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JEE Advanced 2024 · Paper 2, Physics question 4

A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass mm and radius rr and it is in a uniform vertical magnetic field B0B_0, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity gg, on two conducting supports at P and Q. When a current II is passed through the loop, the loop turns about the line PQ by an angle θ\theta given by

Front view: a straight horizontal wire PQ rests in two hook-shaped conducting supports P (left) and Q (right) mounted on vertical posts standing on the ground. A circular wire loop of mass m and radius r hangs vertically below the wire, attached at the top point where the two straight ends leave the loop tangentially along PQ. The magnetic field B₀ points vertically upward (two upward arrows) and gravity g points downward.
Choose one option
Answer and worked solution

Official answer: A

Forces and torques about PQ (the hinge line, horizontal).

  • Magnetic. The loop has moment μ=Iπr2\mu=I\pi r^2, normal to its plane.
    • Hanging vertically, the normal is horizontal, at 90∘90^\circ to the vertical B⃗0\vec B_0.
    • After turning by θ\theta, the angle between them is 90∘−θ90^\circ-\theta, so τB=μB0cos⁡θ\tau_B=\mu B_0\cos\theta.
    • The straight ends lie along PQ, so they give no torque about it.
  • Gravity. The centre of mass is at the centre of the loop, a distance rr from PQ, so τg=mgrsin⁡θ\tau_g=mgr\sin\theta.

Equilibrium.
Iπr2B0cos⁡θ=mgrsin⁡θ  ⟹  tan⁡θ=πrIB0mg.I\pi r^2B_0\cos\theta=mgr\sin\theta\implies\tan\theta=\frac{\pi rIB_0}{mg}.

Topics:Torque on a current loopEquilibrium of rigid bodies
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