A new idea to obtain clean fusion energy

17. Confinement of positive ions and electrons with a static electric and magnetic field

Try to do a small experiment

Back to main page

First I will calculate what magnet is possible to build my self, or to buy.

Let´s assume we have a dc power supply of 100  W.

Electrical cable in Bricodepot.es:  2,5 mm-> 0,22 € / m  (maximum current about 20 A, acc. internet search

Diameter electromagnet (solenoid)  = 60 cm

N = 500 turns, 1 turn = 2 π r = 1,885 m,  500 turn =  942 m  (207 €)

For copper the resistivity ρ = R . A/l  = 1,7 . 10-8  ohm meter  (R = electrical resistance, l = length, A = cross-sectional area)

R = 1,7 . 10-8 . 942 /( 2,5.  10-6 ) = 6.41 ohm

100 W ->  P = I . R  ->  I = P /R  = 100 /  6,41 = 15,6  A

B = μ0 . N/l . I   (see  hyperphysics)

μ0 = 2.10-7 N·A−2

N = amount of windings

l = length solenoid, let´s take 1 m

B= 2.10-7  . 500  . 15,6 / 1 = 1,56 . 10-3 T

rounded- > B = 15 gauss = 15 . 10-4 tesla

Diameter solenoid = 30 cm

N = 942 / ( 2 . π . 0,15) = 999

-> B = 31 gauss

Diameter solenoid = 15 cm

N = 942 / ( 2 . π . 0,075) = 2000

-> B = 2.10-7  . 2000  . 15,6 / 1 =   62   gauss

Fig.1. The magnetic field of an electromagnet with an iron core

r1 = ½.d1

Volume of the iron core = π (r1)2 .  (m1 + m2 + m1 + m2 + g +  π . (r2+.r1 ) . 2 )

Volume of  curved part of iron core  2 . π (r1)2   . π . (r2+.r1 )

d1 = 15 cm
m1 = 1 m
m2 = 0,30 m
r2 = 0,20 m
g = 0,60 m

Volume = π (0,075)2 .  (1 + 0,30 + 1 + 0,30 + 0,60 +  π . (0,20 +0,075 ) . 2 )

= π (0,075)2 . (3,2 +  π . 0,55 ) = 0,087 m3 = 87 . 10-3 m3

ρiron =
7.874 g/cm = 7,874  . 10kg/m3    -> mass  =  685 kg

Only an iron core inside the solonoide:

Volume = π (r1)2 .  m1 = π (0,075)2 .  1 = 17,7 . 10-3 m3 -> mass = 139 kg

We could divide it in 14 discs of 10 kg

Suppose we can create a magnetic field of 1000 gauss

Two negatively charged rings at -80kV.
A vacuum chamber with diameter 15 cm and length 30 cm

Fig. 2.

The electrons are injectet upwards vertically, just above the under ring.

During the short simulation time the electrons stay confined.

If the vacuum chamber would be filled with D2 gas, would this gas be ionized and would the electron cloud attract the D+ ions, more or less towards the centre?

Fig. 3.

The D+ are generated randomly all over the "vacuum chamber".

No magnetic field, and a negative charge in the middle (small ring).

In the following link an electron gun is sold with an energy range: 1 eV to 100 keV and a beam current range: 1 nA to 20 mA   kimballphysics.com/electron-gun-systems

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

According the simulation program the charge of the blue ring at - 200 kV is - 2,38E-6 C.

- 2,38E-6 C / (1.10-9 C) = 2380 s = 40 min

So if you use an electron gun with a current of 1 nA, and the electrons stay in the centre, after 40 minutes we would have a charge equivalent with a charged ring at -200 V

With a electron gun of 20 mA, 20.10times faster, in 0,12 ms.

Fig. 4.

In the centre there is a negative electrical charge of -2,38E-6 C.
The D+ ions are injected upwards vertically just under the under ring, and are somewhat atracted to the charge, and stay somewhat confined.
If the D+ ions are randomly generated, they are also somewhat confined.

Could we generate a electron cloud with, more or less, this charge?
If there is  a little D2 gas in the vacuum chamber, will it be ionized?

If ionized, how many D+ iones will colide with the wall of the vacuum chamber?
Will some D+ iones fuse?

The charged rings can be outside the vacuum chamber, in the air? It seems to be possible, see below.

 Internet: Normally air medium is widely use as an insulating medium in different electrical power equipments and over head lines as its breakdown strength is 30kV/cm. The breakdown electric intensity for air is 3×106 V/m. The maximum charge that can be held by a sphere of radius 1 mm is: 0.33 C   Potential of an electrical charged sphere: V= 9E9. Q /r    (r=radius of sphere, Q=charge of sphere, V=electrical potential of sphere)   V = 9E9 .0,33 /( 1E-3) = 3 E 12 V

Perhaps a kind of van der Graaf Generator can be used to charge the rings?

When operated, the dome of the Van de Graaff generator and anything in contact with it takes on a charge. This charge can be positive or negative depending on how the Van de Graaff generator is designed. Since like charges repel, this can have some entertaining consequences in electrostatics demonstrations.

Pasco.com/  Here a Van de Graaf Generator is sold which can generate  a voltage of about 400 kV (not stated if positive or negative).

Equipment:

Tesla meter  till 2,4 T, 269 €

Amazon cable unipolar  100 m , 50 € , 1000 m -> 500 €

Leroymerlin cable unipolar 2,5 mm2  100 m, 33 €, 1000 m -> 330 €

https://cerlasa.com/  Carpenteria metálico taller Cuerte de Huerva (para tubo acero 1 mtr x 15 cm diameter)

Neutron detector  1295 \$

Vacuum-projects.net  Vacuum chamber 1 mtr x 40 cm diameter 9000 €  in stainless steel 304.

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