JEE Main 2024PhysicsElectromagnetic InductionMediumMCQ

JEE Main 2024 — Electromagnetic Induction Question with Solution

JEE Main 2024 (27 Jan Shift 1)

Question

A rectangular loop of length 2.5 m and width 2 m is placed at 60° to a magnetic field of 4 T. The loop is removed from the field in 10 sec. The average emf induced in the loop during this time is

Choose an option

Show full solutionCorrect option: C
Correct answer
C+1 V

Step-by-step explanation

The formula to calculate the change in the magnetic flux associated with the given loop can be written as

∆ϕ=-BAcosθ=-Blbcosθ   ...1

The formula to calculate the induced emf is given by

ε=-∆ϕ∆t=Blbcosθ∆t   ...2

From equation (2), it follows that

ε=4×2.5×2×cos60°10 V=1 V

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About this question

This is a previous-year question from JEE Main 2024, 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.