Q 12-06-042JEE MainJEE Main 2025 (28 Jan, Shift 2)Easy
A uniform magnetic field of $0.4\ \text{T}$ acts perpendicular to a circular copper disc $20\ \text{cm}$ in radius. The disc has a uniform angular velocity of $10\pi\ \text{rad s}^{-1}$ about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? ($\pi = 3.14$)
Answer: (C) $0.2512\ \text{V}$
Each radius of the disc acts like a rod rotating about one end:
$$\varepsilon = \frac{1}{2}B\omega R^2 = \frac{1}{2}\times0.4\times10\pi\times(0.2)^2 = 0.08\pi = 0.08\times3.14 = 0.2512\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics