A thin disc of radius R has charge Q distributed uniformly on its surface. The disc is rotated about one of its diametric axis with angular velocity `omega`. The magnetic moment of the arrangement is

Options

A. `(QR^(2)omega)/(8)`
B. `(QR^(2)omega)/(4)`
C. `(QR^(2)omega)/(2)`
D. Zero

Answer

To find the magnetic moment of a uniformly charged rotating disc, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a thin disc of radius \( R \) with a total charge \( Q \) distributed uniformly on its surface. The disc is rotating about one of its diametric axes with an angular velocity \( \omega \).

2. **Identify Relevant Concepts**: The magnetic moment \( \mu \) of a rotating charged body can be related to its angular momentum \( L \). The relationship is given by:
\[ \frac{\mu}{L} = \frac{Q}{2m} \]
where \( m \) is the mass of the disc.

3. **Calculate Angular Momentum \( L \)**: The angular momentum \( L \) of the disc can be calculated using the formula:
\[ L = I \omega \]
where \( I \) is the moment of inertia of the disc about the axis of rotation. For a disc rotating about its diameter, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m R^2 \]
Therefore, substituting this into the angular momentum formula gives:
\[ L = \frac{1}{2} m R^2 \omega \]

4. **Substitute Angular Momentum into the Magnetic Moment Equation**: From the relationship between magnetic moment and angular momentum, we can express the magnetic moment \( \mu \) as:
\[ \mu = \frac{Q}{2m} L \]
Substituting the expression for \( L \):
\[ \mu = \frac{Q}{2m} \left( \frac{1}{2} m R^2 \omega \right) \]

5. **Simplify the Expression**: The mass \( m \) cancels out:
\[ \mu = \frac{Q R^2 \omega}{4} \]

6. **Final Result**: Thus, the magnetic moment of the arrangement is:
\[ \mu = \frac{Q R^2 \omega}{4} \]

### Final Answer:
The magnetic moment of the arrangement is \( \mu = \frac{Q R^2 \omega}{4} \).

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