Electromagnetic Induction

JEE Main level, single correct, medium

68 of 153

Question 68 of 153

A metal rod of length LL rotates about one end at origin with a uniform angular velocity ω\omega. The magnetic field radially falls off as B(r)=B0e−λrB(r) = B_0 e^{-\lambda r}; λ\lambda being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :

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Answer: (A) B0ω[1λ2−e−λL(1λ2+Lλ)]B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]

The motional emf dEdE induced in a small element of length drdr at a distance rr from the origin is given by:

dE=B(r)v(r)drdE = B(r) v(r) dr

Since the rod is rotating with angular velocity ω\omega, the velocity of the element is v(r)=ωrv(r) = \omega r.

dE=B0e−λr(ωr)drdE = B_0 e^{-\lambda r} (\omega r) dr

The total emf induced EE is the integral of dEdE from r=0r = 0 to r=Lr = L:

E=∫0LB0ωre−λrdrE = \int_0^L B_0 \omega r e^{-\lambda r} dr

E=B0ω∫0Lre−λrdrE = B_0 \omega \int_0^L r e^{-\lambda r} dr

Using integration by parts:

∫re−λrdr=r(e−λr−λ)−∫1⋅(e−λr−λ)dr\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

=−rλe−λr−1λ2e−λr= -\dfrac{r}{\lambda} e^{-\lambda r} - \dfrac{1}{\lambda^2} e^{-\lambda r}

Evaluating the definite integral from 00 to LL:

∫0Lre−λrdr=[−Lλe−λL−1λ2e−λL]−[0−1λ2e0]\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]

=1λ2−e−λL(1λ2+Lλ)= \dfrac{1}{\lambda^2} - e^{-\lambda L} \left( \dfrac{1}{\lambda^2} + \dfrac{L}{\lambda} \right)

Therefore, the total induced emf is:

E=B0ω[1λ2−e−λL(1λ2+Lλ)]E = B_0 \omega \left[ \dfrac{1}{\lambda^2} - e^{-\lambda L} \left( \dfrac{1}{\lambda^2} + \dfrac{L}{\lambda} \right) \right]

Answer: B0ω[1λ2−e−λL(1λ2+Lλ)]B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]

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