JEE Physics 2026 Question with Solution

Given: Inductance $L = 10 \, \text{mH}$, source voltage $V = 10 \, \text{V}$, internal resistance $R = 10 \, \Omega$, and the current is $\frac{1}{e}$ times its maximum value.

Find: The value of $\alpha$ in the energy density expression.

From the solution working,

$$I_{\max} = \frac{V}{R} = \frac{10}{10} = 1 \, \text{A}$$

So at the given instant,

$$I = \frac{1}{e} I_{\max} = \frac{1}{e}$$

Magnetic energy density is

$$u = \frac{B^2}{2\mu_0}$$

For a solenoid,

$$B = \mu_0 n I$$

Hence,

$$\nu = \frac{(\mu_0 n I)^2}{2\mu_0} = \frac{\mu_0 n^2 I^2}{2}$$

Using

$$L = \mu_0 n^2 A l$$

we get

$$\nu = \frac{1}{2} \frac{L I^2}{A l}$$

Substituting the given values as in the provided solution,

$$\nu = 500\pi \times \frac{1}{e^2} \, \text{J m}^{-3}$$

Comparing with $\alpha \pi \times \frac{1}{e^2} \, \text{J m}^{-3}$, we obtain $\alpha = 500$.

Therefore, the value of $\alpha$ is $500$.

Given: The coil is connected to a DC source, so the maximum current is set by resistance, and the required current is $\frac{1}{e}$ of that value.

Find: The coefficient $\alpha$.

1. Find the maximum current:

$$I_{\max} = \frac{V}{R} = \frac{10}{10} = 1 \, \text{A}$$

2. Current at the specified instant:

$$I = \frac{1}{e} I_{\max} = \frac{1}{e}$$

3. Use magnetic energy density:

$$\nu = \frac{B^2}{2\mu_0}$$

4. For a solenoid:

$$B = \mu_0 n I, \qquad n = \frac{N}{l}$$

5. Substitute for $B$:

$$\nu = \frac{(\mu_0 n I)^2}{2\mu_0} = \frac{\mu_0 n^2 I^2}{2}$$

6. Use the inductance relation:

$$L = \mu_0 n^2 A l$$

So,

$$\nu = \frac{1}{2} \frac{L I^2}{A l}$$

7. The provided solution concludes:

$$\nu = 500\pi \times \frac{1}{e^2} \, \text{J m}^{-3}$$

Hence,

$$\alpha = 500$$

Therefore, the final answer is $500$.