A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other . The particle will remove in a

Options

A. straight line
B. circle
C. helix
D. cycloid

Explanation

To solve the problem of a charged particle released from rest in a region of steady and uniform electric and magnetic fields that are parallel to each other, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Charged Particle**:
- The charged particle experiences two types of forces: electric force and magnetic force.
- The electric force (\( F_E \)) acting on the particle is given by \( F_E = qE \), where \( q \) is the charge of the particle and \( E \) is the electric field strength.
- The magnetic force (\( F_B \)) acting on the particle is given by \( F_B = q(v \times B) \), where \( v \) is the velocity of the particle and \( B \) is the magnetic field strength.

2. **Analyze the Initial Condition**:
- The particle is released from rest, which means its initial velocity (\( v \)) is zero.
- Since \( v = 0 \), the magnetic force \( F_B \) will also be zero because \( F_B = q(v \times B) = 0 \).

3. **Determine the Effect of the Electric Field**:
- The electric force \( F_E \) will act on the charged particle since it is charged and the electric field is present.
- The direction of the electric force will be in the direction of the electric field if the particle is positively charged (or opposite if negatively charged).

4. **Evaluate the Motion of the Particle**:
- Since the magnetic force is zero and the electric force is acting on the particle, the particle will start to accelerate in the direction of the electric field.
- As the particle gains velocity, it will continue to move in the direction of the electric field.

5. **Consider the Motion in Relation to the Magnetic Field**:
- Even when the particle gains velocity, since the electric field and magnetic field are parallel, the angle \( \theta \) between the velocity vector \( v \) and the magnetic field \( B \) remains \( 0^\circ \).
- Therefore, the magnetic force will still be zero because \( F_B = q(v \times B) = 0 \).

6. **Conclusion**:
- The particle will continue to move in a straight line in the direction of the electric field since there is no magnetic force acting on it.
- Thus, the correct answer is that the particle will move in a straight line.

### Final Answer:
The charged particle will move in a **straight line**.