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Unformatted text preview: 12. (a) Eq. 3012 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.
 Fall '08
 SPRUNGER
 Physics, Current

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