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.

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Solution

Q.
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 $60 °$ with the vertical, then the maximum induced EMF between the point of suspension and the point of oscillation is ______ mV. (Take $g = 10 {m/s}^{2}$ ). [2026]

Ans.
(100)

$ε_{\max} = \frac{B ω_{\max} l^{2}}{2} . . . (1)$

Using energy conservation,

$m g ℓ (1 - \cos 60 °) = \frac{1}{2} (m ℓ^{2}) ω_{m}^{2}$

$ω_{m} = \sqrt{\frac{g}{ℓ}} = 10 rad/s$

From eq.(1),

$ε_{\max} = \frac{2 \times 10 \times 0.01}{2} = 0.1 V$

$= 100 mV$

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