JEE Main 2020PhysicsMagnetic Properties of MatterHardMCQ

JEE Main 2020 — Magnetic Properties of Matter Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

A small bar magnet is placed with its axis at 30o with an external magnetic field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C7.2×10−2 J

Step-by-step explanation

τ=MBsinθ=0.018

M=0.018B sin θ=0.0180.06×0.5=0.64 Am2

W=ΔU=Uf−Ui

=−MB cos 180∘−−MB cos0∘

=2MB

=2×0.6×0.06

=0.072 J

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About this question

This is a previous-year question from JEE Main 2020, 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.