An electron having kinetic energy `K` is moving i a circular orbit of radius `R` perpendicular to a uniform magnetic induction. If kinetic energy is douled and magnetic induction tripled, the radius will become

Options

A. `(3R)/(2)`
B. `sqrt(3)/(2)R`
C. `sqrt(2)/(3)R`
D. `sqrt(4)/(3)R`

Answer

To solve the problem, we need to analyze the relationship between the kinetic energy of the electron, the magnetic induction (magnetic field), and the radius of the circular orbit.

### Step-by-Step Solution:

1. **Initial Conditions**:
- Let the initial kinetic energy of the electron be \( K \).
- Let the initial radius of the circular orbit be \( R \).
- Let the initial magnetic induction (magnetic field) be \( B \).

2. **Centripetal Force**:
The centripetal force required to keep the electron moving in a circular path is provided by the magnetic force. The magnetic force acting on the electron is given by:
\[ F = qvB \]
where \( q \) is the charge of the electron, \( v \) is its velocity, and \( B \) is the magnetic field.

3. **Centripetal Acceleration**:
The centripetal force can also be expressed in terms of the mass \( m \) of the electron and its velocity \( v \):
\[ F = \frac{mv^2}{R} \]

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

5. **Solving for Radius**:
Rearranging the equation gives us the expression for the radius \( R \):
\[ R = \frac{mv}{qB} \]

6. **Relating Velocity to Kinetic Energy**:
The kinetic energy \( K \) of the electron is given by:
\[ K = \frac{1}{2} mv^2 \]
From this, we can express \( mv \) in terms of \( K \):
\[ mv = \sqrt{2mK} \]

7. **Substituting into Radius Formula**:
Substituting \( mv \) into the radius formula gives:
\[ R = \frac{\sqrt{2mK}}{qB} \]

8. **Final Conditions**:
Now, according to the problem:
- The kinetic energy is doubled: \( K' = 2K \).
- The magnetic induction is tripled: \( B' = 3B \).

9. **New Radius Calculation**:
Substituting the new kinetic energy and magnetic induction into the radius formula:
\[ R' = \frac{\sqrt{2m(2K)}}{q(3B)} = \frac{\sqrt{4mK}}{3qB} = \frac{2\sqrt{mK}}{3qB} \]

10. **Expressing in Terms of Initial Radius**:
We can relate this to the initial radius \( R \):
\[ R' = \frac{2}{3} R \]

### Final Answer:
The new radius \( R' \) will be:
\[ R' = \frac{2}{3} R \]