A charged particle is moving in a magnetic field of strength B perpendicular to the direction of the field. If q and m denote the charge and mass of the particle respectively, then the frequency of rotation of the particle is

Options

A. `(qB)/(2pi m)`
B. `(qB)/(2pi m^(2))`
C. `(2pi^(2)m)/(qB)`
D. `(2pi m)/(qB)`

Explanation

To find the frequency of rotation of a charged particle moving in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
When a charged particle with charge \( q \) and mass \( m \) moves in a magnetic field \( B \), it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path.

2. **Magnetic Force**:
The magnetic force \( F_m \) acting on the charged particle is given by:
\[ F_m = qvB \]
where \( v \) is the velocity of the particle and \( B \) is the magnetic field strength.

3. **Centripetal Force**:
The centripetal force \( F_c \) required to keep the particle moving in a circular path is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( r \) is the radius of the circular path.

4. **Equating Forces**:
Since the magnetic force provides the necessary centripetal force, we can set these two forces equal to each other:
\[ qvB = \frac{mv^2}{r} \]

5. **Rearranging the Equation**:
We can rearrange this equation to express \( v \):
\[ qvB = \frac{mv^2}{r} \implies qB = \frac{mv}{r} \]

6. **Expressing Velocity in Terms of Frequency**:
The angular velocity \( \omega \) of the particle is related to its linear velocity \( v \) and radius \( r \) by:
\[ v = r\omega \]
The frequency \( f \) is related to angular velocity by:
\[ \omega = 2\pi f \]
Therefore, substituting for \( v \):
\[ qB = \frac{m(r\omega)}{r} = m\omega \]

7. **Substituting for Angular Velocity**:
Now substituting \( \omega = 2\pi f \):
\[ qB = m(2\pi f) \implies f = \frac{qB}{2\pi m} \]

### Final Result:
The frequency of rotation \( f \) of the charged particle in the magnetic field is given by:
\[ f = \frac{qB}{2\pi m} \]