Two particles, each of mass m and charge q, are attached to the two ends of a light rigid rod of len

Q: Why is the equivalent current $2q/T$ and not just $q/T$?

Because there are two charges, each of magnitude $q$, rotating in the same circle. In one complete revolution (time $T$), both charges pass through any given cross-section, making the total charge transferred $2q$.

Q: Can we use the gyromagnetic ratio directly to solve this?

Yes! For any system where the mass and charge are distributed in the exact same way (uniform charge-to-mass ratio), the ratio of magnetic moment to angular momentum is always $q/2m$. Using this fact saves a lot of time in exams.

Q: Why is the area $A = \pi R^2$ and not $2\pi R^2$ since there are two charges?

The area $A$ in the magnetic moment formula $M = iA$ refers to the geometric area of the current loop itself, not the number of charges. Both charges move along the same circular path of radius $R$, so they form a single current loop of area $\pi R^2$.

Q: What if the axis of rotation was at one end of the rod instead of the center?

If the axis was at one end, one charge would be stationary (contributing nothing to $M$ or $L$), and the other would rotate in a circle of radius $2R$. The ratio $M/L$ would still be $q/2m$!

Q: Does the mass of the rod affect the answer?

The problem states it's a 'light rigid rod', meaning its mass is negligible. If the rod had mass but no charge, the angular momentum $L$ would increase, but the magnetic moment $M$ would stay the same, changing the ratio.

Question

Two particles, each of mass $m$ and charge $q$, are attached to the two ends of a light rigid rod of length $2R$. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is

Options:

(A) $\frac{q}{2m}$

(B) $\frac{q}{m}$

(C) $\frac{2q}{m}$

(D) $\frac{q}{\pi m}$

Sigma Percentile · Accepted Answer

### The Rotating Charges

Imagine you are looking at a light rigid rod, spinning smoothly around its center. At each end, there is a particle carrying a charge $q$ and a mass $m$.

This isn't just a mechanical system; it's an electromagnetic one! When charges move, they create an electric current.

To find this equivalent current, we need to ask: how much charge passes a specific point in one complete revolution?

Since there are two charges, a total charge of $\Delta q = 2q$ passes by. The time it takes for one revolution is the time period, $T = \frac{2\pi}{\omega}$.

Therefore, the equivalent current $i$ is:
$$i = \frac{\Delta q}{\Delta t} = \frac{2q}{2\pi/\omega} = \frac{q\omega}{\pi}$$

### Magnetic Moment: The Current Loop

Now that we have a current flowing in a circle, it acts exactly like a current-carrying loop.

Every current loop has a magnetic moment, given by the product of the current and the area it encloses.

The charges are moving in a circle of radius $R$, so the area of our loop is $A = \pi R^2$.

Let's calculate the magnetic moment $M$:
$$M = iA = \left(\frac{q\omega}{\pi}ight)(\pi R^2) = q\omega R^2$$

### Angular Momentum: The Mechanics

Let's switch gears to pure mechanics. The system is rotating, so it possesses angular momentum.

Angular momentum $L$ is the product of the moment of inertia $I$ and the angular velocity $\omega$.

Our system consists of two point masses, each at a distance $R$ from the axis of rotation.

The total moment of inertia is the sum of their individual inertias:
$$I = mR^2 + mR^2 = 2mR^2$$

So, the angular momentum is:
$$L = I\omega = (2mR^2)\omega$$

### The Grand Finale: Gyromagnetic Ratio

We have both pieces of the puzzle. Now, we just need to find their ratio.

Let's divide the magnetic moment by the angular momentum:
$$\frac{M}{L} = \frac{q\omega R^2}{2mR^2\omega}$$

Notice how beautifully the $\omega$ and $R^2$ terms cancel out!

$$\frac{M}{L} = \frac{q}{2m}$$

**This is a profound result.** It tells us that for any system where the charge and mass are distributed uniformly, the ratio of its magnetic moment to its angular momentum is always a constant: $\frac{q}{2m}$.

This constant is known as the **Gyromagnetic Ratio**. Remembering this principle can save you precious time in competitive exams!

Select Answer:

Visualized Solution

$System Setup$

$Equivalent Current$

$Magnetic Moment$

$Angular Momentum$

$Ratio \frac{M}{L}$

$Gyromagnetic Ratio$

The Sigma Insight: Magnetic Moment of Current Loop

The Rotating Charges

Magnetic Moment

Angular Momentum

The Grand Finale

Similar Questions

A particle of charge $q$ and mass $m$ moves in a circular orbit of radius $r$ with angular speed $ω$ . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

An insulating thin rod of length $l$ has a linear charge density $ρ (x) = ρ_{0} \frac{x}{l}$ on it. The rod is rotated about an axis passing through the origin ( $x = 0$ ) and perpendicular to the rod. If the rod makes $n$ rotations per second, then the time averaged magnetic moment of the rod is

A thin circular disk of radius $R$ is uniformly charged with density $σ > 0$ per unit area. The disk rotates about its axis with a uniform angular speed $ω$ . The magnetic moment of the disk is

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 of length $L$ metre carrying a current $i$ ampere is bent in the form of circle. The magnitude of its magnetic moment is ...... in MKS units.

The dipole moment of a circular loop carrying a current $I$ , is $m$ and the magnetic field at the centre of the loop is $B_{1}$ . When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $B_{2}$ . The ratio $\frac{B _{1}}{B _{2}}$ 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

Select Answer:

Visualized Solution

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

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

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

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

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

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