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) $B_0\omega\left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
Concept. For a rotating rod, each element $dr$ moves at $\omega r$: $d\varepsilon=B(r)\,\omega r\,dr$.
Step 1. Integral. $\varepsilon=B_0\omega\displaystyle\int_0^Lre^{-\lambda r}dr$.
Step 2. Evaluate. $\displaystyle\int_0^Lre^{-\lambda r}dr=\frac{1}{\lambda^2}-e^{-\lambda L}\left(\frac{L}{\lambda}+\frac{1}{\lambda^2}\right)$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
