JEE Main 2026 Apr 5 Shift 2, Physics Q11: Motional EMF

JEE Main2026Apr 5, Shift 2Physics

A metal rod of length $L$ rotates about one end at the 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:

A
$B_{0} ω [\frac{1}{λ ^{2}} - e^{- λ L} (\frac{1}{λ ^{2}} + \frac{L}{λ})]$
B
$B_{0} ω [\frac{1}{λ ^{2}} + e^{- λ L} (\frac{1}{λ ^{2}} + \frac{L}{λ})]$
C
$B_{0} ω [\frac{4}{λ ^{2}} - e^{- 2 λ L} (\frac{1}{λ ^{2}} + \frac{2 L}{λ})]$
D
$B_{0} ω [\frac{3}{λ ^{2}} - e^{- 3 λ L} (\frac{3}{λ ^{2}} + \frac{L}{λ})]$

Solution

The motional emf contribution from a small element at radius $r$ moving with speed $ω r$ in field $B (r)$ is $d E = B (r) ω r d r$ .

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

Using integration by parts: $\int r e^{- λ r} d r = - \frac{r}{λ} e^{- λ r} - \frac{1}{λ ^{2}} e^{- λ r}$ .

Evaluating from $0$ to $L$ :

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

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JEE Main 2026 Apr 5 Shift 2, Physics Q11: Motional EMF

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