Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 7 Electrodynamics 7.0 Introduction 7.1 Ohm’s Law 7.2 Faraday’s Law 7.3 Maxwell’s Equations 7.0 Introduction electrostatic 1 E 0 E 0 magnetostatic B 0 B 0 J = conservation of charge J 0 t static ?0 t 0 E ? E ? t t ? B 0 B 0 J ? t ? 0 0 E 0 7.0 (2) Maxwell’s equations. E 0 B 0 Jd B E t B 0 J 0 0 E t displaceme nt current 7.0 (3) = B • E t E d a B d a B da t t Magnetic flux E d Induced electric field (force) B (t ) induce E BE 7.0 (4) • E B t B 0 0 E t Ba Ea induced by Ba Bb induced by Ea Eb induced by Bb Bc induced by Eb e.g. , E , B ~ e i ( kxwt ) E,B field propagate in vacuum 7.0 (5) • E B t B 0 J 0 0 E t J ( x x0,t ) Ba Eb Ba Eb A.C current can generate electromagnetic wave antenna cyclotron mass free electron laser ….. 7.1 Electromotive Force 7.1.1 Ohm’s Law 7.1.2 Electromotive Force 7.1.3 Motional emf 7.1.1 Ohm’s Law • J f Current density conductivity force per unit charge of the medium 1 for perfect conductors resistivity 0 • for f E v B J (E v B) ( a formula based on experience) w for >> v usually true J E k but not in plasma Ohm’s Law 7.1.1 (2) • Total current flowing from one electrode to the other V=I R Ohm’s Law (based on experience) Potential current resistance [ in ohm (Ω) ] Note : for steady current and uniform conductivity 1 E J 0 7.1.1 (3) Ex7.1 uniform I=? R=? uniform V sol: V I JA EA A L in series L1 , L 2 in parallel A1, A 2 L R A R R1 R 2 1 1 1 R R1 R 2 7.1.1 (4) Ex.7.3 Prove the field E is uniform J 0 V=0 A=const V=V0 =const J nˆ 0 at the surfaces on the two ends E nˆ 0 i.e., V 0 n V0 z V ( z ) L V0 E V zˆ L 2V 0 Laplace equation 7.1.1 (5) Ex. 7.2 V I ? ŝ ŝ : line ch arg e density 20s b a V b E d ln ( ) 20 a E V b 1 I J da E da L V 20 [ ln ] L 0 0 a 2L ln ( b ) a V R ln ( b ) 2L a E 7.1.1 (6) The physics of Ohm’s Law and estimation of microscropic the charge will be accelerated by E before a collision time interval of the acceleration is mean free path 2 mfp mfp t min , v a thermal 1 typical case mfp at 2 2 for very strong field and long mean free path 7.1.1 (7) The net drift velocity caused by the directional acceleration is 1 a vave at 2 2v thermal nfq F = qE J n f q v ave 2v thermal m mass of the molecule molecule density e charge free electrons per molecule nfq 2 J E 2mv thermal Joule heating law P VI I 2 R Power is dissipated by collision 7.1.2 Electromotive Force The current is the same all the way around the loop. force f f source E electrostatic electromotive force f d f s d E d 0 ( E 0) b b Vab E d f s d f s d a a fs 0 E V f 0 E f s outside the source Produced by the charge accumulation due to Iin > Iout 7.1.3 motional emf B Fmag ,v qvB Fmag ,v f mag ,v vB q f mag ,v d vBh cause u 7.1.3 (2) f pull uB for equilibrium h f pull d (uB)( ) sin cos v sin u cos = vBh d sin d cos h Work is done by the pull force, not B . dx d d vBh Bh ( ) (Bhx ) dt dt dt magnetic flux 7.1.3 (3) magnetic flux B da Bhx for the loop d dx Bh vBh dt dt d dt flux rule for motional emf 7.1.3 (4) • a general prove d (t dt ) (t ) ribbon B da ribbon da ( v d ) dt d B ( v d ) B ( w d ) dt ( w B) d f mag d f mag d dt 7.1.3 (5) Ex.7.4 =? a 0 f mag ds a 0 wsB ds wBa 2 2 WBa 2 I R 2R f mag v B ws( ŵ B) wsB ŝ 7.2 Electromagnetic Induction 7.2.1 Faraday’s Law 7.2.2 The Induced Electric Field 7.2.3 Inductance 7.2.4 Energy in Magnetic Fields 7.2.1 Faraday’s Law M. Faraday’s experiments Area loop moves Induce I [ v B] B moves induce I [ E ] B induce I [ E ] d emf E d ( E) da dt d B da B da Faraday’s Law (integral form) dt t B Faraday’s Law (differential form) E t 7.2.1 (2) A changing magnetic field induces an electric field. (a) v B, not E drive I (b) & (c) induce E that causes I Lenz’s law : Nature abhors a change in flux ( the induced current will flow in such a direction that the flux it produces tends to cancel the change. ) 7.2.1 (3) Ex.7.5 r̂ zˆ loop Induced (t ) ? sol: K b M nˆ Mˆ B 0 M at center , spread out near the ends max 0 Ma 2 ˆ 7.2.1 (4) Plug in, I induces B Ex. 7.6 B induces Ir Is Ir Plug in , Why ring jump? B vB vB F B F F ring jump. 7.2.2 The Induced Electric Field B E t B 0 J E 0 ( 0) B 0 d E d dt B d 0 I enc 7.2.2 (2) Ex. 7.7 induced E = ? = sol: d d 2 2 dB E d dt dt [s B(t )] s dt E 2s s dB ˆ E B E 2 dt 7.2.2 (3) Ex. 7.8. ẑ B B0 The charge ring is at rest B0 sol: What happens? d 2 dB E d dt a dt torque on d dN r F b (d) E zˆ (bEd) 2 dB 2 dB N dN zˆb E d zˆb[a ] ba dt dt the angular momentum on the wheel 2 0 2 N dt b a d B a bB0 zˆ B 0 7.2.2 (4) Induced E ( s) ? ˆ ẑ I( t ) sol: B 0 I ˆ 2s guasistatic B 7.2.2 (5) = d d 0 I E d dt B da dt 2s' ds' E(s 0 ) E(s) 0 dI s 1 ds ' s0 2 dt s' 0 dI (ln s ln s 0 ) 2 dt 0 dI E(s) [ ln s K] ẑ 2 dt Constant K( s , t ) s << c I dI dt 7.2.3 Inductance 0 d 1 Rˆ B1 I1 I1 2 4 R 2 B1 da2 M 21I1 mutual inductance ( A1 ) da2 A1 d 2 0 I1 d 1 d 2 4 R 0 I1 d 1 A1 4 R 7.2.3 (2) d d 2 M 21 0 1 4 R Neumann formula The mutual inductance is a purely geometrical quantity M 21 M12 M 1 2 if I1 I 2 1 M12 I 2 7.2.3 (3) n2 turns per unit length Ex. 7.10 1 sol: 2 I given n1 turns per unit length assume I too. B1 is too complicated……….. 2 ? Instead, assume I running through solenoid 2 I 2 I1 I 1 n1 1,per turm n1 a 2 B2 B2 0 n 2 I 2 0 a 2 n1 n 2 I 2 0 a 2 n1 n 2 I M 0 a 2 n1 n2 2 (I 2 I1 I) 2 ? M ? 7.2.3 (4) • I1 ( t ) 2 d 2 dI M 1 dt dt changing current I1 in loop1, induces current in loop2 • self inductance I (t ) LI self-inductance (or inductance ) Volt sec [ unit: henries (H) ] 1H 1 A • back emf dI L I will reduce it. dt 7.2.3 (5) Ex. 7.11 N turns b a L(self-inductance)=? sol: N B da 0 NI b1 N h a ds 2 s 0 N 2 Ih b ln ( ) 2 a 0 N 2h b L ln ( ) 2 a 0 NI B 2s 7.2.3 (6) Ex. 7.12 I( t ) ? sol: dI 0 L IR dt I (t ) 0 R 0 R R t ke L particular sol. general sol. if I(0) 0 , k I (t ) 0 R R t 0 L (1 e ) R 0 R (1 e t ) L R time cons tant 7.2.4 Energy in Magnetic Fields In E.S. test charge q From the work done, we find the energy 1 0 2 E W ( V ) d E d in , e 2 2 But , B does no work. WB ? In back emf d dI d 1 WB I L I ( LI 2 ) dt dt dt 2 1 2 1 1 2 WB LI I ( Wk mv ) 2 2 2 sB da s( A) da loop A d WB 1 I A d 1 (A I )d 2 loop 2 loop 7.2.4 (2) In volume 1 WB V ( A J ) d 2 B 1 ( A B) B ( A) A ( B) A ( B )d V 2 0 2 B 1 1 2 B d ( A B )d V V 2 0 2 0 ( A B ) d a 0 s s 1 2 WB B d all space 2 0 1 0 2 Welec (V )d E d 2 2 1 1 2 Wmag ( A J )d B d 2 2 0 7.2.4 (3) Ex. 7.13 WB ? (length ) sol: 0 I B ˆ a< s<b 2s WB dWB 0 I 2 b ds a 4 s 0 I 2 b ln( ) 4 a B0 s <a s >b 0 I 2 ) (2sds ) 2 0 2s 1 ( 1 2 WB L I 2 b L 0 ln ( ) 2 a 7.3 Maxwell’s Equations 7.3.1 Electrodynamics before Maxwell 7.3.2 How to fix Ampere’s Law 7.3.3 Maxwell’s Equations 7.3.4 Magnetic Charge 7.3.5 Maxwell’s Equation in Matter 7.3.6 Boundary Conditions 7.3.1 Electrodynamics before Maxwell E 0 B 0 B E t B 0 J but (Gauss Law ) (no name ) (Faraday' s Law ) (Ampere ' s Law ) B ( E ) ( ) ( B ) 0 t t = ( B ) 0 ( J ) ? Ampere’s Law fails because J 0 0 7.3.1 an other way to see that Ampere’s Law fails for nonsteady current loop 1 2 B d 0 I enc for loop 1 , Ienc 0 for loop 2 , I enc I they are not the same. 7.3.2 How to fix Ampere’s Law continuity equations, charge conservation E J [ 0 E ] ( 0 ) t t t such that, Ampere’s law shall be changed to E B 0 J 0 0 t J d displacement current A changing electric field induces a magnetic field 7.3.2 = E B d a ( J ) d a 0 0 0 t E B d I da 0 enc 0 0 t for the problem in 7.3.1 1 1 Q E between capacitors 0 A 0 E 1 dQ 1 I t 0 A dt 0 A 1 loop1 B d 0 0 0 I 0 I 0 loop 2 B d 0 I 0 0 I 7.3.3 Maxwell’s equations E 0 B 0 Gauss’s law B E Faraday’s law t B 0 J 0 0 E Ampere’s law with Maxwell’s correction t Force law F q(E v B) continuity equation J t ( the continuity equation can be obtained from Maxwell’s equation ) 7.3.3 Since , J produce E , B E 0 B 0 B E 0 t B 0 0 E 0 J t J (r , t ) B E 7.3.4 Magnetic Charge Maxwell equations in free space ( i.e., e 0 , J e 0 ) E 0 B 0 Je With e and If there were m e E 0 B 0 m m Jm t B E 0 symmetric t EB B 00 E B 0 0 E t , the symmetry is broken . ,and J m . B E 0 J m t symmetric E B 0 J e 0 0 t e Je and t So far, there is no experimental evidence of magnetic monopole. 7.3.5 Maxwell’s Equation in Matter bound charge bound current Jb M b P b P t t JP J b 0 no correspond ing polarization current b JP 0 t Q ( b da ) da t t P da J P da t dI surface ch arg e b P 7.3.5 (2) f b f P J J f Jb J P J f M P t 1 Gauss ' s law E ( f P) 0 or D f D 0E P Ampere’s law ( with Maxwell’s term ) B 0 ( J f M P) 0 0 E t t ( B 0 M ) 0 J f 0 ( 0 E P) t 1 H BM H J f D 0 t 7.3.5 (3) In terms of free charges and currents, Maxwell’s equations become B E D f t B 0 H J f D t D, H and E , B are mixed . displacement current one needs constitutive relations : D(E, B) and H(E, B) 7.3.5 (4) for linear dielectric. P 0 xe E or D E f E 0 B 0 M xm H 0 (1 xe ) 1 0 (1 xm ) H B B E t E B J f t 7.3.6 Boundary Condition Maxwell’s equations in integral form s D da Q fenc Over any closed surface S s B da 0 d L E d dt sB da d L H d Ifenc dt sD da D1 , B1 D 2 , B2 for any surface bounded by the S closed loop L 7.3.6 D1 a D2 a f a H1 H 2 K f nˆ = 0 = B2 0 S 0 E1 E2 0 = B1 1E1 2 E2 f = D1 D2 f d E1 E2 B da dt H1 H 2 K f ( n̂ ) ( K f n̂ ) B1 1 2 B2 K f nˆ = 1 = 1