11.4 Confinement of positive ions and electrons with a static electromagnetic field
Let's see what kind of electromagnet we need for our SEM fusor design.
length: 1 mtr
B = μ0 . N/l . I (see hyperphysics)
N= amount of
I = B . l /(μ0 . N)
B = 1 T, l = 1 m, N = 1000 -> I = 1 . 1 / (2.10-7 .1000) = 5000 A , very high..
, amonst others, an electromagnet (coil) is sold: B=
0,13 T , DC power = 5,18 kW (140 A, 37 V), water cooling 12 ltr/min, 360
turns. Not exactly what we need, but just to get an idea. Not enough
magnetic field, consuming a lot of power and also very expensive .
(30 cm gap, 140 A, 11,5 kW, 0,3 T, 16 ltr/min water cooling,
about 60000 $)
Electromagnets are usually in the form of
iron core solenoids, see
I think it would be better to place an electromagnet with an inner core, on top of the fusor and under the fusor (or if we turn it 90 º, at the sides).
In the next link a permanent magnet, with
dimensions 11x9x2 cm and a magnetic field (Remanencia Br) of 1.3 T is
sold for about 120 €, see:
A small disk with the characteristics as in fig.1:
Fig.1. The magnetic field of a small neodymium magnet
If our SEM fusor had a diameter of 20 cm and a length of 0,5 m, the
circumference would be 62 cm.
I think the best idea is to contruct a vacuum chamber with our experiment and to look for an existing magnet in, for example, an university or company. (universidad de Zaragoza?)
Or perhaps here?
(Consider the SEM fusor with a diameter of 45 cm and a length of 100 cm.
Fig.2. The magnetic field of a electromagnet with an iron core
Volume of the iron core = π (0,25)2 . 10 (for the length of the iron core we take 10 m)
= 2 m3 specific mass of iron is about 7 mt/m3 -> mass of the iron core = 14 mt
Price of iron about 1 till 3 € / kilo?
For a coil: B =
μ0 . N/l
µ0= permeability of free space = 4π × 10−7 N·A−2
The iron core causes an increase of the magnetic field inside the coil of about a 1000 times.
N. I = B ( Lcore/μ + Lgap/μ0 ) (assuming that B in the core and in the gap are the same, that no B (flux) is escaping, that the gap is relatively small, L= length)
B = 1 tesla, Lgap = 1,1 m N.I = 1. ( Lcore / (6.3 10−3 ) + 1,1 / (4π × 10−7 ) ) = 8,8.105 Ampere turns. (neglected the Lcore / (6.3 10−3 ) term)
Engineeringtoolbox.com/permeability - > iron 99,8% pure -> μiron = 6.3 10−3 mkg-2C , μiron / μ0 = 5000
The length of the iron core is not so important.
Let's assume the voltage is 12 dc, and we have 1000 turns. Then the current must be 880 A and the power 880*12 = 10,6 kW (about 11 electrical heaters).
Electrical cable in Bricodepot.es: 2,5 mm2 -> 0,22 € / m
the maximum current of such a cable is about 20 A. ->
8,8.105 = N. 20 -> N = 44000 turns.
If we take into account the insulation material around the cable (thickness of it is aproximated) and suppose the area of the cross section is then 3,5 mm2, then we can make over 1 meter length 1 layer with 455 turns.
To get 44000 turns, we need about 100 layers. The thickness of all layers would be around: 22 cm.
For copper the resistivity ρ = R . A/l = 1,68 . 10-8 ohm meter (R = electrical resistance, l = length, A = cross-sectional area)
Diameter vacuum chamber = 50 cm, diameter coils = about 60 cm
R = ρ . l / A = 1,68 . 10-8 . 44000 . 2. π . 0,30 / ( 2,5 . 10-6 ) = 557 ohm
Electrical power generated = I2 . R = 202 . 557 = 224 kW (about 200 electrical heaters of 1000 W)
Quite difficult to realize..