In a region of uniform magnetic induction B = | vedclass

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Question

(C) The magnetic flux linked with the coil is $\phi = NBA \cos(\omega t)$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt}

Vedclass · Accepted Answer

$6$

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(C) The magnetic flux linked with the coil is $\phi = NBA \cos(\omega t)$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = NBA\omega \sin(\omega t)$.The amplitude of the induced $EMF$ is $V_0 = NBA\omega$.The amplitude of the induced current is $i_0 = \frac{V_0}{R} = \frac{NBA\omega}{R}$.Given: $B = 10^{-2} \, T$,$r = 0.3 \, m$,$R = \pi^2 \, \Omega$,$N = 1$,and frequency $f = \frac{200}{60} \, Hz = \frac{10}{3} \, Hz$.Angular velocity $\omega = 2\pi f = 2\pi 	imes \frac{10}{3} = \frac{20\pi}{3} \, rad/s$.Area $A = \pi r^2 = \pi (0.3)^2 = 0.09\pi \, m^2$.Substituting the values: $i_0 = \frac{1 	imes 10^{-2} 	imes 0.09\pi 	imes (20\pi/3)}{\pi^2}$.$i_0 = \frac{10^{-2} 	imes 0.03 	imes 20 	imes \pi^2}{\pi^2} = 10^{-2} 	imes 0.6 = 0.006 \, A = 6 \, mA$.

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