An electron accelerated through a potential difference V passes through a uniform tranverses magnetic field and experiences a force F. If the accelerating potential is increased to 2V, the electron in the same magnetic field will experience a force:

Options

A. (a)F
B. (b)`F//2`
C. (c)`sqrt(2)F`
D. (d)`2F`

Explanation

To solve the problem, we need to analyze how the force experienced by an electron changes when the accelerating potential difference is increased.

### Step-by-Step Solution:

1. **Understanding the Force on a Charged Particle in a Magnetic Field**:
The force \( F \) on a charged particle moving in a magnetic field is given by the equation:
\[ F = B \cdot q \cdot v \]
where:
- \( F \) is the magnetic force,
- \( B \) is the magnetic field strength,
- \( q \) is the charge of the particle,
- \( v \) is the velocity of the particle.

2. **Relating Kinetic Energy to Potential Difference**:
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:
\[ KE = eV \]
where \( e \) is the charge of the electron. The kinetic energy can also be expressed in terms of velocity:
\[ KE = \frac{1}{2} mv^2 \]
Setting these equal gives:
\[ \frac{1}{2} mv^2 = eV \]

3. **Solving for Velocity**:
Rearranging the equation for \( v \):
\[ mv^2 = 2eV \implies v^2 = \frac{2eV}{m} \]
Taking the square root:
\[ v = \sqrt{\frac{2eV}{m}} \]

4. **Substituting Velocity into the Force Equation**:
Now substitute \( v \) back into the force equation:
\[ F = B \cdot q \cdot \sqrt{\frac{2eV}{m}} \]
Since \( q = e \) for an electron, we have:
\[ F = B \cdot e \cdot \sqrt{\frac{2eV}{m}} \]

5. **Increasing the Potential Difference**:
If the potential difference is increased to \( 2V \), we need to find the new force \( F' \):
\[ F' = B \cdot e \cdot \sqrt{\frac{2e(2V)}{m}} = B \cdot e \cdot \sqrt{\frac{4eV}{m}} = B \cdot e \cdot 2 \cdot \sqrt{\frac{eV}{m}} = 2 \cdot F \]

6. **Final Result**:
Therefore, the new force \( F' \) when the potential difference is increased to \( 2V \) is:
\[ F' = \sqrt{2} \cdot F \]

### Conclusion:
When the accelerating potential is increased to \( 2V \), the electron in the same magnetic field will experience a force of \( \sqrt{2}F \).