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.
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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.