Electromagnetic Induction

JEE Main level, numerical answer, medium

69 of 153

Question 69 of 153

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∘60^{\circ} with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ____\_\_\_\_ mV. (Take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2})

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Answer: 100

The maximum induced EMF for a rod of length ℓ\ell rotating in a magnetic field BB is:
εmax=Bωmaxℓ22⋯(1)\varepsilon_{max} = \frac{B \omega_{max} \ell^2}{2} \quad \cdots(1)

Using energy conservation to find ωmax\omega_{max}:
mgℓ(1−cos⁡60°)=12(mℓ2)ωm2mg\ell(1 - \cos 60°) = \frac{1}{2}(m\ell^2)\omega_m^2

ωm=gℓ=100.1=10 rad/s\omega_m = \sqrt{\frac{g}{\ell}} = \sqrt{\frac{10}{0.1}} = 10 \text{ rad/s}

From equation (1):
εmax=2×10×0.012=0.1 V=100 mV\varepsilon_{max} = \frac{2 \times 10 \times 0.01}{2} = 0.1 \text{ V} = 100 \text{ mV}

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