A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentia

Q: Why is the axis of rotation PQ and not the center of the loop?

The problem states the loop is suspended on two conducting supports at P and Q. These supports act as a hinge, so the entire loop swings like a pendulum about the horizontal line connecting P and Q.

Q: How do we know the direction of the magnetic moment?

The magnetic moment vector is always perpendicular to the area of the current loop. If the loop's plane is tilted by $\theta$ from the vertical, its normal (the magnetic moment) is tilted by $\theta$ from the horizontal.

Q: Why is the lever arm for gravity $r \sin\theta$?

In the side view, the center of mass is at a distance $r$ from the hinge. When the loop swings by an angle $\theta$, the horizontal displacement of the center of mass from the vertical line passing through the hinge is exactly $r \sin\theta$.

Q: Why doesn't the magnetic field pull the loop upwards?

A uniform magnetic field exerts a net torque on a current loop, but the net force is always zero. It only causes rotation, not translation. Gravity is the only force providing a net downward pull, which is balanced by the normal forces at the supports.

Q: What happens if we increase the mass of the loop?

Looking at our final formula, $\tan\theta$ is inversely proportional to the mass $m$. A heavier loop will have a stronger gravitational restoring torque, meaning it won't swing as much for the same amount of current. The angle $\theta$ will be smaller.

Question

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 $	heta$ given by

Options:

(A) $	an 	heta = \frac{\pi r I B_0}{mg}$

(B) $	an 	heta = \frac{2\pi r I B_0}{mg}$

(C) $	an 	heta = \frac{\pi r I B_0}{2mg}$

(D) $	an 	heta = \frac{mg}{\pi r I B_0}$

Sigma Percentile · Accepted Answer

### The Setup: A Delicate Balance

Imagine a circular metal loop hanging freely from a horizontal axis, much like a pendulum. In its natural state, gravity pulls it straight down, and it rests perfectly vertically. The center of mass of this circular loop is located exactly at its geometric center, a distance $r$ below the hinge axis $PQ$.

But physics gets interesting when we introduce electricity and magnetism. We pass a current $I$ through the loop and immerse the entire setup in a uniform, vertical magnetic field $\vec{B}_0$. Suddenly, the loop is no longer just a piece of metal; it becomes a magnetic dipole, and it begins to swing!

### The Geometry of the Swing

To truly understand what's happening, we need to look at the system from a side view, looking directly along the axis $PQ$. As the loop swings and settles at a new equilibrium angle $	heta$ with the vertical, two competing torques are at play.

First, let's analyze the restoring torque due to gravity. When the loop tilts by an angle $	heta$, its center of mass shifts horizontally. The perpendicular distance from the hinge to the line of action of the gravitational force $mg$ is the lever arm, which is geometrically $r \sin	heta$. Therefore, the gravitational torque trying to pull the loop back down is:

$$	au_g = mg(r \sin	heta)$$

### Enter the Magnetic Torque

Now, let's look at the magnetic deflecting torque. A current-carrying loop has a magnetic moment $\vec{\mu}$, whose magnitude is the product of the current and the area:

$$\mu = I \cdot \pi r^2$$

This magnetic moment vector points perpendicular to the plane of the loop. If the loop is tilted by an angle $	heta$ from the vertical, its normal vector (the magnetic moment) is tilted by the exact same angle $	heta$ from the horizontal.

Since the external magnetic field $\vec{B}_0$ is perfectly vertical, the angle between the magnetic moment $\vec{\mu}$ and the magnetic field $\vec{B}_0$ is $90^\circ - 	heta$. The magnetic torque is given by the cross product $\vec{\mu} 	imes \vec{B}_0$, which yields:

$$	au_B = \mu B_0 \sin(90^\circ - 	heta) = \mu B_0 \cos	heta$$

### The Battle of Torques

At the new equilibrium position, the loop is perfectly stationary. This means the restoring torque from gravity exactly balances the deflecting torque from the magnetic field. We can set our two torque equations equal to each other:

$$mg r \sin	heta = (I \pi r^2) B_0 \cos	heta$$

Now, the physics is done, and it's time for some elegant algebra. We want to isolate the angle $	heta$. By dividing both sides by $\cos	heta$ and $mg r$, we group the trigonometric terms on the left and the physical constants on the right:

$$\frac{\sin	heta}{\cos	heta} = \frac{I \pi r^2 B_0}{mg r}$$

Simplifying the fraction, we arrive at our beautiful final result:

$$	an	heta = \frac{\pi r I B_0}{mg}$$

This equation tells a compelling story: the tangent of the deflection angle is directly proportional to the current and the magnetic field strength, and inversely proportional to the mass of the loop. A stronger magnet pushes it higher, while a heavier loop resists the swing!

Select Answer:

Visualized Solution

\text{Equilibrium of the Loop}

\text{Restoring Torque due to Gravity}

\text{Magnetic Moment of the Loop}

\text{Deflecting Magnetic Torque}

\text{Equating the Torques}

\text{Solving for } \tan\theta

\text{Final Expression}

\text{Food for Thought}

The Sigma Insight: Magnetic Moment of Current Loop

The Setup

The Geometry of the Swing

Enter the Magnetic Torque

The Battle of Torques

Similar Questions

An infinitely long current-carrying wire and a small current-carrying loop are in the plane of the paper as shown. The radius of the loop is $a$ and distance of its centre from the wire is $d$ ( $d >> a$ ). If the loop applies a force $F$ on the wire, then

A charged particle going around in a circle can be considered to be a current loop. A particle of mass $m$ carrying charge $q$ is moving in a plane with speed $v$ under the influence of magnetic field $B$ . The magnetic moment of this moving particle

A wire carrying current $I$ is bent in the shape ABCDEFA as shown, where rectangle ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths $a$ and $b$ , then the magnitude and direction of magnetic moment of the loop ABCDEFA is

A circular coil having $N$ turns and radius $r$ carries a current $I$ . It is held in the XZ-plane in a magnetic field $B \hat{i}$ . The torque on the coil due to the magnetic field (in N-m) is

A loop carrying current $I$ lies in the $x$ - $y$ plane as shown in the figure. The unit vector $\hat{k}$ is coming out of the plane of the paper. The magnetic moment of the current loop is