A charged particle is moving in a uniform magnetic field and losses 4% of its KE. The radius of curvature of its path change by

Options

A. 0.02
B. 0.04
C. 0.1
D. None of these

Explanation

To solve the problem of how the radius of curvature of a charged particle's path changes when it loses 4% of its kinetic energy in a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between kinetic energy and radius of curvature**:
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.

The kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]

2. **Determine the initial kinetic energy**:
Let the initial kinetic energy be \( K \). Thus, we have:
\[ KE = K = \frac{1}{2} mv^2 \]

3. **Calculate the new kinetic energy after losing 4%**:
After losing 4% of its kinetic energy, the remaining kinetic energy \( K' \) is:
\[ K' = K - 0.04K = 0.96K \]

4. **Relate the new kinetic energy to the new velocity**:
The new kinetic energy can also be expressed in terms of the new velocity \( v' \):
\[ K' = \frac{1}{2} mv'^2 = 0.96K \]

5. **Express the initial and new velocities**:
From the kinetic energy equations, we can express the velocities:
\[ v = \sqrt{\frac{2K}{m}} \quad \text{and} \quad v' = \sqrt{\frac{2K'}{m}} = \sqrt{\frac{2 \times 0.96K}{m}} = \sqrt{0.96} \cdot \sqrt{\frac{2K}{m}} = \sqrt{0.96} \cdot v \]

6. **Find the initial and new radius of curvature**:
The initial radius of curvature \( R_1 \) is:
\[ R_1 = \frac{mv}{qB} \]
The new radius of curvature \( R_2 \) is:
\[ R_2 = \frac{mv'}{qB} = \frac{m(\sqrt{0.96} \cdot v)}{qB} = \sqrt{0.96} \cdot R_1 \]

7. **Calculate the change in radius**:
The change in radius can be expressed as:
\[ \text{Change in radius} = R_1 - R_2 = R_1 - \sqrt{0.96} \cdot R_1 = R_1(1 - \sqrt{0.96}) \]

8. **Calculate the percentage change in radius**:
The percentage change in radius is given by:
\[ \text{Percentage change} = \frac{R_1 - R_2}{R_1} \times 100 = (1 - \sqrt{0.96}) \times 100 \]

9. **Compute the numerical value**:
First, calculate \( \sqrt{0.96} \):
\[ \sqrt{0.96} \approx 0.98 \]
Thus, the percentage change becomes:
\[ \text{Percentage change} \approx (1 - 0.98) \times 100 = 0.02 \times 100 = 2\% \]

### Final Answer:
The radius of curvature changes by approximately **2%**.

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