 Solution

The Lorentz force determines the trajectory: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,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.