JEE Physics 2024 Question with Solution

Given: Mutual inductance is $M = 0.002 \, \text{H}$. Current in the first coil is $i = i_0 \sin \omega t$ with $i_0 = 5 \, \text{A}$ and $\omega = 50\pi \, \text{rad/s}$.

Find: The value of $\alpha$ when the maximum induced emf is $\epsilon_{\max} = \frac{\pi}{\alpha}$.

Using Faraday's law of mutual induction,

$$\epsilon = -M \frac{di}{dt}$$

Now differentiate the current:

$$i(t) = i_0 \sin \omega t = 5 \sin(50\pi t)$$ $$\frac{di}{dt} = 5 \cdot 50\pi \cos(50\pi t) = 250\pi \cos(50\pi t)$$

The maximum value occurs when $\cos(50\pi t) = 1$, so

$$\left(\frac{di}{dt}\right)_{\max} = 250\pi$$

Hence,

$$\epsilon_{\max} = M i_0 \omega = 0.002 \times 250\pi = 0.5\pi = \frac{\pi}{2} \, \text{V}$$

Given that

$$\epsilon_{\max} = \frac{\pi}{\alpha}$$

Equating,

$$\frac\pi2 = \frac{\pi}{\alpha}$$

So,

$$\alpha = 2$$

Therefore, the required numerical value is $2$.

Given: $M = 0.002 \, \text{H}$, $i_0 = 5 \, \text{A}$, and $\omega = 50\pi \, \text{rad/s}$.

Find: $\alpha$ from $\epsilon_{\max} = \frac{\pi}{\alpha}$.

For $i = i_0 \sin \omega t$, the maximum value of $\frac{di}{dt}$ is $i_0\omega$. Hence,

$$\epsilon_{\max} = M i_0 \omega$$

Substitute the values:

$$\epsilon_{\max} = 0.002 \times 5 \times 50\pi = \frac{\pi}{2}$$

So,

$$\frac{\pi}{\alpha} = \frac{\pi}{2}$$

which gives

$$\alpha = 2$$

Therefore, the correct numerical answer is $2$.