JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Electromagnetic Induction

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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}

)

Answer: 100

Solution

Treating the bob as a point mass on the wire, energy conservation from

6 0^{\circ}

to the lowest point gives

m g ℓ (1 - cos 6 0^{\circ}) = \frac{1}{2} (m ℓ^{2}) ω^{2}

, so

ω = \frac{2 g ( 1 - cos 6 0 ^{\circ} )}{ℓ} = \frac{10}{0.1} = 10

rad/s at the lowest point. The wire sweeps like a rod rotating about the point of suspension, for which the motional emf between the pivot and the bob is

ε = \frac{1}{2} B ω ℓ^{2}

. Hence

ε_{m a x} = \frac{1}{2} \times 2 \times 10 \times (0.1)^{2} = 0.1

= 100

mV.

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Integer answer

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JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Electromagnetic Induction

Solution

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