Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields `vecE=E_(0)hatj` and `vecB=B_(0)hatj`. At time `t-0`, this charge has velocity v in the x-y plane, making an angle `theta` with the x- axis. Which of the following option (s) is (are) correct for time `tgt0`?

Options

A. If `theta=0^(@)`, the charge moves in a circular path in the x-z plane
B. If `theta=0^(@)`, the charge undergoes helical motion with constant pitch along the y-axis
C. If `theta=10^(@)`, the charge undergoes helical motion with its pitch increasing with time, along the y-axis
D. If `theta=90^(@)` the charge undergoes linear but accelerated motion along the y-axis

Explanation

To solve the problem of the motion of a positive point charge in the presence of uniform electric and magnetic fields, we can break down the solution step-by-step.

### Step 1: Understand the Given Information
We have a positive point charge \( q \) moving in the x-y plane with an initial velocity \( \vec{v} \) at an angle \( \theta \) with respect to the x-axis. The electric field \( \vec{E} = E_0 \hat{j} \) is directed along the y-axis, and the magnetic field \( \vec{B} = B_0 \hat{j} \) is also directed along the y-axis.

### Step 2: Analyze the Forces Acting on the Charge
The charge experiences two forces:
1. **Electric Force**: The electric force acting on the charge is given by:
\[ \vec{F}_E = q \vec{E} = q E_0 \hat{j} \]
This force acts in the positive y-direction.

2. **Magnetic Force**: The magnetic force is given by the Lorentz force law:
\[ \vec{F}_B = q (\vec{v} \times \vec{B}) \]
Since both \( \vec{v} \) and \( \vec{B} \) are in the x-y plane and along the y-axis, we need to find the angle between \( \vec{v} \) and \( \vec{B} \).

### Step 3: Determine the Angle Between Velocity and Magnetic Field
The angle \( \phi \) between the velocity \( \vec{v} \) and the magnetic field \( \vec{B} \) is given by:
\[ \phi = 90^\circ - \theta \]
This is because \( \vec{B} \) is along the y-axis, and \( \vec{v} \) makes an angle \( \theta \) with the x-axis.

### Step 4: Calculate the Magnetic Force
The magnetic force can be calculated as:
\[ \vec{F}_B = q v B \sin(\phi) = q v B_0 \sin(90^\circ - \theta) = q v B_0 \cos(\theta) \]
This force acts perpendicular to both \( \vec{v} \) and \( \vec{B} \).

### Step 5: Analyze the Motion Based on the Angle \( \theta \)
1. **If \( \theta = 0^\circ \)**: The charge moves in a circular path in the x-y plane due to the magnetic force, and the electric force contributes to an increasing pitch in the z-direction, resulting in a helical motion with increasing pitch.

2. **If \( \theta = 90^\circ \)**: The velocity is entirely in the y-direction, and the magnetic force becomes zero. The charge will experience only the electric force, resulting in linear accelerated motion along the y-axis.

3. **If \( 0 < \theta < 90^\circ \)**: The charge will undergo helical motion due to the combined effects of the electric and magnetic forces. The pitch of the helix will increase due to the constant electric force acting in the y-direction.

### Conclusion
Based on the analysis:
- For \( \theta = 0^\circ \), the charge undergoes helical motion with increasing pitch.
- For \( \theta = 90^\circ \), the charge undergoes linear accelerated motion along the y-axis.
- For \( 0 < \theta < 90^\circ \), the charge undergoes helical motion with increasing pitch.

### Correct Options
Thus, the correct options for the motion of the charge for \( t > 0 \) are:
- **C**: The charge undergoes helical motion with increasing pitch.
- **D**: The charge undergoes linear accelerated motion along the y-axis.