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3. Technical data
Filament voltage:
Anode voltage:
Anode current:
Deflector voltage:
Glass bulb:
Total length:
Gas filling:
4. Operation
To perform experiments using the dual beam tube,
the following equipment is also required:
1 Tube holder S
1 DC power supply 0 – 500 V
or
1 DC power supply 0 – 500 V
1 Helmholtz pair of coils S
1 Analogue multimeter AM50
4.1 Setting up the tube in the tube holder
The tube should not be mounted or removed unless
all power supplies are disconnected.
•
Press tube gently into the stock of the holder
and push until the pins are fully inserted. Take
note of the unique position of the guide pin.
4.2 Removing the tube from the tube holder
•
To remove the tube, apply pressure with the
middle finger on the guide pin and the thumb
on the tail-stock until the pins loosen, then pull
out the tube.
5. Example experiments
5.1 Determination of e/m
An electron of charge e moving at velocity v perpen-
dicularly through a magnetic field B experiences a
force F that is perpendicular to both B and v and the
magnitude of which is given by:
F =
This causes the electron to follow a circular electron
path in a plane perpendicular to B. The centripetal
force for an electron of mass m is
mv
=
F
which implies
7.5 V AC/DC max.
100 V DC max.
30 mA max.
50 V DC max
130 mm dia. approx.
260 mm approx.
Helium at 0.1 torr
pressure
U185001
U33000-115
U33000-230
U185051
U17450
evB
2
=
evB
R
Rearranging the equation gives
If the beam is subjected to a known magnetic field of
magnitude B, and v and R are both calculated then
the ratio e/m can be determined.
The law of conservation of energy means that the
change in kinetic energy plus the change in poten-
tial energy of a charge moving from point 1 to point
2 is equal to zero since no work is performed by
external forces.
⎛
1
2
−
⎜
mv
2
⎝
2
2
The energy of an electron in the dual beam tube is
given by:
By solving for v and replacing it in the equation
the following emerges
The term e/m is the specific charge of an electron
and has the constant value (1.75888 ± 0.0004) x 10
C/kg.
5.1.1 Determination of B
The Helmholtz coils have a diameter of 138 mm and
give rise to a magnetic flux in Helmholtz configura-
tion as given by
=
μ
B
H
= (4.17 x 10
0
2
B
where
I is the current in the Helmholtz coils.
H
The following are also true
e
=
m
I
=
2
I
k
H
5.1.2 Determination of R
Referring to the diagram Fig. 1, the beam emerges
from the electron gun at C travelling along the axis
of the tube. The electron is then deflected in a circu-
lar path with the tube axis forming a tangent. The
2
v
B =
tesla
e
R
m
e
v
=
m
BR
⎞
1
(
)
2
+
−
=
⎟
mv
eU
eU
0
1
2
1
⎠
1
=
2
eU
mv
A
2
e
v
=
m
BR
e
2
U
=
A
2
2
m
B
R
-3
I
)
tesla
H
and
−
=
⋅
6
2
17
.
39
10
I
H
U
⋅
⋅
5
A
1
.
15
10
and
2
2
R
H
U
A
2
R
11
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