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 \(\left.g =10 m / s ^2\right)\)
- A 50
- B 100
- C 150
- D 200
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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