simulation of moving electrons, hydrogen and boron ions in a magnetic field

10b Physics experiment: simulation of moving electrons, hydrogen and boron ions in a magnetic field

Computer simulation of idea 10 (and 11, + ).

Fig. 1 The coordinate system

The program is programmed in Delphi 4.

We use above displayed  coordinate system.

The sides of the cube are s (s=400 pixels = 1 mtr).
The electrons can be injected in any point (we will do this in such a way that they circle around the centre of the cube).

Actually we could inject any particle with any velocity in any point...

A magnetic field is introduced in the program. A positive value means that the magnetic field B is pointed up and is homogeneous.
(we can B also make variable, for example less in the centre and higher at the sides and near the top and the bottom)

In the program we calculate the electric field and the magnetic field in point (x,y,z) due to a charged moving particle in point (xi,yi,zi) (for non relativistic v<<c)  and adding the fixed magnetic field and electric field.

The electric field in a point (x, y, z):  ( See Coulomb's law )

The magnetic field we calculate with Biot-Savart Law. (for v<<c)

Lorentz force

How many electrons produces an electron gun?

A Nanocoulombmeter in combination with a Faraday cup can be used to detect and measure the beams emitted from electron guns and ion guns. (

In the link under an electron gun is sold with an energy range: 1 eV to 100 keV and a beam current range: 1 nA to 20 mA
( )

1.10-9 A = 1.10-9 C/s -> 1.10-9 C/s / (1,6.10-19 C) = 6,3. 109   electrons/s

It seems therefore plausible to have an electron gun that produces about 6.3. 109    electrons/ s
This is  1 electron/1,59 10-10  s

If the energy of an electron is 1 eV, than:

meVe2 = 1 eV = 1,602.10-19 J  => Ve = (2 .  1,602.10-19 / (9,1095 . 10-31) ) = 5,931. 10 m/s

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Suppose you have a boron ion with an energy of 4 MeV. What magnetic field is needed to deflect this ion into a circle with a radius of 0,5 mtr?

B= m .v / (r . q )

― mv2 =  4 .106 . 1,6.10-19 J

=> v = (2 .  4 .106 1,6.10-19 / (11 . 1,67 . 10-27)  )  = 8,35.106  m/s               (boron = 5 proton + 6 neutron,  relativistic effects neglected)

=> B = (11 . 1,67 . 10-27)  . 8,35.106 / (0.5 . 1,6.10-19 ) = 1.9 tesla,  if it's charge is e+    and B = 0,38 tesla if it's charge is 5e+

According en.wikipedia.org/wiki/Orders_of_magnitude_(magnetic_field) this (2 T)  is a magnetic field generated in a modern 60 Hz power transformer.

The remanent field of a neodymium magnet is typically about 1-1.4 tesla,  see en.wikipedia.org/wiki/Neodymium_magnets

To construct a electromagnet of 2 Tesla seems to be a bit difficult, see  instructables.com/answers/2-Tesla-Electromagnet/

https://en.wikipedia.org/wiki/Particle-in-cell

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Experiments with the program:

 (Jan. 2013) (Jan. 2013) (Jan. 2013) (3 Febr. 2013) (6 Febr. 2013) (8 Febr. 2013) (16 Jan. 2014) (17 Jan. 2014) (20 Jan. 2014) ( Febr. 2014- 2016) Computer simulation tenth idea (2014-2016) Eleventh idea   (2016.. 2017)

20 February 2014 - 2016     by  Rinze Joustra        www.valgetal.com