JEE Main 2026 Jan 23 Shift 1, Physics Q21: Motional EMF

JEE Main2026Jan 23, Shift 1Physics

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 $6 0^{\circ}$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ______ mV. (Take $g = 10 m/s^{2}$ )

Your answer

Solution

For a rotating rod about one end in a perpendicular field, the EMF across its length is:

$ε = \frac{B ω ℓ ^{2}}{2}$ , maximum when $ω$ is maximum.

Maximum angular speed from energy conservation (released from $6 0^{\circ}$ ):

$m g ℓ (1 - cos 6 0^{\circ}) = \frac{1}{2} (m ℓ^{2}) ω_{m}^{2} \Rightarrow ω_{m} = \frac{g}{ℓ} = \frac{10}{0.1} = 10 rad/s$

Maximum EMF:

$ε_{m a x} = \frac{2 \times 10 \times ( 0.1 ) ^{2}}{2} = 0.1 V = 100 mV$

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JEE Main 2026 Jan 23 Shift 1, Physics Q21: Motional EMF

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