Textbook 30

# Textbook 30 - Assignment 8 Solutions(a Eq 29-10 gives the...

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Assignment 8 Solutions Chpt 30 23. (a) Eq. 29-10 gives the field at the center of the large loop with R = 1.00 m and current i ( t ). This is approximately the field throughout the area ( A = 2.00 × 10 –4 m 2 ) enclosed by the small loop. Thus, with B = μ 0 i /2 R and i ( t ) = i 0 + kt , where i 0 = 200 A and k = (–200 A – 200 A)/1.00 s = – 400 A/s, we find (a) () ( ) 7 4 00 4 10 H/m 200A (0 ) 1 . 2 6 1 0T 2 2 1.00m i Bt R π× == = = × , (b) ( )( ) 7 4 10 H/m 200A 400A/s 0.500s ( 0.500s) 0, 21 .00m ⎡⎤ ⎣⎦ = and (c) ( )( ) 7 4 4 10 H/m 200A 400A/s 1.00s ( 1.00s) 1.26 10 T, = × or 4 | ( 1.00s)| 1.26 10 T. × (d) Yes, as indicated by the flip of sign of B (t) in (c). (e) Let the area of the small loop be a . Then Φ B Ba = , and Faraday’s law yields 44 42 8 1.26 10 T 1.26 10 T (2.00 10 m ) 1.00 s 5.04 10 V . B d dB a B aa dt dt dt t ε −− Φ Δ ⎛⎞ =− ⎜⎟ Δ ⎝⎠ −× × 1

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30. Noting that | Δ B | = B , we find the thermal energy is 22 thermal 42 2 6 2 63 10 11 ( 2 . 0 01 0m)( 1 7 . 0T ) (5.21 10 )(2.96 10 s) 7.50 10 J. B d tB Pt t A t 2 A B R Rd t R t R ε −− Φ ΔΔ ⎛⎞ ⎛⎞ Δ= = ⎜⎟ t Δ Δ ⎝⎠ ×× = ×Ω × 34. Noting that F net =
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## This homework help was uploaded on 04/11/2008 for the course PHYS 1402 taught by Professor Parsons during the Spring '08 term at Columbia.

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Textbook 30 - Assignment 8 Solutions(a Eq 29-10 gives the...

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