PHY_solution-chapter29

# PHY_solution-chapter29 - 29.1 37 cos 1 37 cos and ° =...

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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: Before: a) ) 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 A credit b) card reader is a search coil. Data is c) 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 At b) 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 From d) 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. 056ddba76af2e97e015697860a997452149a9ecc.doc Page 1 of 26 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 B is getting c) 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 note was uploaded on 02/25/2009 for the course PHY 127 taught by Professor Mr.yang during the Spring '08 term at SUNY Stony Brook.

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PHY_solution-chapter29 - 29.1 37 cos 1 37 cos and ° =...

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