A proton (charge `1.6xx10^(-19) C, mass = 1.67xx10^(-27) kg)` is shot with a speed `8xx10^6 ms^-1` at an angle of `30^@` with the X-axis. A uniform magnetic field `B=0.30 T` exists along the X-axis. Show that path of the proton is a helix. Find the radius and pitch of the helix.

Explanation

To solve the problem step by step, we will analyze the motion of the proton in a magnetic field and derive the radius and pitch of the helical path.

### Step 1: Identify the components of velocity
The proton is shot at an angle of \(30^\circ\) with the X-axis. The speed of the proton is \(v = 8 \times 10^6 \, \text{m/s}\). We can resolve this velocity into two components:
- The component along the magnetic field (X-axis):
\[ v_x = v \cos(30^\circ) = 8 \times 10^6 \cos(30^\circ) \]
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