Experiment 7
The Determination of
The Charge to Mass Ratio of the Electron
A magnetic field produced by Helmholtz coils is used to deflect electrons into circular
paths whose radii are known. By knowing the energy of the electrons and the magnetic field
strength, the ratio of the charge to mass of the electron (e/m) is determined.
Theory
A tube designed specifically to determine the charge to mass ratio of the electron is used
in this experiment. This e/m tube consists of an electron gun composed of a straight filament
surrounded by a coaxial cylindrical anode containing a single axial slit. The beam of electrons is
produced by this gun. (Refer to Figure 1.)The various parts in Figure are labeled as follows:
A)
B)
C)
D)
E)
F)
G)
five crossbars attached to a stiff
wire
a typical path of the beam of
electrons
the cylindrical anode
the distance from the filament to
the far side of the crossbar
the lead wire and support for
anode
the filament
the lead wires and support forthe
filament
Figure I (a) A top view of the e/m tube.
(b) A crossectional view of
the filament assembly.
Electrons emitted from theheated filament F are accelerated by the potential difference
applied between F and the anode C. Some of the electrons emerge from the slits as a narrow
beam. When electrons of a sufficiently high kinetic energy (10.4 eV or more) collide with the
mercury atoms in the tube, a fraction of the atoms will be ionized. On recombination of these
ions with stray electrons, the mercury arc spectrum is enitted with its charactefistic blue color.
This makes the path of the beam of electrons visible as the electrons travel through the mercury
vapor.
The e/m tube is cradled at the center of Helmholtz coils producing a magnetic field that deflects
the electrons into a circular path.
7-1
When a charged I particle such as an electron moves in a magnetic field in a direction at
right angles to the field, it is acted on by a force, the value of which is given by
F = qvB,
(1)
where B is the magnetic field strength, q is the charge on the particle (in this case q = e because
the particle is an electron), and v is the velocity of the particle. This force causes the electron to
move in a circular path in a plane perpendicular to the magnetic field. The radius of this circle is
such that the required centfipetal force is furnished by the force exerted on the electron by the
magnetic field. Therefore,
mv 2
evB 
,
r
(2)
where m is the mass of the electron and r is the radius of its circular path.
If the velocity of the electron is due to its being accelerated through a potential difference
V, then it has a kine.tic energy of
1
mv 2  eV .
2
(3)
e
2V
 2 2.
m B r
(4)
Substituting the value of v from (3) into (2)
Thus, when the accelerating voltage, the field strength, and the radius of the circular path
described by the electron are known, then the value of elm can be computed.
The magnetic field B that causes the electron to move in a circular path is produced by
Helmholtz coils and has a value of
B
8 o NI
125a
,
(5)
where N is the number of tums of wire on each coil, I is the current through the coils, a is the
mean radius of the coils, and m. is the permeability of free space. Substituting (5) into (4) gives
e  3.91a 2  V 


.
m   o2 N 2  I 2 r 2 
7-2
(6)
Equation (6) is the working equation for the apparatus used. The quantity within the first set of
parentheses is constant for any given pair of Helmholtz coils. (For the Helmholtz coils with
which we are working, the number of turns of wire N = 72 and the mean radius of the coils a =
0.33 meters.) The value of r, the radius of the circle in which the electron beam travels, can be
varies by changing either the accelerating voltage or the Helmholtz coil field current.
The current 1, the Helmholtz coil field current that is used to deflect the electrons,
consists of two parts. The first part I, is the current required to cancel out the effect of the earth's
magnetic field. (In this case the earth's magnetic field, although relatively weak, causes a
significant effect on the deflection of the beam. The Helmholtz coils are also tilted so that the
field produced by them is in a direction that is opposite to that of the earth's magnetic field.)
The second part of the current I2is the total current being delivered to the coils. This
includes both I, and the amount of current actually needed to deflect the electrons into their
prescribed paths. Therefore I in (6) is
I = I2 – I1
(7)
Apparatus
o
e/m vacuum tube, with the outside edge of the crossbars located at the following
distances from the filament:
crossbar
o
o
o
1
2
3
4
5
0.065 ± 0.001 meters
0.078 ± 0.001
0.090 ± 0.001
0.103 ± 0.001
0.115 ± 0.001
Helmholtz coils, 72 tums of wire on each coil and a mean radius of 0.33 ±O.Ol m
3 DC power supplies
o magnetic dip needle
3 digi.tal voltmeters
o leads
Procedure
The apparatus is assumed to be aligned properly. The instructor will set the value of the
accelerating voltage V. The only adjustment the student needs to make is to vary the Helmholtz
coil current.
1) Read and record the value of the accelerating voltage.
7-3
2) By adjusting the current to the Helmholtz coils, find the current that causes the beam of
electrons to travel in a straight line. This current is I1.
3) Increase the current to the Helmholtz coils until the electron beam descfibes a circle.
Make fine adjustments to the field current until the outside edge of the electron beam is
tangent to the outside edge of one of the crossbars.
4) Repeat the steps in (3) for the remaining crossbars.
Analysis
Use (6) and (7) to calculate the charge to mass ratio of the electron. One value is to be
found for each crossbar and an average value is found. In a table, list the individual values of
e/m, the average value, the theoretical value, and the percentage error between the average value
and the theoretical value.
Conclusions
Explain what factor(s) caused the greatest error in your determination of the charge to
mass ratio of the ele.ctron and why.
Questions
1) Start with the magnetic field produced by a circular loop, and derive the expression for
the magnetic field produced by Helmholtz coils (5).
2) Why is the outer edge of the beam of electrons used to determine the radius of the path?
7-4