= Solution
The <Lorentz force> determines the trajectory:
$$
\boxed{\frac{d\mathbf p}{dt}=q(\mathbf E+\mathbf v\times\mathbf B),\qquad
\frac{dE}{dt}=q\mathbf E\cdot\mathbf v.}
$$
Thus a <magnetic field> changes <momentum> direction but does no work; an <electric field> can change the particle's <kinetic energy>. In a uniform <electric field> with negligible <magnetic field> and nonrelativistic <speed>, $\mathbf a=q\mathbf E/m$ is constant. A transverse initial <velocity> then gives a <parabola>.
In a uniform <magnetic field>, the parallel <velocity> is unchanged and the perpendicular motion is circular. Nonrelativistically,
$$
r_L=\frac{mv_\perp}{|q|B},\qquad\omega_c=\frac{|q|B}{m};
$$
the combined trajectory is a helix, with circulation reversed when the charge sign reverses. Relativistically $r_L=p_\perp/(|q|B)$ and the gyrofrequency is $|q|B/(\gamma m)$.
For perpendicular uniform fields in the magnetic-dominated case, the drift <velocity> is $\mathbf v_d=\mathbf E\times\mathbf B/B^2$, which obeys $\mathbf E+\mathbf v_d\times\mathbf B=0$. A particle at this <velocity> is undeflected: the crossed-field <velocity selector> selects <speed> $E/B$ in the appropriate direction, <independent> of <mass> and charge. Radiation reaction and collisions are neglected in these ideal trajectories.
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