A proton with velocity `vecv` enters a region of uniform magnetic induction `vecB`, with `vecv` perpendicular to `vecB`and describes a circle of radius R. If an `alpha`- particle enters this region with the same velocity `vecv`, it describes a circle of radius

Options

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

Answer

To solve the problem, we need to find the radius of the circular path described by an alpha particle when it enters a magnetic field with the same velocity as a proton.

### Step-by-step Solution:

1. **Understand the Force Acting on the Particles**:
When a charged particle moves in a magnetic field, it experiences a magnetic force given by:
\[ F = qvB \sin(\theta) \]
Since the velocity \( \vec{v} \) is perpendicular to the magnetic field \( \vec{B} \), \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \). Therefore, the force simplifies to:
\[ F = qvB \]

2. **Centripetal Force**:
This magnetic force acts as the centripetal force required to keep the particle moving in a circular path. The centripetal force 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 gives us:
\[ qvB = \frac{mv^2}{R} \]

4. **Solving for Radius \( R \)**:
Rearranging this equation to solve for the radius \( R \), we get:
\[ R = \frac{mv}{qB} \]

5. **Proton and Alpha Particle Comparison**:
- For a proton:
- Charge \( q_p = e \) (where \( e \) is the elementary charge)
- Mass \( m_p = m \) (mass of the proton)
- Radius \( R = \frac{mv}{eB} \)

- For an alpha particle:
- Charge \( q_{\alpha} = 2e \) (since an alpha particle consists of 2 protons and 2 neutrons)
- Mass \( m_{\alpha} = 4m \) (mass of the alpha particle is approximately 4 times that of a proton)
- Radius \( R' = \frac{(4m)v}{(2e)B} \)

6. **Calculating the Radius of the Alpha Particle**:
Substituting the values for the alpha particle into the radius formula:
\[ R' = \frac{(4m)v}{(2e)B} = \frac{4}{2} \cdot \frac{mv}{eB} = 2 \cdot R \]

7. **Conclusion**:
Thus, the radius of the circular path described by the alpha particle is:
\[ R' = 2R \]

### Final Answer:
The radius of the circular path described by the alpha particle is \( 2R \).