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Physics 143
Measurement of e/m for the Electron
I. General Description of Experiment
It has been proposed that electricity is actually smoke. The evidence to support this
is the well-known property of circuits and motors to fume, spit, spark, and generally
emit gobs of smoke as they cease to work. Obviously, once the smoke (electricity) has
gotten out, the energizer bunny slumps to a sad, silent halt.
Our modern view holds that electricity is the movement of charges, which are a
fundamental property of some types of matter. The way in which those charges
associate themselves with matter is therefore of interest. This experiment is a
conclusive demonstration of one of the most fundamental properties of electrical
charges (specifically, the electron): that the ratio of its charge to its mass is invariant.
This means that charges do not accumulate in various amounts on particles of matter,
but rather that some fundamental type of particle always possesses a precise charge.
You might even call it a quantum (Latin for “how much”) of charge.
We can find this ratio of the electron’s charge to its mass (e/m) by a careful
arrangement of electric (a known potential difference V) and magnetic (B) fields .
II. What’s Up
It was discovered pretty early on that negatively-charged particles as well as light
are emitted from a metal coil if you heat it up enough. In our apparatus we use a hot
filament to “boil” off electrons inside an evacuated glass bulb.
The next thing that happens is the electrons are accelerated away from the filament.
This is accomplished by putting the filament inside a “cathode can”. In the spirit of
clarity, we have labeled the dial and switch corresponding to this the “anode voltage”.
You might remember from last year’s E& M class that a potential difference (a voltage
difference) will act to accelerate or decelerate charges. To get the electrons to move
away from the filament, we put the outside of the can (see figure below) at maybe +20
volts above the voltage at the filament. Therefore the electrons, once they are released
from the filament, start to streak outwards in all directions toward the walls of the can.
Only they don’t all hit the can walls. A small slit in the can (small enough not to
disturb the symmetry of the electric field which has been accelerating the electrons) lets
a few electrons out of the can. We know now almost exactly how much (kinetic)
energy they have! The empty region of space that the electrons travel through (the slit)
is basically at +20 volts, just like the rest of the cathode can. And the energy that the
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charged particles picks up scales linearly with the amount of charge and the potential
difference V according to
Energy = q V
(1)
And we also know what direction these guys are going.
(Figure 1: can, filament, potential difference, and direction of motion)
Figure 1.
This by itself is not enough. If the electrons had huge mass and little charge, they
will be moving at a low velocity. And if they have a small mass and carry lots of
charge, they will be moving very fast. We must “clock” them somehow. One good
method is to use a magnetic field, here generated by Helmholtz coils. Remembering
that
F=q V x B
(2)
(F is force, q is amount of charge, V is the velocity the charges are moving at) we can
predict that the force generated by the electrons’ motion through the B field will
always point perpendicular to the electron’s velocity. It will also point perpendicular
to the B field. (Look up “cross product” if this doesn’t make sense.) So we need to
arrange the B field carefully beforehand. If we do it right, the electrons will follow a
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level, circular path inside the light bulb. And if we know the radius of that circle, we
know something else:
2
F = m V /r
(3)
Finally, if we go back and remember the amount of energy the electrons got from the
potential difference between the filament and the cathode can:
2
1/2 m V = energy = q v
(4)
(where little v in this equation is the voltage difference and capital V is the velocity)
then we can use these few equations and turn them into a formula for finding the
charge-to-mass ratio of the electron. Derive such a formula and show your work in an
appendix to the finished report.
III. The Rack
That imposing frankenstein-looking board full of switches, levers and dials is half
of the apparatus. The other half is the huge light bulb glued between two hula hoops.
We’ve already discussed what is going on inside the glass bulb. You will be able to
“see” the electrons inside the bulb because it is not a perfect vacuum inside; a tiny
amount of mercury vapor in the bulb glows purple when struck by electrons. The
experimental challenges here are to
1. Know what potential difference the electrons experience
2. Measure or compute the magnetic field in the bulb
3. Measure the radius of the circle that the electrons traverse.
The magnetic induction B, supplied by the Helmholtz Coils (those two big ol’
hula hoops), is given theoretically by
B
 0 NIR 2
3/2
d 2  R 2
,
(5)
whereo (the permeability of the vacuum) = 4 x 10-7 weber/amp-meter,
N = number of turns of wire in each Helmholtz Coil,
I = current in amperes,
R = the mean radius of the coils in meters,
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2d = the distance between the coils.
For this apparatus N = 72, 2d = R, and R = 0.33m (see Figure 2).
However, we must remember that the earth and thus room 101 also has a magnetic
field. We will line up the whole apparatus so that the Earth’s field is pointing in the
exact same direction as the field generated by the Helmholtz coils. Then we adjust
the current through the coils (and the B field thereby) until there is zero magnetic
field in the region of the bulb. (How can we tell when the magnetic field
somewhere is zero?) We then correct all future current measurements by this
amount, because the field generated scales linearly with the current. For example,
if you found out it took .2 amps to exactly “zero” the field, and the gauge now
reads .5 amps, it’s just like we walked into a region in deep space where there is
zero magnetic field and cranked the Helmholtz coil current up to .3 amps.
Confused? Go over this part until you understand what you’re doing; it’s
crucial. Ask for help if you need it.
Figure 2.
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IV. Experimental Procedure
Step 1.
Move the apparatus to a position in which the long axis of the e/m tube lies in
the magnetic north-south direction.
Step 2.
Observe the magnetic declination with a dip needle and tilt the coil so that the
plane of the coils makes an angle with the horizontal equal to the complement of
the dip angle. The axis of the coils will then be parallel to the earth's field. Why is
this necessary? Extinguish the bright room lights.
Step 3.
Apply the acceleration potential to the anode by closing the ANODE switch.
This potential may be of any value between 30 and 45 volts. (See Figure 2)
Step 4.
Turn on the filament-current, power-supply (first setting regulator knob on
zero) and increase the current slowly (using the rheostat), watching both filamentcurrent ammeter and the tube, until the electron beam appears. The beam appears
rather abruptly and becomes more intense with increasing current. Increase the
current slightly, if necessary, until the milliammeter indicates a filament-anode
current of 3 or 4 milliamperes and the beam is seen to strike the wall of the tube.
DO NOT LET THE FILAMENT CURRENT EXCEED 4.2 AMPERES AT ANY
TIME. DOING SO MAY BURN OUT THE FILAMENT. DO NOT LET THE
ACCELERATING VOLTAGE EXCEED 45 VOLTS. DO NOT LET THE BEAM
CURRENT EXCEED 5 MILLIAMPERES.
Step 5.
Rotate the tube slightly in its cradle, if necessary, so that the beam is horizontal and
the index bars extend upward from the staff wire. (Note that the electron beam is
deflected slightly downward by the earth's magnetic field.)
Step 6.
Now send a small current through the Helmholtz Coils. If the connections to the
coils have been made correctly, the beam will be seen to straighten or to curve in the
opposite direction.
Step 7.
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Adjust the current in the Helmholtz Coils until the beam is seen to be straight when
a ruler is held close to the tube. The field due to the Helmholtz Coils is now equal to
that of the earth's magnetic field. Record this current as Io. Io should be
redetermined if the accelerating voltage V is changed. Why does Io change?
Step 8.
Increase the field (Helmholtz Coil) current until the electron beam is bent into a
circular path touching the outermost vertical measuring bar (bar No. 5). Make fine
adjustment so that sharp outer edge of beam is tangent to outer edge (i.e., edge farthest
from filament) of bar. Call this current I5. (The outside edge of the beam is used
because it is produced by the electrons with the greatest velocity. The electrons leaving
the negative end of the filament fall through the greatest potential difference between
filament and anode, and have the greatest velocity. It is this potential difference that the
voltmeter measures.) Record the field current required to cause the beam to strike the
next measuring bars. Subtract Io from each of these field currents as a correction for the
effect of the earth's field.
V.
Data Analysis
Take enough data to calculate at least ten different e/m values. It is probably
easiest to use two accelerating voltage V and measure five current values I (one for each
bar) at each voltage. From your measurements determine the mean value, the variance,
and the error of the mean. Compare your result with the presently accepted value 1.76
x 1011 Coulomb/kg.
Bonus Question
Show that the aspect ratio (shape) of the Helmholtz coil gives the most spatially
uniform magnetic field. Hint: We could have put the coils any distance apart. What’s
special about 2d=R?
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Figure 3.