JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

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 :

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
B
$B_0 \omega \left[\dfrac{1}{\lambda^2} + e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
C
$B_0 \omega \left[\dfrac{4}{\lambda^2} - e^{-2\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{2L}{\lambda}\right)\right]$
D
$B_0 \omega \left[\dfrac{3}{\lambda^2} - e^{-3\lambda L}\left(\dfrac{3}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$

Did you get this right?

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:

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

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

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

$\int_0^L r e^{-\lambda r} dr = \left[ -\dfrac{L}{\lambda} e^{-\lambda L} - \dfrac{1}{\lambda^2} e^{-\lambda L} \right] - \left[ 0 - \dfrac{1}{\lambda^2} e^0 \right]$

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

Therefore, the total induced emf is:

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

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

Your note

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Options

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

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

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

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

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:

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

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

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

$\int_0^L r e^{-\lambda r} dr = \left[ -\dfrac{L}{\lambda} e^{-\lambda L} - \dfrac{1}{\lambda^2} e^{-\lambda L} \right] - \left[ 0 - \dfrac{1}{\lambda^2} e^0 \right]$

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

Therefore, the total induced emf is:

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

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

JEE Main 2026 — Physics Electromagnetism

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JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism