JEE Main 2026 — Electromagnetic Induction Question with Solution

The motional emf $dE$ induced in a small element of length $dr$ at a distance $r$ from the origin is given by:

$dE=B(r)v(r)dr$

Since the rod is rotating with angular velocity $\omega$, the velocity of the element is $v(r)=\omega r$.

$dE=B_0{e}^{-\lambda r}(\omega r)dr$

The total emf induced $E$ is the integral of $dE$ from $r=0$ to $r=L$:

$E={\int }_0^L{B}_0\omega r{e}^{-\lambda r}dr$

$E=B_0\omega {\int }_0^{L}r{e}^{-\lambda r}dr$

Using integration by parts:

$\int r{e}^{-\lambda r}dr=r\left(\frac{e^{-\lambda r}}{-\lambda }\right)-\int 1\cdot \left(\frac{e^{-\lambda r}}{-\lambda }\right)dr$

$=-\frac{r}{\lambda }{e}^{-\lambda r}-\frac{1}{{\lambda }^2}{e}^{-\lambda r}$

Evaluating the definite integral from $0$ to $L$:

${\int }_0^{L}r{e}^{-\lambda r}dr=\left[-\frac{L}{\lambda }{e}^{-\lambda L}-\frac{1}{{\lambda }^2}{e}^{-\lambda L}\right]-\left[0-\frac{1}{{\lambda }^2}{e}^0\right]$

$=\frac{1}{{\lambda }^2}-e^{-\lambda L}\left(\frac{1}{{\lambda }^2}+\frac{L}{\lambda }\right)$

Therefore, the total induced emf is:

$E=B_0\omega \left[\frac{1}{{\lambda }^2}-e^{-\lambda L}\left(\frac{1}{{\lambda }^2}+\frac{L}{\lambda }\right)\right]$

Answer: $B_0\omega \left[\frac{1}{{\lambda }^2}-e^{-\lambda L}\left(\frac{1}{{\lambda }^2}+\frac{L}{\lambda }\right)\right]$