 Solution

For the electric-dominated crossed-field orbit, choose a boost $V=c^{2}B/E$ along $x$. Since $|E|>c|B|$, it is subluminal. The Lorentz transformation of electromagnetic fields gives $B_z^{\prime }=0$ and $E_y^{\prime }=E/\Gamma$, where $\Gamma =(1-V^2/c^2{)}^{-1/2}$. The initially stationary electron has $U^{\prime }(0)=(\Gamma c,-\Gamma V,0,0)$, not zero velocity in this frame.

Let $q=-e$, $a=q{E}_y^{\prime }/m_e$ and $s=a\tau /c$. Neglect radiation reaction, as in the prescribed external-field model. Integrating the relativistic Lorentz force in this purely electric frame gives the exact trajectoryIndeed $U^{\prime 0}=\Gamma c\cosh s$, $U^{\prime x}=-\Gamma V$, and $U^{\prime y}=\Gamma c\sinh s$ satisfy both the force equation and $U^{\prime 2}=c^2$. The constant transverse momentum must be retained.

The inverse boost gives a useful exact laboratory parametrization as well:For $E,B>0$, the lab electron starts accelerating toward negative $y$ and bends toward positive $x$, the direction of $\mathbf{E}\times \mathbf{B}$. It never acquires a cyclotron orbit in this electric-dominated case. At late times, $v_x\to V$ and $v_y\to \mathrm{sgn}(q{E}^{\prime })c/\Gamma$, so its speed approaches $c$ along an oblique asymptote. Reversing field signs reverses the appropriate directions. The zero-$B$ limit is the ordinary electric-field hyperbolic motion.