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A straight wire segment 2aB 0i b direction : 0i L d , B 2d -b 0i a R, B 2 R 0i R 0, B 2a The Magnetic Field of a Solenoid 0iR 2 B 3 ; 2( R 2 z 2 ) 2 0 (nidz ) R 2 dB 3 2 2 2 2[ R ( z - d ) ] B 0 niR 2 L/2 2 0 ni 2 [R - L/2 ( dz 2 (z - d ) ] 2 L 2d [R 2 (L 2 d )2 ] If L>>R, B 0 ni 1 3 2 2 L 2-d [R2 (L 2 - d )2 ] Bh 0 nhi 1 ) 2 direction ? Biot-Savart law 00 qidvsrˆrˆ dBB 22 44 rr Magnetic Field Symmetry Moving Electrical charge Current B ds 0I Ampere’s Law A interesting question for interaction force between electric / magnetic field and moving or resting charged particle A charged particle at rest Moving charged particle Whether a charged particle moves or not depends on the chosen frame. Electric field Both electric and magnetic field Whether can we conclude that both electric and magnetic field Depend on the chosen frame? Einstein’s Postulates: The laws of physics are the same in all inertial reference frames. Are these two conclusions contradictory? Is it true? 2017/5/7 [email protected] 4 y In the frame K (at rest) In the frame K Both electric and magnetic force are zero. ’ In the frame K Magnetic force: x r I - - - - - - Fm = qvd B¢ = vd S Electric force: + + + + + + y’ K’ In the frame (with moving electrons) B´ vd I vd - Fm - - l+' = l+ / 1- (vd / c)2 L-' = L / 1- (vd / c)2 l-' = l- 1- (vd / c)2 ql+¢ vd2 FE = qE = q = 2pe0 r 2pe0 rc 2 r We have: I ¢ = l+¢ vd , c2 =1/ (e0m0 ) S + ++ + + + 2017/5/7 L+' = L 1- (vd / c)2 l+¢ - l-¢ x’ - - - m0qvd I ¢ 2p r [email protected] Therefore FE = 5 m0qvd I ¢ 2p r In different inertial reference frame, the results come from the electric or magnetic field is same. Transformation equations of E and B: E x E x E y E z Bx Bx 1 1- b 2 1 1- b 2 ( E y - vBz ) ( E z vBy ) 1 v By ( By 2 Ez ) 2 c 1- b 1 v Bz ( Bz - 2 E y ) 2 c 1- b Magnetic Field Electrical Current Faraday’s Law of Induction Chapter 34 Faraday’s Law of Induction Faraday’s Experiments The experiment about the induction: The common feature: change in flux Faraday’s Law of Induction Magnetic flux ΦB dΦB B dA, ΦB B dA dΦB B dA B cosdA change in B change in θ change in A EMF dΦB dt EMF Faraday’s law Flux change dΦB dt EMF dΦB dt induced current Lentz’ Law Flux (cause) increase decrease Flux (result) induced current decrease increase Conservation of energy considerations: i N S i Mechanic energy V Electric energy Conservation of energy considerations: Electric energy Electric energy Signs in Faraday’s law sign of ddΦΦ BB dd tt sign of dA sign of flux ΦB Motional EMF dΦB d ( BDx ) BDv dt dt d (v B) ds (v B) ds BDv F2 F3 0 From the signs of faraday' law , F1 0 W 0 dΦB the EMF is clockwise. dt the induced current is clockwise. Generators and Motors Generator is a device for converting mechanical work (or other) into electrical work in the load. dΦB d( BA cos t ) BA sin t dt dt alternating current (AC) i R BA R sin t chemical energy electric energy mechanical energy atomic energy electric energy mechanical energy Nuclear reactor at Tihange , Wind-powered electric generator Belgium at Sandia National laboratory. •“Dynamic” Microphones (E.g., some telephones) –Sound oscillating pressure waves oscillating [diaphragm + coil] oscillating magnetic flux oscillating induced emf oscillating current in wire after amplification Sound Induced Electric Field In In conductor Space Changing magnetic field Induced electric fields. E dl i Ei Electrical Electrical Field Current Example r r, R (resistance) are known, Bz=-4.0-5.6t+2.2t2. Find induced current in the loop at t=1s and t=2s. ΦB B dA B dA r 2 B dΦB 2 d -r B r 2 (5.6 - 4.4t ) dt dt 2 r I (5.6 - 4.4t ) R R I t 1 1.2 r 2 R I t 2 -3.2 r 2 R Example E E rE r, R (resistance) are known, Bz=-4.0-5.6t+2.2t2. Find induced current in the loop at t=1s and t=2s. ΦB B dA B dA r 2 B E dΦB 2 d 0 r B E d l Induced Electric Field I I ' 0 dt R r R I t 1 1.2 dt 2 Induced EMF (5.6 - 4.4t ) r 2 R I t 2 -3.2 r 2 R Example ? d (v B) ds d Bvdr , R R 0 0 Bvdr 1 2 Brdr B R 2 suppose oab is a loop, 1 2 ΦB BA B( R ) 2 dΦB 1 2 - BR dt 2 What is the direction of induced EMF? ao Example: As shown, find the emf ε when the rod (length L) in location OM, and ON. i OM : (v B) dl M ω dl 0i 0i 2 l l dl L r0 2 r 4 r 0 0 o N r ON : (v B) dl 0i Direction of EMF ? (r - r0 ) dr 2r O M, ON 0i r0 L ( L - r0 ln ) 2 r0 Exercises P796 23, 24, 27 Problems P798 5, 9 Induced Electric Field In In conductor Space Changing magnetic field dΦB 2REi Ei dl dt Induced Electric Field Ei Electrical Electrical Field Current F qEs F qEi Induced electric fields In the closed loop, the work done on the charge by the induced electric field is W qEi ds W The induced emf in the loop is Ei ds q q d Φ B d -ds 0dA rˆ Electrostatics - B EdsAEs - B Field 2 t 0 r dt 4 dt Induced Electric Fields B Ei ds - t dA 0 Electric fields magnetic field Calculation of Induced Electric Field For the cylindrical field distribution (radius R), dB/dt>0, Ei ds Ei 2r dB dB dB - dt - r dB Ei 2 dt Ei dt dA - dt r 2 (r R) dB dB dB r R, - dA - R 2 dt d t d t dB const. R 2 dB dt Ei (r R) 2r dt r R The betatron dΦB Ei ds ΦB Φo cos t Φo sin t Betatron produces a pulsed rather than a continuous beam. dt Induction furnace dΦB Ei ds dt Change current EMF w i Rt 2 Electric energy Heat current Eddy currents dΦB Ei ds dt Electric energy heat energy Reduce heat energy Transformer Materials are laminated. Magnetic breaking s Magnetic breaking N Damping Oscillation STM Magnetic Damping Induction and Relative Motion In S frame V v vd dWN N (vdt ) FB sin (vdt ) (qVB)(vd /V )(vdt ) (qBv d )(vdt ) (qBv )(vd dt ) qBvds WN qBv ds qBvD WN q BDv Induction and Relative Motion In S frame V v vd dWN N (vdt ) FB sin (vdt ) (qVB)(vd /V )(vdt ) (qBv d )(vdt ) (qBv )(vd dt ) qBvds WN qBv ds qBvD WN q BDv In S’ frame ' -- BDv dBd ( Dx ' ) dt dt N ´ Magnetic Field Induced EMF Motional EMF Induced E Electrical Current dΦB d t d (v B) ds (v B) ds B Ei ds - t dA Magnetic Field Electrical Field Example B 6t - 5 Ei ? when r < R, 2 d (2BrdB B) 2 rEi ii -- r dtdtdt r d2BdB 2rE Ei i--r Ei -3r 2 dt dt When r > R, d B 2 dB i -R dt dt 2 dB 2rEi i -R dt R 2 dB 3R 2 Ei 2r dt r Example 0i dr d B Dv dt 2r dt r i 0i r 0 D ln B Ddr r0 2r 2 r0 d (v B) ds vBds ? i D r v 0 D 0 r direction of 0i vBds Dv 2r r0 r0 DD d B -- 00ii dL v ln(ln( )) ? 22 dt r0 r0 dt r0 D i 0i r0 D 0 L ln( ) B Ldr r0 2 r0 2r Example r dr D i L r0 v direction of d (v B) ds vBds r0 D r0 vBds r0 D r0 0i v dr 2r 0i r0 D v ln( ) 2 r0 Example y Ei As shown, R, L, h, dB/dt are known, find emf in the rod. b B r o θh L R a B 2 dB 2rEi Ei dl - ds -r t dt r dB x Ei 2 dt rh dB h d i Ei dy Ei dy cos dydy dB 0 dt Direction of emf a→b i 2 dt r L h dB b hL dB 2 dy d i -- L a 2 2 dt 2 dt Example y c As shown, R, L, h, dB/dt are known, find emf in the rod. b hLhLdBdB i d i b a 2 2dtdt Loop “a-c-b-a” B o h L R a dB 0 dt 1 2 1 Φ BA B ( R Lh) B acb x 2 2 dΦB R 2 hL dB acb ab ' -( ) dt 2 2 dt acb Direction of emf a→b 2 R dB dB 2 (-R ) 2 dt dt 2 2 ab '- acb hL dB 2 dt Example As shown, R, L, h, dB/dt are known, find emf in the rod. b hLhLdBdB y i d i b a 2 2dtdt Loop “a-b-d-a” B 1 1 o h L d x ΦB BAabd B( R 2 - Lh) c R a dB 0 dt 2 2 dΦB R 2 hL dB ' -( - ) bda ab dt 2 2 dt 2 R dB dB 2 bda (-R ) 2 dt dt 2 2 Direction of emf a→b ab '- acb hL dB 2 dt Example Ei c As shown, R, L, h, dB/dt are known, find emf in the rod. b hLhLdBdB i d i b a 2 2dtdt Loop “a-o-b-a” y B o h L x R a dB 0 dt Ei dl 0 Direction of emf a→b 1 ΦB BAaob B( Lh) 2 dΦB hL dB hL i -( ) 3 dt 2 dt oa ob 0 2 B 2- t 3 emf b→a dB 2 dt 3 Example v v0 b dΦB hL dB hL i1 (b a) dt 2 dt 3 b i 2 (v B) ds -v0 BL -2v0 L (b a) a B o h L R a Loop “a-o-b-a” x 2 B 2- t 3 t 0, B 2 hL i i1 i 2 2v0 L (b a) 3 hL hL d( B) d( ) dΦB hL dB 2 2 i Bdt dt dt 2 dt dh BL hL BL hL hL v0 v0 L dt 2 3 2 3 3 (b a) Example v v0 b oa 0, ob 0 B o h L R a x Exercises P797 29, 31, Problems P799 10, 11, 14