Electromagnetic Induction

JEE Main level, numerical answer, medium

16 of 153

Question 16 of 153

A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B=0.8TB=0.8T. When released the radius of the loop starts shrinking at a constant rate of 2cms–12cm{s}^{–1}. The induced emf in the loop at an instant when the radius of the loop is 10cm10cm will be ______ mVmV.

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Answer: 10

According to the Faraday's law of electromagnetic induction, the emf induced in a conducting loop is given by, e=−dϕdte=-\dfrac{d\phi }{dt}, here, tt is time and ϕ=BA\phi =BA is magnetic flux.
Here, e=∣dϕdt∣=d(BA)dt=Bd(πr2)dte=\left|\dfrac{d\phi }{dt}\right|=\dfrac{d\left(BA\right)}{dt}=B\dfrac{d\left(\pi {r}^{2}\right)}{dt}
=Bπ(2rdrdt)=2×π×0.1×0.8×2×10−2=B\pi \left(2r\dfrac{dr}{dt}\right) =2\times \pi \times 0.1\times 0.8\times 2\times {10}^{-2}
≈10mV\approx 10mV (rounded off)

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