R2 2Rd d2 a2 R2 R 2d p qBR IqB a2 2d r RV The cyclotron formula then giyes Problem 5 2 The general solution is Eq 5 6 y t CI cos u t C2sin u t t C3 z t C2cos u t CI sin u t C4 a y O z O OJ y O E Bj i O O Use these to determine CI C2 C3 and C4 y O 0 CI C3 OJy O u C2 E B E B C2 OJz O 0 C2 C4 0 C4 0 i O 0 CI 0 and hencealso C3 O SoIy t Et B z t 0 1 Does this make sense The magnetic forceisq v x B q E B Bz qE whichexactly cancelsthe electric force since there is no net force theparticle moves in a straight line at constant speed b Assumingit starts from the origin so C3 CI C4 C2 wehavei O 0 CI 0 C3 0 y O 2 C2u 2 C2 2 B C4 y t 2 B sin u t t EEE E z t 2u B cos u t 2u B or y t 2u B 2u t sm u t z t 2u B 1 cos u t Let 3 E 2u B Theny t 3 2u t sin u t z t 3 1 cos u t y 2 3u t 3 sin u t z 3 3 cos u t y 2 3VJt 2 z 3 2 32 This is a circle of radius 3whose center moves to the right at constantspeed Yo 2 3VJt Zo 3 EEEEE c z O y O B ClUJ B CI C3 u Bj C2u B B C2 C4 O 89 90CHAPTER 5 MAGNETOSTATICS EEEE EE y t wB cos wt Bt wB z t wB sm wt y t wB 1 wt cos wt j z t wB sin wt Let 3 EjwB then y 3 1 wi 3cos wt z 3sin wt j y 3 1 wtW z2 32 Thisis a circle of radius 3 whosecenteris atYo 3 1 r vt Zo O z 4 A y 3 c Problem5 3 a From Eq 5 2 F q E v x B 0 E vB Iv 1 qvm b FromEq 5 3 mv qBR FiR Problem5 4 Suppose f flows counterclockwise if not change the sign of the answer The force on the left side toward the left cancels the force on the right side toward the right the force on the top is laB lak a 2 lka2 2 pointing upward and the force on the bottom is laB lka2j2 also upward So the net forceis F IIka2 z 1 Problem5 5 a IK 1 Ibecause the length perpendicular to flowis the circumference 21ra b J 1 J da a s ds de 21ra ds 21raa a 21 J 1 21 1 ss1ra1ras Problem5 6 a v r vr soIK f7r vr 1 b v r vrsin8 IJ pr vrsin8 Iwhere p Qj 4j3 1rR3 Problem5 7 Iv pr dr a r dr V J r dr by the continuityequation Now productrule 5 says V xJ x V J J Vx But Vx x so V xJ x V J Jx Thus Iv V J xdr Iv V xJ dr Iv Jx dr The first term is Is xJ da bythe divergencetheorem and sinceJ is entirely insideV it is zeroonthe surfaceS ThereforeIv V J xdr IvJx dr or combiningthis with the y and z components Iv V J rdr IvJ dr Or referring back to the first line ii J dr Qed Problem5 8 I oll a Use Eq 5 35 wIth z R 82 81 45 and four sIdes B b z R 82 81 andn sides B sin 1r n 91 n LoI 7r I LOII c For small B sm B B So as n t 00 B t 2 R 2R same as Eq 5 38 wIth z 0 Problem5 9 a The straight segments produce no field at P The two quarter circlesgive B I i I out b The two half linesare the same as one infinite line jthe half circlecontributes SoB I 1 I into the page Problem5 10 LaI LOI LoI2a a The forces on the two sIdes cancel At the bottom B 2 F 2 Ia 2 up 7rS7rS7rS La I LoI2a LoI2a2 top B F down The net force IS 2 up 27r S a27r S a7rS S a b Theforceon the bottom is the same as before LoI2 27r up On the left side B LoI z 27rY dF I dl x B I dx x dyy dz z x z dxy dy x But the x component cancels the L12 1 s V3 a 2 1 corresponding term from the right side and Fy dx Here y J3x so 27rs V3Y La 12 s J3 a 2 La12 J3a Fy rq InJ3 rqIn 1 2 The force on the right sIde ISthe same so the net 2y37rs 32y37rS I12 2 J3a Iy force on the triangleisI L 7r 1 J3In1 At the a a 600 s V3 S x z Problem 5 11 UseEq 5 38 for a ring of width dz with I t nI dz LonIJ a2 B 2 3 2 dz But z acotB a2 Z2 a1 B sodz dB and3 2 sm B a2 z2 a So z Zdz B LonI J a2 sin3 B dB LonI J BdB LonI BI h LonI BB a sm cos COS2 COS1 2a3 sin2 B22012 Foraninfinitesolenoid B2 0 BI 7r so cosB2 cosBd 1 1 2 and B I LonI I 92 CHAPTER5 MAGNETOSTATICS Problem5 12 22 Magnetic attractionper unit length Eqs 5 37 and 5 13 1m Po d v 211 Electric field of one wire Eq 2 9 E 2 1 Electric repulsion per unit length on the other wire 1I EO S 1 21 Ie They balance when pov2 or v Puttmgm the numbers 211 EOdEOv EOPO v 1 13 00 X108m s 1 This is precisely the speed of light so in fact you could 8 85x 10 12 411 x 10 7 never get the wires going fast enough the electric force always dominates G Problem5 13 f 0 for s B poI i 2 f or s a 1I S l a l a211 ka331 i s i s b J ks 1 Jda ks 211 s ds 3 k Ienc Jda ks 211 s ds 00211 a00 PoI S2A 33 cjJ for s a 211 kss211 a 13 for s a SoIB 3aPolA cjJ for s a 211 s Problem5 14 By the right hand rule the field points in the ydirection for z 0 and in the y di rection for z IB PoJzyI a z a Ienc polaJ soIB poJa for z a ampedanloop I oJay 1m z a 1z y l Problem5 15 The field inside a solenoid is ponI and outside it is zero The outer solenoid s field points to the left 2 whereasthe inner one pointsto the right z So i IB poI n1 n2 z I ii IB poln2z I iii IB 0 1 Problem5 16 From Ex 5 8 the top plate produces a field poK 2 aiming out of the page for points above it and into the page for points below The bottom plate produces a field poK 2 aiming into the page for points above it and out of the page for points below Above and below bothplates the two fields cancel between the plates they add up to poK pointing in a IB poO V in Ibetweem the plates IB 0Ielsewhere b The Lorentz force law says F J Kx B da so the force per unit area is f K x B Here K TV to the right and B the field of the lower plate is poO v 2 into the page SoI1m PO0 2V2 2 up 93 c The electric field of the lower plate is O 2Eo the electric force per unit area on the upper plate is IIe O 2 2Eo down ITheybalanceif POV2 I Eo orIv 1 ftOiIO cI the speed of light as in Prob 5 12 Problem5 17 We might as well orient the axes so the field point r lies on the y axis r 0 y 0 Considera sourcepoint at x y z on loop 1 x x y y y z z dl dx X dy y dl X X dx x Y dy y y z 0 z dy X z dx y y y dx x dy z z Pol dl x Pol z dy X z dx y y y dx x dy Z dB1 471 1 3471 x 2 y y 2 z 2 3 2 Nowconsider the symmetrically placed source element onz loop 2 at x y z Since z changes sign while every thingelse is the same the x andycomponentsfrom dB1 and dB2cancel leaving only a z component qed With this Ampere s law yields immediately B ponI Z inside the solenoid 0 outside r y thesame as for a circular solenoid Ex 5 9 For the toroid N 271 s n the number of turns per unit length so Eq 5 58 yields B pon1 inside and zero outside consistentwith the solenoid Note N 271 s n applies only ifthe toroid is large in circumference so that s is essentially constant over the cross section Problem 5 18 IIt doesn t matter IAccording to Theorem 2 in Sect 1 6 2 JJ da is independentof surface for any given boundary line provided that J is divergenceless which it is for steady currents Eq 5 31 Problem 5 19 a charge charge atoms moles grams e N d where PvolumeatommolegramvolumeM e charge of electron N Avogadro s number M atomic mass of copper d density of copper 1 6X 10 19C 6 0X io23 mole 764gm mole 9 0gm cm3 p 1 6x 10 19 6 0 x 1023 11 4X104C cm3 1 1 11II b J 7I S2 pv v 7I S2p 71 2 5x 10 3 1 4 x 104 9 1X10 3 cm s or about 33 cm hr This isastonishingly small literallyslower than a snail s pace Po 1112 471 X 10 7 II c From Eq 5 37 1m 271 d 271 2 X 10 7 N cm 94CHAPTER5 MAGNETOSTATICS d E fe A1A2 ltI2 c2 Lo ltI2 C2fm where 211 fOd211 fOdV2 211 fOdv2211 dV2 f C2 3 0X 1010 2 C l yfO LO 3 00 x 108m s Herei 2 9 1X 10 3 11 1 x 1025 1 fe 1 1 x 1025 2 x 10 7 12 X 1018N cm 1 Problem5 20 Ampere s law says V x B LoJ Together with the continuity equation 5 29 this gives V V x B LoV J Loop ot whichis inconsistentwith div curl Ounless p is constant magnetostatics The other Maxwell equations are OK V x E 0 V V x E 0 and as forthe two divergenceequations there is no relevant vanishing second derivative the other one is curl grad which doesn t involve the divergence Problem5 21 At this stage I d expect no changes in Gauss s law or Ampere s law T4e divergence of B would take the formIV B o OPm IwherePm is the densityof magneticcharge and 0 0 is some constant analogousto 0I and Lo The curl of E becomesIV x E 3oJm IwhereJm is the magnetic current density representingthe flow of magnetic charge and 3o is another constant Presumably magnetic charge is conserved so Pm and Jm satisfy a continuity equation V Jm oPm ot As for the Lorentz force law one might guess something of the form qm B v x E where qm is the magneticcharge But this is dimensionallyimpossible since E has the same units as vB Evidently we need to divide v x E by somethingwith the dimensions of velocity squared The naturalcandidate is c 1 01 0 IF q IE v x B qm B vx E I n this fo m the magnetic aoalog to Coulomb law reads F 0 0qml m2 f so to determine 0 0we would first introduce arbitrarily a unit of magnetic charge 411 r then measure the force between unit charges at a given separation For further details and an explanation of the minus sign in the force law see Prob 7 35 Problem5 22 A LofI Z dz LoIZt2dz 411 Iz 411 JZ1yz2 s2 LolA I v2 2 Z2 Loll Z2 V Z2 2 S2 A znZZ8 nz 411 Zl411 Zl V Zl 2 82 v Z Z B V x A oA p LoI 18 18 p 08411 Z2 V Z2 2 82 V Z2 2 82Zl v zd2 82 V Zl 2 82 LoI8 Z2 V Z2 2 821 Zl v zd2 821 p 411 Z2 2 Z2 2 82 V Z2 2 82z zd2 82 v zd2 82 LoI8 Z2 1 Zl 1 p LoI Z2 Zl j 411 82V Z2 2 82V Zl 2 82 411 8 V Z2 2 82v zd2 82 Zl Z2 or sInce smlh and sm 2 v zd2 82V Z2 2 82 4 LOI sin 2 sin l p I as in Eq 5 35 11 8 95 Problem5 23 1a k 11 a k A k B VxA a sk z z J VxB a cp zcp sssfLofLossfLoS Problem5 24 V A V r x B B Vx r r V x B 0 since V x B 0 B is uniform and 11 V x r 0 Prob 1 62 V x A 2Vx r x B 2 B V r r V B r V B B V r But r V B 0 andV B 0 since B is uniform and V r 1 1 1 3 Finally B V r Bx x By y Bz z xx yy zz Bxx Byy Bzz B So VxA B 3B B qed Problem5 25 a A points in the same direction as I and is a function only of s the distance from the wire In cylindrical coordinates then A A s Z so B V x A a a A fL 2o1 the field of an infinite wire Therefore s7rS a a A fL 2oI and A r fL 2o1 In s a z the constant a is arbitrary you could use 1 but then the units s 7r aAzaAz fLoI lookfishy V A a o vx A a cp cp B zS27rS b Here Ampere s law givesfB dl B 27rs fLolenc fLoJ 7rS2 fLo 7r 27rS2 fL 2 J LoIs aAfLoI sfLoI22 B 21 R2 cpoas 21 R2 A 41 R2 s b z Here b IS agalllarbItrary exceptthatSlllce A mustbe continuous at R In R a 2 R2 b2 whichmeans that wemust picka and bsuch that 2In R b 1 b R 2 I ll use a b R Then fLoI 2 A 47rR2 S R2 Z fLoI 21 In s R Z for s R for s R Problem 5 26 K Kx B xfLoKY plus for z 0 2 A is parallelto K and depends only on z so A A z x xyzaAK B VxA 1a ax a ay a az 8y xfL y A z 00z JA IzixI ill do the job orthis plus any constant z y x Problem 5 27 a V A v dr V V J J V But the first term is zero because J r isafunctionofthe sourcecoordinates not the fieldcoordinates And since r r V V So 96CHAPTER5 MAGNETOSTATICS V J V ButV V J J V andV J 0inmagnetostatics Eq 5 31 So J J d h Po J v J Po f J d f V V and hence by theIvergence t eorem V A 47r dr 47r a where the integral is now over the surface surrounding all the currents But J 0onthis surface soV A 0 b V x A JV x dr j V x J J x V dr But V x J 0 since J is not 1 4po j JX4 a function of f and V 2 Eq 1 101 so V x A dr B IJ IJ 47rIJ c 72A J 72 dr But 72 J 72 once again J is a constant as far as differenti ation with respect to r is concerned and 72 47r 53 1t Eq 1 102 c So 72A Po j J r 4m53 1t dr poJ r 47r Problem5 28 Pol fB dl lbVU dl U b U a by the gradient theorem so U b fU a qed For an infinite straight wire B Pol Jy IU P2oIe Iwould do the job in the sense that 27rS7r VU Pol V e Pol e Jy B But whene advances by 27r this function does not return to its initial 27r27r s lie value it works say for 0 e 27r but at 27rit jumps back to zero Problem5 29 Use Eq 5 67 withR f and a pdt A powpsinB lr i d Powp B lR d f rr rsm f rr 3r203r POWP 1 r5 r22 APowp R2r2 A 3smBr25 2 R r p rsmB3 5 p pow p Ia R2r2 A1a 2 R2r2 A B V x A sm Br sm B r r sm B 2rsmBaB3 5r or35 R2r2 A R22r2 A Q powp3 5cosBr 3 5smB But p 4 3 7rR3 so powQ 3r2 A 6r2 A 1 cosBr 1 sm B 47rR5R25R2 Problem5 30 a a z FY WZ X Y Z f FY X Y Z dX Cl y z a y Fz Wy x y z f Fz x y z dx C2 y z These satisfy ii and iii for any C1 and C2 it remains to choose these functions so as to satisfy i 97 l X 8Fy x y z d 8Cl l x 8Fz x y z d 8C2 F B 8Fx8Fy8Fz 0 8x 8x xx y z ut so 0YuY0zuZuXuYuZ t8Fx x y z 8Cl8C2 X 8Fx x y z 108x dx 8y 8z Fxx y z NowJo8x dx Fxx y z FxO y z so 8 1 8 2 Fx O y z We may as well pick C2 0 Cl y z lY Fx O y z dy andwe redone with Wx 0 Wy lx Fz x y z dx Wz lY Fx O y z dy lxFy x y z dx b V x W 8Wz 8Wy x 8Wx 8Wz y 8Wy 8Wx Z8y8z8z8x8x8y F r8Fy x y z d X8Fz x y z d X0 y z Jo8yx Jo8zxx 0 Fy x y z y Fz x y z 0 z But V F 0 sothe x termis Fx O y Z lx 8Fx y z dX Fx O y z Fx x y z Fx O y z soV x W F W 8Wx8Wy8Wz 0 l x 8Fz x y z d l Y8Fx 0 y z d l X8Fy x y z d t O v x Y x I 8x8y8z08y08z08z ingeneral Xx2 y Xy2 c Wy Jo x dx 2 Wz Jo y dy Jo Zdx 2 zx Iw Y zx 1V x w xyZ 8 8x8 8y8 8z 0 x2 2 y2 2 zx yx zy xz F Problem5 31 a At the surface of the solenoid Babove 0 Bbelow J LonI z J LoK z ii s so K x ii K Z EvidentlyEq 5 74 holds b In Eq 5 67 both expressions reduce to J LoR2VJa 3 sin iJat the surface so Eq 5 75 is satisfied aA I J LoR4VJa 2Sin 1 2J LoRv Ja 8A I J LoRv Ja 8hI f d f a 3 P 3sm P I 3sm P 0 teet S1e 0 r R rRurR Eq 5 76 is J LoRVJa sin iJ Meanwhile K av a x r aVJRsin iJ so the right side of Eq 5 76 is l0 J1J R sin iJ and the equation is satisfied Problem 5 32 Because Aabove Abelowat every point on the surface it follows that and are the same above andbelow any discontinuity is confined to the normal derivative B B 8AYabove 8AYbelow x 8Axabove 8Axbelow Y But Eq574says this equals abovebelow8z8z8z8z loK y 80 8Atove 8A OW and 8A 7ove 8A e ow J LoK Thusthe normalderivativeofthe com f ffid KI 8Aabove8Abelow K ponent0A parallel to K su ers alscontmmty J Lo or more compact y 8n 8n J Lo Problem5 33 Sameidea as Prob 3 33 Write m m f f m 0 0 mcos f msin O Fig 5 54 Then 3 m f f m 3m cos f mcos f msin O 2mcos f msin O and Eq 5 87 Eq 5 86 Qed 98CHAPTER5 MAGNETOSTATICS Problem5 34 a m Ia lhrR2i 1 b B I I 2 2cos0 f sin 08 c On the z axis 0 0 r z f i for z 0 soIB iI for z w 2 for all four Jz2 w 2 2 sides multiply by 4 IB LoIW2i IFor 211 z2 w2 4 Jz2 w2 2 z w B L 2 i The field of a dipole 1m IW2 Ifor points on the z axis Eq 5 86 with r z f i 0 0 is B Lomi 211 z3 Problem5 38 Themobile charges do pull in toward the axis but the resulting concentrationof negative charge sets up an electric field that repels away further accumulation Equilibrium is reached when the electric repulsion on a mobile charge q balances the magnetic attraction F q E v x B 0 E v x B Say the current z RsinO z W 2 99 isin the z direction J p vz where p and v are both negative fB dl B 27rS l1oJ7rS2 B 110P2 vs J E da E27rsl p p 7rs21 E 21 p p ss fOfO 2 O p p ss vz x 110P2 VS J Op V2ss p p p fOI10V2 p Evidently p p 1 orp 2p In this naive model the mobile negative charges fill a smallerinner cylinder leaving a shell of positive stationary charge at the outside But since v c the effect isextremely small Problem5 39 a If positive charges flow to the right they are deflectedIdown Iand the bottom plate acquires a positive charge b qvB qE E vB V Et IvEt Iwith thebottom at higherpotential c If negative charges flow to theleft they are also deflected down and the bottom plate acquires a negative charge The potential difference is still the same but this time the top plate is at the higher potential Problem5 40 From Eq 5 17 F I J dl x B But B is constant in this case so it comes outside the integral F I Jdl x B andJ dl w the vector displacement from the point at which the wire first enters the field to thepoint whereit leaves Sincewand B are perpendicular F IBw and F is perpendicular to w Problem5 41 The angular momentum acquired by the particle as it moves out from the center to the edge is L dt N dt r x F dt r x q v x B dt q r x dl x B q r B dl B r dl But r i perpendicularto B so r B 0 and r dl r dr d r r d r2 rdr lj27r 27rrdr SoL 2 foRB27rr dr 2 Bda It followsthatIL Iwhere JB da is the total flux In particular if 0 then L 0 and the charge emerges with zero angular momentum which means it is goingalong a radial line qed Problem5 42 From Eq 5 24 F J K x Bave da Here K av v wR sin BJ da R2 sin BdBd4J and Bave Bin Bout From Eq 5 68 100 CHAPTER5 MAGNETOSTATICS Bin J 100 R VZ J 100 R V cosBf sinBO FromEq 5 67 Bout V XA V X J 10R4 VO sinB J 1oR4 VO Sin2B f SinB 0 3r23rsmBoBr2r orr J 10R4 VO 2 B BO J 1oR VO 2 B BO R 3cosr sm 3 cosr smsmce r 3r B J 1oR VO 4 B BO aye cos r SIn 6 J 10R VO J 10 KX Baye O VRsin B 6fjJx 4cosBf sinBO 6 O VR 4cosBO sinBf sinB Picking out the z component of0 namely sinB and of f namely cosB wehave K x Baye z 0 O VR 2sin2BcosB so Fz o O VR 2 R2fsin3 BcosBdBdcj o O VR2 2 211 Ci 4B C2 orIF O VR2 2z 1 Problem5 43 F B J 10qeqm J 10qeqm a ma qeV X v x r a v x r 411 r411 mr 1 dIddvdv b BecauseaJ v a v O Buta v 2dt v v 2dt v2 vdt So dt 0 qed dQ J 1oqeqm d r 0 J 1oqeqm X J J 1oqeqm Vr dr c m v x vm r x a rv x r dt411 dtr411 r3411 rr2 dt J 1oqeqm r2v r v rJ J 1oqeqm f v f f 2 r V O v 411 r3rr2 dt411 rrr2rr J 1oqeqm d i Q fjJ Q z fjJ mer x v fjJ 411 f fjJ But z fjJ f fjJ 0 so r x v fjJ O But r r f and v dl r f rB 0 r sin B where dots denote differentiation with respect to time so dt f0 r x v I rOOI r2sinB 0 r2B rrB rsinB Therefore r x v r2B 0 so B is constant qed ii Q f Q z f mer x v f J 1o qm f f But z f cosB and r x v J r r x v f 0 so Q B J 1oqeqm Q J 1oqeqm A d B t Q d cos 4 or 4B nsmceIScons ant so too IS qe11 11 cos iii Q O Q z O mer x v 0 J 1o qm f O But z O sinB f O 0 and r x v O r2 sinB f Q B 2 Bl lQk th k QJ 1oqeqm rom1 so sIn mrsm f f 2 2 WI 4 B mrrm1I mcbs 2 22 2 k2222k22k2 sin 2B e v2 r2 r B2 r sin Bcj but B 0 and cj 2 so r v r sin B4 v 2 rrr 101 dr 2 1 2 V2 ksin r 2 2 Vr 2 2 dr Vr 2 2 defy 2 k2 r4 rksm defyrvksm dr 1 vr vr f J dcjJ cjJ cjJo sec 1 k sec cjJ cjJo sm k or k 2 2 smsmsm rvr sm AI n r efy cjJ cjJ where A 4 cos 0 sm1fmv Problem5 44 Put the field point on the x axis so r s 0 0 Then J lo K x 4 A B da da RdcjJdz K K P K sincjJx coscjJy s RcoscjJ X RsincjJy zz XyZ sin cjJcos cjJ0 8 R cos cjJ R sin cjJ z K z coscjJ