A particle having a mass of `10^(-2) kg` carries a charge of `5 xx 10^(-8) C`. The particle is given an initial horizontal velocity of `10^(5) msec^(-1)` in the presence of electric field `vec(E)` and magnetic field `vec(B)`. To keep the particle moving in a horizontal direction, it is necessary that
A. `vec(B)` should be perpendicular to the direction of velocity and `vec(E)` should be along the direction of velocity
B. Both `vec(B)` and `vec(E)` should be along the direction of velocity
C. Both `vec(B)` and `vec(E)` are mutually perpendicular and perpendicular to the direction of velocity
D. `vec(B)` should be along the direction of velocity and `vec(E)` should be perpendicular to the direction of velocity

Options

A. `A` and `B`
B. `C` and `D`
C. `B` and `C`
D. `B` and `D`

Explanation

To determine the conditions under which a charged particle maintains a horizontal motion in the presence of electric and magnetic fields, we can analyze the forces acting on the particle.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle**:
- The particle has a charge \( q = 5 \times 10^{-8} \, C \).
- The particle is moving with a velocity \( \vec{v} = 10^5 \, m/s \).
- The electric force \( \vec{F_E} \) acting on the particle due to the electric field \( \vec{E} \) is given by:
\[ \vec{F_E} = q \vec{E} \]
- The magnetic force \( \vec{F_B} \) acting on the particle due to the magnetic field \( \vec{B} \) is given by:
\[ \vec{F_B} = q (\vec{v} \times \vec{B}) \]

2. **Condition for Horizontal Motion**:
- For the particle to maintain horizontal motion, the net force acting on it must be zero. This means that the electric force must balance the magnetic force:
\[ \vec{F_E} + \vec{F_B} = 0 \]
- This implies:
\[ q \vec{E} + q (\vec{v} \times \vec{B}) = 0 \]
- Rearranging gives:
\[ \vec{E} = -(\vec{v} \times \vec{B}) \]

3. **Analyzing Directions**:
- The velocity \( \vec{v} \) is in the horizontal direction.
- The magnetic force \( \vec{F_B} \) is given by the cross product \( \vec{v} \times \vec{B} \). For \( \vec{F_B} \) to be directed opposite to \( \vec{E} \), \( \vec{B} \) must be perpendicular to \( \vec{v} \).
- Therefore, \( \vec{B} \) should be perpendicular to the direction of velocity \( \vec{v} \).

4. **Direction of Electric Field**:
- Since \( \vec{E} \) must act in the opposite direction to \( \vec{F_B} \) to maintain horizontal motion, it can be concluded that \( \vec{E} \) should also be perpendicular to \( \vec{B} \) and not aligned with \( \vec{v} \).

5. **Conclusion**:
- The correct condition for the particle to keep moving in a horizontal direction is that both \( \vec{B} \) and \( \vec{E} \) are mutually perpendicular and perpendicular to the direction of velocity.

### Final Answer:
**C. Both \( \vec{B} \) and \( \vec{E} \) are mutually perpendicular and perpendicular to the direction of velocity.**