JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2026
- Easy
- Numerical
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 \(\left.g =10 m / s ^2\right)\)
Pick an option to check your answer.
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),
\(\varepsilon_{\max }=\frac{2 \times 10 \times 0.01}{2}=0.1 V\)
\(=100 mV\)
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),
\(\varepsilon_{\max }=\frac{2 \times 10 \times 0.01}{2}=0.1 V\)
\(=100 mV\)
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