A magnetic field of 2 times 10^{-2} , T acts | vedclass

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Question

(A) The induced emf is given by Faraday's law: $|e| = N \frac{|\Delta \Phi|}{\Delta t}$.Here,$N = 50$,$A = 100 \, cm^2 = 100 	imes 10^{-4} \, m^2 = 1

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(A) The induced emf is given by Faraday's law: $|e| = N \frac{|\Delta \Phi|}{\Delta t}$.Here,$N = 50$,$A = 100 \, cm^2 = 100 	imes 10^{-4} \, m^2 = 10^{-2} \, m^2$,$B_1 = 2 	imes 10^{-2} \, T$,$B_2 = 0 \, T$,and $e = 0.1 \, V$.The change in magnetic flux is $\Delta \Phi = A(B_2 - B_1) = 10^{-2} 	imes (0 - 2 	imes 10^{-2}) = -2 	imes 10^{-4} \, Wb$.The magnitude of induced emf is $|e| = N \frac{|\Delta \Phi|}{t}$.Substituting the values: $0.1 = 50 	imes \frac{2 	imes 10^{-4}}{t}$.$t = \frac{50 	imes 2 	imes 10^{-4}}{0.1} = \frac{100 	imes 10^{-4}}{0.1} = \frac{10^{-2}}{10^{-1}} = 0.1 \, s$.

$A$ magnetic field of $2 \times 10^{-2} \, T$ acts at right angles to a coil of area $100 \, cm^2$ with $50$ turns. The average emf induced in the coil is $0.1 \, V$,when it is removed from the field in time $t$. The value of $t$ is $... \, sec$.

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