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 will analyze the motion of both the proton and the alpha particle in a magnetic field and derive the radius of the circular path for each.

### Step-by-Step Solution:

1. **Understanding the Motion of the Proton:**
- A proton enters a uniform magnetic field \( \vec{B} \) with a velocity \( \vec{v} \) that is perpendicular to \( \vec{B} \).
- The magnetic force acting on the proton provides the centripetal force required for circular motion.

2. **Magnetic Force and Centripetal Force:**
- The magnetic force \( F \) on a charged particle moving in a magnetic field is given by:
\[ F = q v B \sin \theta \]
Since \( \vec{v} \) is perpendicular to \( \vec{B} \), \( \sin \theta = 1 \).
Thus, the force becomes:
\[ F = q v B \]
- For circular motion, this magnetic force acts as the centripetal force:
\[ F = \frac{mv^2}{R} \]
- Equating the two forces gives:
\[ q v B = \frac{mv^2}{R} \]

3. **Deriving the Radius for the Proton:**
- Rearranging the equation to solve for \( R \):
\[ R = \frac{mv}{qB} \]
- For a proton, the charge \( q \) is equal to \( e \) (the elementary charge), and we denote the radius of the proton's circular path as \( R_p \):
\[ R_p = \frac{mv}{eB} \]

4. **Analyzing the Alpha Particle:**
- An alpha particle has a charge of \( 2e \) (twice the charge of a proton) and a mass of \( 4m \) (four times the mass of a proton).
- Using the same formula for the radius of circular motion, we can write:
\[ R_{\alpha} = \frac{(4m)v}{(2e)B} \]
- Simplifying this expression:
\[ R_{\alpha} = \frac{4mv}{2eB} = \frac{2mv}{eB} \]

5. **Relating the Radius of the Alpha Particle to the Proton:**
- We already found that:
\[ R_p = \frac{mv}{eB} \]
- Therefore, we can express \( R_{\alpha} \) in terms of \( R_p \):
\[ R_{\alpha} = 2R_p \]

6. **Final Result:**
- Since we are given that the radius of the proton's path is \( R \):
\[ R_{\alpha} = 2R \]

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