 Solution

Use a Lorentz boost along positive $x$ with $v=E/B$. Since $E<cB$, its dimensionless speed $\beta =E/(cB)$ is less than one. The given transformation of the electric field and magnetic field yields $E_y^{\prime }=\gamma (\beta )(E-vB)=0$ andAll other electric components vanish. In this frame a magnetic field does no work, so the particle's speed $u<c$ and Lorentz factor ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$ are constant. In proper time, the Lorentz force equations are $d({\gamma }_u{u}_x^{\prime })/d\tau =(q{B}^{\prime }/m){\gamma }_u{u}_y^{\prime }$ and $d({\gamma }_u{u}_y^{\prime })/d\tau =-(q{B}^{\prime }/m){\gamma }_u{u}_x^{\prime }$. Thus, choosing the phase and origin,This is the signed proper-time cyclotron frequency; the frequency with respect to $t^{\prime }$ is $\omega /{\gamma }_u$. The inverse Lorentz boost givesIn particular the average drift speed is $\beta c=E/B$ along $\mathbf{E}\times \mathbf{B}$.

For $u>0$, $q\ne 0$, set $s=\omega \tau$ and $d=\beta c/u$. The dimensionless spatial curve isThe drift between successive cycles is $2\pi \gamma (\beta )d$. When $2\pi \beta \ll u/c$, this is small compared with the oscillation size: the curve consists of nearly closed loops with slow rightward drift. When $2\pi \beta \gg u/c$, the longitudinal drift is large, producing a progressing oscillatory curve. More precisely $d\tilde{x}/ds=\gamma (\beta )(d+\cos s)$: loops occur for $d<1$, cusps at $d=1$, and monotone progression for $d>1$. The stated large-drift inequality compares the drift over a full cycle, so an example of that regime with $d>1$ is shown; that comparison alone does not replace the exact loop criterion.

Reversing $q$ reverses the cyclotron orientation; the drift remains along positive $x$. If $u=0$ the trajectory is pure straight drift and the dimensionless coordinates are undefined. If $q=0$ the particle is inertial and the displayed $1/\omega$ parametrization must be replaced by its inertial limit.