A rod of length $60 \mathrm{~cm}$ rotates with a uniform angular velocity $20 \mathrm{rad} \mathrm{s}^{-1}$ about its perpendicular bisector, in a uniform magnetic filed $0.5 T$. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is _____V.

Solution

$\begin{aligned} & \because \mathrm{V}_0-\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{B} \omega \ell^2}{2} \\ & \mathrm{~V}_0-\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{B} \omega \ell^2}{2} \\ & \therefore \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \therefore \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=0\end{aligned}$

Hence, the correct answer is 0.