A square-shaped coil of area 70 , cm^2 having | vedclass

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(D) Given: $N = 600$,$A = 70 	imes 10^{-4} \, m^2$,$B = 0.4 \, T$,$f = \frac{500}{60} \, Hz$.Angular velocity $\omega = 2 \pi f = 2 \pi 	imes \frac{

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$44$

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(D) Given: $N = 600$,$A = 70 	imes 10^{-4} \, m^2$,$B = 0.4 \, T$,$f = \frac{500}{60} \, Hz$.Angular velocity $\omega = 2 \pi f = 2 \pi 	imes \frac{500}{60} = \frac{50 \pi}{3} \, rad/s$.The instantaneous emf is given by $e = N A B \omega \sin 	heta$,where $	heta$ is the angle between the area vector $\vec{A}$ and the magnetic field $\vec{B}$.Since the plane of the coil is inclined at $60^{\circ}$ with the field,the angle between the normal to the plane (area vector) and the field is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.$e = 600 	imes (70 	imes 10^{-4}) 	imes 0.4 	imes (\frac{50 \pi}{3}) 	imes \sin(30^{\circ})$.$e = 600 	imes 0.007 	imes 0.4 	imes \frac{50}{3} 	imes \frac{22}{7} 	imes 0.5$.$e = 4.2 	imes 0.4 	imes \frac{50}{3} 	imes \frac{22}{7} 	imes 0.5$.$e = 1.68 	imes \frac{50}{3} 	imes \frac{22}{7} 	imes 0.5 = 44 \, V$.

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