A portion is fired from origin with velocity `vec(v) = v_(0) hat(j)+ v_(0) hat(k)` in a uniform magnetic field `vec(B) = B_(0) hat(j)`. In the subsequent motion of the proton

Options

A. its Z co-ordinate can never be negative
B. its x co-ordinate can never be positive
C. its x and z co-ordinates cannot be zero at the same time
D. its y co-ordinate will be proportional to its time of flight

Answer

To solve the problem, we need to analyze the motion of a positron fired from the origin with a specific velocity in a uniform magnetic field. Let's break down the solution step by step.

### Step 1: Understand the Initial Conditions
The positron is fired from the origin with a velocity vector given by:
\[ \vec{v} = v_0 \hat{j} + v_0 \hat{k} \]
This indicates that the positron has equal components of velocity in the y-direction (j) and z-direction (k).

### Step 2: Identify the Magnetic Field
The uniform magnetic field is given by:
\[ \vec{B} = B_0 \hat{j} \]
This means the magnetic field is directed along the y-axis.

### Step 3: Determine the Angle Between Velocity and Magnetic Field
The angle between the velocity vector and the magnetic field vector can be calculated. The velocity has components in both the j and k directions, while the magnetic field is only in the j direction. The angle θ between the velocity vector and the magnetic field vector can be found using the dot product:
\[ \cos(\theta) = \frac{\vec{v} \cdot \vec{B}}{|\vec{v}| |\vec{B}|} \]
Calculating the magnitudes:
\[ |\vec{v}| = \sqrt{(v_0)^2 + (v_0)^2} = v_0 \sqrt{2} \]
\[ |\vec{B}| = B_0 \]
Now, the dot product:
\[ \vec{v} \cdot \vec{B} = (v_0 \hat{j} + v_0 \hat{k}) \cdot (B_0 \hat{j}) = v_0 B_0 \]
Thus,
\[ \cos(\theta) = \frac{v_0 B_0}{v_0 \sqrt{2} B_0} = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ \]

### Step 4: Analyze the Motion
Since the angle between the velocity and the magnetic field is neither 0° nor 90°, the motion of the positron will be helical. The positron will move in a circular path in the plane perpendicular to the magnetic field while also moving linearly along the direction of the magnetic field.

### Step 5: Determine the Components of Motion
1. **Linear Motion**: The component of velocity parallel to the magnetic field (v_y) will result in linear motion along the y-axis:
\[ v_y = v_0 \]
The position in the y-direction as a function of time (t) will be:
\[ y(t) = v_0 t \]

2. **Circular Motion**: The component of velocity perpendicular to the magnetic field (v_z) will result in circular motion in the x-z plane. The radius of the circular motion can be determined by the Lorentz force:
\[ F = q(\vec{v} \times \vec{B}) \]
The circular motion will occur in the x-z plane, and the position will depend on the radius and angular frequency.

### Step 6: Conclusion
The overall motion of the positron is helical, with:
- Linear motion along the y-axis.
- Circular motion in the x-z plane.

### Final Answer
The correct option regarding the motion of the positron is that it will execute helical motion, with its y-coordinate being proportional to time and the x-coordinate being negative.