JEE Main 2026 (5 Apr, Shift 2) Physics Q36: Faraday's Law

Question

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 $B(r)=B_0\,e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

Accepted Answer

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

Answer

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

Answer

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

Answer

$B_0\omega\left[\frac{4}{\lambda^2}-e^{-2\lambda L}\left(\frac{1}{\lambda^2}+\frac{2L}{\lambda}\right)\right]$

Answer

$B_0\omega\left[\frac{3}{\lambda^2}-e^{-3\lambda L}\left(\frac{3}{\lambda^2}+\frac{L}{\lambda}\right)\right]$

JEE Main 5 April 2026, Shift 2, Physics Q36

Same idea in other shifts