In a region of uniform magnetic induction B = | vedclass

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(C) The magnetic flux through the coil at any time $t$ is given by $\phi = NBA \cos(\omega t)$,where $\omega$ is the angular frequency.The induced ele

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$6$

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(C) The magnetic flux through the coil at any time $t$ is given by $\phi = NBA \cos(\omega t)$,where $\omega$ is the angular frequency.The induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt} = NBA\omega \sin(\omega t)$.The amplitude of the induced $EMF$ is $e_0 = NBA\omega$.The amplitude of the induced current is $i_0 = \frac{e_0}{R} = \frac{NBA\omega}{R}$.Given: $B = 10^{-2} \, T$,$r = 0.3 \, m$,$A = \pi r^2 = \pi(0.3)^2 = 0.09\pi \, m^2$,$R = \pi^2 \, \Omega$,and frequency $f = \frac{200}{60} = \frac{10}{3} \, Hz$.Angular frequency $\omega = 2\pi f = 2\pi 	imes \frac{10}{3} = \frac{20\pi}{3} \, rad/s$.Assuming $N=1$ (single turn coil):$i_0 = \frac{1 	imes 10^{-2} 	imes (0.09\pi) 	imes (20\pi/3)}{\pi^2} = \frac{10^{-2} 	imes 0.03 	imes 20 	imes \pi^2}{\pi^2} = 0.006 \, A = 6 \, mA$.

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