Thomson’s Charge-to-Mass Experiment (EB3)
Lab Exercise (“dry lab”)
The discovery of cathode rays in the mid-1800s created a lot of excitement in the physics
community. Sir William Crookes did some clever experiments in 1875 to determine several
qualitative properties of cathode rays, but he was not able to determine whether the rays were a
form of light or a stream of charged particles. Finally, in 1897 Sir Joseph John Thomson created
an experimental design and apparatus to show that cathode rays are negatively charged particles.
Complete the Analysis and Evaluation sections of the laboratory report. The evidence has
been gathered for you. If possible, use a spreadsheet to record and calculate required quantities,
create a best-fit straight line and determine the slope of the line.
Purpose
The purpose of this experiment is to create a value of the e/m ratio for cathode rays.
Problem
What is the charge-to-mass ratio of the negatively charged particles in a cathode ray?
Experimental Design
A cathode ray is produced that travels perpendicular
to a net, external magnetic field (Figure 1). In each
trial of the experiment, the electric current in the pair
of wire loops (Helmholtz coils, Figure 2) is
manipulated for a constant accelerating voltage of
the beam. The radius of curvature of the beam is
measured as the responding variable. Three different
trials with three different accelerating voltages of the
cathode ray are conducted.
Figure 1: e/m Tube
Materials
-
-
e/m tube with pegs set at 3.25 cm, 3.90 cm, 4.50
cm, 5.15 cm and 5.75 cm (Figure 1)
pair of Helmholtz coils with magnetic field
aligned anti-parallel to Earth’s magnetic field
(Figure 2)
variable power supplies for the cathode ray tube
and the Helmholtz coils
A
I
18o
Procedure
1. Adjust the accelerating voltage to about 20 V


BE
BH
with the coil current set to zero. Record the
specific voltage.
Figure 2: Alignment of Helmholtz Coils
2. Keeping the accelerating voltage constant,
increase the electric current in the coils until the


beam aligns with the outside of the first (green) peg on the crossbar. Record the electric
current.
3. Repeat step 2 for each of the other pegs.
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Thomson’s Charge-to-Mass Experiment (EB3)
4. Repeat steps 1 to 3 using accelerating voltages of about 30 V and about 40 V.
Evidence/Analysis
Accelerating
Voltage, V
Helmholtz
Current, I
(A)
Peg
Radius, r
(V)
1
( 102 m)
3.25
2
3.90
20.5
3
4.50
2.01
4
5.15
1.75
5
5.75
1
3.25
2
3.90
3
4.50
2.34
4
5.15
2.09
5
5.75
1
3.25
2
3.90
3
4.50
2.69
4
5.15
2.39
5
5.75
2.69
(approx. 20 V)
2.29

2V
r
½
(V / m)
BH
(T)


1.61
3.19
30.1
(approx. 30 V)
2.70
1.88
3.64
40.5
(approx. 40 V)
3.09
2.15
(x-axis values)
(y-axis values)
According to the law of conservation of energy, the change in potential energy of the electron
equals the increase in its kinetic energy at the anode (Figure 1).
Eqn 1
Ep

Ek
qeV

1
2
mv 2
(assuming vi = 0)
According to Newton’s second law, the net force acting perpendicular to the motion of the
electrons causes a centripetal acceleration. The net force is the magnetic force from the net
 magnetic field.
external
Fnet

 mac
B  qe v 
Eqn 2
mv 2
r
Rearrange both equations to solve for v2 and then set these two expressions equal to each other.

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Thomson’s Charge-to-Mass Experiment (EB3)
2
2qeV
m

(Eqn 1)
B qe2 r 2
m2
(Eqn 2)
Simplifying the terms and rearranging this equation,

Eqn 3
qe

m
2V
 2
B r 2
The net external magnetic field is the vector addition of the magnetic field of the Helmholtz coils
and Earth’s magnetic field (Figure 2). The general symbol for electric charge, q, is replaced by the
 symbol, e, for an elementary charge.
alternate
Eqn 4
e
m
2V

2
 

2
BH  B E  r



According to Ampere’s law, the magnetic field of the Helmhotz coils is calculated as follows:

Eqn 5

BH
0 = 4  107 Tm/A
 8  N 
  0 I
R 125 
where
N = 72
R = 0.33 m
I = measured Helmholtz current
Eqn
4 is rearranged into the form of a general mathematical equation for a straight line and the
charge-to-mass ratio (e/m) is calculated from the slope of the best-fit straight line for a graph of

2V
BH versus
.
r


Eqn 6
BH

(y

m 2V

e
r
mx


BE
 b)

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