the effect of the
magnetic force we know e/mv. Combining these results we can find both
e/m and v.

[Illustration: FIG. 13.]

The method by which this determination is carried out in practice is
illustrated in fig. 13. The cathode rays start from the electrode C in
a highly exhausted tube, pass through two small holes in the plugs A
and B, the holes being in the same horizontal line. Thus a pencil of
rays emerging from B is horizontal and produces a bright spot at the
far end of the tube. In the course of their journey to the end of the
tube they pass between the horizontal plates E and D, by connecting
these plates with an electric battery a vertical electric field is
produced between E and D and the phosphorescent spot is deflected. By
measuring this deflection we determine e/mv^2. The tube is now placed
in a uniform magnetic field, the lines of magnetic force being
horizontal and at right angles to the plane of the paper. The magnetic
force makes the rays describe a circle in the plane of the paper, and
by measuring the vertical deflection of the phosphorescent patch at
the end of the tube we can determine the radius of this circle, and
hence the value of e/mv. From the two observations the value of e/m
and v can be calculated.

Another method of finding e/m for the negative ion which is applicable
in many cases to which the preceding one is not suitable, is as
follows: Let us suppose that the ion starts from rest and moves in a
field where the electric and magnetic forces are both uniform, the
electric force X being parallel to the axis of x, and the magnetic
force Z parallel to the axis of z; then if x, y, are the co-ordinates
of the ion at the time t, the equations of motion of the ion are--

d^2x dy
m ---- = Xe - He --,
dt^2 dt

d^2y dx
m ---- = He --.
dt^2 dt

The solution of these equations, if x, y, dx/dt, dy/dt all vanish when
t = 0, is

Xm / / e \ \
x = ---- {1 - cos( -- Ht ) }
eH^2 \ \ m / /

Xm /e / e \ \
y = ---- {-- Ht - sin( -- Ht ) }.
eH^2 \m \ m / /

These equations show that the path of the ion is a cycloid, the
generating circle of which has a diameter equal to 2Xm/eH^2, and rolls
on the line x = 0.

Suppose now that we have a number of ions starting from the plane x =