A rectangular coil of N turns and of length a and width b is rotated at frequency $$f$$ in a uniform magnetic field $$\overrightarrow {B}$$, as indicated in Figure . The coil is connected to co-rotating cylinders, against which metal brushes slide to make contact. (a) Show that the
emf induced in the coil is given (as a function of time t) by
$$\mathscr{E}=2\pi fNab\text{ }\sin{2\pi
ft}=\mathscr{E_0}\text{}\sin{2\pi ft}.$$
This is the principle of the commercial alternating-current generator. (b) What value of Nab gives an emf with $$\mathscr{E_0}=150V$$ when the loop is rotated at 60.0 rev/s in a uniform magnetic field of 0.500 T?
(a) It should be emphasized that the result, given in terms of $$\sin (2 \pi f t)$$, could as easily be given in terms of $$\cos (2 \pi f t)$$ or even $$\cos (2 \pi f t+\phi)$$ where $$\phi$$ is a phase constant . The angular position $$\theta$$ 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 $$B A \cos \theta, B A
\sin \theta$$ or $$B A \cos (\theta+\phi) .$$ Here our choice is such that $$\Phi_{B}=B
A \cos \theta .$$ since the coil is rotating steadily, $$\theta$$ increases linearly with time. Thus, $$\theta=\omega t$$ (equivalent to $$\theta=2 \pi f t$$) if $$\theta$$ is understood to be in radians (and $$\omega$$ would be the angular velocity). since the area of the rectangular coil is $$A=a b$$,
Faraday's law leads to
$$\varepsilon=-N \dfrac{d(B A \cos
\theta)}{d t}=-N B A \dfrac{d \cos (2 \pi f t)}{d t}=N B a b 2 \pi f \sin (2
\pi f t)$$
which is the desired result, shown in the problem statement. The second way this is written $$\left(a_{0} \sin (2
\pi f)\right)$$ is meant to emphasize that the voltage output is sinusoidal (in its time dependence) and has an amplitude of $$\varepsilon_{0}=2 \pi f N a b B$$
(b) We solve
$$\varepsilon_{0}=150 \mathrm{V}=2
\pi f \mathrm{NabB}$$
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 $$N a b=0.796 \mathrm{m}^{2}$$