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 particle of mass \( m \) and charge \( q \) that is accelerated by a potential difference \( V \) and then enters a uniform magnetic field \( B \) perpendicularly, executing a circular arc of radius \( R \).

### Step-by-Step Solution:

1. **Kinetic Energy from Potential Difference**:
The work done on the charge \( q \) when it is accelerated through a potential difference \( V \) is given by:
\[ KE = qV \]
where \( KE \) is the kinetic energy of the particle.

2. **Kinetic Energy in Terms of Mass and Velocity**:
The kinetic energy can also be expressed in terms of mass \( m \) and velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]

3. **Equating the Two Expressions for Kinetic Energy**:
Setting the two expressions for kinetic energy equal gives:
\[ qV = \frac{1}{2} mv^2 \]

4. **Solving for Velocity**:
Rearranging this equation to find \( v^2 \):
\[ v^2 = \frac{2qV}{m} \]
Taking the square root to find \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

5. **Magnetic Force and Circular Motion**:
When the charged particle enters the magnetic field, it experiences a magnetic force given by the Lorentz force:
\[ F = qvB \]
This force provides the centripetal force required for circular motion:
\[ F = \frac{mv^2}{R} \]

6. **Setting the Forces Equal**:
Equating the magnetic force to the centripetal force:
\[ qvB = \frac{mv^2}{R} \]

7. **Rearranging for \( \frac{q}{m} \)**:
Rearranging the equation to isolate \( \frac{q}{m} \):
\[ qvB = \frac{mv^2}{R} \implies \frac{q}{m} = \frac{v}{BR} \]

8. **Substituting for Velocity**:
Now substitute the expression for \( v \) from step 4 into this equation:
\[ \frac{q}{m} = \frac{\sqrt{\frac{2qV}{m}}}{BR} \]

9. **Squaring Both Sides**:
To eliminate the square root, square both sides:
\[ \left(\frac{q}{m}\right)^2 = \frac{2qV}{m} \cdot \frac{1}{B^2R^2} \]

10. **Final Rearrangement**:
Rearranging gives:
\[ \frac{q}{m} = \frac{2V}{B^2R^2} \]

### Final Result:
Thus, the ratio \( \frac{q}{m} \) is given by:
\[ \frac{q}{m} = \frac{2V}{B^2R^2} \]