JEE Main 2023PhysicsElectromagnetic InductionHardNumerical

JEE Main 2023 — Electromagnetic Induction Question with Solution

JEE Main 2023 (29 Jan Shift 1)

Question

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

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Show full solutionCorrect answer: 10
Correct answer
10

Step-by-step explanation

According to the Faraday's law of electromagnetic induction, the emf induced in a conducting loop is given by, e=-dϕdt, here, t is time and ϕ=BA is magnetic flux.

Here, e=dϕdt=dBAdt=Bdπr2dt

=Bπ2rdrdt=2×π×0.1×0.8×2×10-2

 ≈10 mV (rounded off)

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About this question

This is a previous-year question from JEE Main 2023, covering the Electromagnetic Induction chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.