Motion in a uniform electric or magnetic field
We want now to describe—mainly in a qualitative way—the motions of charges in various circumstances. Most of the interesting phenomena in which charges are moving in fields occur in very complicated situations, with many, many charges all interacting with each other. For instance, when an electromagnetic wave goes through a block of material or a plasma, billions and billions of charges are interacting with the wave and with each other. We will come to such problems later, but now we just want to discuss the much simpler problem of the motions of a single charge in a given field. We can then disregard all other charges—except, of course, those charges and currents which exist somewhere to produce the fields we will assume. We should probably ask first about the motion of a particle in a uniform electric field. At low velocities, the motion is not particularly interesting—it is just a uniform acceleration in the direction of the field. However, if the particle picks up enough energy to become relativistic, then the motion gets more complicated. But we will leave the solution for that case for you to play with. Next, we consider the motion in a uniform magnetic field with zero electric field. We have already solved this problem—one solution is that the particle goes in a circle. The magnetic force $q\FLPv\times\FLPB$ is always at right angles to the motion, so $d\FLPp/dt$ is perpendicular to $\FLPp$ and has the magnitude $vp/R$, where $R$ is the radius of the circle: \begin{equation*} F=qvB=\frac{vp}{R}. \end{equation*}  The radius of the circular orbit is then \begin{equation} \label{Eq:II:29:1} R=\frac{p}{qB}. \end{equation}  That is only one possibility. If the particle has a component of its motion along the field direction, that motion is constant, since there can be no component of the magnetic force in the direction of the field. The general motion of a particle in a uniform magnetic field is a constant velocity parallel to $\FLPB$ and a circular motion at right angles to $\FLPB$—the trajectory is a cylindrical helix (Fig. 29–1). The radius of the helix is given by Eq. (29.1) if we replace $p$ by $p_\perp$, the component of momentum at right angles to the field.