A rod of length $l$ rotates with a small but uniform angular velocity $\omega$ about its perpendicul

Question

A rod of length $l$ rotates with a small but uniform angular velocity $\omega$ about its perpendicular bisector. A uniform magnetic field $B$ exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is

Professor PC Thomas Classes · Answer

$d \varepsilon=B \omega x d x$\ $\varepsilon=\int d \varepsilon=\int_{0}^{l/2} B \omega x d x=\frac{1}{8} B \omega l^2$