A coil of 45 turns and radius 4 cm is placed | vedclass

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(C) Given:Number of turns $N = 45$Radius $r = 4 \ cm = 0.04 \ m$Area $A = \pi r^2 = \pi 	imes (0.04)^2 = 16\pi 	imes 10^{-4} \ m^2$Initial magnetic

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$0.72$

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(C) Given:Number of turns $N = 45$Radius $r = 4 \ cm = 0.04 \ m$Area $A = \pi r^2 = \pi 	imes (0.04)^2 = 16\pi 	imes 10^{-4} \ m^2$Initial magnetic field $B_1 = 0 \ T$Final magnetic field $B_2 = 0.70 \ T$Time interval $\Delta t = 220 \ s$The magnetic flux $\phi = B \cdot A \cdot \cos(	heta)$. Since the plane is perpendicular to the field,the angle $	heta = 0^\circ$,so $\cos(0^\circ) = 1$.Change in flux $\Delta \phi = A(B_2 - B_1) = 16\pi 	imes 10^{-4} 	imes (0.70 - 0) = 11.2\pi 	imes 10^{-4} \ Wb$.According to Faraday's law,the induced emf $\varepsilon = -N \frac{\Delta \phi}{\Delta t}$.Magnitude of emf $|\varepsilon| = \frac{45 	imes 11.2\pi 	imes 10^{-4}}{220}$.$|\varepsilon| = \frac{504\pi 	imes 10^{-4}}{220} \approx \frac{1583.36 	imes 10^{-4}}{220} \approx 7.2 	imes 10^{-4} \ V = 0.72 \ mV$.

$A$ coil of $45$ turns and radius $4 \ cm$ is placed in a uniform magnetic field such that its plane is perpendicular to the direction of the field. If the magnetic field increases from $0$ to $0.70 \ T$ at a constant rate in a time interval of $220 \ s$,then the induced emf in the coil is (in $mV$)

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