To determine the conditions under which a charged particle would continue to move with a constant velocity, we need to analyze the effects of electric and magnetic fields on the particle. Let's break down the problem step by step.
### Step 1: Understanding Constant Velocity
A charged particle moves with a constant velocity when the net force acting on it is zero. According to Newton's first law of motion, if no net external force acts on an object, it will maintain its state of motion (constant velocity).
### Step 2: Analyzing the Options
We have four options to analyze regarding the presence of electric field (E) and magnetic field (B):
1. **Option A: E = 0 and B ≠ 0**
- In this case, there is a magnetic field present, but no electric field. The magnetic force acting on the charged particle is given by \( F_m = q(v \times B) \), which acts perpendicular to the velocity. This means the particle will undergo circular motion, changing direction but not speed. Thus, the magnitude of the velocity remains constant, but the direction changes. Therefore, this option does not satisfy the condition of constant velocity.
2. **Option B: E ≠ 0 and B ≠ 0**
- Here, both electric and magnetic fields are present. The electric force \( F_e = qE \) acts in the direction of the electric field, while the magnetic force \( F_m = q(v \times B) \) acts perpendicular to the velocity. If the magnitudes of these forces are equal and opposite (\( F_e = F_m \)), the net force will be zero, allowing the particle to move with constant velocity. Thus, this option can satisfy the condition.
3. **Option C: E ≠ 0 and B = 0**
- In this scenario, there is an electric field present, but no magnetic field. The electric force \( F_e = qE \) will act on the charged particle, causing it to accelerate. Since there is no balancing magnetic force, the particle cannot maintain a constant velocity. Therefore, this option does not satisfy the condition.
4. **Option D: E = 0 and B = 0**
- In this case, both fields are absent. Since there are no forces acting on the charged particle, it will continue to move with the same velocity (both magnitude and direction). Thus, this option satisfies the condition for constant velocity.
### Conclusion
From the analysis, the conditions under which a charged particle would continue to move with a constant velocity are:
- Option B (E ≠ 0 and B ≠ 0, with forces balanced)
- Option D (E = 0 and B = 0).
Thus, the correct options are B and D.
