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 12 Electromagnetic Induction 第十二章 电磁感应 §12-1 Nonelectrostatic Force, Source & Electromotive Force 非静电力 电源 电动势 1. Nonelectrostatic Force A B When S is turned off, the charges on A and B are constant, the E-field between A and B is constant. SS R When S is turned on, there is a current in circle, the charges on A and B decrease, E-field q=0, E=0, I=0 In order to sustain a constant current, some kind of nonelectrostatic force ( F)n must be supplied to move positive charges from B (low potential) to A (high potential) resisting Fe A B Fn Fe R The device which supplies Fn is called source. Nonelectrostatic Field Fn En q ---nonelectrostatic force on per unit positive charge. 2. Electromotive Force 电动势 The work done by nonelectrostatic force for moving per unit positive charge from B to A within the source is defined as A B En dl ---emf B A scalar High potential Low potential A B + _ R R If Fn exists in a closed loop, then En dl l S N I The emf in a complete circuit equals to the line integral of the nonelectrostatic force round the circuit . B 3. Complete circuit Ohm law 全电路欧姆定律 A B + _ r Source: emf = , internal resistance = r In a time interval dt, emf moves dq from B to A R i dq = idt So the work done by emf is dW=dq = idt Meanwhile, the resistances R and r consume energy Pdt = (i2R + i2r) dt thermal energy According to the principle of conservation of energy, dW = Pdt idt = (i2R + i2r) dt i.e. i A B + _ a i Vab Rr Vab = iR r R Vab = - ir b ---Complete circuit Ohm law §12-2 Faraday’s Law of Induction 电磁感应定律 1. Induction phenomena S magnet moves N Ii Ii B Ii wire moves loop rotates v B B Conclusion:An induced current appears in a closed conductor loop, only when the B-flux that pass through the loop is changed. 2. Lenz’s Law Determine the direction of an induced current. The induced emf and current are in such a direction as to oppose the change that produces them.(感应电动势和电流的方向总是在 抵抗产生他们的变化的方向上) S B B Bi and B have opposite direction I ’s direction is as Fig. N I Bi I B 3. Faraday’s law of induction For SI d B dt d B dt The induced emf in a circuit is equal to the rate at which the B-flux through that circuit is changing with time. 1 d B Induced current I R R dt The induced charge passing through the loop 1 q Idt ( B1 B 2 ) R Note : for a coil of N turns d 1 d 2 1 2 N ( dt dt ) d N N ( i ) ---total flux i dt i 1 If i j for any i, j i 1 N--number of flux linkages d dt [Example] Alternating current (AC) generator A coil of area S,N turns rotating in a uniform M-field. Angular velocity is ,and =0 when t=0. Find =? B N NBS cos NBS cos t d NBS sin t dt Let m NBS m sin t —alternating emf [Example] A long, straight wire carries a variable current I (t). Find the induced emf in the rectangular loop. dI 0 、a 、b 、 l are known dt ab 0 I ldr B B ds a 2 r dr S r I ds B l 0 Il a b ln 2 a a b 0 l a b dI d B ln 2 a dt dt Direction:inverse-clockwise §123 Motional electromotive force 动生电动势 d d B dt dt B d S S B is a constant,S is a variable ----motional emf S a constant,B is a variable ----induced emf Motional emf: a+ ++ En v B F The charge +q in the rod When the rod is moved with , v +q suffers Lorentz force F qv B --nonelectrostatic force --- b nonelectrostatic field is F En v B q a b (v B) dl Notes: ’s direction is along ba d l v is the instantaneous velocity of dl . B is the M-field in the spot that dl locates. This formula can be used to calculate the motional emf induced by any shape of moving conductor in any M-field. Question: a b (v B) dl is the work done by Lorentz force to positive unit charge. From another formula F qv B , We get F v . Lorentz force does not do any work for q. How can we explain their conflict? q has two motions in the rod: F qv B F qv B a FR F v vR v F b The total work done by resultant force is FR v R ( F F ) ( v v ) F v F v F v F v =0 Positive work =0 =0 negative work [Example] l 、B 、v、 are known, find =? b solution: ( v B) dl a v B b l dl a v vB cos dl a 2 b vBl sin Ub > Ua [Example]:l、B 、 are known, find v B B v L o l A dl =? A o ( v B) dl L vBdl 0 1 2 B ldl BL 0 2 “-”represents ’s direction is opposite dl’s direction, i.e. AO. UO > UA L Another solution:using L B d B dt Make a subsidiary line, then the M-flux through the fan area is 1 B B L L 2 1 2 1 2 d d B BL BL dt 2 dt 2 A metal disk with radius R is rotating around its center in the uniform M-field. B o R The motional emf from O to any point P of its edge is P 1 2 B R 2 [Example] I 、d、L、v are known, find =? B I x d I 0 Solution B 2 x v dx a b L vB Ua> Ub b a v B dl d L d 0I vdx 2 x 0 Iv d L ln 2 d §12-4 Induced electric field 1. Maxwell ’s hypothesis A changing M-field produces an nonelectrostatic field -- induced electric field E n. En B En En B >0 t En 2. Induced emf L En dl or l En dl 3 . The relation between induced emf and changing M-field En dl L d B B dS B dS S dt t t B En dl S dS L t S —the area surrounded by L. 3 . The differences between electrostatic field and induced electric field Electrostatic field is set up by static charges. E-line originates on +q, terminates on –q. It is never closed. Induced electric field E n is produced by the changing of M-field. E n-line is closed, neither start, nor end. --non-vortex field (无旋场) E d l 0 L E-potential can be introduced to describe electrostatic field. L En dl --vortex field (有旋场) Potential can not be introduced to describe induced electric field. [Example] A uniform M-field fills a cylindrical B volume of R. And 0. t Find the distributions of E n E n-lines are concentrical rings R L B r . En P r<R:Choose a closed circle L passing point P L En dl Endl En 2 r L d B B B 2 r dS S t t dt B use En dl S dS L t B 2 En 2 r r t r B En “ ”represents that the 2 t direction of E n is determined by Lenz’s Law L En B r>R: R r L P En dl En 2 r B B 2 S t dS t R En r R B En 2r t 2 §12-5 Applications of induced electric field 1. Betatron 感应加速器 A device that can accelerate electrons to high nenrgies by induced electric field. S N B(r,t) When an electron enters the changing M-field, Fm ev B F m En 1 dB , Fe eEn En r 2 dt Fe eE n eE n Tangent acceleration: a t m t e dB t e r dt rBt speed:v t at dt 0 0 2m 2m dt when B(t) , v(t) . ( r keeps a constant. ) 2.Vortex Current 涡电流 (1) formation Alternating current I changing B E vortex ~I n current Iv dB dt En Iv ~I (2) application Using the hot effect of current, vortex current can be used on electromagnetic stove or induced stove to produce heat energy. The energy loss of vortex current should be decreased as less as possible on the transformer systems. I ~ §12-5 Self-induction & Mutual-induction 自感和互感 1. Self-induction I B I Induced emf M-field B varies B I Varies I B N B I LI L --Self inductance of the circle. It is determined by the size, the shape, the number of turns, and the magnetic properties of the material around the circuit. It has nothing with I. The SI unit of L is Henry(H)亨利 Self-induced emf :when I is changing, dI L L dt Self inductance: L I or L L dI dt [Example]:A long and thin straight solenoid with l S r N. Find its L=? rS l N Solution Assume the solenoid carries current I, then B r 0 nI r 0 N l I NBS 0 r N S 0 r lSN 2 L 2 I l l 2 L n V 2 V --the volume of the solenoid. [Example]:Find L=? for the unit length of coaxial cable. (a,b,0 are known) a b Then the distributions of B is B I Solution assume it carries I , I B 0 (r<a) 0I 2 r (a<r<b) 0 (r>b) 0 I B d B h dr a 2 r b a b B h1 I 0 I b ln 2 a The self inductance for unit length cable is I r dr B 0 b L ln I 2 a 2. Mutual induction I1 I2 1 B1 N112 12 2 B2 N 2 21 21 Mutual induced emf : when the currents I1, I2 change, d12 1 dt d21 2 dt Circle 1: 12 I 2 12 M12 I 2 M12 :the mutual inductance of loop 1 relative to 2. It is determined by the sizes, the shapes, the number of turns, and the magnetic properties of the material around the loops, the relative position of the two loops. It has nothing with I2. The SI unit of M is Henry(H) dI 2 1 M12 dt Circle 2: 21 I1 21 M 21I1 M 21 :the mutual inductance of loop 2 relative to 1. dI1 2 M 21 dt It was proved that M12 M 21 M dI1 2 M dt dI 2 1 M dt Calculate M: or 12 21 M I2 I1 M 1 dI 2 dt 2 dI1 dt [Example]:A long and thin straight solenoid with l S r and two coils N1 N 2 . Find their M=? Solution N2 S r N1 l Assume the coil 1carries current I1, then the M-field produced by I1 is B1 r 0 nI1 r 0 N1 l I1 The number of magnetic flux linkages through the coil 2 is 0 r N1 N 2 S 21 N 2 B1 S2 I1 l 21 0 r N1 N 2 S M I1 l If the coils 2 carries current I2,M=? §12-7 Energy of the M-field R L 。。 K Turn on the switch K , the current in the circle 0 i(t), At any time, L iR di L iR dt di dt iR L R t di dt 1 e L i t 0 iR 0 L R When t,i t I (constant) R i t t During the current changes from 0 to ,Mfield changes from 0 to . The work done by the battery: di W dq idt iR L idt dt t I i Rdt Lidi 2 0 0 Energy of M-field Joule energy dissipated on R i.e. the part of work is transferred in the energy of M-field. Wm I 0 1 2 Lidi LI 2 Energy of M-field is 1 2 Wm LI 2 The energy density of M-field (the density of magnetic energy) is dWm wm dV For long, thin straight solenoid, L n V , 2 B nI 2 1 2 1 B V Wm LI 2 2 The density of magnetic energy is 2 dWm 1 B 1 1 2 H BH wm 2 2 2 dV The energy of M-field is dWm wmdV Wm dWm V wm dV the volume filled by M-field [Example]:Find Wm=? for the unit length of coaxial cable. (I,a,b,0 are known) The distributions of B is a b 0 B 0I 2 r 0 h1 I I (r<a) (a<r<b) (r>b) choose a section of the cable with h=1, on the region of r<a and r>b , we have W m 0 1B I 0 a<r<b: w m 2 2 2 0 8 r 2 0 I dWm wm dV 2 2 1 2rdr 8 r 2 b I2 0 I b 0 Wm dWm ln 2 rdr 2 2 a 8 r 4 a 2 2 1 2 Or usingW m LI : 2 b 0 We have known that L ln 2 a 2 0 I b Wm ln 4 a