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Unformatted text preview: 22. (a) First, we observe that a large portion of the ﬁgure contributes ﬂux which “cancels out.” The ﬁeld
(due to the current in the long straight wire) through the part of the rectangle above the wire is
out of the page (by the righthand rule) and below the wire it is into the page. Thus, since the
height of the part above the wire is b − a, then a strip below the wire (where the strip borders the
long wire, and extends a distance b − a away from it) has exactly the equalbutopposite ﬂux which
cancels the contribution from the part above the wire. Thus, we obtain the nonzero contributions
to the ﬂux:
a
µ0 ib
a
µ0 i
ΦB = B dA =
(b dr ) =
ln
.
2πr
2π
b−a
b −a
Faraday’s law, then, (with SI units and 3 signiﬁcant ﬁgures understood) leads to
E =
=
= dΦB
d µ0 ib
a
=−
ln
dt
dt 2π
b−a
µ0 b
a
di
µ0 b
−
ln
=−
ln
2π
b − a dt
2π
−µ0 b(9t − 10)
a
ln
.
2π
b−a − a
b−a d
dt 92
t − 10t
2 With a = 0.120 m and b = 0.160 m, then, at t = 3.00 s, the magnitude of the emf induced in the
rectangular loop is
E = 4π × 10−7 (0.16)(9(3) − 10)
0.12
ln
2π
0.16 − 0.12 = 5.98 × 10−7 V . di
(b) We note that dt > 0 at t = 3 s. The situation is roughly analogous to that shown in Fig. 315(c).
From Lenz’s law, then, the induced emf (hence, the induced current) in the loop is counterclockwise. ...
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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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