A charged particle is moving in a uniform magnetic field in a circular path. Radius of circular path is `R`. When energy of particle is doubled, then new radius will be

Options

A. `Rsqrt(2)`
B. `Rsqrt(3)`
C. `2R`
D. `3R`

Answer

To solve the problem, we need to analyze the relationship between the kinetic energy of a charged particle moving in a magnetic field and the radius of its circular path.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Let the mass of the charged particle be \( m \).
- Let the charge of the particle be \( q \).
- Let the initial velocity of the particle be \( v \).
- The initial kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
- The radius \( R \) of the circular path in a magnetic field \( B \) is given by the formula:
\[ R = \frac{mv}{qB} \]

2. **Doubling the Kinetic Energy**:
- According to the problem, the kinetic energy is doubled. Therefore, the new kinetic energy \( KE' \) is:
\[ KE' = 2 \times KE = 2 \times \frac{1}{2} mv^2 = mv^2 \]
- The new kinetic energy can also be expressed in terms of the new velocity \( v' \):
\[ KE' = \frac{1}{2} mv'^2 \]
- Setting these equal gives:
\[ mv^2 = \frac{1}{2} mv'^2 \]

3. **Finding the New Velocity**:
- Rearranging the equation to solve for \( v' \):
\[ mv'^2 = 2mv^2 \]
\[ v'^2 = 2v^2 \]
\[ v' = \sqrt{2}v \]

4. **Calculating the New Radius**:
- Now, we can find the new radius \( R' \) using the new velocity \( v' \):
\[ R' = \frac{mv'}{qB} \]
- Substituting \( v' = \sqrt{2}v \):
\[ R' = \frac{m(\sqrt{2}v)}{qB} = \sqrt{2} \cdot \frac{mv}{qB} = \sqrt{2}R \]

5. **Conclusion**:
- The new radius of the circular path when the energy of the particle is doubled is:
\[ R' = \sqrt{2}R \]

### Final Answer:
The new radius will be \( \sqrt{2}R \).