JEE Main 2025 — Electromagnetic Induction Question with Solution

To determine the potential difference between the center and the rim of the disc (often called a Faraday disc or unipolar generator), we use the formula for motional EMF in a rotating disc:

$$V=\frac{1}{2}B\omega R^2$$

where:

$$B$$ is the magnetic field strength,

$$\omega$$ is the angular velocity,

$$R$$ is the radius of the disc.

Given:

$$B=0.4\text{T}$$,

$$\omega =10\pi \text{rad/s}$$,

$$R=20\text{cm}=0.2\text{m}$$,

we substitute these values into the formula:

First calculate $$R^2$$:

$$R^2=(0.2\text{m}{)}^2=0.04{\text{m}}^2$$

Substitute into the formula:

$$V=\frac{1}{2}\times 0.4\times (10\pi )\times 0.04$$

Multiply step-by-step:

$$0.4\times 10\pi =4\pi$$,

Then, $$4\pi \times 0.04=0.16\pi$$,

Finally, multiply by $$\frac{1}{2}$$:

$$V=\frac{1}{2}\times 0.16\pi =0.08\pi$$

Substitute $$\pi \approx 3.14$$:

$$V\approx 0.08\times 3.14=0.2512\text{V}$$

Thus, the potential difference developed between the axis and the rim is approximately $$0.2512\text{V}$$, which corresponds to Option C.