The magnetic field in a certain region of space is given by `B=8.35 xx10^(-2) hati T.`A proton is shot into the field with velocity `v=(2xx10^(5)hati+4xx10^(5)hatj)m//s.` the proton follows a helical path in the field. The distance moved by proton in the x-direction during the period of one revolution in the yz-plane will be ( mass of proton`=1.67 xx 10^(-27) kg)`

Options

A. `0.053 m`
B. `0.136 m`
C. `0.157 m`
D. `0.236 m`

Answer

To solve the problem step by step, we need to find the distance moved by the proton in the x-direction during one complete revolution in the yz-plane.

### Step 1: Identify the Given Values
- Magnetic field \( B = 8.35 \times 10^{-2} \, \hat{i} \, \text{T} \)
- Velocity of the proton \( v = (2 \times 10^{5} \, \hat{i} + 4 \times 10^{5} \, \hat{j}) \, \text{m/s} \)
- Mass of the proton \( m = 1.67 \times 10^{-27} \, \text{kg} \)
- Charge of the proton \( q = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Calculate the Time Period of the Proton's Motion
The time period \( T \) for a charged particle moving in a magnetic field is given by the formula:
\[ T = \frac{2\pi m}{qB} \]

Substituting the values:
\[ T = \frac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(8.35 \times 10^{-2})} \]

Calculating the denominator:
\[ qB = (1.6 \times 10^{-19})(8.35 \times 10^{-2}) = 1.336 \times 10^{-20} \]

Now substituting back:
\[ T = \frac{2\pi (1.67 \times 10^{-27})}{1.336 \times 10^{-20}} \]
Calculating \( T \):
\[ T \approx \frac{3.34 \times 10^{-27}}{1.336 \times 10^{-20}} \approx 2.5 \times 10^{-7} \, \text{s} \]

### Step 3: Determine the Parallel Component of Velocity
The velocity \( v \) has two components:
- \( v_x = 2 \times 10^{5} \, \text{m/s} \) (in the x-direction)
- \( v_y = 4 \times 10^{5} \, \text{m/s} \) (in the y-direction)

The parallel component of velocity \( v_{\parallel} \) in the x-direction is:
\[ v_{\parallel} = v_x = 2 \times 10^{5} \, \text{m/s} \]

### Step 4: Calculate the Distance Moved in the x-Direction
The distance moved in the x-direction during one complete revolution (pitch) can be calculated using:
\[ \text{Distance} = v_{\parallel} \times T \]

Substituting the values:
\[ \text{Distance} = (2 \times 10^{5}) \times (2.5 \times 10^{-7}) \]
Calculating:
\[ \text{Distance} = 5.0 \times 10^{-2} \, \text{m} = 0.05 \, \text{m} \]

### Step 5: Final Result
The distance moved by the proton in the x-direction during the period of one revolution in the yz-plane is:
\[ \text{Distance} = 0.05 \, \text{m} \]