and it is at right angles both to the field and to the velocity. As usual, the force is equal to the rate of change of momentum with time. If the field is directed upward out of the paper, the momentum of the particle and the force on it are as shown in Fig. 34-4. Since the force is at right angles to the velocity, the kinetic energy, and therefore the speed, remains constant. All the magnetic field does is to ge the direction of motion. In a short time At, the momentum vector changes at right angles to itself by an amount Ap= FAL, and therefore p turns through an angle 40= Ap/p= qUB At/p, since F= quB. But in this same time the particle has gone a distance As v AL. Evidently, the two lines AB and CD will intersect at a point O such that OA =OC = R, where As =R A0. Combining this with the find r△0/△ P in a circular (or helical)path in a uniform magnetic field qub/p Since this same argument can be applied during the next instant, the next, and so on,we conclude that the particle must be moving in a circle of radius R, with angu- The result that the momentum of the particle is equal to a charge times the radius times the magnetic field is a very important law that is used a great deal It is important for practical purposes because if we have elementary particles which all have the same charge and we observe them in a magnetic field, we can measure the radii of curvature of their orbits and, knowing the magnetic field, thus deter- mine the momenta of the particles. If we multiply both sides of Eq (34.7)by c, and express q in terms of the electronic charge, we can measure the momentum in units of the electron volt. In those units our formula is 3×105(q/q)BR where B, R, and the speed of light are all expressed in the mks system, the latter numerically. The mks unit of magnetic field is called a weber per square meter. There is an older unit which is still in common use, called a gauss. One weber/m is equal to 10* gauss. To give an idea of how big magnetic fields are, the strongest magnetic field that one can usually make in iron is about 1.5x 10* gauss; beyond that, the advantage of using iron disappears. Today, electromagnets conducting wire are able to produce steady fields of over 10 gauss strength--that is, 10 mks units. The field of the earth is a few tenths of a gauss at the equator. Returning to Eq. (34.9), we could imagine the synchrotron running at a billion electron volts, so pc would be 10 for a billion electron volts. (We shall come back to the energy in just a moment. Then, if we had a b corresponding to, say, 10,000 gauss,which is a good substantial field, one mks unit, then we see that R would have to be 3.3 meters. The actual radius of the Caltech synchrotron is 3.7 meters, the field is a little bigger, and the energy is 1.5 billion, but it is the same idea. S now we have a feeling for why the synchrotron has the size it has We have calculated the momentum, but we know that the total energy, including the rest energy, is given by W= vpac2 +m c4, and for an electron the rest energy corresponding to mc2 is 0.511 X 106 ev, so when pe is 10 ev we can neglect mc2, and so for all practical purposes w=pc when the speeds are relativistic. It is practically the same to say the energy of an electre billion electron volts as to say the momentum times c is a billion electron volts. If w 10%ev, it is easy to show that the speed differs from the speed of light by but one We turn now to the radiation emitted by such a particle. a particle moving on a circle of radius 3. 3 meters, or 20 meters circumference, goes around once in roughly the time it takes light to go 20 meters. So the wavelength that should be emitted by such a particle would be 20 meters-in the shortwave radio region But because of the piling up effect that we have been discussing(Fig. 34-3), and because the distance by which we must extend the radius to reach the speed c is