A charged particle moves in a uniform magnetic field perpendicular to it, with a radius of curvature 4 cm. On passing through a metallic sheet it loses half of its kinetic energy. Then, the radius of curvature of the particle is

Options

A. `2 cm`
B. `4 cm`
C. `8 cm`
D. `2sqrt2 cm`

Explanation

To solve the problem step by step, we will use the relationship between the radius of curvature of a charged particle moving in a magnetic field and its kinetic energy.

### Step 1: Understand the relationship between radius of curvature and kinetic energy
The radius of curvature \( R \) of a charged particle moving in a magnetic field is given by the formula:
\[ R = \frac{mv}{QB} \]
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle,
- \( Q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

### Step 2: Express velocity in terms of kinetic energy
The kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
From this, we can express the velocity \( v \) as:
\[ v = \sqrt{\frac{2KE}{m}} \]

### Step 3: Substitute velocity in the radius of curvature formula
Substituting the expression for \( v \) into the radius of curvature formula:
\[ R = \frac{m \sqrt{\frac{2KE}{m}}}{QB} = \frac{\sqrt{2m \cdot KE}}{QB} \]

### Step 4: Establish the relationship between initial and final kinetic energy
Let \( KE_1 \) be the initial kinetic energy and \( KE_2 \) be the final kinetic energy after the particle passes through the metallic sheet. According to the problem, the particle loses half of its kinetic energy:
\[ KE_2 = \frac{1}{2} KE_1 \]

### Step 5: Relate the radii of curvature before and after the energy loss
Using the relationship derived earlier, we can express the ratio of the radii of curvature in terms of kinetic energy:
\[ \frac{R_1}{R_2} = \frac{\sqrt{KE_1}}{\sqrt{KE_2}} \]
Substituting \( KE_2 \):
\[ \frac{R_1}{R_2} = \frac{\sqrt{KE_1}}{\sqrt{\frac{1}{2} KE_1}} = \frac{\sqrt{KE_1}}{\sqrt{KE_1} \cdot \frac{1}{\sqrt{2}}} = \sqrt{2} \]

### Step 6: Solve for the final radius of curvature
Given that \( R_1 = 4 \, \text{cm} \):
\[ \frac{4}{R_2} = \sqrt{2} \]
Cross-multiplying gives:
\[ 4 = R_2 \sqrt{2} \]
Thus, solving for \( R_2 \):
\[ R_2 = \frac{4}{\sqrt{2}} = 2\sqrt{2} \, \text{cm} \]

### Final Answer
The radius of curvature of the particle after passing through the metallic sheet is:
\[ R_2 = 2\sqrt{2} \, \text{cm} \]