A charged particle moves in a circular path in a uniform magnetic field. If its speed is reduced, then its tiem period will

Options

A. increase
B. decrease
C. remain same
D. none of these

Explanation

To solve the problem, we need to analyze the motion of a charged particle in a magnetic field and derive the relationship between its speed, radius of the circular path, and the time period of the motion.

### Step-by-Step Solution:

1. **Understanding the Forces**:
A charged particle moving in a circular path in a magnetic field experiences a magnetic force that acts as the centripetal force required for circular motion. The magnetic force \( F \) on a charged particle is given by:
\[ F = qvB \]
where \( q \) is the charge of the particle, \( v \) is its speed, and \( B \) is the magnetic field strength.

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

3. **Equating Forces**:
For circular motion, the magnetic force must equal the centripetal force:
\[ qvB = \frac{mv^2}{r} \]

4. **Solving for Radius**:
Rearranging the equation gives us:
\[ r = \frac{mv}{qB} \]
This equation shows that the radius \( r \) of the circular path depends on the speed \( v \) of the particle.

5. **Calculating the Time Period**:
The time period \( T \) of the motion is the time taken to complete one full circle. The distance traveled in one complete revolution is the circumference of the circle, which is \( 2\pi r \). Therefore, the time period \( T \) can be expressed as:
\[ T = \frac{\text{Distance}}{\text{Speed}} = \frac{2\pi r}{v} \]

6. **Substituting for Radius**:
Substituting the expression for \( r \) into the equation for \( T \):
\[ T = \frac{2\pi \left(\frac{mv}{qB}\right)}{v} \]

7. **Simplifying the Time Period**:
The \( v \) in the numerator and denominator cancels out:
\[ T = \frac{2\pi m}{qB} \]
This shows that the time period \( T \) does not depend on the speed \( v \) of the particle.

8. **Conclusion**:
Since the time period \( T \) is independent of the speed \( v \), if the speed of the charged particle is reduced, the time period will remain the same.

### Final Answer:
The time period will remain the same.