JEE Main 2024PhysicsElectromagnetic InductionMediumNumerical

JEE Main 2024 — Electromagnetic Induction Question with Solution

JEE Main 2024 (29 Jan Shift 1)

Question

A square loop of side 10 cm and resistance 0.7 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, the magnitude of induced emf is x×10-3 V. The value of x is _______.

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

The area vector for the square loop is given by

A→=(0.1)2 j^ m2

According to the above diagram, the magnetic field is given by

B→=0.22i^+0.22j^

Hence, the magnitude of induced emf can be calculated as follows

e=ΔϕΔt=B→·A→-01=0.22i^+0.22j^·(0.1)2 j^ -01 V=2×10-3 V

Hence, x=2.

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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.