3B Scientific Teltron Dual Beam Tube D - Manual (Page 2)

2
3. Technical data
Filament voltage: max. 7.5 V AC/DC
Anode voltage:
max. 100 V DC
Anode current: max. 30 mA
Deflector voltage: max. 50 V DC
Glass bulb:
130 mm dia. approx.
Total length: 260 mm approx.
Gas filling: Helium at 0.1 torr pressure
4. Operation
To perform experiments using the dual beam
tube, the following equipment is also required:
1 Tube holder D
1008507
1 DC power supply 500 V (@230 V) 1003308
or
1 DC power supply 500 V (@115 V) 1003307
1 Helmholtz pair of coils D
1000644
1 Analogue multimeter AM50 1003073
4.1 Setting up the tube in the tube holder
The tube should not be mounted or removed
unless all power supplies are disconnected.
• Push the jaw clamp sliders on the stanchion
of the tube holder right back so that the jaws
open.
• Push the bosses of the tube into the jaws.
• Push the jaw clamps forward on the stan-
chions to secure the tube within the jaws.
4.2 Removing the tube from the tube holder
• To remove the tube, push the jaw clamps
right back again and take the tube out of the
jaws.
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:
evB 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
evB
R
mv
F
= =
2
which implies
R
m
e
v
B
= tesla
Rearranging the equation gives
BR
v
m
e =
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.
( )
0
2
1
2
1
1 2
2
1
2
2
= − + ⎟
⎠
⎞
⎜
⎝
⎛
−
eU eU mv mv
The energy of an electron in the dual beam tube
is given by:
2
2
1
mv eU
A
=
By solving for v and replacing it in the equation
BR
v
m
e =
the following emerges
2 2
2
R B
U
m
e
A =
The term e/m is the specific charge of an elec-
tron and has the constant value (1.75888 ±
0.0004) x 10
11 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
H B
0
μ = = (4.17 x 10 -3 ) I
H
tesla
and
2 6 2 10 39 . 17
H
I B
− ⋅ =
where I
H
is the current in the Helmholtz coils.
The following are also true
5
2 2
10 15 . 1 ⋅ ⋅ =
R I
U
m
e
H
A and
2
2
R
U
k I A
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 circular path with the tube axis
forming a tangent. The centre of this circle is at