Two coils have mutual inductance $0.002 \mathrm{~H}$. The current changes in the first coil according to the relation $\mathrm{i}=\mathrm{i}_0 \sin \omega \mathrm{t}$, where $\mathrm{i}_0=5 \mathrm{~A}$ and $\omega=50 \pi$ rad/s. The maximum value of emf in the second coil is $\frac{\pi}{\alpha} \mathrm{~V}$. The value of $\alpha$ is _______.
Answer (integer)
2
Solution
<p>$$\begin{aligned}
& \phi=\mathrm{Mi}=\mathrm{Mi}_0 \sin \omega \mathrm{t} \\
& \mathrm{EMF}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=-0.002\left(\mathrm{i}_0 \omega \cos \omega \mathrm{t}\right) \\
& \mathrm{EMF}_{\max }=\mathrm{i}_0 \omega(0.002)=(5)(50 \pi)(0.002) \\
& \mathrm{EMF}_{\max }=\frac{\pi}{2} \mathrm{~V}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Electromagnetic Induction · Topic: Self and Mutual Inductance
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