A particle of mass m carrying charge q is accelerated by a potential difference V. It enters perpendicularly in a region of uniform magnetic field B and executes circular arc of radius R, then `q/m` equals

Options

A. `(2V)/(B^(2)R^(2))`
B. `V/(2BR)`
C. `(VB)/(2R)`
D. `(mV)/(BR)`

Answer

To solve the problem, we need to find the ratio \( \frac{q}{m} \) for a charged particle of mass \( m \) and charge \( q \) that is accelerated by a potential difference \( V \) and enters a magnetic field \( B \) perpendicularly, executing a circular arc of radius \( R \).

### Step-by-Step Solution:

1. **Understanding the Forces**:
When a charged particle enters a magnetic field perpendicularly, it experiences a magnetic force given by the Lorentz force formula:
\[ F = qvB \]
where \( v \) is the velocity of the particle and \( B \) is the magnetic field strength.

2. **Centripetal Force**:
The particle moves in a circular path, so the magnetic force acts as the centripetal force. The centripetal force required to keep the particle moving in a circle of radius \( R \) is given by:
\[ F = \frac{mv^2}{R} \]

3. **Setting the Forces Equal**:
Since the magnetic force provides the centripetal force, we can set them equal to each other:
\[ qvB = \frac{mv^2}{R} \]

4. **Rearranging the Equation**:
We can rearrange this equation to express \( \frac{q}{m} \):
\[ qvB = \frac{mv^2}{R} \implies q = \frac{mv^2}{RvB} \]
Dividing both sides by \( m \):
\[ \frac{q}{m} = \frac{v}{RB} \]

5. **Finding Velocity from Potential Difference**:
The particle is accelerated by a potential difference \( V \). The kinetic energy gained by the particle is equal to the work done on it by the electric field:
\[ qV = \frac{1}{2}mv^2 \]
Rearranging this gives:
\[ v^2 = \frac{2qV}{m} \]

6. **Substituting Velocity into the Ratio**:
Now we substitute \( v^2 \) back into our expression for \( \frac{q}{m} \):
\[ \frac{q}{m} = \frac{v}{RB} = \frac{\sqrt{\frac{2qV}{m}}}{RB} \]

7. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ \left(\frac{q}{m}\right)^2 = \frac{2qV}{mR^2B^2} \]

8. **Final Expression for \( \frac{q}{m} \)**:
Rearranging gives:
\[ \frac{q}{m} = \frac{2V}{R^2B^2} \]

### Conclusion:
Thus, the final result for the ratio \( \frac{q}{m} \) is:
\[ \frac{q}{m} = \frac{2V}{R^2B^2} \]