A proton of mass `1.67xx10^(-27)` kg charge `1.6xx10^(-19) C` is projected in xy-plane with a speed of `2xx10^(6) m//s` at an angle of `60^(@)` to the X-axis. If a uniform magnetic field of 0.14 T is applies along the Y-axis, then the path of the proton is

Options

A. a circle of radiuis 0.2 m and time period `pixx10^(-7)s`
B. a circle of radius 0.1 m and time period `2pixx10^(-7)s`
C. a helix of radius 0.07 m and time period `0.5xx10^(-6)s`
D. a helix of radius 0.14 m and time period `1.0xx10^(-7)s`

Answer

To solve the problem step by step, we will analyze the motion of the proton in the presence of a magnetic field.

### Step 1: Identify the components of velocity
The proton is projected at a speed of \( V = 2 \times 10^6 \, \text{m/s} \) at an angle of \( 60^\circ \) to the x-axis. We need to find the components of the velocity in the x and y directions.

- The component of velocity in the x-direction:
\[ V_x = V \cos(60^\circ) = 2 \times 10^6 \times \frac{1}{2} = 1 \times 10^6 \, \text{m/s} \]

- The component of velocity in the y-direction:
\[ V_y = V \sin(60^\circ) = 2 \times 10^6 \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^6 \, \text{m/s} \]

### Step 2: Determine the angle with respect to the magnetic field
The magnetic field is applied along the y-axis. The angle between the velocity vector and the magnetic field is \( 30^\circ \) (since \( 90^\circ - 60^\circ = 30^\circ \)).

### Step 3: Identify the components of velocity relative to the magnetic field
- The component of velocity perpendicular to the magnetic field (y-direction):
\[ V_{\perp} = V_y = \sqrt{3} \times 10^6 \, \text{m/s} \]

- The component of velocity parallel to the magnetic field (x-direction):
\[ V_{\parallel} = V_x = 1 \times 10^6 \, \text{m/s} \]

### Step 4: Determine the radius of the helical path
The radius of the circular motion in the magnetic field can be calculated using the formula:
\[ r = \frac{m V_{\perp}}{q B} \]
Where:
- \( m = 1.67 \times 10^{-27} \, \text{kg} \) (mass of the proton)
- \( q = 1.6 \times 10^{-19} \, \text{C} \) (charge of the proton)
- \( B = 0.14 \, \text{T} \) (magnetic field strength)

Substituting the values:
\[ r = \frac{1.67 \times 10^{-27} \times \sqrt{3} \times 10^6}{1.6 \times 10^{-19} \times 0.14} \]

Calculating the numerator:
\[ \text{Numerator} = 1.67 \times 10^{-27} \times \sqrt{3} \times 10^6 \approx 2.89 \times 10^{-21} \, \text{kg m/s} \]

Calculating the denominator:
\[ \text{Denominator} = 1.6 \times 10^{-19} \times 0.14 \approx 2.24 \times 10^{-20} \, \text{C T} \]

Now, calculating the radius:
\[ r \approx \frac{2.89 \times 10^{-21}}{2.24 \times 10^{-20}} \approx 0.129 \, \text{m} \]

### Step 5: Determine the time period of the motion
The time period \( T \) of the circular motion is given by:
\[ T = \frac{2 \pi m}{q B} \]
Substituting the values:
\[ T = \frac{2 \pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.14} \]

Calculating:
\[ T \approx \frac{3.34 \times 10^{-27}}{2.24 \times 10^{-20}} \approx 1.49 \times 10^{-7} \, \text{s} \]

### Conclusion
The proton moves in a helical path due to the combination of its velocity components and the magnetic field. The radius of the helical path is approximately \( 0.129 \, \text{m} \) and the time period is approximately \( 1.49 \times 10^{-7} \, \text{s} \).