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(C) Given:Number of turns $N = 20$Area $A = 25 \, cm^2 = 25 	imes 10^{-4} \, m^2$Resistance $R = 100 \, \Omega$Rate of change of magnetic field $\fra

Vedclass · Accepted Answer

$0.5$

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(C) Given:Number of turns $N = 20$Area $A = 25 \, cm^2 = 25 	imes 10^{-4} \, m^2$Resistance $R = 100 \, \Omega$Rate of change of magnetic field $\frac{dB}{dt} = 1000 \, T/s$Since the magnetic field is perpendicular to the plane of the coil,the angle $	heta$ between the area vector and the magnetic field is $0^\circ$.According to Faraday's law,the induced electromotive force $(EMF)$ is given by $e = -N \frac{d\phi}{dt} = -N A \cos 	heta \frac{dB}{dt}$.The magnitude of the induced $EMF$ is $|e| = N A \frac{dB}{dt} \cos 0^\circ = N A \frac{dB}{dt}$.Substituting the values:$|e| = 20 	imes (25 	imes 10^{-4}) 	imes 1000 = 20 	imes 25 	imes 10^{-1} = 50 	imes 10^{-1} = 5 \, V$.The current $i$ in the coil is given by $i = \frac{|e|}{R}$.$i = \frac{5}{100} = 0.05 \, A$. Wait,recalculating: $20 	imes 25 	imes 10^{-4} 	imes 1000 = 20 	imes 25 	imes 10^{-1} = 50 	imes 0.1 = 5 \, V$. Then $i = 5 / 100 = 0.05 \, A$. Re-evaluating the provided solution: The provided solution states $0.5 \, A$. Let's check: $20/100 * 1000 * 25 * 10^{-4} = 0.2 * 1000 * 0.0025 = 200 * 0.0025 = 0.5 \, A$. The calculation $20 	imes 25 	imes 10^{-4} 	imes 1000 = 500000 	imes 10^{-4} = 50$. $50 / 100 = 0.5 \, A$. Correct.

$A$ rectangular coil of $20$ turns and an area of cross-section $25 \, cm^2$ has a resistance of $100 \, \Omega$. If a magnetic field which is perpendicular to the plane of the coil changes at the rate of $1000 \, T/s$,the current in the coil is $....... \, A$.

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