Two coils have a mutual inductance of 0.004 | vedclass

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(D) The induced e.m.f. in the second coil is given by the formula $|e_s| = M \frac{dI}{dt}$.Given $I = I_0 \sin \omega t$,we differentiate with respec

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(D) The induced e.m.f. in the second coil is given by the formula $|e_s| = M \frac{dI}{dt}$.Given $I = I_0 \sin \omega t$,we differentiate with respect to time $t$:$\frac{dI}{dt} = I_0 \omega \cos \omega t$.Substituting this into the e.m.f. equation:$|e_s| = M I_0 \omega \cos \omega t$.The maximum value of the induced e.m.f. occurs when $\cos \omega t = 1$,so $|e_s|_{\max} = M I_0 \omega$.Substituting the given values: $M = 0.004 \ H$,$I_0 = 10 \ A$,and $\omega = 50 \pi \ rad \ s^{-1}$:$|e_s|_{\max} = 0.004 	imes 10 	imes 50 \pi = 2 \pi \ V$.

Two coils have a mutual inductance of $0.004 \ H$. The current changes in the first coil according to the equation $I = I_0 \sin \omega t$,where $I_0 = 10 \ A$ and $\omega = 50 \pi \ rad \ s^{-1}$. The maximum value of e.m.f. in the second coil in volt is (in $\pi$)

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