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Unformatted text preview: 11.6 Problems Problem 30.15
In ﬁgure 33, a stiﬀ wire bent into a semicircle of radius a = 2, cm is rotated at constant angular
speed f = 40 rev/s in a uniform 20 mT magnetic ﬁeld. What are the (a) frequency and (b)
amplitude of the emf induced in the loop? Figure 33: Figure for problem 30.15
The induced emf is generated by the changing magnetic ﬂux through the circuit. In this case,
the changing magnetic ﬂux is not due to the magnetic ﬁeld changing but rather the amount of area
though which the ﬁeld passes is changing. Since the area oscillates back and forth at the frequency
f , the frequency of the emf must be the same.
To ﬁnd the amplitude of the emf we must do some calculations. We want to write down the
ﬂux through the loop as a function of time. Lets return to the deﬁnition of ﬂux.
ΦB = B · dA = BA1 + B dA2 cos θ where θ is the angle between the magnetic ﬁeld and the normal to the changing surface area (in
or out will both work just ﬁne). The ﬁrst term is the magnetic ﬂux through the square part of the
loop which does not change. Since it does not change, it will not contribute to the emf and can
therefore be neglected.
Integrating over dA2 gives
ΦB = B ( πa2 ) cos θ
Since the loop is turning, θ is a function of time. If the loop turns at a rate of f , then theta changes
at a rate of
ω = 2πf
Hence, the magnetic ﬂux through the circuit as a function of time is
ΦB = 1
Bπa2 cos(2πf t)
2 Finally, from Faraday’s law the induced emf will be
E (t) = −Bπ 2 a2 f sin(2πf t)
from which we can read oﬀ the amplitude as
|E| = Bπ 2 a2 f 44 ...
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- Spring '08