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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 Power supply 500 V (115 V, 50/60 Hz) 1003307
or
1 Power supply 500 V (230 V, 50/60 Hz) 1003308
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 per-
pendicularly through a magnetic field B experi-
ences a force F that is perpendicular to both B
and v and the magnitude of which is given by:
F =
evB
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
2
mv
=
=
F
evB
R
which implies
v
B =
tesla
e
R
m
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
1014525
1000611
1003073
Rearranging the equation gives
e
v
=
m
BR
If the beam is subjected to a known magnetic
field of magnitude B, and v and R are both calcu-
lated 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
potential energy of a charge moving from point 1
to point 2 is equal to zero since no work is per-
formed by external forces.
⎛
⎞
1
1
2
2
⎜
−
⎟
mv
mv
2
1
2
2
⎝
⎠
The energy of an electron in the dual beam tube
is given by:
1
2
=
eU
mv
A
2
By solving for v and replacing it in the equation
e
v
=
m
BR
the following emerges
2
e
U
=
A
2
2
m
B
R
The term e/m is the specific charge of an elec-
tron and has the constant value (1.75888 ±
11
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
configuration as given by
=
μ
= (4.17 x 10
B
H
0
−
2
6
2
=
⋅
17
.
39
10
B
I
H
where
I is the current in the Helmholtz coils.
H
The following are also true
e
U
=
⋅
⋅
. 1
15
10
A
2
2
m
I
R
H
U
2
=
A
I
k
H
2
R
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 circular path with the tube axis
forming a tangent. The centre of this circle is at
B and it lies in the plane of DCD' about 2 mm
behind the plane of EE'.
2
2
2
=
+
AB
BC
AC
2
(
)
+
−
=
0
eU
eU
2
1
-3
)
tesla
I
H
and
5
and
−
⋅
2
BC
DC
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