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Magnetic Properties of Matter question

2025 · 3 Apr · Shift 2 · Q55
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Magnetic Properties of Matter question

2025 · 3 Apr · Shift 2 · Q55

JEE MainPhysicsMagnetic Properties of MatterMCQ+4 / −1
A magnetic dipole experiences a torque of 803 N m80 \sqrt{3} \mathrm{~N} \mathrm{~m}803​ N m when placed in uniform magnetic field in such a way that dipole moment makes angle of 60∘60^{\circ}60∘ with magnetic field. The potential energy of the dipole is :
  1. A
     −403 J\text { }-40 \sqrt{3} \mathrm{~J} −403​ J
  2. B
    -80 J
  3. C
    80 J
  4. D
    -60 J
View written solutionFree

Correct answer: B

  1. Use the torque formula for a magnetic dipole

For a magnetic dipole in a uniform magnetic field,

τ=MBsin⁡θ\tau = MB\sin\thetaτ=MBsinθ

where MMM is the magnetic dipole moment, BBB is the magnetic field, and θ\thetaθ is the angle between them.

Given:

τ=803 N m,θ=60∘\tau = 80\sqrt{3}\ \text{N m}, \qquad \theta = 60^\circτ=803​ N m,θ=60∘

So,

MBsin⁡60∘=803MB\sin 60^\circ = 80\sqrt{3}MBsin60∘=803​ MB(32)=803MB\left(\frac{\sqrt{3}}{2}\right)=80\sqrt{3}MB(23​​)=803​ MB=160MB = 160MB=160
  1. Use the potential energy formula

Potential energy of a magnetic dipole in a magnetic field is

U=−MBcos⁡θU = -MB\cos\thetaU=−MBcosθ

Substitute MB=160MB=160MB=160 and θ=60∘\theta=60^\circθ=60∘:

U=−160cos⁡60∘U = -160\cos 60^\circU=−160cos60∘ U=−160(12)U = -160\left(\frac{1}{2}\right)U=−160(21​) U=−80 JU = -80\ \text{J}U=−80 J
  1. Match with the options

The correct option is:

B: −80 J\boxed{\text{B: } -80\ \text{J}}B: −80 J​
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