A charged particle is released from rest in a region of steady and uniform electric and magnetic fields where \[{\text{E}}\parallel {\text{B}}\]. The particle will follow
A. Circular path
B. Elliptical path
C. Helical path
D. Straight path

Hint: Recall the expression for the electric force and magnetic force on the charged particle. First assume the effect of electric force on the charged particle. Then according to the expression for the magnetic force, determine the effect of the magnetic field on the charged particle.

Complete step by step answer:
We have given that the charged particle is released in the uniform electric and magnetic field. Therefore, the electric and magnetic force will act on it to change its direction in the field.
We know that the force on a particle of charge q placed in the uniform electric field is given as,
\[{F_e} = qE\]
Here, E is the electric field.
As soon as the particle is released, due to the electric force \[{F_e}\] on the charged particle, the charge particle will follow the straight line path in the direction of the electric field.
We also know that the force on a particle of charge q placed in the uniform magnetic field is given as,
\[{F_m} = qvB\sin \theta \]
Since we have given that the electric field and magnetic field are parallel to each other, the velocity of the charge is also parallel to the magnetic field. Therefore, the angle between the velocity and the magnetic field is zero. We substitute 0 for \[\theta \] in the above equation.
\[{F_m} = qvB\sin \left( {0^\circ } \right)\]
\[ \Rightarrow {F_m} = 0\]
Therefore, the magnetic force on the charged particle will be zero. So, the particle will follow the path of the electric field that is the straight line path.

Note:
Students should note that the magnetic force on the particle is given by the cross product of its velocity and magnetic field and not the dot product. In this question, we don’t know the charge on the particle. If the charge is positive, it follows the direction of the electric field and if the charge is negative, it follows the path in the opposite direction of the electric field.