Two particles , each of mass ` m` and charge `q`, are attached to the two ends of a light rigid rod of length ` 2 R` . 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. `q//2m`
B. `q//m`
C. `2q//m`
D. `q//pim`

Explanation

To solve the problem, we need to find the ratio of the magnetic moment of the system to its angular momentum about the center of the rod. Let's break down the solution step by step.

### Step 1: Understanding the System
We have two particles, each with mass \( m \) and charge \( q \), attached to the ends of a rigid rod of length \( 2R \). The rod is rotating with a constant angular speed \( \omega \) about an axis perpendicular to the rod and passing through its center.

### Step 2: Calculate the Current
The current \( I \) generated by the rotating charges can be calculated using the formula:
\[ I = q \cdot f \]
where \( f \) is the frequency of rotation. The frequency can be expressed in terms of angular speed \( \omega \) as:
\[ f = \frac{\omega}{2\pi} \]
Thus, the current becomes:
\[ I = q \cdot \frac{\omega}{2\pi} \]

### Step 3: Calculate the Area of Rotation
Each charge moves in a circular path with radius \( R \). The area \( A \) of the circular path is given by:
\[ A = \pi R^2 \]

### Step 4: Calculate the Magnetic Moment
The magnetic moment \( \mu \) of the system can be calculated using the formula:
\[ \mu = I \cdot A \]
Substituting the values of \( I \) and \( A \):
\[ \mu = \left(q \cdot \frac{\omega}{2\pi}\right) \cdot (\pi R^2) \]
This simplifies to:
\[ \mu = \frac{q \omega R^2}{2} \]

### Step 5: Calculate the Angular Momentum
The angular momentum \( L \) of the system can be calculated using the moment of inertia \( I \) and angular speed \( \omega \):
\[ L = I \cdot \omega \]
For our system, the moment of inertia \( I \) for two point masses at a distance \( R \) from the center is:
\[ I = mR^2 + mR^2 = 2mR^2 \]
Thus, the angular momentum becomes:
\[ L = (2mR^2) \cdot \omega \]

### Step 6: Calculate the Ratio of Magnetic Moment to Angular Momentum
Now, we find the ratio of the magnetic moment \( \mu \) to the angular momentum \( L \):
\[ \text{Ratio} = \frac{\mu}{L} = \frac{\frac{q \omega R^2}{2}}{2mR^2 \omega} \]
The \( R^2 \) and \( \omega \) terms cancel out:
\[ \text{Ratio} = \frac{q}{4m} \]

### Final Answer
Thus, the ratio of the magnitudes of the magnetic moment of the system to its angular momentum about the center of the rod is:
\[ \frac{q}{4m} \]