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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Magnetic Properties of Matter

JEE Main 2024 · 4 April · Evening Shift · Physics · Magnetic Properties of Matter

The magnetic moment of a bar magnet is 0.5 Am^2 . It is suspended...

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:

Options

  1. A8×10−2 J8 \times 10^{-2} \mathrm{~J}
  2. B4×10−2 J4 \times 10^{-2} \mathrm{~J}
  3. CZero
  4. D16×10−2 J16 \times 10^{-2} \mathrm{~J}

Correct answer

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

Step-by-step solution

At stable equilibrium U=−mBcos⁡0∘=−mBAtunstableequilibrium\mathrm{U}=-\mathrm{mB} \cos 0^{\circ}=-\mathrm{mB} At unstable equilibriumU=−mBcos⁡180∘=+mBW=ΔU W.D. =2mB=2(0.5)8×10−2=8×10−2J\begin{aligned} & \mathrm{U}=-\mathrm{mB} \cos 180^{\circ}=+\mathrm{mB} \\ & \mathrm{W}=\Delta \mathrm{U} \\ & \text{ W.D. }=2 \mathrm{mB} \\ & =2(0.5) 8 \times 10^{-2}=8 \times 10^{-2} \mathrm{ J} \end{aligned}$$
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