A simple pendulum made of mass 10 g and a metallic wire of length 10… | JEE Main 2026 (23 Jan, Shift 1)

Q11JEE Main 2026 · 23 JAN · SHIFT 1Difficult

Motional EMF

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 $g = 10\text{ m/s}^2$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 100

Speed at the lowest point: $v = \sqrt{2gl(1 - \cos 60^\circ)} = \sqrt{2 \times 10 \times 0.1 \times 0.5} = 1\text{ m/s}$
Maximum angular speed $\omega = \dfrac{v}{l} = \dfrac{1}{0.1} = 10\text{ rad/s}$
Wire rotating about one end: $e = \tfrac{1}{2}B\omega l^2 = \tfrac{1}{2} \times 2 \times 10 \times (0.1)^2$
$e = 0.1\text{ V} = 100\text{ mV}$

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Motional EMF question from JEE Main 2026 (23 Jan, Shift 1)

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