A charged particle of mass `m` and charge `q` is accelerated through a potential differences of `V` volts. It enters of uniform magnetic field `B` which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius

Options

A. `((sqrtVm)/(2qB^2))`
B. `(2Vm)/(qB^2)`
C. `sqrt((2Vm)/q)(1/B)`
D. `sqrt((Vm)/q) (q/B)`

Explanation

To find the radius of the circular path of a charged particle moving in a uniform magnetic field, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
A charged particle with mass \( m \) and charge \( q \) is accelerated through a potential difference \( V \). It then enters a magnetic field \( B \) that is perpendicular to its motion.

2. **Kinetic Energy from Potential Difference**:
The work done on the charged particle by the electric field when it is accelerated through the potential difference \( V \) is given by:
\[ W = qV \]
This work done is converted into kinetic energy (KE) of the particle:
\[ KE = \frac{1}{2} mv^2 = qV \]

3. **Solving for Velocity**:
From the kinetic energy equation, we can express the velocity \( v \) of the particle:
\[ \frac{1}{2} mv^2 = qV \implies mv^2 = 2qV \implies v^2 = \frac{2qV}{m} \implies v = \sqrt{\frac{2qV}{m}} \]

4. **Momentum of the Particle**:
The momentum \( p \) of the particle is given by:
\[ p = mv \]
Substituting the expression for \( v \):
\[ p = m \sqrt{\frac{2qV}{m}} = \sqrt{2mqV} \]

5. **Radius of Circular Motion**:
The radius \( R \) of the circular path of a charged particle moving in a magnetic field is given by the formula:
\[ R = \frac{mv}{qB} \]
Substituting the expression for \( v \):
\[ R = \frac{m \sqrt{\frac{2qV}{m}}}{qB} = \frac{\sqrt{2mqV}}{qB} \]

6. **Simplifying the Expression**:
We can simplify the expression for \( R \):
\[ R = \frac{\sqrt{2m}}{B} \cdot \sqrt{\frac{qV}{q}} = \frac{\sqrt{2mV}}{qB} \]

### Final Result:
The radius of the circular path of the charged particle is:
\[ R = \frac{\sqrt{2mV}}{qB} \]