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Physics · Electromagnetic Induction

JEE Main 2026 — 23 January, Morning Shift — Question 44

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∘60^{\circ} with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ____\_\_\_\_ mV.(\mathrm{mV} .\left(\right. Take g=10 m/s2)\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

εmax =Bωmax ℓ22\varepsilon_{\text {max }}=\frac{\mathrm{B} \omega_{\text {max }} \ell^{2}}{2}

Using energy conservation, mg⁡ℓ(1−cos⁡60∘)=12( mℓ2)ωm2\operatorname{mg} \ell\left(1-\cos 60^{\circ}\right)=\frac{1}{2}\left(\mathrm{~m} \ell^{2}\right) \omega_{\mathrm{m}}^{2} ωm=gℓ=10rad/s\omega_{\mathrm{m}}=\sqrt{\frac{\mathrm{g}}{\ell}}=10 \mathrm{rad} / \mathrm{s} From eq.(1), εmax =2×10×0.012=0.1 V\varepsilon_{\text {max }}=\frac{2 \times 10 \times 0.01}{2}=0.1 \mathrm{~V} =100mV=100 \mathrm{mV}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Motional EMF

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