When an electron is accelerated through a potential difference `V`, it experience a force `F` through as uniform transverse magnetic field. If the potential difference is increased to `2 V,` the force experoenced by the electron in the same magnetic field is

Options

A. 2F
B. `2sqrt2F`
C. `sqrt2F`
D. `4F`

Explanation

To solve the problem, we need to analyze the relationship between the force experienced by an electron in a magnetic field and the potential difference through which it is accelerated.

### Step-by-Step Solution:

1. **Understanding the Force on the Electron**:
The force \( F \) experienced by a charged particle (like an electron) moving in a magnetic field is given by the equation:
\[ F = q \cdot v \cdot B \cdot \sin(\theta) \]
where:
- \( q \) is the charge of the electron,
- \( v \) is the velocity of the electron,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

2. **Relating Potential Difference to Kinetic Energy**:
When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field:
\[ KE = qV \]
This kinetic energy can also be expressed in terms of velocity:
\[ KE = \frac{1}{2} mv^2 \]
Setting these two expressions for kinetic energy equal gives:
\[ qV = \frac{1}{2} mv^2 \]

3. **Solving for Velocity**:
Rearranging the equation for velocity \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

4. **Substituting Velocity into the Force Equation**:
Now, substituting \( v \) back into the force equation:
\[ F = q \cdot \left(\sqrt{\frac{2qV}{m}}\right) \cdot B \cdot \sin(\theta) \]
This simplifies to:
\[ F = k \cdot \sqrt{V} \]
where \( k \) is a constant that includes \( q \), \( B \), \( \sin(\theta) \), and other constants.

5. **Considering the New Potential Difference**:
If the potential difference is increased to \( 2V \), we can find the new force \( F' \):
\[ F' = k \cdot \sqrt{2V} \]

6. **Relating the New Force to the Original Force**:
We can express \( F' \) in terms of \( F \):
\[ F' = k \cdot \sqrt{2} \cdot \sqrt{V} = \sqrt{2} \cdot F \]

7. **Conclusion**:
Therefore, the force experienced by the electron when the potential difference is increased to \( 2V \) is:
\[ F' = F \sqrt{2} \]

### Final Answer:
The force experienced by the electron in the same magnetic field when the potential difference is increased to \( 2V \) is \( F \sqrt{2} \).