 # Magnetic Deflection of Electrons

If you have not already done so please read the General Introduction to Experiments EF-1 and EF-2.

In this experiment we consider the effect of a magnetic field on the motion of electrons. Just as the electric field represents the interaction of two charged particles at rest, the magnetic field represents the interaction of charged particles resulting from relative motion. The electric field can be defined as the force per unit charge, and the magnetic field in terms of the force on a unit current element.

A magnetic field can be produced by a current in a conductor. The field, in turn, exerts a force on a particle of charge q moving through it according to the Lorentz force equation

Thus an electron moving through a magnetic field is accelerated by a force with magnitude F proportional to the component of velocity perpendicular to the field, in a direction always perpendicular to both the field  and the instantaneous velocity . This relation between the directions of  and  has an immediate and important consequence: The magnetic field force never does work on the particle, since the particle always moves in a direction perpendicular to the force acting on it. For this reason such a particle moves with constant kinetic energy and thus with constant speed. The direction of the velocity, of course, can change and in this experiment you will observe the deflection of an electron beam by a magnetic field oriented perpendicular to the direction of the beam.

Consider the situation shown in the next figure. Electrons with charge  emerge from the electron gun with a speed v determined by the energy relation

just as in Experiment EF-1. Now the beam enters a region of length l in which there is a uniform magnetic field B (the source of which will be discussed later) oriented perpendicular to the plane of the figure, pointing out of the page. The resulting magnetic force has a magnitude F = evB and is always perpendicular to the velocity. Further, since the acceleration produced by this force is at each instant perpendicular to , only the direction of  changes.

Now the conditions just described are those needed to produce circular motion at constant speed. The radius R of the circular arc can be obtained from the equation for centripetal acceleration. Equating the centripetal and magnetic forces we have

After leaving the region of the magnetic field, the electrons again travel in a straight line, deflected by an angle  from the original axial direction. The length of the circular arc is S and hence . If  is small and R is large the arc length S will be approximately equal to l and the angle  will be small, giving

Also, from the figure, the transverse displacement a at the point of emergence from the field is

The beam strikes the screen at a point displaced a distance D from the undeflected beam position. The total displacement is given by

When the above expressions for  and a are substituted into this equation, the results are somewhat complicated. It can be simplified considerably by using the fact that  is small so that we can use the approximations  and . We then find the total deflection D to be

Using the energy relation, Eq. (1), we obtain

As this expression shows, the beam deflection is proportional to the magnetic field B. It depends inversely on the square root of the accelerating potential; in contrast to experiement EF-1 where the deflection varied as  itself, not its square root. The difference is that here we have an additional velocity dependence because of the nature of the magnetic force.

The magnetic field will be produced by a pair of solenoidal coils arranged as shown below. In the figure, the CRT face is shown end-on, while the solenoid coils are shown as if cut by a plane containing their longitudinal axis. Point C in the figure is at the center of the CRT screen and on the axis of the solenoids. The angles  and  are between lines from C to the ends of the coil windings and the axis. .5in

For a solenoid of infinite length the magnetic field is uniform along the axis and across the space inside the solenoid. The field strength is given by  where n is the number of turns per unit length of the coil, I is the current, and the constant . The expression for the field strength for one finite coil is given by

where the angles  and  are as indicated in the figure. They can be determined from the dimensions of the apparatus. Of course, the total field strength for two coils is twice the value calculated from Eqn. 3.

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