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Physics · Magnetism and Matter

JEE Main 2024 — 4 April, Shift 2 — Question 32

The magnetic moment of a bar magnet is 0.5Am20.5 \mathrm{Am}^{2}. It is suspended in a uniform magnetic field of 8×10−2 T8 \times 10^{-2} \mathrm{~T}. The work done in rotating it from its most stable to most unstable position is :

  1. Option A:

    16×10−2 J16 \times 10^{-2} \mathrm{~J}

  2. Option B:

    8×10−2 J8 \times 10^{-2} \mathrm{~J}

    Correct
  3. Option C:

    4×10−2 J4 \times 10^{-2} \mathrm{~J}

  4. Option D:

    Zero

Answer: B

Step-by-step solution

At stable equilibrium

U=−mBcos⁡0∘=−mB\mathrm{U}=-\mathrm{mB} \cos 0^{\circ}=-\mathrm{mB}

At unstable equilibrium

U=−mBcos⁡180∘=+mB\mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB}

W=ΔU\mathrm{W}=\Delta \mathrm{U}

W.D. =2mB=2 \mathrm{mB}

=2(0.5)8×10−2=8×10−2 J=2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{~J}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 2
Subject
Physics
Chapter
Magnetism and Matter
Topic
Natural Magnetism: Bar Magnets
The magnetic moment of a bar magnet is 0.5 Am 2 . It is suspended in… | JEE Main 2024 PYQ with Solution · DhiX AI