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, we need to find the distance moved by the proton in the x-direction during the period of one revolution in the yz-plane. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Motion of the Proton
The proton is moving in a magnetic field, which causes it to follow a helical path. The velocity of the proton has components in both the x and y directions, while the magnetic field is directed along the x-axis. The motion in the yz-plane will be circular due to the magnetic field.

### Step 2: Determine the Time Period of One Revolution
The time period \( T \) for one complete revolution of a charged particle in a magnetic field can be calculated using the formula:
\[ T = \frac{2\pi m}{qB} \]
Where:
- \( m \) is the mass of the proton (\( 1.67 \times 10^{-27} \) kg),
- \( q \) is the charge of the proton (\( 1.6 \times 10^{-19} \) C),
- \( B \) is the magnetic field strength (\( 8.35 \times 10^{-2} \) T).

### Step 3: Substitute the Values into the Formula
Substituting the known values into the formula for \( T \):
\[ T = \frac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(8.35 \times 10^{-2})} \]

### Step 4: Calculate the Time Period
Calculating the denominator:
\[ qB = (1.6 \times 10^{-19})(8.35 \times 10^{-2}) = 1.336 \times 10^{-20} \]
Now substituting back into the time period formula:
\[ T = \frac{2\pi (1.67 \times 10^{-27})}{1.336 \times 10^{-20}} \approx \frac{3.34 \times 10^{-27}}{1.336 \times 10^{-20}} \approx 2.5 \times 10^{-7} \text{ seconds} \]

### Step 5: Calculate the Distance Moved in the x-direction
The distance moved in the x-direction during the time period \( T \) can be calculated using the formula:
\[ \text{Distance} = V_x \cdot T \]
Where \( V_x \) is the x-component of the velocity, given as \( 2 \times 10^5 \) m/s.

Substituting the values:
\[ \text{Distance} = (2 \times 10^5) \cdot (2.5 \times 10^{-7}) \approx 0.05 \text{ meters} \]

### Step 6: Final Calculation
Now, we can finalize the calculation:
\[ \text{Distance} = 0.05 \text{ meters} = 5 \times 10^{-2} \text{ meters} \]

### Conclusion
The distance moved by the proton in the x-direction during the period of one revolution in the yz-plane is approximately \( 0.157 \) meters.