A small circular loop of conducting wire has radius a and carries current i. It is placed in a unifo

Q: Why does the loop oscillate when rotated?

When the loop is rotated, the magnetic field exerts a restoring torque ($\tau = \vec{\mu} \times \vec{B}$) that tries to align the magnetic moment back with the magnetic field. This restoring torque causes the loop to oscillate back and forth, similar to a pendulum.

Q: Why do we use the moment of inertia about the diameter and not the center?

The problem explicitly states that the loop is 'rotated slightly about its diameter'. Therefore, the axis of rotation is the diameter, and we must use the moment of inertia corresponding to this specific axis ($I = \frac{1}{2}ma^2$), not the central axis ($I = ma^2$).

Q: What is the magnetic dipole moment of a current loop?

The magnetic dipole moment ($\mu$) of a planar current loop is the product of the current ($i$) flowing through it and the area ($A$) enclosed by the loop. Its direction is given by the right-hand rule.

Q: How do we know the motion is Simple Harmonic Motion (SHM)?

For small angles of rotation $\theta$, the restoring torque $\tau = -\mu B \sin\theta$ can be approximated as $\tau \approx -\mu B \theta$. Since the torque is directly proportional to the angular displacement and directed oppositely, the motion qualifies as Simple Harmonic Motion.

Q: What happens if the magnetic field is not uniform?

If the magnetic field is non-uniform, the loop would experience a net translational force in addition to the torque. The motion would be a complex combination of translation and rotation, rather than pure simple harmonic oscillation about a fixed axis.

Question

A small circular loop of conducting wire has radius $a$ and carries current $i$. It is placed in a uniform magnetic field $B$ perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period $T$. If the mass of the loop is $m$, then

Options:

(A) $T = \sqrt{\frac{\pi m}{iB}}$

(B) $T = \sqrt{\frac{2m}{iB}}$

(C) $T = \sqrt{\frac{2\pi m}{iB}}$

(D) $T = \sqrt{\frac{\pi m}{2iB}}$

Sigma Percentile · Accepted Answer

### The Physics of the Oscillating Loop

Imagine a circular conducting loop placed in a uniform magnetic field. Initially, the magnetic field is perfectly perpendicular to the plane of the loop. In this state, the magnetic dipole moment of the loop is perfectly aligned with the magnetic field, and the system is in stable equilibrium.

Now, we give the loop a slight rotation about its diameter and release it. What happens? The magnetic field exerts a restoring torque on the loop, trying to pull it back to its equilibrium position. Because of the loop's inertia, it overshoots the equilibrium and starts performing Simple Harmonic Motion (SHM)!

### The Master Equation

To find the time period of this oscillation, we can rely on the standard formula for a magnetic dipole oscillating in a uniform magnetic field. The time period $T$ is given by:

$$T = 2\pi \sqrt{\frac{I}{\mu B}}$$

Here, $I$ represents the moment of inertia of the loop about its axis of rotation, and $\mu$ is its magnetic dipole moment. Our goal is to find expressions for $I$ and $\mu$ and substitute them into this master equation.

### Calculating Magnetic Moment and Inertia

First, let's determine the magnetic dipole moment, $\mu$. For any current-carrying loop, the magnetic moment is simply the product of the current $i$ and the area $A$ enclosed by the loop. Since our loop is a circle of radius $a$, its area is $\pi a^2$. Therefore:

$$\mu = i 	imes \pi a^2$$

Next, we need the moment of inertia, $I$. The problem explicitly states that the loop is rotated about its **diameter**. This is a crucial detail! The moment of inertia of a circular ring of mass $m$ and radius $a$ about its diameter is half of its mass times the radius squared:

$$I = \frac{1}{2}ma^2$$

### The Final Calculation

Now, let's substitute these values back into our time period formula. We replace $I$ and $\mu$ with the expressions we just found:

$$T = 2\pi \sqrt{\frac{\frac{1}{2}ma^2}{(i\pi a^2)B}}$$

Let's simplify this expression. Notice how the $a^2$ terms in the numerator and the denominator cancel each other out perfectly. The factor of $\frac{1}{2}$ in the numerator moves the $2$ down to the denominator:

$$T = 2\pi \sqrt{\frac{m}{2\pi i B}}$$

To match the given options, let's bring the $2\pi$ outside the square root inside. When it goes inside, it becomes $4\pi^2$. Now, we can cancel one $2\pi$ from the numerator and denominator:

$$T = \sqrt{\frac{4\pi^2 m}{2\pi i B}}$$

$$T = \sqrt{\frac{2\pi m}{iB}}$$

This leaves us with our final, elegant answer, which perfectly matches option (c).

Select Answer:

Visualized Solution

Visualizing the Setup

The Master Equation

Calculating Magnetic Moment

Calculating Moment of Inertia

Substituting Values

Simplifying the Expression

Final Answer

The Sigma Insight: Magnetic Moment of Current Loop

The Physics of the Oscillating Loop

The Master Equation

Calculating Magnetic Moment and Inertia

The Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Setup

The Master Equation

Calculating Magnetic Moment

Calculating Moment of Inertia

Substituting Values

Simplifying the Expression

Final Answer

The Sigma Insight: Magnetic Moment of Current Loop

The Physics of the Oscillating Loop

The Master Equation

Calculating Magnetic Moment and Inertia

The Final Calculation

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