= 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 $E'_y=\gamma(\beta)(E-vB)=0$ and
$$
\boxed{\beta=\frac{E}{cB},\qquad B'=B\sqrt{1-\beta^2}=\sqrt{B^2-E^2/c^2}}.
$$
All 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 $d(\gamma_u u_x')/d\tau=(qB'/m)\gamma_u u_y'$ and $d(\gamma_u u_y')/d\tau=-(qB'/m)\gamma_u u_x'$. Thus, choosing the phase and origin,
$$
ct'=\gamma_uc\tau,\qquad x'=\frac{\gamma_uu}{\omega}\sin\omega\tau,\qquad y'=\frac{\gamma_uu}{\omega}\cos\omega\tau,
\qquad\boxed{\omega=\frac{qB'}m}.
$$
This is the signed proper-time <cyclotron frequency>; the frequency with respect to $t'$ is $\omega/\gamma_u$. The inverse <Lorentz boost> gives
$$
\boxed{\begin{aligned}
ct&=\gamma_u\gamma(\beta)\left(c\tau+\frac{\beta u}{\omega}\sin\omega\tau\right),\\
x&=\gamma_u\gamma(\beta)\left(\beta c\tau+\frac{u}{\omega}\sin\omega\tau\right),\\
y&=\frac{\gamma_uu}{\omega}\cos\omega\tau.
\end{aligned}}
$$
In particular the average drift speed is $\beta c=E/B$ along $\mathbf E\times\mathbf B$.
For $u>0$, $q\ne0$, set $s=\omega\tau$ and $d=\beta c/u$. The dimensionless spatial curve is
$$
\boxed{\widetilde x=\gamma(\beta)(ds+\sin s),\qquad\widetilde y=\cos s}.
$$
The 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 $d\widetilde x/ds=\gamma(\beta)(d+\cos s)$: 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-1-relativistic-drift.png]
{title=The relativistic drift of a charged particle}
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.
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