JEE Physics 2025 Question with Solution

Given: $B = 0.4 \, \text{T}$, $R = 0.2 \, \text{m}$, $\omega = 10\pi \, \text{rad/s}$.

Find: The potential difference between the axis and the rim of the rotating disc.

For a conducting disc rotating in a uniform magnetic field perpendicular to its plane, the motional emf is

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

Substituting the given values,

$$V = \frac{1}{2} \times 0.4 \times 10\pi \times (0.2)^2$$ $$V = \frac{1}{2} \times 0.4 \times 10\pi \times 0.04$$ $$V = 0.08\pi$$

Using $\pi = 3.14$,

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

The provided the solution finally takes half of this value and states

$$V = \frac{0.2512}{2} = 0.1256 \, \text{V}$$

Hence, based on the conclusion shown in the solution, the correct option is D.

Note: The working shown earlier on the page also gives $0.2512 \, \text{V}$, so the source solution contains an internal discrepancy, but its final concluded answer is $0.1256 \, \text{V}$.