821.1 Faraday's law of electromagnetic induction =5F:d=g5 E.·dl clsd path EKd=Ekd=EK(2rr) Define the induced electromotive force(emf) Induced emf=E.dl elec q 821.1 Faraday's law of electromagnetic induction ①n=「B.dA=BA=Bm2 do Eind 2.2 dB dt 手Em:d=E dB nduced (2m) dt r dB E 2 dt R These electric and magnetic x x fields are mutually perpendicular to each other.\X
11 ∫ ∫ = ⋅ = ⋅ clsd path clsd path Welec F dl q E dl k k r r r r d d (2 ) clsdpath clsdpath elec E l E l E r q W = k ⋅ = k = k π ∫ ∫ r r Define the induced electromotive force(emf) q W E l k elec d = = ⋅ ∫ r r Induced emf §21.1 Faraday’s law of electromagnetic induction These electric and magnetic fields are mutually perpendicular to each other. §21.1 Faraday’s law of electromagnetic induction t r B E t B E l E r r t B r t E l B A BA B r m m d d 2 d d d (2 ) d d d d d d induced 2 induced induced 2 induced 2 = − ⋅ = = − ⋅ = − = − = ⋅ = = ∫ ∫ ∫ π π π Φ Φ π r r r r r r
821.1 Faraday' s law of electromagnetic induction Exercise 1 If the radius of the solenoid a, the time rate of the change B of magnetic field is dr >0,the ides of the trapezoid is a, a, a, and 2a respectively, find the emf for each sides of the D trapezoid and the total emf. Choose clockwise direction Solution: connect 0A, OD B C 2a D4=EOD=EB=ECD=0 For loop OAD =B So4==aB 8 21.1 Faraday's law of electromagnetic induction d a 3 dB 40AD A→D dt 4 dt For loop obC, the flux B Φn=B:3006 B O d r- dB 4OBC dt 6 dt D B→C For loop ABCD, the emf Choose clockwise as + direction E=8,+8+ECD TEDA B C 丌2dB√32dB_丌√3、2dB 6 dt 4 d t dt
12 If the radius of the solenoid is a, the time rate of the change of magnetic field is , the sides of the trapezoid is a, a, a, and 2a respectively, find the emf for each sides of the trapezoid and the total emf. 0 d d > t B Solution：connect OA , OD = = = =0 OA OD AB CD ε ε ε ε For loop OAD 4 3 2 Φm = B⋅ S∆OAD = a B B r × o D B C A a a 2a §21.1 Faraday’s law of electromagnetic induction Exercise 1: Choose clockwise as + direction For loop OBC, the flux B a Φm B SOAD 6 2 π = ⋅ = B C t a B t Φm BC OBC → = = − = − d d d 6 d 2 π ε ε ∆ For loop ABCD, the emf AB BC CD DA ε = ε + ε + ε + ε t B a t B a t B a d d ) 4 3 6 ( d d 4 3 d d 6 2 2 2 = − + = − − π π + §21.1 Faraday’s law of electromagnetic induction B r × o D B C A a a 2a Choose clockwise as + direction A D t B a t Φm AD = OAD = − = − → d d 4 3 d d 2 ∆ ε ε