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(C) The peak induced electromotive force $(EMF)$ is given by $e_0 = N B A \omega$.Given: $N = 1$,$B = 10^{-2} \ T$,$r = 0.3 \ m$,$R = \pi^2 \ \Omega$,

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$6$

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(C) The peak induced electromotive force $(EMF)$ is given by $e_0 = N B A \omega$.Given: $N = 1$,$B = 10^{-2} \ T$,$r = 0.3 \ m$,$R = \pi^2 \ \Omega$,and frequency $
u = \frac{200}{60} = \frac{10}{3} \ Hz$.The angular velocity is $\omega = 2\pi 
u = 2\pi 	imes \frac{10}{3} = \frac{20\pi}{3} \ rad/s$.The area $A = \pi r^2 = \pi (0.3)^2 = 0.09\pi \ m^2$.Thus,$e_0 = 1 	imes 10^{-2} 	imes 0.09\pi 	imes \frac{20\pi}{3} = 10^{-2} 	imes 0.03\pi 	imes 20\pi = 0.6 	imes 10^{-2} 	imes \pi^2 \ V$.The peak current $i_0 = \frac{e_0}{R} = \frac{0.6 	imes 10^{-2} 	imes \pi^2}{\pi^2} = 0.6 	imes 10^{-2} \ A = 6 	imes 10^{-3} \ A = 6 \ mA$.

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