Rotating Conductors & AC Generator question from MHT CET 2026 (15 May, Shift 1)

Answer: (c)

1. For a rod rotating about a pivot, the emf between the pivot and a point at distance $r$ is $\frac{1}{2}B\omega r^2$.
2. With $\vec B$ into the page and counter-clockwise rotation, $q\vec v \times \vec B$ pushes positive charge towards O, so O is at the higher potential.
3. $V_O - V_A = \frac{1}{2}B\omega L^2$, $V_O - V_C = \frac{1}{2}B\omega(3L)^2 = \frac{9}{2}B\omega L^2$
4. $V_A - V_C = (V_O - V_C) - (V_O - V_A) = \frac{9}{2}B\omega L^2 - \frac{1}{2}B\omega L^2 = 4B\omega L^2$
5. So the answer is (c) $V_A - V_C = 4B\omega L^2$.