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

A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T . The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of \(60^{\circ}\) with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ___________ mV. Take \(\left.g =10 m / s ^2\right)\)

  1. A 50
  2. B 100
  3. C 150
  4. D 200
Verified Solution

Answer & Solution

Correct Answer

(B) 100

Step-by-step Solution

Detailed explanation

\(\varepsilon_{\max }=\frac{B \omega_{\max } \ell^2}{2}\) \(\quad\) ...(1) Using energy conservation, \(\operatorname{mg} \ell\left(1-\cos 60^{\circ}\right)=\frac{1}{2}\left(m \ell^2\right) \omega_{ m }^2\) \(\omega_{ m }=\sqrt{\frac{ g }{\ell}}=10 rad / s\) From eq.(1),…
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