JEE Main 2023PhysicsElectromagnetic InductionEasyNumerical

JEE Main 2023 — Electromagnetic Induction Question with Solution

JEE Main 2023 (12 Apr Shift 1)

Question

A conducting circular loop is placed in a uniform magnetic field of 0.4 T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1 mm s-1. The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm will be _______ μV.

Enter your answer

Show full solutionCorrect answer: 50
Correct answer
50

Step-by-step explanation

The data given is 

B=0.4 TR=2×10-2 m

It is given that dRdt=10-3 m s-1

Area of the loop is A=πR2

So, dAdt=2πRdRdt

The flux is,

ϕ=2πRdRdt×B

So, the magnitude of emf is 

ε=2×227×2×10-2×10-3×0.4

=5.028×10-5 V=50.28 μV≈50  μV

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Electromagnetic Induction chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

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.