Following charge has maximum frequency of rotation in uniform transverse magnetic field :-

Options

A. a proton
B. an alpha particle
C. an electron
D. a neutron

Answer

To solve the problem of determining which charged particle has the maximum frequency of rotation in a uniform transverse magnetic field, we will follow these steps:

### Step 1: Understand the Motion of Charged Particles in a Magnetic Field
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts perpendicular to its velocity. This causes the particle to undergo circular motion. The frequency of this motion can be derived from the relationship between the magnetic force and the centripetal force required for circular motion.

### Step 2: Write the Expression for Magnetic Force
The magnetic force \( F \) on a charged particle moving with velocity \( v \) in a magnetic field \( B \) is given by:
\[ F = qvB \sin(\theta) \]
Where:
- \( q \) is the charge of the particle,
- \( v \) is the velocity,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

Since the motion is transverse, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \), thus:
\[ F = qvB \]

### Step 3: Relate Magnetic Force to Centripetal Force
For circular motion, the centripetal force \( F_c \) required to keep the particle moving in a circle of radius \( r \) is given by:
\[ F_c = \frac{mv^2}{r} \]
Where:
- \( m \) is the mass of the particle.

Setting the magnetic force equal to the centripetal force gives:
\[ qvB = \frac{mv^2}{r} \]

### Step 4: Solve for the Radius of the Circular Path
Rearranging the equation to find the radius \( r \):
\[ r = \frac{mv}{qB} \]

### Step 5: Find the Time Period of Rotation
The time period \( T \) for one complete revolution is the circumference of the circle divided by the velocity:
\[ T = \frac{2\pi r}{v} = \frac{2\pi}{qB} \cdot \frac{mv}{v} = \frac{2\pi m}{qB} \]

### Step 6: Calculate the Frequency
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{qB}{2\pi m} \]

### Step 7: Compare Frequencies for Different Particles
Now we can analyze the frequency for different charged particles:
1. **Proton**: Charge \( q = e \) and mass \( m = m_p \)
\[ f_p = \frac{eB}{2\pi m_p} \]

2. **Alpha Particle**: Charge \( q = 2e \) and mass \( m = 4m_p \)
\[ f_{\alpha} = \frac{2eB}{2\pi (4m_p)} = \frac{eB}{4\pi m_p} \]

3. **Electron**: Charge \( q = -e \) and mass \( m = m_e \)
\[ f_e = \frac{eB}{2\pi m_e} \]

### Step 8: Determine Which Frequency is Maximum
Now we need to compare the frequencies:
- The frequency of the proton is \( \frac{eB}{2\pi m_p} \).
- The frequency of the alpha particle is \( \frac{eB}{4\pi m_p} \).
- The frequency of the electron is \( \frac{eB}{2\pi m_e} \).

Since \( m_p \) (mass of proton) is much greater than \( m_e \) (mass of electron), the frequency of the electron will be the highest due to its much smaller mass.

### Conclusion
The charged particle with the maximum frequency of rotation in a uniform transverse magnetic field is the **electron**.