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. (c ) and (d)
B. (b) and (c )
C. (b) and (d)
D. (a) and (c )

Answer

To solve the problem, we need to analyze the conditions under which a charged particle moves in a horizontal direction in the presence of electric and magnetic fields.

### Step-by-Step Solution:

1. **Understanding the Forces:**
- A charged particle experiences two forces when in an electric field \( \vec{E} \) and a magnetic field \( \vec{B} \):
- The electric force \( \vec{F_E} = q \vec{E} \)
- The magnetic force \( \vec{F_B} = q (\vec{v} \times \vec{B}) \)
- Here, \( q \) is the charge of the particle, \( \vec{v} \) is its velocity, and \( \times \) denotes the cross product.

2. **Condition for Horizontal Motion:**
- For the particle to continue moving horizontally, the net force acting on it must be zero in the vertical direction. This means that the vertical components of the electric and magnetic forces must balance each other out.

3. **Case 1: Both Forces Parallel to Velocity**
- If both \( \vec{E} \) and \( \vec{B} \) are in the same direction as the velocity \( \vec{v} \), the magnetic force \( \vec{F_B} \) will be zero because the angle \( \theta \) between \( \vec{v} \) and \( \vec{B} \) is zero (i.e., \( \sin(0) = 0 \)). Thus, only the electric force acts, which will increase the speed but maintain the horizontal direction.
- This corresponds to option B.

4. **Case 2: Forces Perpendicular to Each Other**
- If \( \vec{B} \) is perpendicular to \( \vec{v} \) and \( \vec{E} \) is also perpendicular to \( \vec{v} \), we can arrange the forces such that:
- Let \( \vec{B} \) be in the negative z-direction and \( \vec{E} \) in the negative y-direction while \( \vec{v} \) is in the positive x-direction.
- The magnetic force \( \vec{F_B} \) will act upwards (positive y-direction) due to the right-hand rule, and if we set the electric force \( \vec{F_E} \) downwards (negative y-direction), we can balance the forces:
\[ \vec{F_E} + \vec{F_B} = 0 \]
- This corresponds to option C, where both fields are mutually perpendicular and perpendicular to the direction of velocity.

5. **Conclusion:**
- Both options B and C are valid conditions for the particle to maintain horizontal motion.

### Final Answer:
- The correct options are B and C.