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Magnetic Properties of Matter question

2024 · 4 Apr · Shift 2 · Q66
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  5. /2024 · 4 Apr · Shift 2 · Q66

Magnetic Properties of Matter question

2024 · 4 Apr · Shift 2 · Q66

JEE MainPhysicsMagnetic Properties of MatterMCQ+4 / −1
The magnetic moment of a bar magnet is 0.5 Am20.5 \mathrm{~Am}^20.5 Am2. It is suspended in a uniform magnetic field of 8×10−2 T8 \times 10^{-2} \mathrm{~T}8×10−2 T. The work done in rotating it from its most stable to most unstable position is:
  1. A
    4×10−2 J4 \times 10^{-2} \mathrm{~J}4×10−2 J
  2. B
    16×10−2 J16 \times 10^{-2} \mathrm{~J}16×10−2 J
  3. C
    8×10−2 J8 \times 10^{-2} \mathrm{~J}8×10−2 J
  4. D
    Zero
View written solutionFree

Correct answer: C

  1. Potential energy of a magnetic dipole in a uniform magnetic field

A bar magnet with magnetic moment mmm placed in a uniform magnetic field BBB has potential energy

U=−mBcos⁡θU=-mB\cos\thetaU=−mBcosθ

where θ\thetaθ is the angle between m⃗\vec mm and B⃗\vec BB.

  1. Identify stable and unstable positions

  • Most stable position: θ=0∘\theta=0^\circθ=0∘ Umin⁡=−mBU_{\min}=-mBUmin​=−mB

  • Most unstable position: θ=180∘\theta=180^\circθ=180∘ Umax⁡=+mBU_{\max}=+mBUmax​=+mB

  1. Work done to rotate from stable to unstable position

The external work done equals the increase in potential energy:

W=Umax⁡−Umin⁡W=U_{\max}-U_{\min}W=Umax​−Umin​

So,

W=(+mB)−(−mB)=2mBW=(+mB)-(-mB)=2mBW=(+mB)−(−mB)=2mB

  1. Substitute the given values

Given:

m=0.5 A m2,B=8×10−2 Tm=0.5\,\text{A m}^2, \qquad B=8\times 10^{-2}\,\text{T}m=0.5A m2,B=8×10−2T

Therefore,

W=2×0.5×8×10−2W=2\times 0.5 \times 8\times 10^{-2}W=2×0.5×8×10−2

W=1×8×10−2W=1\times 8\times 10^{-2}W=1×8×10−2

W=8×10−2 JW=8\times 10^{-2}\,\text{J}W=8×10−2J

  1. Match with the options

8×10−2 J8\times 10^{-2}\,\text{J}8×10−2J corresponds to Option C.

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