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) = Bo e-λ r; λ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is: (a) Boω[(1)/(λ²) - e-λ L((1)/(λ²) + (L)/(λ))] (b) Boω[(1)/(λ²) + e-λ L((1)/(λ²) + (L)/(λ))] (c) Boω[(4)/(λ²) - e-2λ L((1)/(λ²) + (2L)/(λ))] (d) B₀ω[(3)/(λ²) - e-3λ L((3)/(λ²) + (L)/(λ))]

Answer: (a). Element dr at distance r moves with speed ω r: de = B(r) ω r dr. e = Boω∫₀L r e-λ rdr Integrating by parts: ∫₀L re-λ rdr = (1)/(λ²) - e-λ L((L)/(λ) + (1)/(λ²)) So the answer is (a) Boω[(1)/(λ²) - e-λ L((1)/(λ²) + (L)/(λ))].

A metal rod of length L rotates about one end at origin with a… | JEE Main 2026 (5 Apr, Shift 2)

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Rotating Conductors & AC Generator

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

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Answer: (a)

Element $dr$ at distance $r$ moves with speed $\omega r$ : $de = B(r)\,\omega r\,dr$ .
$e = B_o\omega\int_0^L r e^{-\lambda r}dr$
Integrating by parts: $\int_0^L re^{-\lambda r}dr = \frac{1}{\lambda^2} - e^{-\lambda L}\left(\frac{L}{\lambda} + \frac{1}{\lambda^2}\right)$
So the answer is (a) $B_o\omega\left[\frac{1}{\lambda^2} - e^{-\lambda L}\left(\frac{1}{\lambda^2} + \frac{L}{\lambda}\right)\right]$ .

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