 Solution

The Lorentz force determines the motion of a charged particle:At nonrelativistic speeds, $\mathbf{p}=m\mathbf{v}$. The electric field does work at rate $q\mathbf{E}\cdot \mathbf{v}$, while the magnetic part of the Lorentz force is perpendicular to velocity and does no work. A uniform electric field gives constant acceleration $q\mathbf{E}/m$; with a transverse initial velocity, the trajectory is a parabola.

In a uniform magnetic field, the parallel velocity is constant, while the perpendicular component undergoes circular motion. The angular cyclotron frequency, orbit radius and helical pitch areThe sign of $q$ determines the sense of rotation. For perpendicular uniform electric fields and magnetic fields, the guiding centre has the electric drift velocityIndeed, $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$, so the remaining motion is cyclotron rotation about that drifting centre, with possible motion along the magnetic field. The drift is independent of electric charge sign and mass. At relativistic energies, retain $\mathbf{p}=\Gamma m\mathbf{v}$ and the corresponding relativistic orbit radius rather than assuming constant nonrelativistic acceleration.