A conducting rod of length $L$ rotates with angular velocity $\omega$ about one end in a uniform magnetic field $B$ perpendicular to the plane of rotation. The induced emf between the ends is?

Solution

1. Consider a small segment of the rod of length $dr$ at a distance $r$ from the pivot point.
2. The velocity of this segment is $v=\omega r$, so the induced emf across the segment is $$\begin{array}{rl}dϵ& =Bvdr\\ & =B\omega rdr\end{array}$$.
3. Integrate this expression from $r=0$ to $r=L$ to find the total induced emf: $$\begin{array}{rl}ϵ& ={\int }_0^{L}B\omega rdr\\ & =B\omega {\left[\frac{r^2}{2}\right]}_0^L\end{array}$$.
4. Evaluating the integral yields $ϵ=\frac{1}{2}B\omega L^2$.
5. Hence the answer is (B).