Question 5.2: Cycloid Motion. A more exotic trajectory occurs if we includ...

Let’s think it through qualitatively, first. Initially, the particle is at rest, so the magnetic force is zero, and the electric field accelerates the charge in the z-direction. As it picks up speed, a magnetic force develops which, according to Eq. 5.1, pulls the charge around to the right. The faster it goes, the stronger F_{mag} becomes; eventually, it curves the particle back around towards the y axis. At this point the charge is moving against the electrical force, so it begins to slow down—the magnetic force then decreases, and the electrical force takes over, bringing the particle to rest at point a, in Fig. 5.7. There the entire process commences anew, carrying the particle over to point b, and so on.

Now let’s do it quantitatively. There being no force in the x-direction, the position of the particle at any time t can be described by the vector (0, y(t), z(t)); the velocity is therefore

F_{mag}=Q(V\times B).        (5.1)

where dots indicate time derivatives. Thus

and hence, applying Newton’s second law,

Or, treating the \hat{y} and \hat{z} components separately,

For convenience, let

\omega =\frac{QB}{m}.                            (5.4)

(This is the cyclotron frequency, at which the particle would revolve in the absence of any electric field.) Then the equations of motion take the form

\ddot{y}=\omega \dot{z},\ddot{z}=\omega \left(\frac{E}{B}-\dot{y} \right).                               (5.5)

Their general solution^{4} is

\begin{cases}y(t) = C_1 \cos \omega t + C_2 \sin \omega t + (E/B)t + C_3, \\z(t) = C_2 \cos \omega t − C_1 \sin \omega t + C_4.\end{cases}            (5.6)

But the particle started from rest (\dot{y}(0)=\dot{z}(0)=0), at the origin (y(0) = z(0) =0); these four conditions determine the constants C_{1},C_{2},C_{3},     and    C_{4}:

y(t)\frac{E}{\omega B}(\omega t-\sin\omega t) , z(t)=\frac{E}{\omega B}(1-\cos\omega t) .         (5.7)

In this form, the answer is not terribly enlightening, but if we let

R\equiv \frac{E}{\omega B},                   (5.8)

and eliminate the sines and cosines by exploiting the trigonometric identity \sin^{2} \omega t + \cos^{2} \omega t = 1, we find that

(y − R\omega t)^{2} + (z − R)^{2} = R^{2}.        (5.9)

This is the formula for a circle, of radius R, whose center (0, Rωt, R) travels in the y-direction at a constant speed

u=\omega R=\frac{E}{B} .                   (5.10)

The particle moves as though it were a spot on the rim of a wheel rolling along the y axis. The curve generated in this way is called a cycloid. Notice that the overall motion is not in the direction of E, as you might suppose, but perpendicular to it.