\text{A positive point charge with velocity }\overrightarrow{v}=5\hat{i}\text{ enters a region having electric field }\overrightarrow{E}=4\hat{j}\text{ and magnetic field }\overrightarrow{B}=-6\hat{k}.\text{ Which one of the following statements}

\text{is correct for the force on the charge as it enters the region?}

The correct option is: The force will be along the electric field but perpendicular to the magnetic field

To understand why this is correct, let us analyze the forces acting on the positive point charge as it enters the region with both electric and magnetic fields.
The total force on a charge $q$ moving with velocity $\overrightarrow{v}$ in an electric field $\overrightarrow{E}$ and a magnetic field $\overrightarrow{B}$ is given by the Lorentz force formula:

$$\overrightarrow{F}={\overrightarrow{F}}_{\text{electric}}+{\overrightarrow{F}}_{\text{magnetic}}=q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})$$

Let us evaluate each component of this force using the values given in the problem:
- The charge is positive, so $q>0$.
- Velocity: $\overrightarrow{v}=5\hat{i}$
- Electric field: $\overrightarrow{E}=4\hat{j}$
- Magnetic field: $\overrightarrow{B}=-6\hat{k}$

1. Electric Force:
The electric force is given by:

$${\overrightarrow{F}}_{\text{electric}}=q\overrightarrow{E}=q\left(4\hat{j}\right)=4q\hat{j}$$

Since $q>0$, this force is directed along the positive y-axis, which is the direction of the electric field $\overrightarrow{E}$.

2. Magnetic Force:
The magnetic force is given by:

$${\overrightarrow{F}}_{\text{magnetic}}=q(\overrightarrow{v}\times \overrightarrow{B})=q\left[\right(5\hat{i})\times (-6\hat{k}\left)\right]$$

Using the vector cross product rule for unit vectors where $\hat{i}\times \hat{k}=-\hat{j}$, we have:

$${\overrightarrow{F}}_{\text{magnetic}}=q[-30(\hat{i}\times \hat{k}\left)\right]=q[-30(-\hat{j}\left)\right]=30q\hat{j}$$

This force is also directed along the positive y-axis (direction of $\hat{j}$).

3. Net Force:
Combining the two components, the total force acting on the charge is:

$$\overrightarrow{F}=4q\hat{j}+30q\hat{j}=34q\hat{j}$$

Since the net force vector points in the direction of $\hat{j}$:
- It is parallel to (along) the electric field $\overrightarrow{E}=4\hat{j}$.
- It is perpendicular to the magnetic field $\overrightarrow{B}=-6\hat{k}$ because the dot product of $\hat{j}$ and $\hat{k}$ is zero ($\hat{j}·\hat{k}=0$).
Therefore, the force will be along the electric field but perpendicular to the magnetic field.