A charged particle with charge `q` enters a region of constant, uniform and mututally orthogonal fields `vec(E) and vec(B)` with a velocity `vec(v)` perpendicular to both `vec(E) and vec(B)`, and comes out without any change in magnitude or direction of `vec(v)`. Then

Options

A. `vecv=vecB xx vecE//E^2`
B. `vecv=vecE xx vecB//B^2`
C. `vecv=vecB xx vecE//B^2`
D. `vecv=vecE xx vecB//E^2`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in the presence of electric and magnetic fields. We are given that the particle enters a region where both fields are mutually orthogonal and that its velocity is perpendicular to both fields. The particle exits this region without any change in the magnitude or direction of its velocity.

### Step-by-Step Solution:

1. **Understanding Forces on the Charged Particle**:
- When a charged particle with charge \( q \) moves in an electric field \( \vec{E} \) and a magnetic field \( \vec{B} \), it experiences two forces:
- The electric force \( \vec{F_E} = q \vec{E} \)
- The magnetic force \( \vec{F_B} = q \vec{v} \times \vec{B} \)

2. **Condition for No Change in Velocity**:
- For the particle to exit the region without any change in the magnitude or direction of its velocity, the net force acting on it must be zero:
\[ \vec{F_E} + \vec{F_B} = 0 \]
- This implies:
\[ q \vec{E} + q (\vec{v} \times \vec{B}) = 0 \]
- Dividing through by \( q \) (assuming \( q \neq 0 \)):
\[ \vec{E} + (\vec{v} \times \vec{B}) = 0 \]
- Rearranging gives:
\[ \vec{E} = -(\vec{v} \times \vec{B}) \]

3. **Magnitude of Forces**:
- The magnitudes of the forces must be equal:
\[ |\vec{E}| = |\vec{v} \times \vec{B}| \]
- The magnitude of the magnetic force can be expressed as:
\[ |\vec{v} \times \vec{B}| = vB \sin(\theta) \]
- Since \( \vec{v} \) is perpendicular to \( \vec{B} \), \( \sin(\theta) = 1 \):
\[ |\vec{v} \times \vec{B}| = vB \]

4. **Equating Forces**:
- From the earlier relationship, we have:
\[ |\vec{E}| = vB \]

5. **Finding the Velocity Relation**:
- Rearranging the equation gives us:
\[ v = \frac{E}{B} \]
- This shows that the velocity \( v \) of the particle is directly proportional to the electric field \( E \) and inversely proportional to the magnetic field \( B \).

### Conclusion:
The relationship derived indicates that for the charged particle to maintain its velocity while traversing through the fields, the velocity must satisfy the equation:
\[ v = \frac{E}{B} \]