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JEE Mains · Physics · STD 12 -6. Electromagnetic induction

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 :

  1. A \(B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]\)
  2. B \(B_0 \omega \left[\dfrac{1}{\lambda^2} + e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]\)
  3. C \(B_0 \omega \left[\dfrac{4}{\lambda^2} - e^{-2\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{2L}{\lambda}\right)\right]\)
  4. D \(B_0 \omega \left[\dfrac{3}{\lambda^2} - e^{-3\lambda L}\left(\dfrac{3}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(B_0 \omega \left[\dfrac{1}{\lambda^2} - e^{-\lambda L}\left(\dfrac{1}{\lambda^2} + \dfrac{L}{\lambda}\right)\right]\)

Step-by-step Solution

Detailed explanation

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 \(v(r) = \omega r\).…
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