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3B Scientific Teltron Dual Beam Tube S User Manual

Page 2

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3. Technical data

Filament voltage:

7.5 V AC/DC max.

Anode voltage:

100 V DC max.

Anode current:

30 mA max.

Deflector voltage:

50 V DC max

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 S

1014525

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

1000611

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.

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:

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

)

H

I

tesla

and

2

6

2

10

39

.

17

H

I

B

=

where

H

I

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

B and it lies in the plane of DCD’ about 2 mm

behind the plane of EE’.

DC

BC

AC

BC

AB

+

=

2

2

2

2