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 ×10–4m2) enclosed by the small loop. Thus, with B= µ0i/2Rand i(t) = i0+ kt, where i0= 200 A and k= (–200 A – 200 A)/1.00 s = – 400 A/s, we find (a) ()( )7400410 H/m200A(0)1.2610T,22 1.00miBtR−−π×==== ×(b) ( )( )7410 H/m200A400A/s0.500s(0.500s)0,21.00m−−⎡⎤⎣⎦=and (c) ( )( )74410 H/m200A400A/s 1.00s(1.00s)1.26 10 T,−−−=−×or 4| (1.00s)| 1.26 10 T.
This is the end of the preview. Sign up
access the rest of the document.