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Unformatted text preview: 13. (a) It should be emphasized that the result, given in terms of sin(2π ft), could as easily
be given in terms of cos(2π ft) or even cos(2π ft + φ) where φ is a phase constant as
discussed in Chapter 15. The angular position θ of the rotating coil is measured from
some reference line (or plane), and which line one chooses will affect whether the
magnetic flux should be written as BA cosθ, BA sinθ or BA cos(θ + φ). Here our choice is
such that Φ B = BA cosθ . Since the coil is rotating steadily, θ increases linearly with time.
Thus, θ = ωt (equivalent to θ = 2π ft) if θ is understood to be in radians (and ω would be
the angular velocity). Since the area of the rectangular coil is A=ab , Faraday’s law leads
d ( BA cosθ )
d cos ( 2π ft )
ε = −N
= − NBA
= N Bab2π f sin ( 2π ft )
which is the desired result, shown in the problem statement. The second way this is
written (ε0 sin(2π ft)) is meant to emphasize that the voltage output is sinusoidal (in its
time dependence) and has an amplitude of ε0 = 2π f N abB.
(b) We solve ε0 = 150 V = 2π f N abB when f = 60.0 rev/s and B = 0.500 T. The three unknowns are N, a, and b which occur in
a product; thus, we obtain N ab = 0.796 m2. ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
- Spring '08