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 magnetic field, we can follow these steps:

### Step 1: Understand the Forces Acting on the Particle
When a charged particle moves in a magnetic field, it experiences a magnetic force given by:
\[ F_B = q \cdot v \cdot B \cdot \sin(\theta) \]
Since the magnetic field is perpendicular to the direction of motion, \(\theta = 90^\circ\) and \(\sin(90^\circ) = 1\). Therefore, the magnetic force simplifies to:
\[ F_B = q \cdot v \cdot B \]

### Step 2: Relate the Magnetic Force to Centripetal Force
For a particle moving in a circular path, the magnetic force acts as the centripetal force. The centripetal force is given by:
\[ F_c = \frac{m \cdot v^2}{r} \]
where \(m\) is the mass of the particle, \(v\) is its speed, and \(r\) is the radius of the circular path.

Setting the magnetic force equal to the centripetal force, we have:
\[ q \cdot v \cdot B = \frac{m \cdot v^2}{r} \]

### Step 3: Solve for the Radius \(r\)
Rearranging the equation to solve for \(r\):
\[ r = \frac{m \cdot v}{q \cdot B} \]

### Step 4: Determine the Speed \(v\) of the Particle
The particle is accelerated through a potential difference \(V\), which gives it kinetic energy. The kinetic energy gained by the particle is equal to the work done on it by the electric field:
\[ KE = q \cdot V \]
The kinetic energy can also be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
Setting these equal gives:
\[ q \cdot V = \frac{1}{2} m v^2 \]

### Step 5: Solve for \(v\)
Rearranging the equation for \(v\):
\[ v^2 = \frac{2qV}{m} \]
Taking the square root:
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 6: Substitute \(v\) Back into the Radius Equation
Now substitute \(v\) back into the equation for \(r\):
\[ r = \frac{m \cdot \sqrt{\frac{2qV}{m}}}{q \cdot B} \]

### Step 7: Simplify the Expression for \(r\)
This simplifies to:
\[ r = \frac{m}{qB} \cdot \sqrt{\frac{2qV}{m}} \]
\[ r = \frac{1}{B} \cdot \sqrt{\frac{2mV}{q}} \]

### Final Expression
Thus, the radius of the circular path is given by:
\[ r = \frac{1}{B} \cdot \sqrt{\frac{2mV}{q}} \]