A coil of circular cross-section having 1000 | vedclass

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Question

(A) Given: Number of turns $N = 1000$.Face area $A = 4 \, cm^2 = 4 	imes 10^{-4} \, m^2$.Change in magnetic field $\Delta B = 10^{-2} \, Wb \, m^{-2}

Vedclass · Accepted Answer

$400$

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(A) Given: Number of turns $N = 1000$.Face area $A = 4 \, cm^2 = 4 	imes 10^{-4} \, m^2$.Change in magnetic field $\Delta B = 10^{-2} \, Wb \, m^{-2}$.Time interval $\Delta t = 0.01 \, s = 10^{-2} \, s$.The induced $e.m.f.$ is given by Faraday's law: $e = N \frac{\Delta \phi}{\Delta t} = N A \frac{\Delta B}{\Delta t} \cos 	heta$.Since the axis is parallel to the magnetic field,the angle $	heta$ between the area vector and the magnetic field is $0^\circ$,so $\cos 0^\circ = 1$.Substituting the values: $e = 1000 	imes (4 	imes 10^{-4} \, m^2) 	imes \frac{10^{-2} \, Wb \, m^{-2}}{10^{-2} \, s}$.$e = 1000 	imes 4 	imes 10^{-4} = 0.4 \, V$.Converting to $mV$: $0.4 \, V = 400 \, mV$.

$A$ coil of circular cross-section having $1000$ turns and $4 \, cm^2$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{-2} \, Wb \, m^{-2}$ in $0.01 \, s$. The $e.m.f.$ induced in the coil is....$mV$

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