A metal rod of length L rotates about one end at origin with a uniform angular velocity ω . The magnetic field radially falls off as B(r) = B 0 e -λ r λ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is:

B 0ω/λ 2[1 - e -λ L (1+λ L)]

Physics · Electromagnetic Induction

JEE Main 2026 — 5 April, Evening Shift — Question 11

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:

Option A:
$\frac{B_0\omega}{\lambda^2}[1 - e^{-\lambda L}(1+\lambda L)]$
Correct
Option B:
$\frac{B_0\omega}{\lambda^2}[1 - e^{-\lambda L}]$
Option C:
$\frac{B_0\omega}{4\lambda^2}[1 - e^{-2\lambda L}]$
Option D:
$\frac{B_0\omega}{3\lambda^2}[1 - e^{-3\lambda L}]$

Answer: A

Step-by-step solution

$e = \int_0^L B(r) \omega r dr = B_0\omega \int_0^L r e^{-\lambda r} dr = B_0\omega \left[ \frac{1}{\lambda^2} - e^{-\lambda L}\left(\frac{1}{\lambda^2}+\frac{L}{\lambda}\right)\right]$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Motional EMF

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Answer: A

Step-by-step solution

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