JEE Main 2026 · 5 Apr · Shift 2 · Physics Q36

← JEE Main 2026 · 5 Apr · Shift 2

PhysicsSingle correct+4 / -1Question 36 of 75

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 $B(r)=B_oe^{-\lambda r}$ ; $\lambda$ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

B_o\omega\left[\dfrac1{\lambda^2}-e^{-\lambda L}\left(\dfrac1{\lambda^2}+\dfrac L\lambda\right)\right]

B_o\omega\left[\dfrac1{\lambda^2}+e^{-\lambda L}\left(\dfrac1{\lambda^2}+\dfrac L\lambda\right)\right]

B_o\omega\left[\dfrac4{\lambda^2}-e^{-2\lambda L}\left(\dfrac1{\lambda^2}+\dfrac{2L}\lambda\right)\right]

B_o\omega\left[\dfrac3{\lambda^2}-e^{-3\lambda L}\left(\dfrac3{\lambda^2}+\dfrac L\lambda\right)\right]

Answer and worked solution

Official answer: A

Element of the rod. A piece dr at distance r moves at speed ωr, so its emf is $B(r)\,\omega r\,dr$ .

Integrate.
$\varepsilon=B_o\omega\int_0^Lre^{-\lambda r}\,dr=B_o\omega\left[\frac1{\lambda^2}-e^{-\lambda L}\left(\frac1{\lambda^2}+\frac L\lambda\right)\right].$

(A)

Topics:Motional EMF

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JEE Main 2026 · 5 Apr · Shift 2, Physics question 36