A proton of mass `m` and charge `q` enters a magnetic field `B` with a velocity `v` at an angle `theta` with the direction of `B`.The radius of curvature of the resulting path is

Options

A. `(mv)/(qB)`
B. `(mv sin theta)/(qB)`
C. `(mv)/(qB sin theta)`
D. `(mv cos theta)/(qB)`

Explanation

To find the radius of curvature of a proton moving in a magnetic field at an angle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Parameters:**
- Mass of the proton: \( m \)
- Charge of the proton: \( q \)
- Velocity of the proton: \( v \)
- Magnetic field strength: \( B \)
- Angle between the velocity and the magnetic field: \( \theta \)

2. **Break Down the Velocity into Components:**
- The velocity \( v \) can be resolved into two components:
- Perpendicular to the magnetic field: \( v_{\perp} = v \sin \theta \)
- Parallel to the magnetic field: \( v_{\parallel} = v \cos \theta \)

3. **Determine the Magnetic Force:**
- The magnetic force acting on the proton is given by:
\[ F = q(v_{\perp} \times B) = q(v \sin \theta) B \]
- Since \( v_{\perp} \) is perpendicular to the magnetic field, we can simplify this to:
\[ F = q v \sin \theta B \]

4. **Relate the Magnetic Force to Centripetal Force:**
- For a charged particle moving in a circular path, the magnetic force provides the centripetal force required for circular motion:
\[ F = \frac{m v_{\perp}^2}{r} \]
- Substituting \( v_{\perp} = v \sin \theta \):
\[ F = \frac{m (v \sin \theta)^2}{r} \]

5. **Set the Forces Equal:**
- Equating the magnetic force and the centripetal force:
\[ q v \sin \theta B = \frac{m (v \sin \theta)^2}{r} \]

6. **Rearranging for Radius \( r \):**
- Rearranging the equation to solve for \( r \):
\[ r = \frac{m (v \sin \theta)}{q B} \]
- This can be simplified to:
\[ r = \frac{mv}{qB \sin \theta} \]

### Final Result:
The radius of curvature \( r \) of the proton's path in the magnetic field is given by:
\[ r = \frac{mv}{qB \sin \theta} \]