Which of the following particles will take the least time to complete a circle when projected with the same velocity perpendicular to a uniform magnetic field?

Options

A. `He^(+)`
B. `Li^(+)`
C. Electron
D. Proton

Explanation

To solve the problem of which particle takes the least time to complete a circle when projected with the same velocity perpendicular to a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion in a Magnetic Field**:
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as a centripetal force, causing the particle to move in a circular path.

2. **Identify the Forces**:
The magnetic force \( F \) acting on a charged particle with charge \( q \) moving with velocity \( v \) in a magnetic field \( B \) is given by:
\[ F = qvB \sin(\theta) \]
Since the velocity is perpendicular to the magnetic field, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \). Thus, the force simplifies to:
\[ F = qvB \]

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

4. **Equating Forces**:
Setting the magnetic force equal to the centripetal force gives:
\[ qvB = \frac{mv^2}{r} \]
Rearranging this equation, we find:
\[ \frac{mv}{r} = qB \]
Thus,
\[ r = \frac{mv}{qB} \]

5. **Angular Frequency**:
The angular frequency \( \omega \) of the circular motion can be expressed as:
\[ \omega = \frac{qB}{m} \]

6. **Time Period**:
The time period \( T \) for one complete revolution is given by:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi m}{qB} \]
This shows that the time period \( T \) is directly proportional to the mass \( m \) of the particle and inversely proportional to its charge \( q \) and the magnetic field \( B \).

7. **Comparing Different Particles**:
In this problem, we are given four particles (Electron, Proton, Helium, Lithium) with the same charge \( q \) (which is the elementary charge \( e \) for all these particles) but different masses:
- Mass of Electron \( m_e \)
- Mass of Proton \( m_p \)
- Mass of Helium \( m_{He} = 4m_p \)
- Mass of Lithium \( m_{Li} = 7m_p \)

8. **Determine the Particle with the Least Mass**:
Since the time period \( T \) is directly proportional to the mass \( m \), the particle with the least mass will have the least time period. The masses in increasing order are:
- Electron \( m_e \) (least)
- Proton \( m_p \)
- Helium \( 4m_p \)
- Lithium \( 7m_p \)

9. **Conclusion**:
Therefore, the particle that takes the least time to complete a circle is the **Electron**.

### Final Answer:
The particle that takes the least time to complete a circle is the **Electron**.