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PhyHardAdv 2024 • Paper 2

Equilibrium Angle of Circular Wire Loop in Magnetic Field

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

Question Diagram 1

Options

A

tan⁡θ=πrIB0/(mg)\tan \theta = \pi r I B_0 / (m g)

Correct
B

tan⁡θ=2πrIB0/(mg)\tan \theta = 2 \pi r I B_0 / (m g)

C

tan⁡θ=πrIB0/(2mg)\tan \theta = \pi r I B_0 / (2 m g)

D

tan⁡θ=mg/(πrIB0)\tan \theta = m g / (\pi r I B_0)

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceTorque

Step-by-Step Solution

To find the angle θ\theta by which the circular loop turns about the line PQPQ, we analyze the equilibrium of torques acting on the loop about the axis of rotation PQPQ.

1. Gravitational Torque (τg\tau_g)

The circular loop has mass mm and radius rr. Since the upper tangent of the loop lies along the horizontal axis PQPQ, the center of mass of the loop is at a perpendicular distance rr from the axis PQPQ.

When the loop rotates about PQPQ by an angle θ\theta, the center of mass moves such that the line connecting PQPQ to the center of mass makes an angle θ\theta with the vertical. The magnitude of the restoring gravitational torque about the line PQPQ is: τg=mgrsin⁡θ\tau_g = m g r \sin \theta

2. Magnetic Torque (τm\tau_m)

The area vector A⃗\vec{A} of the circular loop has magnitude A=πr2A = \pi r^2 and is directed normal to the plane of the loop.

Let PQPQ be along the xx-axis and the vertical direction be along the zz-axis, so the magnetic field is B⃗0=B0k^\vec{B}_0 = B_0 \hat{k}.

Initially, when the loop hangs vertically, the normal vector points along the horizontal yy-axis (j^\hat{j}). When the loop rotates by an angle θ\theta about the xx-axis, the normal vector becomes: n^=j^cos⁡θ+k^sin⁡θ\hat{n} = \hat{j} \cos \theta + \hat{k} \sin \theta

The magnetic dipole moment of the loop is: μ⃗=IA⃗=Iπr2(j^cos⁡θ+k^sin⁡θ)\vec{\mu} = I \vec{A} = I \pi r^2 (\hat{j} \cos \theta + \hat{k} \sin \theta)

The magnetic torque acting on the loop is given by τ⃗m=μ⃗×B⃗0\vec{\tau}_m = \vec{\mu} \times \vec{B}_0: τ⃗m=Iπr2(j^cos⁡θ+k^sin⁡θ)×(B0k^)=Iπr2B0cos⁡θ i^\vec{\tau}_m = I \pi r^2 (\hat{j} \cos \theta + \hat{k} \sin \theta) \times (B_0 \hat{k}) = I \pi r^2 B_0 \cos \theta \, \hat{i}

Thus, the magnitude of the magnetic torque about the line PQPQ is: τm=Iπr2B0cos⁡θ\tau_m = I \pi r^2 B_0 \cos \theta

3. Rotational Equilibrium

For the loop to be in equilibrium at an angle θ\theta, the magnitude of the gravitational torque must equal the magnitude of the magnetic torque: τg=τm\tau_g = \tau_m mgrsin⁡θ=Iπr2B0cos⁡θm g r \sin \theta = I \pi r^2 B_0 \cos \theta

Rearranging the terms to find tan⁡θ\tan \theta: sin⁡θcos⁡θ=Iπr2B0mgr\frac{\sin \theta}{\cos \theta} = \frac{I \pi r^2 B_0}{m g r}

tan⁡θ=πrIB0mg\tan \theta = \frac{\pi r I B_0}{m g}

Thus, the correct option is (A).

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Equilibrium Angle of Circular Wire Loop in Magnetic Field | Physics PYQ Solution - JEE Challenger