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

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(A) According to Faraday's law of electromagnetic induction, the magnitude of induced e.m.f. is given by $|e| = N \frac{|\Delta \phi|}{\Delta t}$.Here

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$0.1$

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

$A$ 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 e.m.f. induced in the coil is $0.1 \,V$, when it is removed from the field in time $t$. The value of $t$ is (in seconds): (in $s$)

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