Minimally Sufficient

21 02 2009

So the goal of this first machine is only to achieve first fusion. This could mean a single solitary neutron in a detector.

Towards that end I need to make calculations for a minimally sufficient design, so it’s _expected_ to produce fusion. Off the top of my head these are some of the relevant inputs:

radius of the core
ampturns of each coil
current going to the electron emitters
positive potential of the magrid skin
radius of the vacuum chamber
quality of the vacuum

I need to make calculations to take those inputs and calculate:

well depth
B-field strength
sufficient for fusion?

OK. So where to start? The WB6 report. Starting on page 4

Its ability to trap electrons inside the device is measured by an overall trapping factor, called Gmj, which is the ratio of the electron lifetime in the machine environment with B fields turned on to that with no B fields.  This overall e- current trapping factor is composed of two terms; one due to internal trapping by diamagnetic electron confinement leading to cusp confinement, and one due to electron trapping/flow through coil corner positions, which act somewhat like line cusps.

The first of these represents the effect of B field expansion under electron pressure, and is called the Wiffle-Ball factor Gwb.For a truncated cube geometry (used in all machines tested to date) this factor is Gwb  = (BR)^2/110Eo, where B is the magnetic field strength (in G) on-axis of  the main faces, R is the radius of the device (in cm) from its center to the midplane of the field coils, and Eo is the depth of the electric potential well (in eV) resulting from the injection of the energetic electrons that drive the device.  Typically the well depth is about 0.7-0.9 of the electron injection energy (Ei), depending on the exact geometry of the device and of the injection system.  In WB-6 well depth was about 0.8 of injection energy.

So there are some unknowns for a truncated dodecahedral configuration. I’m thinking a good place to start would be to express these calculations in ruby for the truncated cube, just to get my head around the ideas. Ouch my head hurts. 


So first we need to convert our magnetic field from Ampere-turn to Gauss. But first I want to understand Ampere-turn better:

The ampere-turn (AT) is the MKS unit of magnetomotive force

Lets take a closer look at magnetomotive force:

Magnetomotive force is any physical cause that produces magnetic flux. If a magnetic field (measured in teslas) passes through a cross sectional area A (measured in square meters), it produces a flux given by the equation MMF×A = flux (in webers).


Magnetic flux, represented by the Greek letter Φ (phi), is a measure of quantity of magnetism, taking into account the strength and the extent of a magnetic field. The SI unit of magnetic flux is the weber (in derived units: volt-seconds), and the unit of magnetic field is the weber per square meter, or tesla.

OK. One tesla is equal to 104 gauss. So we need to go from  Ampere-turn to Tesla.

The tesla (symbol T) is the SI derived unit of magnetic field B (which is also known as “magnetic flux density” and “magnetic induction”). The tesla is equal to one weber per square meter.

Now if I can keep these all in my head at once, I can connect them together. 

What is the relationship between magnetomotive force and  magnetic flux? How do I go from  Ampere-turn to webers?

magnetomotive force =  magnetic flux *  reluctance of the magnetic circuit. So now I need to determine the reluctance of the magnetic circuit.

Electromagnetic theory… whoa. 

Magnetic reluctance or “magnetic resistance”, is analogous to resistance in an electrical circuit (although it does not dissipate magnetic energy). In likeness to the way an electric field causes an electric current to follow the path of least resistance, a magnetic field causes magnetic flux to follow the path of least magnetic reluctance. It is a scalarextensive quantity, akin to electrical resistance.

Ok. MSimon directed me to this page which has more direct calculations. So for a loop of R = 0.03462 M and 400 ampturns, we get a a B field of 0.007259 T or 72.59601 G. This page raises the question: Are we interested in the B-field at the center of the current loop, or on the axis of the current loop (the center of the core)? I would guess the center of the core, since that’s where the electrons are. At this location we have B-field = 0.00046324 T or 4.632465 G. MSimon also mentioned that the WB6 had a B-field of 0.1 T, which would require the 14X the coil I currently have coming. This assumes we are measuring the B-field at the center of the coil, and not the center of the core. Correct? 

I took the javascript from this page and converted it to ruby, and checked the results against the page:


u = 2*Math::PI * (10.0**-7) * ((Unit(“tesla”) * Unit(“m”))/Unit(“ampere”))

puts b_field =  u*(ic*turns)/(torus_radius >> Unit(“meter”))


This gives us a B-field of 0.0072596 T


Gwb  = (BR)^2/110Eo, where B is the magnetic field strength (in G) on-axis of  the main faces, R is the radius of the device (in cm) from its center to the midplane of the field coils, and Eo is the depth of the electric potential well (in eV) resulting from the injection of the energetic electrons that drive the device.

Before I could continue I needed to define electronvolt in ruby-units. Fortunately this is pretty easy:


class Unit < Numeric


    {‘<eV>’=>[%w{eV electron_volt electronvolt},  1.60217653e-19, :energy, %w{joule} ]}




Now we have an expression in ruby for the gwb:


puts gwb = (((b_field >> Unit(‘G’)) * (torus_midplane_radius >> Unit(‘cm’)))**2) / 110*eo

So for the decawell unit running at 1000 eV potential, we get this gwb with a rather comical unit:
3.02178e+06 eV*cm^2*G^2





2 responses

21 02 2009
M. Simon

Amp turns and gauss and Teslas explained

As for “Webers” – you are on your own.

BTW WB-6 was 1,000 gauss – .1 Tesla.

22 02 2009
M. Simon

I believe GWB is a unitless number. A ratio.

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