Two coils have mutual inductance 0.002 H. Th | vedclass

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(C) The magnetic flux linked with the second coil is given by $\phi = Mi = M i_0 \sin \omega t$. The induced $emf$ in the second coil is given by Fara

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$2$

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(C) The magnetic flux linked with the second coil is given by $\phi = Mi = M i_0 \sin \omega t$. The induced $emf$ in the second coil is given by Faraday's law: $emf = -\frac{d\phi}{dt} = -M \frac{di}{dt}$. Substituting the given expression for current: $emf = -M \frac{d}{dt}(i_0 \sin \omega t) = -M i_0 \omega \cos \omega t$. The maximum value of the induced $emf$ is $|emf|_{max} = M i_0 \omega$. Given $M = 0.002 \ H$,$i_0 = 5 \ A$,and $\omega = 50 \pi \ rad/s$: $|emf|_{max} = (0.002) 	imes (5) 	imes (50 \pi) = 0.01 	imes 50 \pi = 0.5 \pi = \frac{\pi}{2} \ V$. Comparing this with $\frac{\pi}{\alpha} \ V$,we get $\alpha = 2$.

Two coils have mutual inductance $0.002 \ H$. The current changes in the first coil according to the relation $i = i_0 \sin \omega t$,where $i_0 = 5 \ A$ and $\omega = 50 \pi \ rad/s$. The maximum value of $emf$ in the second coil is $\frac{\pi}{\alpha} \ V$. The value of $\alpha$ is . . . . . . .

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