A charged particle travels along a circle under the influence of...

For a charged particle to move in a circle, its speed must remain constant. This implies that the net work done by the fields on the particle must be zero at all times.

The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity $\vec{v}$, so it does no work.

The power delivered by the electric field is $P = q\vec{E} \cdot \vec{v}$. For the speed to be constant, we must have $q\vec{E} \cdot \vec{v} = 0$ at all times. Since the particle is moving in a circle, its velocity vector continuously changes direction in the plane of the circle. For the dot product to be zero at all points on the circle, the electric field $\vec{E}$ must be perpendicular to the plane of the circle.

If $\vec{E}$ is perpendicular to the plane of the circle, it will exert a constant force $q\vec{E}$ perpendicular to the plane. To keep the particle in the circular plane, the net force in the perpendicular direction must be zero. The magnetic force perpendicular to the plane is given by the component of $q(\vec{v} \times \vec{B})$ along $\vec{E}$. Since $\vec{v}$ continuously changes direction, this component of the magnetic force will vary sinusoidally and cannot cancel the constant electric force $q\vec{E}$.

Therefore, the electric field must be zero ($E = 0$).

To provide the necessary centripetal force for circular motion, a magnetic field must be present. Thus, $B \neq 0$.