 Solution

In vacuum, the source-free Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-{\partial }_{t}B$ and $\nabla \times B=c^{-2}{\partial }_{t}E$. Put $\vartheta =\omega (t-z/c)$ and denote the field amplitude by $E_0$. The corresponding travelling-wave magnetic field isBoth divergences vanish because the fields have no longitudinal component and no dependence on the transverse coordinates. Direct differentiation gives $\nabla \times E=(\omega E_0/c)(\cos \vartheta ,\sin \vartheta ,0)=-{\partial }_{t}B$ and $\nabla \times B=(\omega E_0/c^2)(-\sin \vartheta ,\cos \vartheta ,0)=c^{-2}{\partial }_{t}E$.

Introducing potentials by $B=\nabla \times A$ and $E=-\nabla \varphi -{\partial }_{t}A$ automatically satisfies the magnetic-divergence and Faraday equations. The remaining Maxwell equations become equations for these potentials, with the usual gauge freedom. For $\omega \ne 0$ take $\varphi =0$ andDifferentiation verifies both potential relations.

The electromagnetic force is $Q(E+v\times B)$. Confinement also supplies a normal reaction force. For motion in a plane of constant $z$, both $v$ and $B$ are tangent to the plane, so their cross product is normal and is balanced by that reaction. It does not enter the in-plane equation $m\ddot{r}=Q{E}_0(\cos \vartheta ,\sin \vartheta )$.

At the fixed plane, the solution with no uniform drift isIt is circular, withOther initial data can add a constant drift velocity; the question only asks for the possibility of circular motion.