P31_012

# P31_012 - 12. (a) Eq. 30-12 gives the ﬁeld at the center...

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Unformatted text preview: 12. (a) Eq. 30-12 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 m2 ) enclosed by the small loop. Thus, with B = µ0 i/2R and i(t) = i0 +kt (where i0 = 200 A and k = (−200 A−200 A)/1.00 s = −400 A/s), we ﬁnd B t=0 = t=0.500 s = B |t=1.00 s = B µ0 i0 (4π × 10−7 H/m)(200 A) = = 1.26 × 10−4 T , 2R 2(1.00 m) (4π × 10−7 H/m)[200 A − (400 A/s)(0.500 s)] =0, 2(1.00 m) (4π × 10−7 H/m)[200 A − (400 A/s)(1.00 s)] = −1.26 × 10−4 T . 2(1.00 m) (b) Let the area of the small loop be a. Then ΦB = Ba, and Faraday’s law yields E = = d(Ba) dB ∆B dΦB =− = −a = −a dt dt dt ∆t −1.26 × 10−4 T − 1.26 × 10−4 T −(2.00 × 10−4 m2 ) 1.00 s − = 5.04 × 10−8 V . ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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