= Solution
For the <magnetic-dominated crossed-field orbit>, choose $V=E/B$ along $x$, which is subluminal when $|E|<c|B|$. Then $E'_y=0$ and $B'_z=B/\Gamma$. The initial <velocity> in this frame is $(-V,0,0)$, and the constant particle <Lorentz factor> is $\gamma'=\Gamma$ because a <magnetic field> does no work.
Define the signed relativistic <cyclotron frequency> $\Omega=qB'_z/(\Gamma m_e)$, retaining $q=-e$. With $t'$ as parameter, the exact primed trajectory is
$$
\boxed{x'=-\frac V\Omega\sin(\Omega t'),\qquad
y'=\frac V\Omega[1-\cos(\Omega t')],\qquad z'=0.}
$$
It is a circle of radius $|V/\Omega|$ with centre $(0,V/\Omega)$; its initial <velocity> is $(-V,0,0)$ as required. Transforming back gives
$$
\boxed{x=\Gamma V\left[t'-\frac{\sin(\Omega t')}{\Omega}\right],\qquad
y=\frac V\Omega[1-\cos(\Omega t')],\qquad
t=\Gamma\left[t'-\frac{V^2}{c^2\Omega}\sin(\Omega t')\right].}
$$
In the laboratory this is a stretched cycloidal trajectory with cusps where the <electron> is instantaneously at rest. Its average drift is $\langle v_x\rangle=V=E/B$, along $\mathbf E\times\mathbf B$ for the positive field orientation. Its <energy> oscillates as the <electric field> exchanges <energy> with it; it does not accelerate without bound. If $E=0$, the initially stationary <electron> remains stationary. At $|E|=c|B|$, neither of these finite boosts exists; that separate null-field limit is not one of the two strict cases requested.
\Image[/past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2017-crossed-field-trajectories.png]
{title=Exact electron trajectories in electric-dominated and magnetic-dominated crossed fields}
{description=Original curves use $q=-e$ and an <electron> released from rest in the laboratory. Adapted inertial frames eliminate the <magnetic field> or <electric field>, respectively. The lab magnetic-dominated curve has cycloidal cusps; the electric-dominated curve has an oblique relativistic asymptote.}
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