A particle of charge 'q' and mass 'm' move in a circular orbit of radius 'r' with frequency 'v' the ratio of the magnetic moment to angular momentum is:

Options

A. `(2qv)/(m)`
B. `(qv)/(2m)`
C. `(q)/(2mr)`
D. `(q)/(2m)`

Answer

To find the ratio of the magnetic moment (μ) to the angular momentum (L) of a charged particle moving in a circular orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- The **magnetic moment (μ)** of a charged particle moving in a circular path is given by:
\[ \mu = \frac{q \cdot I \cdot A}{c} \]
where \(I\) is the current, \(A\) is the area, and \(c\) is the speed of light. However, for a single charge moving in a circle, we can simplify it to:
\[ \mu = \frac{q \cdot v \cdot r}{2} \]
- The **angular momentum (L)** of the particle is given by:
\[ L = m \cdot v \cdot r \]

2. **Relating Frequency and Angular Velocity**:
- The frequency \(v\) is related to the angular velocity \(\omega\) by the equation:
\[ \omega = 2\pi v \]
- Therefore, the linear velocity \(v\) can be expressed as:
\[ v = \omega \cdot r \]

3. **Substituting for Magnetic Moment**:
- Substituting \(v\) in the expression for magnetic moment:
\[ \mu = \frac{q \cdot (\omega \cdot r) \cdot r}{2} = \frac{q \cdot \omega \cdot r^2}{2} \]

4. **Substituting for Angular Momentum**:
- The angular momentum can be expressed as:
\[ L = m \cdot v \cdot r = m \cdot (\omega \cdot r) \cdot r = m \cdot \omega \cdot r^2 \]

5. **Finding the Ratio of Magnetic Moment to Angular Momentum**:
- Now, we can find the ratio \(\frac{\mu}{L}\):
\[ \frac{\mu}{L} = \frac{\frac{q \cdot \omega \cdot r^2}{2}}{m \cdot \omega \cdot r^2} \]
- The \(r^2\) and \(\omega\) terms cancel out:
\[ \frac{\mu}{L} = \frac{q}{2m} \]

### Final Result:
Thus, the ratio of the magnetic moment to angular momentum is:
\[ \frac{\mu}{L} = \frac{q}{2m} \]