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 magnitudes of the magnetic moment of the system to its angular momentum about the center of the rod. Let's break it down 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 about a perpendicular axis through its center.

### Step 2: Calculate the Magnetic Moment
The magnetic moment \( M \) of a system can be calculated using the formula:
\[ M = I \cdot A \]
where \( I \) is the current and \( A \) is the area swept by the charges.

1. **Calculate the Current \( I \)**:
The total charge \( Q \) that is effectively moving is \( 2q \) (since there are two charges). The time period \( T \) for one complete rotation is given by:
\[ T = \frac{2\pi}{\omega} \]
Therefore, the current \( I \) is:
\[ I = \frac{\text{Total charge}}{\text{Time period}} = \frac{2q}{T} = \frac{2q \cdot \omega}{2\pi} = \frac{q \omega}{\pi} \]

2. **Calculate the Area \( A \)**:
The area swept by the charges as they rotate is a circle with radius \( R \):
\[ A = \pi R^2 \]

3. **Calculate the Magnetic Moment \( M \)**:
Substituting \( I \) and \( A \) into the magnetic moment formula:
\[ M = I \cdot A = \left(\frac{q \omega}{\pi}\right) \cdot \left(\pi R^2\right) = q \omega R^2 \]

### Step 3: Calculate the Angular Momentum
The angular momentum \( L \) of the system can be calculated using the formula:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia of the system.

1. **Calculate the Moment of Inertia \( I \)**:
The moment of inertia \( I \) for two point masses at a distance \( R \) from the center is:
\[ I = mR^2 + mR^2 = 2mR^2 \]

2. **Calculate the Angular Momentum \( L \)**:
Now substituting \( I \) into the angular momentum formula:
\[ L = (2mR^2) \cdot \omega \]

### Step 4: Calculate the Ratio of Magnetic Moment to Angular Momentum
Now, we can find the ratio \( \frac{M}{L} \):
\[ \frac{M}{L} = \frac{q \omega R^2}{2mR^2 \omega} \]
Cancelling \( \omega \) and \( R^2 \) from the numerator and denominator, we get:
\[ \frac{M}{L} = \frac{q}{2m} \]

### 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:
\[ \frac{M}{L} = \frac{q}{2m} \]

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