Q74 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)
Correct answer: 100
Solution
The maximum induced EMF for a rod of length $\ell$ rotating in a magnetic field $B$ is:
$\varepsilon_{max} = \frac{B \omega_{max} \ell^2}{2} \quad \cdots(1)$
Using energy conservation to find $\omega_{max}$:
$mg\ell(1 - \cos 60°) = \frac{1}{2}(m\ell^2)\omega_m^2$
$\omega_m = \sqrt{\frac{g}{\ell}} = \sqrt{\frac{10}{0.1}} = 10 \text{ rad/s}$
From equation (1):
$\varepsilon_{max} = \frac{2 \times 10 \times 0.01}{2} = 0.1 \text{ V} = 100 \text{ mV}$
$\varepsilon_{max} = \frac{B \omega_{max} \ell^2}{2} \quad \cdots(1)$
Using energy conservation to find $\omega_{max}$:
$mg\ell(1 - \cos 60°) = \frac{1}{2}(m\ell^2)\omega_m^2$
$\omega_m = \sqrt{\frac{g}{\ell}} = \sqrt{\frac{10}{0.1}} = 10 \text{ rad/s}$
From equation (1):
$\varepsilon_{max} = \frac{2 \times 10 \times 0.01}{2} = 0.1 \text{ V} = 100 \text{ mV}$
Hence, the correct answer is 100.