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 \, \text{A} $ and $ \omega = 50\pi \, \text{rad/s} $. The maximum value of emf in the second coil is $ \frac{\pi}{\alpha} $. The value of $ \alpha $ is ______.

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The mutual inductance between two coils is given as $M = 0.002 \, \text{H}$. The current $i$ in the first coil changes according to $i = i_0 \sin \omega t$, where $i_0 = 5 \, \text{A}$ and $\omega = 50\pi \, \text{rad/s}$. We need to determine the maximum emf ($\epsilon$) induced in the second coil and find the value of $\alpha$ when $\epsilon_{\text{max}} = \frac{\pi}{\alpha}$. According to Faraday's Law of electromagnetic induction, the emf induced in the second coil is given by: $\epsilon = -M \frac{di}{dt}$  Where $ \frac{di}{dt} $ is the rate of change of current in the first coil. Now, let's calculate $\frac{di}{dt}$: $i(t) = i_0 \sin \omega t = 5 \sin (50\pi t)$ $\frac{di}{dt} = 5 \cdot 50\pi \cos (50\pi t)$ $\frac{di}{dt} = 250\pi \cos (50\pi t)$ The maximum value of $\cos (50\pi t)$ is 1, so the maximum rate of change of current is $250\pi$. Substituting into the emf formula, we get: $\epsilon_{\text{max}} = -0.002 \cdot 250\pi = -0.5\pi$ Ignoring the negative sign, $\epsilon_{\text{max}} = 0.5\pi$. Given $\epsilon_{\text{max}} = \frac{\pi}{\alpha}$, we equate and solve: $0.5\pi = \frac{\pi}{\alpha}$ $\alpha = \frac{\pi}{0.5\pi} = 2$ The computed value of $\alpha$ is 2, which falls within the expected range [2, 2]. Thus, the value of $\alpha$ is 2 .

Correct Answer: 2

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