A proton moves at a speed `v = 2xx10^(6) m//s` in a region of constant magnetic field of magnitude B = 0.05 T. The direction of the proton when it enters this field is `theta = 30^(@)` to the field. When you look along the direction of the magnetic field, then the path is a circle projected on a plane perpendicular to the magnetic field. How far will the proton move along the direction of B when two projected circles have been completed?

Options

A. `4.35 m`
B. `0.209 m`
C. `2.82 m`
D. `2.41 m`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the components of the proton's velocity
The proton is moving at a speed \( v = 2 \times 10^6 \, \text{m/s} \) at an angle \( \theta = 30^\circ \) to the magnetic field. We need to find the components of the velocity:
- The component of velocity parallel to the magnetic field:
\[ v_{\parallel} = v \cos \theta \]
- The component of velocity perpendicular to the magnetic field:
\[ v_{\perpendicular} = v \sin \theta \]

### Step 2: Calculate the time period of the circular motion
The time period \( T \) for one complete revolution of a charged particle in a magnetic field is given by:
\[ T = \frac{2\pi m}{qB} \]
where:
- \( m \) is the mass of the proton (\( m \approx 1.67 \times 10^{-27} \, \text{kg} \)),
- \( q \) is the charge of the proton (\( q \approx 1.6 \times 10^{-19} \, \text{C} \)),
- \( B \) is the magnetic field strength (\( B = 0.05 \, \text{T} \)).

### Step 3: Calculate the distance traveled along the magnetic field during two revolutions
The distance traveled along the direction of the magnetic field when two projected circles are completed is given by:
\[ \text{Distance} = v_{\parallel} \times \text{Total Time} \]
The total time for two revolutions is:
\[ \text{Total Time} = 2T \]
Thus, the distance can be expressed as:
\[ \text{Distance} = v \cos \theta \times 2T \]

### Step 4: Substitute the values and calculate
1. Calculate \( v_{\parallel} \):
\[ v_{\parallel} = v \cos \theta = (2 \times 10^6) \cos(30^\circ) = (2 \times 10^6) \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^6 \, \text{m/s} \]

2. Calculate the time period \( T \):
\[ T = \frac{2\pi m}{qB} = \frac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(0.05)} \approx 4.19 \times 10^{-8} \, \text{s} \]

3. Calculate the total time for two revolutions:
\[ \text{Total Time} = 2T \approx 2 \times 4.19 \times 10^{-8} \approx 8.38 \times 10^{-8} \, \text{s} \]

4. Finally, calculate the distance:
\[ \text{Distance} = v_{\parallel} \times \text{Total Time} = (\sqrt{3} \times 10^6) \times (8.38 \times 10^{-8}) \approx 0.035 \, \text{m} \approx 3.5 \, \text{cm} \]

### Final Result
The proton will move approximately **3.5 cm** along the direction of the magnetic field when two projected circles have been completed.

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