If a charged particle goes unaccelerated in a region containing electric and magnetic fields,

Options

A. `vec(E)` must be parallel to `vec(B)`
B. `vec(V)` must be perpendicular to Electric field
C. `vec(V)` must be perpendicular to `vec(B)`
D. E must be equal to to vB

Explanation

To solve the problem of a charged particle moving unaccelerated in a region with electric and magnetic fields, we will analyze the forces acting on the particle and derive the necessary conditions.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Charged Particle**:
- A charged particle in an electric field \( \mathbf{E} \) experiences an electric force given by:
\[ \mathbf{F}_E = q \mathbf{E} \]
- In a magnetic field \( \mathbf{B} \), the charged particle with velocity \( \mathbf{v} \) experiences a magnetic force given by:
\[ \mathbf{F}_B = q (\mathbf{v} \times \mathbf{B}) \]

2. **Condition for Unaccelerated Motion**:
- For the particle to be unaccelerated, the net force acting on it must be zero:
\[ \mathbf{F}_E + \mathbf{F}_B = 0 \]
- This implies:
\[ q \mathbf{E} + q (\mathbf{v} \times \mathbf{B}) = 0 \]
- Dividing through by \( q \) (assuming \( q \neq 0 \)):
\[ \mathbf{E} + (\mathbf{v} \times \mathbf{B}) = 0 \]
- Rearranging gives:
\[ \mathbf{E} = -(\mathbf{v} \times \mathbf{B}) \]

3. **Analyzing the Relationship Between the Vectors**:
- The equation \( \mathbf{E} = -(\mathbf{v} \times \mathbf{B}) \) indicates that the electric field \( \mathbf{E} \) is perpendicular to the plane formed by the velocity \( \mathbf{v} \) and the magnetic field \( \mathbf{B} \).
- Since the cross product \( \mathbf{v} \times \mathbf{B} \) results in a vector that is perpendicular to both \( \mathbf{v} \) and \( \mathbf{B} \), we conclude that:
- \( \mathbf{E} \) is perpendicular to \( \mathbf{B} \)
- \( \mathbf{E} \) is also perpendicular to \( \mathbf{v} \)

4. **Conclusion**:
- The electric field \( \mathbf{E} \) must be perpendicular to both the velocity \( \mathbf{v} \) and the magnetic field \( \mathbf{B} \) for the charged particle to remain unaccelerated.
- Thus, the correct answer to the question is that the electric field is perpendicular to the magnetic field and the velocity of the charged particle.