Question

A rod of length l rotates with a small but uniform angular velocity ω 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
(a) zero
(b) $\frac{1}{8}\mathrm{\omega }B{l}^2$
(c) $\frac{1}{2}\mathrm{\omega }B{l}^2$
(d) Bωl².

Answer 1

(b) $\frac{1}{8}\mathrm{\omega }B{l}^2$


Let us consider a small element dx at a distance x from the centre of the rod rotating with angular velocity ω about its perpendicular bisector. The emf induced in the rod because of this small element is given by
$de=Bvl=B\omega xdx$

The emf induced across the centre and end of the rod is given by
$$\begin{gathered} \int de={\int }_0^{l/2}B\omega xdx \\ \Rightarrow E=B\omega {\left[\frac{x^2}{2}\right]}_0^{l/2} \\ \Rightarrow E=\frac{1}{8}B\omega l^2 \end{gathered}$$