A proton moving with a constant velocity passes through a region of space without any changing its velocity. If `E` and `B` represent the electric and magnetic fields, respectively. Then, this region of space may have

Options

A. `E=0,B=0`
B. `E=0, B!=0`
C. `E!=0,B=0`
D. `E!=0,B!=0`

Explanation

To solve the problem, we need to analyze the conditions under which a proton (or any charged particle) can move through a region of space without changing its velocity. This involves understanding the forces acting on the proton due to electric and magnetic fields.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- A charged particle like a proton experiences forces when it moves through electric and magnetic fields. The electric force (\( F_E \)) is given by \( F_E = qE \), where \( q \) is the charge of the proton and \( E \) is the electric field. The magnetic force (\( F_B \)) is given by \( F_B = q(v \times B) \), where \( v \) is the velocity of the proton and \( B \) is the magnetic field.

2. **Condition for Constant Velocity**:
- For the proton to maintain a constant velocity, the net force acting on it must be zero. This means that the electric force must be equal in magnitude and opposite in direction to the magnetic force:
\[ F_E + F_B = 0 \quad \Rightarrow \quad F_E = -F_B \]

3. **Direction of Forces**:
- If the electric field \( E \) is directed downwards, the electric force \( F_E \) will also act downwards. For the magnetic force \( F_B \) to act upwards, the velocity \( v \) of the proton must be perpendicular to both the electric field and the magnetic field.

4. **Using the Right-Hand Rule**:
- To determine the direction of the magnetic force, we can use the right-hand rule. If we point our thumb in the direction of the velocity \( v \) (which is perpendicular to both fields), and our fingers in the direction of the magnetic field \( B \), then the force \( F_B \) will be directed out of our palm.

5. **Equating Forces**:
- Since \( F_E = qE \) and \( F_B = qvB \), we can set them equal to each other:
\[ qE = qvB \]
- Dividing both sides by \( q \) (assuming \( q \neq 0 \)):
\[ E = vB \]

6. **Conclusion**:
- The region of space can have both electric and magnetic fields present, and they must be perpendicular to each other, as well as to the velocity of the proton. Therefore, the proton can move with a constant velocity in a region where both electric and magnetic fields exist, as long as the forces balance each other.

### Final Answer:
The region of space may have both electric field \( E \) and magnetic field \( B \) such that \( E = vB \), with all three vectors (velocity, electric field, and magnetic field) being mutually perpendicular.

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