A proton of mass `1.67 xx 10^(-27)` kg and charge `1.6 xx 10^(-19)` C is projected with a speed of `2 xx 10^(6)` m/s at an angle of `60^(@)` to the X–axis. If a uniform magnetic field of 0.104 tesla is applied along Y–axis, the path of proton is

Options

A. A circle of radius = 0.2 m and time period `pi xx 10^(-7) s`
B. A circle of radius = 0.1 m and time period `2pi xx 10^(-7) s`
C. A helix of radius = 0.1 m and time period `2pi xx 10^(-7) s`
D. A helix of radius = 0.2 m and time period `4 pi xx 10^(-7) `s

Explanation

To solve the problem of a proton moving in a magnetic field, we will follow these steps:

### Step 1: Understand the Problem
We have a proton with a mass \( m = 1.67 \times 10^{-27} \) kg and charge \( q = 1.6 \times 10^{-19} \) C. It is projected with a speed \( v = 2 \times 10^{6} \) m/s at an angle of \( 60^\circ \) to the X-axis. A uniform magnetic field \( B = 0.104 \) T is applied along the Y-axis. We need to determine the path of the proton.

### Step 2: Identify the Components of Velocity
The velocity of the proton can be broken down into two components:
- The component parallel to the magnetic field (Y-axis):
\[ v_{\parallel} = v \cos(60^\circ) = 2 \times 10^{6} \times \frac{1}{2} = 1 \times 10^{6} \text{ m/s} \]
- The component perpendicular to the magnetic field (X-axis):
\[ v_{\perpendicular} = v \sin(60^\circ) = 2 \times 10^{6} \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^{6} \text{ m/s} \]

### Step 3: Determine the Magnetic Force
The magnetic force acting on the proton is given by the formula:
\[ F = q v_{\perpendicular} B \]
Substituting the values:
\[ F = (1.6 \times 10^{-19}) \times (\sqrt{3} \times 10^{6}) \times (0.104) \]

### Step 4: Calculate the Radius of Circular Motion
The radius \( r \) of the circular motion due to the magnetic force can be calculated using the formula:
\[ r = \frac{mv_{\perpendicular}}{qB} \]
Substituting the known values:
\[ r = \frac{(1.67 \times 10^{-27}) \times (\sqrt{3} \times 10^{6})}{(1.6 \times 10^{-19}) \times (0.104)} \]

### Step 5: Determine the Path of the Proton
Since there are two components of motion:
- The \( v_{\parallel} \) component causes the proton to move linearly along the Y-axis.
- The \( v_{\perpendicular} \) component causes the proton to move in a circular motion in the X-Z plane.

Thus, the overall path of the proton will be a helical path, as it moves in a circular motion while simultaneously translating along the Y-axis.

### Conclusion
The path of the proton is a **helical path**.

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