Two coils have a mutual inductance of 0.005 | vedclass

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(B) Given: Mutual inductance $M = 0.005 \ H$,current $I = I_0 \sin \omega t$,$I_0 = 10 \ A$,and $\omega = 100 \pi \ rad/s$.The induced emf $(e)$ in th

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$5 \pi \ V$

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(B) Given: Mutual inductance $M = 0.005 \ H$,current $I = I_0 \sin \omega t$,$I_0 = 10 \ A$,and $\omega = 100 \pi \ rad/s$.The induced emf $(e)$ in the second coil is given by the formula:$e = M \frac{dI}{dt}$Substituting the expression for current:$e = M \frac{d}{dt} (I_0 \sin \omega t) = M I_0 \omega \cos \omega t$The maximum value of the induced emf $(e_{\max})$ occurs when $\cos \omega t = 1$:$e_{\max} = M I_0 \omega$Substituting the given values:$e_{\max} = 0.005 	imes 10 	imes 100 \pi$$e_{\max} = 0.05 	imes 100 \pi = 5 \pi \ V$Therefore,the maximum induced emf is $5 \pi \ V$.

Two coils have a mutual inductance of $0.005 \ H$. The current changes in the first coil according to the equation $I = I_0 \sin \omega t$,where $I_0 = 10 \ A$ and $\omega = 100 \pi \ rad/s$. The maximum value of the induced emf in the second coil is:

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