Answer:
The charged particle will follow a circular path.
Explanation:
The expression for the magnetic force due to the charged particle is as follows;
[tex]F=qvBsin\theta[/tex]
Here, q is the charge, B is the magnetic field, v is the velocity, [tex]\theta[/tex] is the angle between velocity and magnetic force and F is the magnetic force.
If the direction of velocity has both perpendicular and parallel components to the direction magnetic field then the parallel component moves the charged particle in a straight line and the perpendicular component moves the charged particle in a circular direction. Then, the trajectory followed by the charged particle is helix.
In the given problem, a charged particle moves in the uniform magnetic field in such a way velocity vector of the charged particle is perpendicular to the magnetic field vector.
Here, [tex]sin\theta=90^{\circ}[/tex] and the magnetic force is constant.
Then, the magnitude of magnetic force becomes, F= qvB. In this case, the force will be maximum. If the charged particle moves along the direction of magnetic field then the F=0 as [tex]sin\theta=0^{\circ}[/tex] .
Therefore, the path followed by the charged particle is circular path.