A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure,

Q: Why is the angle between the magnetic moment and the magnetic field 90 degrees?

The magnetic field is given as parallel to the plane of the coil. The magnetic moment vector, by definition, is always perpendicular to the plane of the coil. Therefore, the angle between a vector in the plane and a vector perpendicular to the plane is exactly $90^\circ$.

Q: Why doesn't the coil rotate while the capacitor is discharging?

The problem specifies that the discharge happens in a 'very short time'. Due to the mechanical inertia of the coil, it cannot instantly change its position. It only acquires angular momentum (a 'kick') during this infinitesimal time, and begins to visibly rotate only after the discharge is complete.

Q: What is the physical meaning of the integral of current over time?

Current $i$ is the rate of flow of charge, $dq/dt$. When you integrate $i$ with respect to time ($dt$), you are summing up all the tiny packets of charge $dq$ that flowed. The total sum is the total charge $Q$ that left the capacitor.

Q: Does the resistance of the coil affect the final angular momentum?

Surprisingly, no! A higher resistance would make the discharge take slightly longer and lower the peak current, but the total area under the current-time graph ($\int i dt$) will always equal the total charge $Q$. Since the angular momentum depends only on this integral, it is independent of the resistance.

Q: How is this setup different from a standard electric motor?

A standard motor receives a continuous supply of current, resulting in a continuous torque that drives sustained rotation. This setup receives a single, instantaneous burst of charge, resulting in an angular impulse that causes the coil to swing once, much like a ballistic pendulum.

Question

A circular coil of radius $R$ and $N$ turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge $Q$ through a switch. The coil is in a horizontal uniform magnetic field $B_o$ parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time)-

Options:

(A) $\frac{\pi}{2} N Q B_o R^2$

(B) $\pi N Q B_o R^2$

(C) $2\pi N Q B_o R^2$

(D) $4\pi N Q B_o R^2$

Sigma Percentile · Accepted Answer

This problem is a beautiful intersection of electromagnetism and rotational mechanics. It perfectly illustrates the concept of an **angular impulse** delivered by a transient magnetic torque. Let's break down the physics step-by-step.

### Analyzing the Setup

We are given a circular coil of radius $R$ and $N$ turns, hanging vertically. A uniform magnetic field $B_o$ is applied horizontally, meaning it is perfectly parallel to the plane of the coil.

When the switch is closed, the capacitor discharges, sending a sudden, short-lived surge of current $i$ through the coil. This current transforms the coil into a magnetic dipole. The magnetic moment $M$ of this dipole is given by the product of the number of turns, the current, and the area of the coil:

$$M = N i A = N i \pi R^2$$

By the right-hand rule, the direction of this magnetic moment vector $\mathbf{M}$ is perpendicular to the plane of the coil.

### The Master Equation: Magnetic Torque

Whenever a magnetic dipole is placed in an external magnetic field, it experiences a torque. This torque is mathematically defined as the cross product of the magnetic moment and the magnetic field:

$$\boldsymbol{	au} = \mathbf{M} 	imes \mathbf{B_o}$$

Because the coil is in a vertical plane and the magnetic field is horizontal and parallel to that plane, the angle $	heta$ between the magnetic moment vector $\mathbf{M}$ (which is perpendicular to the plane) and the magnetic field $\mathbf{B_o}$ is exactly $90^\circ$.

Therefore, the magnitude of the torque simplifies beautifully:

$$	au = M B_o \sin(90^\circ) = M B_o = (N i \pi R^2) B_o$$

### The Angular Impulse

The problem states that the discharge happens in a "very short time," meaning the coil doesn't have time to rotate while the current is flowing. Instead, it receives a sharp rotational "kick"—an **angular impulse**.

From Newton's Second Law for rotation, torque is the rate of change of angular momentum:

$$	au = \frac{dL}{dt} \implies dL = 	au dt$$

To find the total angular momentum $\Delta L$ gained by the coil, we integrate this torque over the duration of the discharge:

$$\Delta L = \int 	au dt = \int (N i \pi R^2 B_o) dt$$

### Final Calculation

Notice that $N$, $\pi$, $R^2$, and $B_o$ are all constants. We can pull them out of the integral, leaving us with the integral of the current with respect to time:

$$\Delta L = N \pi R^2 B_o \int i dt$$

What is $\int i dt$? By definition, current is the rate of flow of charge ($i = \frac{dq}{dt}$). Therefore, the integral of current over the entire discharge time is simply the total charge $Q$ that was initially stored in the capacitor!

Substituting $\int i dt = Q$ into our equation, we arrive at the final expression for the angular momentum gained:

$$\Delta L = \pi N Q B_o R^2$$

This elegant result is the foundational principle behind a **ballistic galvanometer**, an instrument designed to measure the quantity of charge passing through it by observing the maximum deflection of a suspended coil.

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Visualized Solution

$Visualizing the Setup$

$Magnetic Moment of the Coil$

$Torque on the Coil$

$Angular Impulse$

$Integrating the Current$

$Final Angular Momentum$

$The Way Forward$

The Sigma Insight: Magnetic Moment of Current Loop

Analyzing the Setup

The Master Equation

The Angular Impulse

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Moment of the Coil

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque on the Coil

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Impulse

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrating the Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Angular Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward