Q61 / 75 · JEE Main 2026 (05 April Shift 2) · Physics
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 :
- Correct answerA$B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
- 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$B_0 \omega \left[\dfrac{3}{\lambda^2} - e^{-3\lambda L}\left(\dfrac{3}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
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: $B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
$dE = B(r) v(r) dr$
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: $B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]$
Hence, the correct answer is option: 'A'.