Chapter_29_Solutions - 29.1 B f = NBA and Bi = NBAcos 53.0...

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29.1: Φ B f = NBA , and Φ B i = NBAcos 53.0 °⇒ ∆Φ B = NBA(1 cos 53.0 ° ) ε =− ∆Φ B t cos 53.0 ° ) t (80)(1.10 T)(0.400 m)(0.25 m)(1 cos 53.0 ° ) 0.0600 s = 58.4 V.
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29.2: a) Before: Φ B = NBA = (200)(6.0 × 10 5 T)(12 × 10 4 m 2 ) = 1.44 × 10 5 T m 2 ;after :0 b) ε = ∆Φ B t = NBA t = (200)(6.0 × 10 5 T)(1.2 × 10 3 m 2 ) 0.040 s = 3.6 × 10 4 V.
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29.3: a) ε = ∆Φ B t = NBA t = IR = Q t R QR = NBA Q = NBA R . b) A credit card reader is a search coil. c) Data is stored in the charge measured so it is independent of time.
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29.4: From Exercise (29.3), Q = NBA R = (90)(2.05 T)(2.20 × 10 4 m 2 ) 6.80 Ω+ 12.0 = 2.16 × 10 3 C.
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29.5: From Exercise (29.3), Q = NBA R B = QR NA = (3.56 × 10 5 C)(60.0 Ω+ 45.0 ) (120)(3.20 × 10 4 m 2 ) = 0.0973 T.
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29.6: a) ε = Nd Φ B dt = NA d dt (B) = NA d dt (0.012 T s)t + (3.00 × 10 5 Ts 4 )t 4 ( ) = NA (0.012 T s) + (1.2 × 10 4 4 )t 3 () = 0.0302 V + (3.02 × 10 4 Vs 3 )t 3 . b) At t = 5.00 s = 0.0302 V + (3.02 × 10 4 2 )(5.00 s) 3 =+ 0.0680 V . I = R = 0.0680 V 600 = 1.13 × 10 4 A.
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29.7: a) ε =− d Φ B dt d dt NAB 0 1 cos 2 π t T ⎥ =− 2 NAB 0 T sin 2 t T for 0 < t < T; zero otherwise. b) = 0att = T 2 . c) max = 2 NAB 0 T occurs at t = T 4 and t = 3T 4 . d) From 0 < t < T 2 ,B is getting larger and points in the + z direction. This gives a clockwise current looking down the z axis. From T 2 t < T, B is getting smaller but still points in the direction. This gives a counterclockwise current. + z
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29.8: a) ε ind = d Φ B dt = d dt (B 1 A) ind = Asin60 ° dB dt = Asin 60 ° d dt (1.4 T)e 0.057s 1 t ( ) = ( π r 2 )(sin 60 ° )(1.4 T)(0.057s 1 )e 0.057s 1 t = (0.75 m) 2 (sin 60 ° )(1.4 T)(0.057 s 1 )e 0.057s 1 t =0 .12 V e 0.057 s 1 t b) = 1 10 0 = 1 10 (0.12 V) 1 10 (0.12 V) = 0.12 V e 0.057 s 1 t ln(1 10) =− 0.057 s 1 t t = 40.4 s 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.
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29.9: a) c = 2 π r and A = r 2 so A = c 2 /4 Φ B = BA = (B 4 )c 2 ε = d Φ B dt = B 2 c dc dt At t = 9.0 s, c = 1.650 m (9.0s)(0.120 s) = 0.570 m = (0.500 T)(1 2 )(0.570 m)(0.120 m s) = 5.44 mV b) Flux is decreasing so the flux of the induced current Φ ind is and I is clockwise.
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29.10: According to Faraday’s law (assuming that the area vector points in the positive z -direction) ε =− ∆Φ t 0 (1.5 T) π (0.120 m) 2 2.0 × 10 3 s =+ 34 V (counterclockwise)
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29.11: Φ B = BAcos φ ; is the angle between the normal to the loop and , so G B ° = 53 ε = d Φ B dt = (Acos )(dB dt) = (0.100 m) 2 cos 53 ° (1.00 × 10 3 Ts ) = 6.02 × 10 6 V
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29.12: a) ε = d Φ B dt = d dt (NBAcos ω t) = NBA sin t and 1200 rev min = 20 rev s, so: max = NBA = (150)(0.060 T) π (0.025 m) 2 (440 rev min)(1 min 60sec)(2 rad rev) = 0.814 V. b) Average = 2 max = 2 0.814 V = 0.518 V.
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29.13: From Example 29.5, ε av = 2N ω BA π = 2(500)(56 rev s)(2 rad rev)(0.20 T)(0.10 m) 2 = 224 V
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29.14: ε =− d Φ B dt d dt (NBAcos ω t) = NBA sin t max = NBA = max NBA = 2.40 × 10 2 V (120)(0.0750 T)(0.016 m) 2 = 10.4 rad /s.
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29.15:
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29.16: a) If the magnetic field is increasing into the page, the induced magnetic field must oppose that change and point opposite the external field’s direction, thus requiring a counterclockwise current in the loop.
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This homework help was uploaded on 04/01/2008 for the course PHYSICS 240 taught by Professor Davewinn during the Fall '08 term at University of Michigan.

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Chapter_29_Solutions - 29.1 B f = NBA and Bi = NBAcos 53.0...

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