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Electromagnetism Zhu Jiongming Department of Physics Shanghai Teachers University Electromagnetism Chapter 1 Chapter 2 Chapter 3 Chapter 4 Electric Field Conductors Dielectrics Direct-Current Circuits Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Magnetic Field Electromagnetic Induction Magnetic Materials Alternating Current Electromagnetic Waves Chapter 6 Electromagnetic Induction §1. Electromagnetic Induction §2. Lenz’s Law §3. Motional Electromotive Force (emf) §4. Induced emf Induced Electric Field §5. Self Induction §6. Mutual Induction §7. Eddy Current §8. Transient State of RL Circuit §9. Transient State of RC Circuit §10. Magnetic Energy §1. Electromagnetic Induction 1. Experiment A current can be induced in a loop as the magnetic flux through the loop changes with time 2. Faraday’s Law of Induction d K dt A Induced emf = K time rate of change of flux in SI unit(V、Wb、s) K = 1 d | | dt ( Direction --- Lenz’s Law) §2. Lenz’s Law 1. Two statements ● Induced current always tends to oppose the flux change ● Magnetic force on induced current always tends to oppose the motion of conductor 2. Expression of Faraday’s Law 3. Examples 1. Two Statements (1) Induced current always tends to oppose the flux change —— flux of induced current ● in the opposite direction for > 0 ● in the same direction for < 0 N A S N A S 1. Two Statements (2) Magnetic force on induced current always tends to oppose the motion of conductor I ∵ v towards right ∴ fAmpere left( fAmpere= Idl B ) R f安 v ∴ I counterclockwise ( or: > 0 ∴ flux of I in opposite direction ) Work and Energy: Work by external force I 2Rt ( heat energy ) (constant velocity,kinetic energy not changed) If fAmpere towards right,energy conservation violated 2. Expression of Faraday’s Law Positive directions for and : d right hand rule then dt 0 d 0 dt 0 0 d 0 dt 0 0 d 0 dt 0 0 d 0 dt 0 Example (p.276 / 6-2-1) A small N turns coil A of radius r is placed coaxially in a long solenoid S of n turns per unit length. Current I in S decreases from 1.5A to -1.5A at a steady rate in 0.05 s. Find emf in A. Sol.:B = 0nI NBr 2 N 0 nr 2 I d 2 dI 0 nNr dt dt dI 1.5 1.5 60 dt 0.05 600 nNr 2 0 Direction: same with I and not changed Exercises p.276 / 6 - 2 - 2, 3 §3. Motional emf d dt where : B dS S Three ways to change the flux : change the area ( move the loop ) —— motional emf change the field B —— induced emf change both of them §3. Motional emf 1. Motional emf and Lorentz Force 2. Calculation of Motional emf 3. Examples 4. Alternating-Current Generator 1. Motional emf and Lorentz Force Magnetic force on electrons in conductor ab a I d f = - ev B (downward) I counterclockwise direction (caused by emf on ab ) f emf = work on unit charge c b by Lorentz force as moving it from b to a a a a f dl (v B) dl vBdl vBl b b b e ∵ vl is the rate at which the area changes ∴ vBl is the rate at which the flux changes , = d/dt v Motional emf General formula for Motional emf : (v B) dl motional suitable also for unclosed conductor,but no I, only emf v o B in the formula not change with time 2. Calculation of Motional emf Two ways to calculate Motional emf : (1) motional (v B ) dl where v、B might not be uniform,so could not be brought out in front of the integral sign d where : B dS (2) S dt If not closed,add an imaginary v curve to make it closed or get d/dt by the area swept per unit time Example (p.229/ [Ex.1]) (1) A straight wire of length L is rotating at angular velocity in a uniform magnetic field B. Find emf ab and potential difference Uab . b Sol.:(1) take dl at a distance l from a dl v B same direction with dl , v = l l a b L 1 v 2 ab a (v B) dl 0 Bl dl BL 2 ( 0, direction :apotential,negative b) a: low charges ab b: high potential,positive charges U ba ab 1 U ab ab BL2 2 Example (p.229/ [Ex.1])(2) (2) assume rotating d in d t the area swept is 1 dS L Ld 2 a 1 2 d BdS BL d 2 d 1 2 d 1 2 BL BL dt 2 dt 2 b L d direction:imaginary loop counterclockwise, ab Ld 4. Alternating-Current Generator Loop plane is perpendicular to magnetic field at t = 0 ( S is in the same direction with B , = 0 at t = 0 ) B dS BS cos t S d BS sin t dt BS sin t B N S Exercises p.277 / 6- 3 - 1, 2 §4. Induced emf Induced Field 1. Induced Electric Field 2. Properties of Induced Electric Field 3. Induced Electric Field Due to the Changing Magnetic Field in a Solenoid 4. Examples 1. Induced Electric Field Loop does not move,magnetic field is changing d B changing flux =B·dS changing 0 dt induced current force F on q F/ q = E Coulomb's law Coulomb's electric field E C changing magnetic field induced electric field E I Total electric field E = E C + E I or d EC dl dt 2. Properties of Induced Electric Field Potential field E dl 0 Source field E dS q / d d E dl dt dt B dS B dS 0 Eddy field t Sourceless field E dS 0 EC : L S EI : L C C in 0 I S S S I Total electric field E dS q S in / 0 E = EC + EI B LE dl S t dS 3. Induced Electric Field in a Solenoid symmetry(infinite long),E perpendicular to the axis —— no axial component symmetry(circle), E I —— no radial component E S 感 I is in the plane S dS 0 Conclusion: E I has only tangent component , E I lines are concentric circles Example 1 (p.236/ [Ex.1]) In a long solenoid of radius R, dB/dt is known, find EI . Sol.: EI dl EI dl EI dl EI 2r L L L B B dS dS S t S t dB r 2 (r R) dB dt d S dB dt S R 2 (r R) dt r dB Inside EI 2 dt R 2 dB outside EI 2r dt R r Example 2(p.237/ [Ex.2]) In a long solenoid of radius R, dB/dt, h, L are known. Find MN . N E dl I M MN Sol.: o L r dB r h cosdl ( r cos h ) 0 2 dt L M N 1 dB h dB L dl hL 2 dt 2 dt 0 1 Another way: OMN = BS hLB 2 emfs are zero on OM and ON d 1 dB MN OMN hL dt 2 dt Exercises p.279 / 6- 4 - 2 §5. Self Induction 1. Self Induction 2. Self Inductance 3. Examples 1. Self Induction IB if d 0 dt then B I d 0 dt An induced emf appears in any coil in which the current is changing —— Self-induced emf 2. Self Inductance tightly wound coil:every turn as a loop,the same flux S passes through all the turns N - turn coil:flux linkage S = NS S S B I Definition: S = L I L:Self inductance d S dS dI S N L dt dt dt 1 Weber Unit: 1 Henry 1 Ampere Example (p.245/ [Ex.]) A long solenoid of volume ,n turns per unit length. Find it’s L . Sol.:B = 0nI S = N S = nl S = 0n2IS l = 0n2 I = LI L = 0n2 Exercises p.280 / 6- 5 - 1, 2, 3 §6. Mutual Induction 1. Mutual Induction 2. Mutual Inductance 3. Two Coils in Series ● connect in the same direction ● connect in the opposite direction 1. Mutual Induction I1 11 ,12 I1 11 12 12 : flux through coil 2 by the field of I1 d11 Self induced emf dt 2 d12 1 Mutual induced emf dt in the same way 21 : flux through coil 1 by the field of I2 total flux through coil 1 1= 11 21 + for I1 , I2 in same direction - for opposite direction B 2. Mutual Inductance flux linkage:11 = N111 , 21 = N121 , 1 = N11 1= 11 21 ( 1= 11 21 ) 11 = L1 I1 , 21 = M21 I2 M21: flux through coil 1 by the field of I2 I2 M12: flux through coil 2 by the field of I1 I1 can be prove: M12 = M21 = M Mutual Inductance d1 dI1 dI 2 1 ( L1 M ) M :interaction dt dt dt of two coils d2 dI 2 dI1 and 2 ( L2 M ) dt dt dt 3. Two Coils in Series Connect in the same direction Connect in the opposite direction Connect in the Same Direction 1= 11 + 21 2= 22 + 12 dI dI I 1 ( L1 M ) dt dt dI ( L1 M ) dt dI and 2 ( L2 M ) dt dI 1 2 ( L1 L2 2M ) dt L L1 L2 2M I Connect in the Opposite Direction 1= 11 - 21 2= 22 - 12 dI 1 ( L1 M ) dt dI 2 ( L2 M ) dt I dI 1 2 ( L1 L2 2M ) dt L L1 L2 2M I Exercises p.280 / 6- 6 - 1, 2 §7. Eddy Current Applications Harmfulness Electromagnetic Damping Skin Effect §8. Transient State of RL Circuit 1. Connect RL Circuit with an emf 2. Remove the emf from RL Circuit R L Steady state:I = 0 (K open) and I =/R(K close) Transient State :i :0 I K capitals: I、U steady quantities little: i、u varying quantities Ohm’s Law, Kirchhoff’s Rules are workable , as long as the quasi-steady condition satisfied 1. Connect RL Circuit with an emf (1) Equation:( Kirchhoff’s loop rule) L R d i S iR S L S dt di i L iR dt —— deferential equation of i Initial condition :i 0 at t 0 K di General solution : L iR dt R t di R dt i Ae L i / R L R R ln( i / R) t K(constant ) L Connect RL Circuit with an emf (2) Spetial solution :i A 0 at t 0 R A R R t i (1 e L ) R I (1 e (t 0) R t L ) R L i 自 K i (t) I R 0 t Connect RL Circuit with an emf (3) i(t ) I (1 e R t L i (t) ) I Discussion: R 0.63I (1) if t ,then i I , actually need only little time 0 1 2 t (2) Inductive time constant R R the time constant is, t The less e L 0 rate governed by L rapidly the current the more L goes to it’s equilibrium value. Definition : ( time constant ) R when t , i I (1 e1 ) 0.63I 2. Remove the emf from RL Circuit(1) L R The switch closes at t = 0,consider the uper part of the circuit ( loop ) 自 i di S iR S L dt di L iR dt R’ di R dt at t 0 i A I 0 i L R ln i t K(constant ) R R' L i Ae R t L i(t ) I 0e R t L Remove the emf from RL Circuit(2) i (t ) I 0 e R t L i (t) I0 Discussion: (1) if t ,then i 0 , 0.37I0 actually need only little time 0 1 2 t (2) time constant : L The less the time constant is, the more rapidly the current R goes to it’s equilibrium value 0 . i(t ) I e t / 0 when t , i I 0e 0.37 I 0 1 Exercises p.280 / 6- 8 - 1, 5 §9. Transient State of RC Circuit 1. Connect RC Circuit with an emf 2. Remove the emf from RC Circuit 3. Summary 1. Connect RC Circuit with an emf(1) Charging a capacitor:uc:0 Equation(Kirchhoff’s loop rule) R C uR uC uR uC i dq q CuC uR iR i dt duC K RC uC dt duC dt uC RC uC 0 at t 0 1 ln( uC ) tK A 0 RC t / RC t / RC u ( t ) ( 1 e ) C uC Ae Connect RC Circuit with an emf(2) uC (t ) (1 e Discussion: t / RC ) uC (t) (1) if t ,then uC , t actually need only little time 0 (2) Capacitive time constant: = RC The less the , the more rapidly it goes to equilibrium . R smaller,current larger Need less time C smaller,need less charge 2. Remove the emf from RC Circuit The switch 2 1 at t = 0 R (discharging a capacitor) u R uC 0 i duC RC uC 0 dt duC dt uC RC uC (t) t / RC uC Ae at t 0 , uC A uC (t ) e t / RC e t / 0 C 1 2 K t 3. Summary Steps to treat transient process : differential equation general solution (in exponential fashion) initial condition constant Features : C voltage can not change suddenly L current can not change suddenly energy(L、C can store energy,need time) C: W 1 Cu 2 L: WM 1 Li 2 E 2 2 ( R does not store energy,I, U change suddenly ) Exercises p.281 / 6- 9 - 1, 2 §10. Magnetic Energy 1. Magnetic Energy of a Self-Inductor 2. Magnetic Energy of Mutual-Inductors 1. Magnetic Energy of a Self-Inductor Steady:I constant, = IR Power of emf I = I2R L Thermal energy transient : di L iR dt idt: idt = i 2R dt + Lidi R K work of emf = thermal energy + energy to build up field in L (this is magnetic energy) i :0 I Wm I 0 1 2 1 Lidi LI ( compare :WE CU 2 ) 2 2 ( I lager ,B stronger, Wm lager ) 2. Magnetic Energy of Mutual-Inductors di1 di2 1 L1 M i1R1 ① dt dt di2 di1 2 L2 M i2 R2 ② dt dt ①×i1dt + ②× i2dt: 1i1dt 2i2dt L1 M K1 R1 L2 K2 R2 1 2 2 2 i1 R1dt i2 R2dt L1i1di1 L2i2di2 M (i1di2 i2di1 ) work of emfs = thermal energy in Rs+ magnetic energy Wm in Ls I1I 2 Wm L1i1di1 L2i2di2 M d(i1i2 ) 0 0 0 1 1 2 2 L1 I1 L2 I 2 MI1I 2 2 2 I1 I2 Exercises p.282 / 6- 11 - 2