A metal rod of length $L$ rotates about on end at origin with a uniform angular velocity $\omega$. The magnetic field radially falls off as $B(r)=B_o e^{-\lambda r}$; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:

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

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

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

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

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Electromagnetic Induction

A metal rod of length

L

rotates about on end at origin with a uniform angular velocity

ω

. The magnetic field radially falls off as

B (r) = B_{o} e^{- λ r}

;

λ

being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:

(A) $B_{o} ω [\frac{1}{λ ^{2}} - e^{- λ L} (\frac{1}{λ ^{2}} + \frac{L}{λ})]$ ✓ Correct
(B) $B_{o} ω [\frac{1}{λ ^{2}} + e^{- λ L} (\frac{1}{λ ^{2}} + \frac{L}{λ})]$
(C) $B_{o} ω [\frac{4}{λ ^{2}} - e^{- 2 λ L} (\frac{1}{λ ^{2}} + \frac{2 L}{λ})]$
(D) $B_{o} ω [\frac{3}{λ ^{2}} - e^{- 3 λ L} (\frac{3}{λ ^{2}} + \frac{L}{λ})]$

Answer: (A)

An element of the rod at distance

r

from the axis moves with speed

ω r

in a field

B (r) = B_{0} e^{- λ r}

, contributing an emf

d ε = B (r) (ω r) d r = B_{0} ω r e^{- λ r} d r

. The total emf is

ε = B_{0} ω \int_{0}^{L} r e^{- λ r} d r

. Integrating by parts,

\int_{0}^{L} r e^{- λ r} d r = [- \frac{r}{λ} e^{- λ r} - \frac{1}{λ ^{2}} e^{- λ r}]_{0}^{L} = \frac{1}{λ ^{2}} - e^{- λ L} (\frac{1}{λ ^{2}} + \frac{L}{λ})

. The induced emf is

B_{0} ω [\frac{1}{λ ^{2}} - e^{- λ L} (\frac{1}{λ ^{2}} + \frac{L}{λ})]

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option