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Unformatted text preview: 7. The primary diﬀerence between this and the situation described in Sample Problem 31-1 is in the quantity
A. The area through which there is magnetic ﬂux is not the area of the short coil, in this case, but is the
area of the solenoid (there is no ﬁeld outside an ideal solenoid). Actually, because of the current (which
we calculate here) in the short coil, there is a very small amount of ﬁeld outside the solenoid (caused
by that current) – but it may be disregarded in this calculation. The values are as indicated in Sample
Problem 31-1 except that A = πD2 /4 (where D = 0.032 m) and N = 120 for the short coil. Thus, we
ﬁnd ΦB,i = 3.3 × 10−5 Wb, and the magnitude of the induced emf is 0.16 V. Ohm’s law then yields
0.16 V/5.3 Ω = 0.030 A. ...
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- Fall '08