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

Options

A. w and q
B. w, q and m
C. q and m
D. w and m

Answer

To solve the problem, we need to find the ratio of the magnitude of the magnetic moment (μ) to that of the angular momentum (L) of a particle of charge \( q \) and mass \( m \) moving in a circular orbit of radius \( r \) with angular speed \( \omega \).

### Step-by-Step Solution:

1. **Calculate Angular Momentum (L)**:
The angular momentum \( L \) of a particle moving in a circular path is given by the formula:
\[ L = mvr \]
where \( v \) is the linear velocity of the particle. The linear velocity \( v \) can be expressed in terms of angular speed \( \omega \):
\[ v = \omega r \]
Therefore, substituting \( v \) into the angular momentum formula:
\[ L = m(\omega r)r = m \omega r^2 \]

2. **Calculate Magnetic Moment (μ)**:
The magnetic moment \( \mu \) for a current loop can be defined as:
\[ \mu = nIA \]
where \( n \) is the number of turns (which is 1 in this case), \( I \) is the current, and \( A \) is the area of the loop. The area \( A \) of the circular path is:
\[ A = \pi r^2 \]
To find the current \( I \), we use the definition of current:
\[ I = \frac{q}{T} \]
where \( T \) is the time period for one complete revolution. The time period \( T \) can be expressed as:
\[ T = \frac{2\pi r}{v} = \frac{2\pi r}{\omega r} = \frac{2\pi}{\omega} \]
Substituting \( T \) back into the current formula:
\[ I = \frac{q}{\frac{2\pi}{\omega}} = \frac{q \omega}{2\pi} \]
Now substituting \( I \) and \( A \) into the magnetic moment formula:
\[ \mu = 1 \cdot \left(\frac{q \omega}{2\pi}\right) \cdot \pi r^2 = \frac{q \omega r^2}{2} \]

3. **Find the Ratio of Magnetic Moment to Angular Momentum**:
Now, we can find the ratio of the magnetic moment \( \mu \) to the angular momentum \( L \):
\[ \frac{\mu}{L} = \frac{\frac{q \omega r^2}{2}}{m \omega r^2} \]
Simplifying this expression:
\[ \frac{\mu}{L} = \frac{q \omega r^2}{2m \omega r^2} = \frac{q}{2m} \]

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