A charge particle is moving in a uniform magnetic field ib a circular path. The energy of the particle is doubled . If the initial radius of he circular path was `R`, the radius of the new circular path after the energy is doubled will be

Options

A. `R/2`
B. `sqrt(2)R`
C. 2R
D. `R/sqrt(2)`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Motion of the Charged Particle**:
A charged particle moving in a uniform magnetic field describes a circular path due to the Lorentz force acting as the centripetal force. The magnetic force acting on the particle is given by:
\[ F = qvB \]
where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field strength.

2. **Centripetal Force**:
The centripetal force required to keep the particle in circular motion is given by:
\[ F = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle and \( r \) is the radius of the circular path.

3. **Equating Forces**:
Setting the magnetic force equal to the centripetal force, we have:
\[ qvB = \frac{mv^2}{r} \]

4. **Solving for Velocity**:
Rearranging the equation to solve for \( v \):
\[ v = \frac{qBr}{m} \]

5. **Kinetic Energy of the Particle**:
The kinetic energy \( E \) of the particle is given by:
\[ E = \frac{1}{2}mv^2 \]
Substituting \( v \) from the previous step:
\[ E = \frac{1}{2}m\left(\frac{qBr}{m}\right)^2 = \frac{1}{2}\frac{q^2B^2r^2}{m} \]

6. **Energy Proportionality**:
From the expression for energy, we can see that the energy \( E \) is directly proportional to \( r^2 \):
\[ E \propto r^2 \]

7. **Doubling the Energy**:
If the energy is doubled, we can express this as:
\[ E_2 = 2E_1 \]
Therefore, we have:
\[ 2E_1 \propto r_2^2 \]

8. **Setting Up the Ratio**:
Since \( E_1 \propto r_1^2 \), we can write:
\[ \frac{E_2}{E_1} = \frac{r_2^2}{r_1^2} \]
Substituting \( E_2 = 2E_1 \):
\[ 2 = \frac{r_2^2}{r_1^2} \]

9. **Finding the New Radius**:
Rearranging gives:
\[ r_2^2 = 2r_1^2 \]
Taking the square root:
\[ r_2 = r_1\sqrt{2} \]

10. **Final Result**:
If the initial radius \( r_1 = R \), then the new radius \( r_2 \) is:
\[ r_2 = R\sqrt{2} \]

### Conclusion:
The radius of the new circular path after the energy is doubled will be \( R\sqrt{2} \).

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