A metal rod of length $L$ rotates about one end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(r) = B_0e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :
Answer: (A) $B_0\omega\left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
An element $dr$ at distance $r$ moves with speed $\omega r$, so $d\varepsilon = B(r)\,\omega r\,dr$.
$$\varepsilon = B_0\omega\int_0^L re^{-\lambda r}dr = B_0\omega\left[-\frac{re^{-\lambda r}}{\lambda} - \frac{e^{-\lambda r}}{\lambda^2}\right]_0^L = B_0\omega\left[\frac{1}{\lambda^2} - e^{-\lambda L}\left(\frac{1}{\lambda^2} + \frac{L}{\lambda}\right)\right]$$
Solution by Sreeraj P, M.Sc Physics