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

Options

A. ` omega and q `
B. ` omega , q and m `
C. ` q and m `
D. ` omega and m `

Explanation

To solve the problem, we need to find the ratio of the magnetic moment (μ) to the angular momentum (L) of a charged particle moving in a circular orbit. Let's go through the steps systematically.

### Step 1: Understand the definitions
- **Magnetic Moment (μ)**: For a charged particle moving in a circular path, the magnetic moment is given by:
\[ \mu = I \cdot A \]
where \(I\) is the current and \(A\) is the area of the circular path.

- **Angular Momentum (L)**: The angular momentum of a particle moving in a circular path is given by:
\[ L = m \cdot v \cdot r \]
where \(m\) is the mass of the particle, \(v\) is its linear velocity, and \(r\) is the radius of the circular path.

### Step 2: Calculate the current (I)
The current \(I\) due to a charge \(q\) moving in a circular path can be expressed as:
\[ I = \frac{q}{T} \]
where \(T\) is the time period of one complete revolution.

### Step 3: Find the time period (T)
The time period \(T\) can be calculated using the relationship between linear velocity and angular velocity:
\[ v = r \cdot \omega \]
The time period \(T\) is given by:
\[ T = \frac{2\pi r}{v} = \frac{2\pi r}{r \cdot \omega} = \frac{2\pi}{\omega} \]

### Step 4: Substitute \(T\) into the expression for current (I)
Now substituting \(T\) back into the expression for current:
\[ I = \frac{q}{T} = \frac{q \cdot \omega}{2\pi} \]

### Step 5: Calculate the area (A)
The area \(A\) of the circular path is:
\[ A = \pi r^2 \]

### Step 6: Calculate the magnetic moment (μ)
Now substituting \(I\) and \(A\) into the expression for magnetic moment:
\[ \mu = I \cdot A = \left(\frac{q \cdot \omega}{2\pi}\right) \cdot (\pi r^2) = \frac{q \cdot \omega \cdot r^2}{2} \]

### Step 7: Calculate the angular momentum (L)
Now, we can calculate the angular momentum:
\[ L = m \cdot v \cdot r = m \cdot (r \cdot \omega) \cdot r = m \cdot r^2 \cdot \omega \]

### Step 8: Find the ratio of magnetic moment to angular momentum
Now, we find the ratio \(\frac{\mu}{L}\):
\[ \frac{\mu}{L} = \frac{\frac{q \cdot \omega \cdot r^2}{2}}{m \cdot r^2 \cdot \omega} \]
The \(r^2\) and \(\omega\) terms cancel out:
\[ \frac{\mu}{L} = \frac{q}{2m} \]

### Conclusion
The ratio of the magnitude of the magnetic moment to that of the angular momentum depends on the charge \(q\) and the mass \(m\) of the particle. Thus, the final answer is:
\[ \frac{\mu}{L} = \frac{q}{2m} \]