A charged particle, having charge `q_(1)` accelerated through a potential difference V enters a perpendicular magnetic field in which it experiences a force F. It V is increased to 5V, the particle will experience a force

Options

A. F
B. 5F
C. `(F)/(5)`
D. `sqrt5F`

Answer

To solve the problem step by step, we will analyze the relationship between the potential difference, the velocity of the charged particle, and the force experienced by it in a magnetic field.

### Step 1: Relate the kinetic energy to the potential difference
The work done on the charged particle by the potential difference \( V \) is equal to the kinetic energy gained by the particle. This can be expressed as:

\[ \text{Work done} = q_1 V = \text{Kinetic Energy} = \frac{1}{2} m v^2 \]

### Step 2: Rearrange the equation to find velocity
From the equation above, we can rearrange it to find the velocity \( v \):

\[ q_1 V = \frac{1}{2} m v^2 \]

Multiplying both sides by 2 gives:

\[ 2 q_1 V = m v^2 \]

Now, solving for \( v \):

\[ v^2 = \frac{2 q_1 V}{m} \]

Taking the square root of both sides:

\[ v = \sqrt{\frac{2 q_1 V}{m}} \]

### Step 3: Write the expression for the magnetic force
The force \( F \) experienced by the charged particle in a magnetic field is given by:

\[ F = q_1 v B \sin \theta \]

Since the particle enters the magnetic field perpendicularly, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \), so:

\[ F = q_1 v B \]

### Step 4: Substitute the expression for velocity into the force equation
Substituting the expression for \( v \) into the force equation:

\[ F = q_1 \left(\sqrt{\frac{2 q_1 V}{m}}\right) B \]

This simplifies to:

\[ F = q_1 B \sqrt{\frac{2 q_1 V}{m}} \]

### Step 5: Analyze the effect of increasing the potential difference
Now, if the potential difference \( V \) is increased to \( 5V \), we can denote the new force as \( F' \):

\[ F' = q_1 B \sqrt{\frac{2 q_1 (5V)}{m}} = q_1 B \sqrt{\frac{10 q_1 V}{m}} \]

### Step 6: Relate the new force to the original force
From the original force \( F \):

\[ F = q_1 B \sqrt{\frac{2 q_1 V}{m}} \]

We can see that:

\[ F' = \sqrt{5} \cdot F \]

### Conclusion
Thus, when the potential difference is increased to \( 5V \), the new force experienced by the particle will be:

\[ F' = \sqrt{5} F \]

### Final Answer
The particle will experience a force that is \( \sqrt{5} \) times the original force \( F \).

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