A coil of area 100 , cm^2 has 500 turns. A ma | vedclass

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(B) The induced $e.m.f.$ $(e)$ is given by Faraday's law of induction: $e = -N \frac{d\phi}{dt} = -N \frac{A \cos 	heta (B_2 - B_1)}{\Delta t}$.Given

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$5$

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(B) The induced $e.m.f.$ $(e)$ is given by Faraday's law of induction: $e = -N \frac{d\phi}{dt} = -N \frac{A \cos 	heta (B_2 - B_1)}{\Delta t}$.Given:Number of turns $(N)$ = $500$Area $(A)$ = $100 \, cm^2 = 100 	imes 10^{-4} \, m^2 = 10^{-2} \, m^2$Initial magnetic field $(B_1)$ = $0.1 \, Wb/m^2$Final magnetic field $(B_2)$ = $0 \, Wb/m^2$Time interval $(\Delta t)$ = $0.1 \, s$Angle $(	heta)$ = $0^\circ$ (since the field is perpendicular to the coil, the normal to the coil is parallel to the field).Substituting the values:$e = - \frac{500 	imes (0 - 0.1) 	imes 10^{-2} 	imes \cos 0^\circ}{0.1}$$e = - \frac{500 	imes (-0.1) 	imes 10^{-2}}{0.1} = 5 \, V$.Thus, the magnitude of the induced $e.m.f.$ is $5 \, V$.

$A$ coil of area $100 \, cm^2$ has $500$ turns. $A$ magnetic field of $0.1 \, Wb/m^2$ is perpendicular to the coil. The field is reduced to zero in $0.1 \, s$. The induced $e.m.f.$ in the coil is $...... \, V$.

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