JEE Main 2023PhysicsElectromagnetic InductionMediumMCQ

JEE Main 2023 — Electromagnetic Induction Question with Solution

JEE Main 2023 (24 Jan Shift 1)

Question

A conducting loop of radius 10π cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is:

Choose an option

Show full solutionCorrect option: B
Correct answer
Bemf=10mV

Step-by-step explanation

Given: B=0.5 T  at t=0 & B=0, at t=0.5 s.

Magnetic field will decrease linearly as given below. We can say, value of magnetic field at t=0.25 will be B=0.25.

Therefore,

εind =ΔϕΔt=ΔBAΔt=AΔBΔt

=π×10π2×10-4×0.250.25

=10-2×1

=0.01 V=10 mV

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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.