JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Electromagnetic Induction

The emf developed across a rod rotating about one end builds up along its length because each part moves at a different speed. An element at distance $x$ from the pivot moves with speed $v=\omega x$, so it contributes $d\epsilon =Bvdx=B\omega xdx$. Adding the contributions from the pivot to the free end, $\epsilon ={\int }_0^{L}B\omega xdx=B\omega \frac{L^2}{2}$. The induced emf across the rotating rod is $\frac{1}{2}B{L}^2\omega$.