A beam of protons with a velocity of `4 X 10 ^5 ms^(-1)` enters a uniform magnetic field of 0.3 T. The velocity makes an angle of `60^@` with the magnetic field. Find the radius of the helicla path taken by the proton beam and the pitch of the helix.

Options

A. `0.036`
B. `0.012m`
C. `0.024m`
D. `0.048m`

Answer

To solve the problem, we need to find the radius of the helical path and the pitch of the helix for a beam of protons moving in a magnetic field. Here are the steps to calculate both:

### Step 1: Identify the given values
- Velocity of protons, \( V = 4 \times 10^5 \, \text{m/s} \)
- Magnetic field strength, \( B = 0.3 \, \text{T} \)
- Angle with the magnetic field, \( \theta = 60^\circ \)
- Mass of a proton, \( m = 1.67 \times 10^{-27} \, \text{kg} \)
- Charge of a proton, \( q = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Resolve the velocity into components
The velocity \( V \) can be resolved into two components:
- Perpendicular component: \( V_{\perp} = V \sin \theta \)
- Parallel component: \( V_{\parallel} = V \cos \theta \)

Calculating these components:
- \( V_{\perp} = 4 \times 10^5 \sin(60^\circ) = 4 \times 10^5 \times \frac{\sqrt{3}}{2} \approx 3.464 \times 10^5 \, \text{m/s} \)
- \( V_{\parallel} = 4 \times 10^5 \cos(60^\circ) = 4 \times 10^5 \times \frac{1}{2} = 2 \times 10^5 \, \text{m/s} \)

### Step 3: Calculate the radius of the helical path
The formula for the radius \( r \) of the helical path is given by:
\[ r = \frac{m V_{\perp}}{q B} \]

Substituting the values:
\[ r = \frac{(1.67 \times 10^{-27} \, \text{kg}) \times (3.464 \times 10^5 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (0.3 \, \text{T})} \]

Calculating this:
\[ r = \frac{(1.67 \times 10^{-27}) \times (3.464 \times 10^5)}{(1.6 \times 10^{-19}) \times (0.3)} \approx \frac{5.78 \times 10^{-22}}{4.8 \times 10^{-20}} \approx 1.204 \times 10^{-2} \, \text{m} \approx 0.012 \, \text{m} \]

### Step 4: Calculate the pitch of the helix
The pitch \( P \) of the helix can be calculated using the formula:
\[ P = V_{\parallel} \times T \]
where \( T \) is the time period of one complete revolution, given by:
\[ T = \frac{2 \pi m}{q B} \]

Calculating \( T \):
\[ T = \frac{2 \pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19}) \times (0.3)} \approx \frac{3.34 \times 10^{-27}}{4.8 \times 10^{-20}} \approx 6.958 \times 10^{-8} \, \text{s} \]

Now substituting \( V_{\parallel} \) and \( T \) into the pitch formula:
\[ P = (2 \times 10^5) \times (6.958 \times 10^{-8}) \approx 1.3916 \times 10^{-2} \, \text{m} \approx 0.0139 \, \text{m} \]

### Final Answers
- Radius of the helical path, \( r \approx 0.012 \, \text{m} \)
- Pitch of the helix, \( P \approx 0.0139 \, \text{m} \)