ch30-p023 - 23. (a) Eq. 29-10 gives the field at the center...

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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.
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