Q 12-06-053JEE MainJEE Main 2024 (27 Jan, Shift 1)Easy
Two coils have mutual inductance $0.002\ \text{H}$. The current changes in the first coil according to the relation $i = i_0\sin\omega t$, where $i_0 = 5\ \text{A}$ and $\omega = 50\pi\ \text{rad s}^{-1}$. The maximum value of emf in the second coil is $\dfrac{\pi}{\alpha}\ \text{V}$. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 2
$$\varepsilon = -M\frac{di}{dt} = -Mi_0\omega\cos\omega t \;\Rightarrow\; \varepsilon_{max} = Mi_0\omega$$
$$\varepsilon_{max} = 0.002\times5\times50\pi = \frac{\pi}{2}\ \text{V} \;\Rightarrow\; \alpha = 2$$
Solution by Sreeraj P, M.Sc Physics