JEE Main 2023PhysicsMagnetic Properties of MatterHardNumerical

JEE Main 2023 — Magnetic Properties of Matter Question with Solution

JEE Main 2023 (10 Apr Shift 1)

Question

The current required to be passed through a solenoid of 15 cm length and 60 turns in order to demagnetise a bar magnet of magnetic intensity 2.4×103 A m–1 is ________ A.

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Show full solutionCorrect answer: 6
Correct answer
6

Step-by-step explanation

The magnetic intensity is

H=Bμ0-M

For M=0, it can be written

H=Bμ0=ni

The data given is 

H=2.4×103 A m-1l=15×10-2 mN=60

Using the relation, n=Nl and H=ni, the value of the current is,

i=Hn=2.4×103×15×10-260=6 A

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About this question

This is a previous-year question from JEE Main 2023, covering the Magnetic Properties of Matter 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.