JEE Physics 2025 Question with Solution

Given: An infinitely long wire has linear charge density $\lambda = 2 \times 10^{-9} \, \text{C/m}$. The cube has side length $\sqrt{3} \, \text{cm}$ and the wire passes through two maximally displaced corners, so it lies along the body diagonal of the cube.

Find: The net electric flux through the cube.

Using Gauss's law,

$$\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}$$

The length of wire inside the cube is the body diagonal. For a cube of side $a$, body diagonal is $a\sqrt{3}$. Here $a = \sqrt{3} \, \text{cm}$, so

$$l = a\sqrt{3} = \sqrt{3} \times \sqrt{3} \, \text{cm} = 3 \, \text{cm} = 0.03 \, \text{m}$$

Therefore, enclosed charge is

$$Q_{\text{enc}} = \lambda l = 2 \times 10^{-9} \times 0.03 = 6 \times 10^{-11} \, \text{C}$$

Now,

$$\Phi = \frac{6 \times 10^{-11}}{8.85 \times 10^{-12}} \approx 6.78 \, \text{N m}^2\text{/C}$$

Since $2.16\pi \approx 6.79$, the flux matches $2.16\pi \, \text{N m}^2\text{/C}$.

Therefore, the correct option is A.

Given: The wire passes through two corners of the cube that are maximally displaced from each other.

Find: The segment length of the wire enclosed by the cube.

The two maximally displaced corners of a cube are opposite vertices. The straight line joining them is the body diagonal.

If the side of the cube is $a = \sqrt{3} \, \text{cm}$, then the body diagonal is

$$\sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$$

Substituting $a = \sqrt{3} \, \text{cm}$,

$$l = \sqrt{3} \cdot \sqrt{3} \, \text{cm} = 3 \, \text{cm}$$

This is the wire length enclosed inside the cube. Then Gauss's law gives the same result:

$$\Phi = \frac{\lambda l}{\varepsilon_0} = \frac{2 \times 10^{-9} \times 0.03}{8.85 \times 10^{-12}} \approx 2.16\pi$$

Hence, the correct option is A.