Particle acceleration by an induced electric field; the betatron
We have said that the electromotive force generated by a changing magnetic field can exist even without conductors; that is, there can be magnetic induction without wires. We may still imagine an electromotive force around an arbitrary mathematical curve in space. It is defined as the tangential component of $\FLPE$ integrated around the curve. Faraday’s law says that this line integral is equal to minus the rate of change of the magnetic flux through the closed curve, Eq. (17.3).  As an example of the effect of such an induced electric field, we want now to consider the motion of an electron in a changing magnetic field. We imagine a magnetic field which, everywhere on a plane, points in a vertical direction, as shown in Fig. 17–4. The magnetic field is produced by an electromagnet, but we will not worry about the details. For our example we will imagine that the magnetic field is symmetric about some axis, i.e., that the strength of the magnetic field will depend only on the distance from the axis. The magnetic field is also varying with time. We now imagine an electron that is moving in this field on a path that is a circle of constant radius with its center at the axis of the field. (We will see later how this motion can be arranged.) Because of the changing magnetic field, there will be an electric field $\FLPE$ tangential to the electron’s orbit which will drive it around the circle. Because of the symmetry, this electric field will have the same value everywhere on the circle. If the electron’s orbit has the radius $r$, the line integral of $\FLPE$ around the orbit is equal to minus the rate of change of the magnetic flux through the circle. The line integral of $\FLPE$ is just its magnitude times the circumference of the circle, $2\pi r$. The magnetic flux must, in general, be obtained from an integral. For the moment, we let $B_{\text{av}}$ represent the average magnetic field in the interior of the circle; then the flux is this average magnetic field times the area of the circle. We will have \begin{equation*} 2\pi rE=\ddt{}{t}(B_{\text{av}}\cdot\pi r^2). \end{equation*}  Since we are assuming $r$ is constant, $E$ is proportional to the time derivative of the average field: \begin{equation} \label{Eq:II:17:4} E=\frac{r}{2}\,\ddt{B_{\text{av}}}{t}. \end{equation}  The electron will feel the electric force $q\FLPE$ and will be accelerated by it. Remembering that the relativistically correct equation of motion is that the rate of change of the momentum is proportional to the force, we have \begin{equation} \label{Eq:II:17:5} qE=\ddt{p}{t}. \end{equation}  For the circular orbit we have assumed, the electric force on the electron is always in the direction of its motion, so its total momentum will be increasing at the rate given by Eq. (17.5). Combining Eqs. (17.5) and (17.4), we may relate the rate of change of momentum to the change of the average magnetic field: \begin{equation} \label{Eq:II:17:6} \ddt{p}{t}=\frac{qr}{2}\,\ddt{B_{\text{av}}}{t}. \end{equation}  Integrating with respect to $t$, we find for the electron’s momentum \begin{equation} \label{Eq:II:17:7} p=p_0+\frac{qr}{2}\,\Delta B_{\text{av}}, \end{equation}  where $p_0$ is the momentum with which the electrons start out, and $\Delta B_{\text{av}}$, is the subsequent change in $B_{\text{av}}$. The operation of a betatron—a machine for accelerating electrons to high energies—is based on this idea. To see how the betatron operates in detail, we must now examine how the electron can be constrained to move on a circle. We have discussed in Chapter 11 of Vol. I the principle involved. If we arrange that there is a magnetic field $\FLPB$ at the orbit of the electron, there will be a transverse force $q\FLPv\times\FLPB$ which, for a suitably chosen $\FLPB$, can cause the electron to keep moving on its assumed orbit. In the betatron this transverse force causes the electron to move in a circular orbit of constant radius. We can find out what the magnetic field at the orbit must be by using again the relativistic equation of motion, but this time, for the transverse component of the force. In the betatron (see Fig. 17–4), $\FLPB$ is at right angles to $\FLPv$, so the transverse force is $qvB$. Thus the force is equal to the rate of change of the transverse component $p_t$ of the momentum: \begin{equation} \label{Eq:II:17:8} qvB=\ddt{p_t}{t}. \end{equation} When a particle is moving in a circle, the rate of change of its transverse momentum is equal to the magnitude of the total momentum times $\omega$, the angular velocity of rotation (following the arguments of Chapter 11, Vol. I): \begin{equation} \label{Eq:II:17:9} \ddt{p_t}{t}=\omega p, \end{equation} where, since the motion is circular, \begin{equation} \label{Eq:II:17:10} \omega=\frac{v}{r}. \end{equation} Setting the magnetic force equal to the transverse acceleration, we have \begin{equation} \label{Eq:II:17:11} qvB_{\text{orbit}}=p\,\frac{v}{r}, \end{equation} where $B_{\text{orbit}}$ is the field at the radius $r$.  As the betatron operates, the momentum of the electron grows in proportion to $B_{\text{av}}$, according to Eq. (17.7), and if the electron is to continue to move in its proper circle, Eq. (17.11) must continue to hold as the momentum of the electron increases. The value of $B_{\text{orbit}}$ must increase in proportion to the momentum $p$. Comparing Eq. (17.11) with Eq. (17.7), which determines $p$, we see that the following relation must hold between $B_{\text{av}}$, the average magnetic field inside the orbit at the radius $r$, and the magnetic field $B_{\text{orbit}}$ at the orbit: \begin{equation} \label{Eq:II:17:12} \Delta B_{\text{av}}=2\,\Delta B_{\text{orbit}}. \end{equation} The correct operation of a betatron requires that the average magnetic field inside the orbit increases at twice the rate of the magnetic field at the orbit itself. In these circumstances, as the energy of the particle is increased by the induced electric field the magnetic field at the orbit increases at just the rate required to keep the particle moving in a circle.  The betatron is used to accelerate electrons to energies of tens of millions of volts, or even to hundreds of millions of volts. However, it becomes impractical for the acceleration of electrons to energies much higher than a few hundred million volts for several reasons. One of them is the practical difficulty of attaining the required high average value for the magnetic field inside the orbit. Another is that Eq. (17.6) is no longer correct at very high energies because it does not include the loss of energy from the particle due to its radiation of electromagnetic energy (the so-called synchrotron radiation discussed in Chapter 34, Vol. I). For these reasons, the acceleration of electrons to the highest energies—to many billions of electron volts—is accomplished by means of a different kind of machine, called a synchrotron.