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Magnetism question

2024 · Shift 2 · Q38
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  5. /2024 · Shift 2 · Q38

Magnetism question

2024 · Shift 2 · Q38

JEE AdvancedPhysicsMagnetismMCQ+3 / −1
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 mmm and radius rrr and it is in a uniform vertical magnetic field B0B_0B0​, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity ggg, on two conducting supports at P\mathrm{P}P and Q\mathrm{Q}Q. When a current III is passed through the loop, the loop turns about the line PQ\mathrm{PQ}PQ by an angle θ\thetaθ given by JEE Advanced 2024 Paper 2 Online Physics - Magnetism Question 3 English
  1. A
    tan⁡θ=πrIB0/(mg)\tan \theta=\pi r I B_0 /(m g)tanθ=πrIB0​/(mg)
  2. B
    tan⁡θ=2πrIB0/(mg)\tan \theta=2 \pi r I B_0 /(m g)tanθ=2πrIB0​/(mg)
  3. C
    tan⁡θ=πrIB0/(2mg)\tan \theta=\pi r I B_0 /(2 m g)tanθ=πrIB0​/(2mg)
  4. D
    tan⁡θ=mg/(πrIB0)\tan \theta=m g /\left(\pi r I B_0\right)tanθ=mg/(πrIB0​)
View written solutionFree

Correct answer: A

  1. Identify the axis of rotation and the torques

The loop can rotate about the horizontal line PQPQPQ, which is tangent to the circular loop at its top point.

When current flows, the loop experiences a magnetic torque due to the uniform vertical magnetic field B0B_0B0​.

At equilibrium,

τmag=τg.\tau_{\text{mag}} = \tau_{g}.τmag​=τg​.
  1. Magnetic moment of the loop

The wire forms a complete circular loop of radius rrr, so its area is

A=πr2.A = \pi r^2.A=πr2.

Hence the magnetic moment is

μ=IA=Iπr2.\mu = I A = I\pi r^2.μ=IA=Iπr2.

The magnetic field is vertical. Initially the plane of the loop is vertical, so the magnetic moment is horizontal. After turning by angle θ\thetaθ about PQPQPQ, the angle between μ⃗\vec \muμ​ and B⃗\vec BB becomes 90∘−θ90^\circ-\theta90∘−θ.

Therefore magnetic torque is

τmag=μB0sin⁡(90∘−θ)=μB0cos⁡θ\tau_{\text{mag}} = \mu B_0 \sin(90^\circ-\theta)= \mu B_0 \cos\thetaτmag​=μB0​sin(90∘−θ)=μB0​cosθ

so

τmag=Iπr2B0cos⁡θ.\tau_{\text{mag}} = I\pi r^2 B_0\cos\theta.τmag​=Iπr2B0​cosθ.
  1. Gravitational torque about tangent axis PQPQPQ

The center of mass of the circular loop is at its geometric center, a distance rrr from the tangent axis PQPQPQ.

When the loop rotates by angle θ\thetaθ, the perpendicular distance of the weight mgmgmg from the axis becomes

rsin⁡θ.r\sin\theta.rsinθ.

Thus the gravitational torque is

τg=mgrsin⁡θ.\tau_g = mgr\sin\theta.τg​=mgrsinθ.
  1. Equilibrium condition

Set magnetic torque equal to gravitational torque:

Iπr2B0cos⁡θ=mgrsin⁡θ.I\pi r^2 B_0\cos\theta = mgr\sin\theta.Iπr2B0​cosθ=mgrsinθ.

Divide by rcos⁡θr\cos\thetarcosθ:

πrIB0=mgtan⁡θ.\pi r I B_0 = mg\tan\theta.πrIB0​=mgtanθ.

Hence,

tan⁡θ=πrIB0mg.\tan\theta = \frac{\pi r I B_0}{mg}.tanθ=mgπrIB0​​.
  1. Match with options

This corresponds to:

Option A:

tan⁡θ=πrIB0mg\tan \theta=\frac{\pi r I B_0}{m g}tanθ=mgπrIB0​​
  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.

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