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 \( r \) of the path of a proton moving in a magnetic field at an angle \( \theta \), we can follow these steps:

### Step 1: Understand the Motion in a Magnetic Field
When a charged particle (like a proton) moves through a magnetic field, it experiences a magnetic force that acts perpendicular to both the velocity of the particle and the magnetic field direction. This force causes the particle to move in a circular path.

### Step 2: Resolve the Velocity
The velocity \( v \) of the proton can be resolved into two components:
- The component parallel to the magnetic field: \( v \cos \theta \)
- The component perpendicular to the magnetic field: \( v \sin \theta \)

Only the perpendicular component \( v \sin \theta \) contributes to the magnetic force that causes circular motion.

### Step 3: Calculate the Magnetic Force
The magnetic force \( F \) acting on the proton can be expressed using the formula:
\[ F = q(v \sin \theta)B \]
where:
- \( q \) is the charge of the proton,
- \( B \) is the magnetic field strength,
- \( v \sin \theta \) is the component of the velocity perpendicular to the magnetic field.

### Step 4: Relate the Magnetic Force to Centripetal Force
For circular motion, the magnetic force provides the necessary centripetal force. The centripetal force \( F_c \) required to keep the proton in circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the proton,
- \( r \) is the radius of curvature.

### Step 5: Set the Forces Equal
Setting the magnetic force equal to the centripetal force gives us:
\[ q(v \sin \theta)B = \frac{mv^2}{r} \]

### Step 6: Solve for the Radius \( r \)
Rearranging the equation to solve for \( r \):
\[ r = \frac{mv^2}{q(v \sin \theta)B} \]
This simplifies to:
\[ r = \frac{mv \sin \theta}{qB} \]

### Final Result
Thus, the radius of curvature of the resulting path is:
\[ r = \frac{mv \sin \theta}{qB} \]