A coil has 1,000 turns and 500 , cm^2 as its | vedclass

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(B) Given: Number of turns $N = 1000$,Area $A = 500 \, cm^2 = 500 	imes 10^{-4} \, m^2 = 0.05 \, m^2$,Magnetic field $B = 2 	imes 10^{-5} \, Wb/m^2$

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$10$

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(B) Given: Number of turns $N = 1000$,Area $A = 500 \, cm^2 = 500 	imes 10^{-4} \, m^2 = 0.05 \, m^2$,Magnetic field $B = 2 	imes 10^{-5} \, Wb/m^2$,Time $\Delta t = 0.2 \, s$.Initially,the plane of the coil is perpendicular to the magnetic field,so the angle between the area vector and the magnetic field is $	heta_1 = 0^o$.After rotating by $180^o$,the angle becomes $	heta_2 = 180^o$.The induced $e.m.f.$ is given by Faraday's law: $e = -N \frac{\Delta \phi}{\Delta t} = -N \frac{BA(\cos 	heta_2 - \cos 	heta_1)}{\Delta t}$.Substituting the values: $e = -\frac{1000 	imes 2 	imes 10^{-5} 	imes 0.05 	imes (\cos 180^o - \cos 0^o)}{0.2}$.$e = -\frac{1000 	imes 10^{-7} 	imes (-1 - 1)}{0.2} = -\frac{10^{-3} 	imes (-2)}{0.2} = \frac{2 	imes 10^{-3}}{0.2} = 10^{-2} \, V$.$10^{-2} \, V = 10 \, mV$.

$A$ coil has $1,000$ turns and $500 \, cm^2$ as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{-5} \, Wb/m^2$. The coil is rotated through $180^o$ in $0.2 \, s$. The average $e.m.f.$ induced in the coil,in milli-volts,is

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