W9_parallel_resonance.eps

Week 6: Moving Charges and Magnetic Force 205

e−

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E

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L D

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Figure 69: Joseph John Thomson’s apparatus for measuring the ratio of the charge on the electron
to its mass (improved by the addition of a magnetic velocity selector). This was a critical experiment
in determining the structure of the atom, for which Thomson received the Nobel Prize (only the
sixth such prize awarded in physics).

A cartoon schematic of Thomson’s apparatus is shown above, although I had a very hard time
finding any useful picture of how he applied the magnetic field in his second series of experiments
and the magnetic part of the apparatus may be incorrect.

Let’s see how Thomson used his apparatus to measuree/mfor the electron. First, he cooked
up some electrons using a wire heated by joule heating until electrons “boiled off”. These slow
electrons passed through collimating slits and fell across a potential difference maintained between
the plates containing the slits to speed them up to a roughly consistent velocity. The electrons, each
with (approximate) velocityv 0 then entered the region between to capacitor plates built right into
the tube. The downward electric field then produced an upward, constant upward acceleration and
hence deflection of the electrons (where we can completely ignore gravity in the experiment as the
electrical acceleration was vastly greater) as they traversed the plate lengthL. On the far side they
emerged from the field, travelled in a straight line for anxdistance ofD, and then struck the glass
of the screen, where they made a glowing spot.

By measuring thetotal distanceof upward deflection of the spot from the center of the screen
(where they struck when theE-field was off) and the point where they struck when theE-field was
turned on to some known value, Thomson could reason backwards to the ratio ofe/mas follows.

First, we analyze the constant acceleration motion of the electronwhile it is between the plates:

Fx = 0 (418)

Fy = eE=may (419)

from which we find (using 2D kinematics from the first semester – theproblem isidenticalto
analyzing trajectories with a constant gravitational acceleration):

x(t) = v 0 t (420)

vx(t) = v 0 (421)

y(t) =

1

2 ayt

(^2) =eE
2 mt
(^2) (422)
vy(t) = ayt=
eE
m
t (423)
We can easily find thetimethe electron is between the plates:
t 1 =

L

v 0 (424)