# chap29 - 29.1: ) . 37 cos 1 ( . 37 cos and , °- = ∆Φ...

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Unformatted text preview: 29.1: ) . 37 cos 1 ( . 37 cos and , °- = ∆Φ ⇒ ° = Φ = Φ NBA NBA NBA B B B i f V. 5 . 29 s 0600 . ) . 37 cos 1 )( m 25 . )( m 400 . )( T 10 . 1 )( 80 ( ) . 37 cos 1 ( = ⇒ °-- = ∆ °-- = ∆ ∆Φ- = ⇒ ε ε t NBA t B 29.2: a) Before: ) m 10 12 )( T 10 . 6 )( 200 ( 2 4 5-- × × = = Φ NBA B : after ; m T 10 44 . 1 2 5 ⋅ × =- b) . V 10 6 . 3 s 040 . ) m 10 2 . 1 )( T 10 . 6 )( 200 ( 4 2 3 5--- × = × × = ∆ = ∆ ∆Φ = t NBA t B ε 29.3: a) . R NBA Q NBA QR R t Q IR t NBA t B = ⇒ = ⇒ ∆ = = ∆ = ∆ ∆Φ = ε b) A credit card reader is a search coil. c) Data is stored in the charge measured so it is independent of time. 29.4: From Exercise (29.3), . C 10 16 . 2 . 12 80 . 6 ) m 10 20 . 2 ( T) 05 . 2 )( 90 ( 3 2 4-- × = Ω + Ω × = = R NBA Q 29.5: From Exercise (29.3), . T 0973 . ) m 10 20 . 3 )( 120 ( ) . 45 . 60 )( C 10 56 . 3 ( 2 4 5 = × Ω + Ω × = = ⇒ =-- NA QR B R NBA Q 29.6: a) ( 29 4 4 5 ) s T 10 00 . 3 ( s) T 012 . ( ) ( t t dt d NA B dt d NA dt Nd B- × + = = = Φ ε ( 29 . ) s V 10 02 . 3 ( V 0302 . ) s T 10 2 . 1 ( ) s T 012 . ( 3 3 4 3 4 4 t t NA-- × + = × + = ⇒ ε b) At V 0680 . ) s 00 . 5 )( s V 10 02 . 3 ( V 0302 . s 00 . 5 3 2 4 + = × + = ⇒ =- ε t . A 10 13 . 1 600 V 0680 . 4- × = Ω = = ⇒ R I ε 29.7: a) for 2 sin 2 2 cos 1 - = -- = Φ- = T t T NAB T t NAB dt d dt d B π π π ε otherwise. zero ; T t < < . b) 2 at T t = = ε c) . 4 3 and 4 at occurs 2 max T t T t T πNAB = = = ε d) From B t T , 2 < < is getting larger and points in the z + direction. This gives a clockwise current looking down the z- axis. From B T t T , 2 < is getting smaller but still points in the z + direction. This gives a counterclockwise current. 29.8: a) ) ( 1 ind A B dt d dt d = = Φ B ε ( 29 t s 057 . ind 1 ) T 4 . 1 ( 60 sin 60 sin-- ° = ° = e dt d A dt dB A ε t s 057 . 1 2 1 ) s 057 . )( T 4 . 1 )( 60 )(sin (--- ° = e r π t s e 1 057 . 1 2 ) s 057 . )( T 4 . 1 )( 60 (sin ) m 75 . (--- ° = π = t s e 1 057 . V 12 .-- b) ) V 12 . ( 10 1 10 1 = = ε ε t s e 1 057 . V 12 . ) V 12 . ( 10 1-- = s 4 . 40 s 057 . ) 10 1 ( ln 1 = →- =- t t c) B is getting weaker, so the flux is decreasing. By Lenz’s law, the induced current must cause an upward magnetic field to oppose the loss of flux. Therefore the induced current must flow counterclockwise as viewed from above. 29.9: a) π π π 4 / so and 2 2 2 c A r A r c = = = 2 ) 4 ( c B BA B π = = Φ dt dc c π B dt d ε B = Φ = 2 At m 570 . ) s 120 . )( s . 9 ( m 650 . 1 , s . 9 =- = = c t mV 44 . 5 ) s m 120 . )( m 570 . )( 2 1 )( T 500 . ( = = π ε b) Flux ⊗ is decreasing so the flux of the induced current ⊗ Φ is ind and I is clockwise....
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## This homework help was uploaded on 04/08/2008 for the course PHY 205 taught by Professor Nearing during the Fall '07 term at University of Miami.

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chap29 - 29.1: ) . 37 cos 1 ( . 37 cos and , °- = ∆Φ...

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