Electromagnetic Induction

JEE Main level, numerical answer, medium

83 of 153

Question 83 of 153

A conducting circular loop is placed in a uniform magnetic field of 0.4T0.4T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1mms−11mm{s}^{-1}. The magnitude of induced emf in the loop at an instant when the radius of the loop is 2cm2cm will be _______ μV\mu V.

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Answer: 50

The data given is
B=0.4TR=2×10−2mB=0.4T R=2\times {10}^{-2}m
It is given that dRdt=10−3ms−1\dfrac{dR}{dt}={10}^{-3}m{s}^{-1}
Area of the loop is A=πR2A=\pi {R}^{2}
So, dAdt=2πRdRdt\dfrac{dA}{dt}=2\pi R\dfrac{dR}{dt}
The flux is,
ϕ=2πRdRdt×B\phi =2\pi R\dfrac{dR}{dt}\times B
So, the magnitude of emf is
ϵ=2×227×2×10−2×10−3×0.4\epsilon =2\times \dfrac{22}{7}\times 2\times {10}^{-2}\times {10}^{-3}\times 0.4
=5.028×10−5V=50.28μV≈50μV=5.028\times {10}^{-5}V=50.28\mu V\approx 50\mu V

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