Mindyard | 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 𝑚 and radius 𝑟 and it is in a uniform vertical magnetic field 𝐵 0 , as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity 𝑔, on two conducting supports at P and Q. When a current 𝐼 is passed through the loop, the loop turns about the line PQ by an angle 𝜃 given by tan ⁡ θ = π r I B 0 / ( m g ) , tan ⁡ θ = 2 π r I B 0 / ( m g ) , tan ⁡ θ = π r I B 0 / ( 2 m g ) , tan ⁡ θ = m g / ( π r I B 0 )

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 𝑚 and radius 𝑟 and it is in a uniform vertical magnetic field 𝐵₀, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity 𝑔, on two conducting supports at P and Q. When a current 𝐼 is passed through the loop, the loop turns about the line PQ by an angle 𝜃 given by

Options

$\tan θ = π r I B_{0} / (m g)$

$\tan θ = 2 π r I B_{0} / (m g)$

$\tan θ = π r I B_{0} / (2 m g)$

$\tan θ = m g / (π r I B_{0})$

Correct Answer :

Option A

$\tan θ = π r I B_{0} / (m g)$

tan θ = π r I B_0 / (m g)

Solution :

To find the angle $θ$ by which the loop turns, we analyze the balance of torques acting on the loop about the axis of rotation (the horizontal line PQ).

From the diagram, the following components are given:
- Radius of the circular loop = $r$
- Mass of the loop = $m$
- Uniform vertical magnetic field pointing upwards = $B_{0}$
- Current through the loop = $I$

1. Magnetic Torque:
The magnetic dipole moment $\vec{M}$ of a circular loop carrying current $I$ is given by:

$M = I A = I (π r^{2})$

Initially, the loop hangs vertically downwards, meaning the plane of the loop is vertical and the magnetic moment vector $\vec{M}$ (which is perpendicular to the loop's plane) is horizontal. The magnetic field $B_{0}$ is vertical.
When the loop rotates about the line PQ by an angle $θ$ , the normal to the loop tilts by an angle $θ$ with respect to the horizontal. Therefore, the angle between the magnetic moment $\vec{M}$ and the vertical magnetic field $\vec{B_{0}}$ becomes $90 ° - θ$ .
The magnitude of the torque due to the magnetic field is:

$τ_{B} = M B_{0} \sin (90 ° - θ) = M B_{0} \cos θ$

Substituting the expression for $M$ :

$τ_{B} = π r^{2} I B_{0} \cos θ$

2. Gravitational Torque:
The center of mass of the circular loop is at its geometric center, located a distance of $r$ from the axis of rotation PQ. When the loop is tilted by an angle $θ$ , the perpendicular distance from the line PQ to the line of action of the gravitational force $m g$ is $r \sin θ$ .
The torque due to gravity about the axis PQ is:

$τ_{g} = m g r \sin θ$

3. Equilibrium:
For rotational equilibrium, the magnetic torque must balance the gravitational torque:

$τ_{B} = τ_{g}$

$π r^{2} I B_{0} \cos θ = m g r \sin θ$