Electron interactions with the magnetic field

[tombo]

The way I see it is:
1. The force on a coil from the coil on the opposite side is reduced because the conductive ball of B-field-excluding plasma in the center shields them from each other. (pushes their lines away from each other)
Likewise the far sides of the 4 adjacent coils. (depending on the plasma ball diameter)
My first cut would be to approximate the field seen by a coil as the current in the near halves (or so) of the 4 adjacent coils. Modeling them as one coil at maybe twice the spacing of the actual nearest approach.
No I have not gotten around to it yet.

2. Any force on a coil from induced plasma currents would reduce the force from the other coils.

3. Any force on a coil from induced plasma currents would (I think) be equal to the physical pressure of the plasma itself. So, how much force would we see from a pressure that really is still a pretty hard vacuum. Not much I think.

4. Consider the bottom coil of the 6 coil structure before the plasma is formed:
The fields on it from the top coil and the top parts of the 4 side coils cancel each other.
Also the fields from the sides of the 4 side coils (where they come together) cancel each other.
All horizontal forces cancel by symmetry. (That works for the whole coil, but this assumption needs to be revisited later when calculating hoop stresses within the coil. My first look tells me that they mitigate hoop stresses because the nearer currents predominate and are opposite to the current in the coil in question.)
What is left is the field from the horizontal components of the bottom portions of the 4 side coils. And, as a second order term some portion of the horizontal current component of the 4 triangular virtual coils at the top corners must be considered (but might also cancel).

Conclusion: I think a conservative worst case model would be to calculate the force on the bottom coil from an identical (parallel) coil located at the height of the bottoms of the 4 side coils.
A closer approximation would be to place it at the height of the rms average of the height of the bottom semicircles of the 4 side coils.
(I could have done the integral in college but...)

[MSimon]

2. Any force on a coil from induced plasma currents would reduce the force from the other coils.

I see it as more like pressure in a balloon. Thus no force cancellation or addition.

But I may be wrong.

[hanelyp]

The way I see it is:
3. Any force on a coil from induced plasma currents would (I think) be equal to the physical pressure of the plasma itself. So, how much force would we see from a pressure that really is still a pretty hard vacuum. Not much I think.

PV = NRT
Low density, but VERY HIGH temperature. As Dr. Buzzard described the wiffleball, beta=1. In other words plasma pressure == magnetic field pressure.

[icarus]

It is simply a matter of equivalency of singularities that produce the integral resultants of the fields. There will be a change to the forces on the coils as surely as there is a change to the field between the coils and the wiffleball when the wiffleball forms.

For the purposes of forces on the physical coils, using the image system to calculate the new forces is the simplest way. Imagine if there was a physical set of coils that is equivalent to the image system; stating the obvious, it would produce exactly the field external to the wiffleball (regardless of what is physically happening inside the wiffleball). It is like a center of mass calculation, or center of vorticity using singular vortices calculation. For this type of spherical diamagnetic inclusion, located concentrically to the coils, the forces seen by the physical coils will be determined purely by the radius of the sphere, regardless of the currents internal to the wiffleball surface, which are by all accounts unknown anyway.

The beta=1 boundary of the wiffleball obviously has the magnetic field strength and plasma pressure in balance. As the radius of the wiffleball increases, due to an increase in plasma pressure, the forces on the coils will surely change due to the magnetic field being pushed back? or not?

It's just a matter of performing the the magnetic force calculation on the coils in two ways, one with and one without the image system coils included. Would have been a neat little undergrad project in my day. An analytical expression having coil reaction forces as some function of the wiffleball radius and using that excess force over the case without the image system to determine plasma pressure force (divided by wiffleball surface area gives plasma pressure).

[MSimon]

The way I see it is:
3. Any force on a coil from induced plasma currents would (I think) be equal to the physical pressure of the plasma itself. So, how much force would we see from a pressure that really is still a pretty hard vacuum. Not much I think.

PV = NRT
Low density, but VERY HIGH temperature. As Dr. Buzzard described the wiffleball, beta=1. In other words plasma pressure == magnetic field pressure.

That was what I was getting at. You said it better.

[MSimon]

It is simply a matter of equivalency of singularities that produce the integral resultants of the fields. There will be a change to the forces on the coils as surely as there is a change to the field between the coils and the wiffleball when the wiffleball forms.

For the purposes of forces on the physical coils, using the image system to calculate the new forces is the simplest way. Imagine if there was a physical set of coils that is equivalent to the image system; stating the obvious, it would produce exactly the field external to the wiffleball (regardless of what is physically happening inside the wiffleball). It is like a center of mass calculation, or center of vorticity using singular vortices calculation. For this type of spherical diamagnetic inclusion, located concentrically to the coils, the forces seen by the physical coils will be determined purely by the radius of the sphere, regardless of the currents internal to the wiffleball surface, which are by all accounts unknown anyway.

The beta=1 boundary of the wiffleball obviously has the magnetic field strength and plasma pressure in balance. As the radius of the wiffleball increases, due to an increase in plasma pressure, the forces on the coils will surely change due to the magnetic field being pushed back? or not?

It's just a matter of performing the the magnetic force calculation on the coils in two ways, one with and one without the image system coils included. Would have been a neat little undergrad project in my day. An analytical expression having coil reaction forces as some function of the wiffleball radius and using that excess force over the case without the image system to determine plasma pressure force (divided by wiffleball surface area gives plasma pressure).

I don't see how there can be any extra forces developed in a fluid that is not very viscous. Even if it was at air density.

[icarus]

I don't see how there can be any extra forces developed in a fluid that is not very viscous.

I don't understand what extra forces in the fluid you are referring to here. I'm looking for an expression that will relate the average plasma pressure of the wiffleball to its radius. I believe it can be got readily by comparing the magnetic reaction forces on the Polywell coils (all radial by symmetry) of the cases with and without a spherical wiffleball, as described by our method of images solution discussed previously.

Hydrostatic pressure can be thought of as a measure of energy density contained in the molecular motions of the fluid. Think of the extra coil magnetic reaction forces, when the wiffleball forms, as being due to confining the plasma energy, similar to stresses developed by a pressure vessel.

Regardless of how the magnetic field distortion described by the wiffleball is caused, i.e. it could be a solid diamagnetic lump of something, or a blob of superfluid, it doesn't matter. The only thing that is relevant for calculating the coil magnetic reaction forces, due to the presence of the wiffleball field distortion, is a system of singularities that will be consistent with that shape, e.g. the image system, (which also happily happens to be the simplest;)

[MSimon]

icarus,

I was responding to an earlier comment that thought the magnetic forces in the plasma would augment the forces caused by the fields generated by the coils.

I agree with you. The forces can only be those generated by the magnets and distributed by the plasma.

[alexjrgreen]

We should be able then to relate wiffle-ball radius and electron pressure in this way.

Imagine an inflating balloon (the wiffle ball) between two spring-loaded boards (the magnetic field). As the pressure in the balloon goes up the springs give to equalise the load. The extra force is resisted by the structure (the casing of the magrid).

Since the forces must be equal at the boundary of the wiffle ball, you can either calculate electron pressure or magnetic field pressure.

[KeithChard]

The forces on a Magrid coil can be calculated as being due to the sum of the the magnetic fields due to the other five coils plus the sum of the fields due to all the image coils. This calculation can be done without knowledge of the energy and total number of electrons that have been introduced to the wiffleball. The force on the Magrid coil, due to the image coils, will be equal to the electron plasma pressure integrated over the sector of the wiffleball that correponds to that coil. This gives us the required total energy of the electrons that is necessary to form the wiffleball assumed for doing the magnetic field calculations of the image sysem. The trouble is that at this stage of our knowledge we can only do the calculations very approximately because we are only using a simplistic model for the coil image system. It may be accurate enough for stressing the coil supports, but we have some way to go before we can quantify the exact shape and size of the wiffleball and also to get a realistic structure for the image current flows. Perhaps it does not really matter at this stage because the experimental programme has successfully set up the wiffleball without us having a full understanding! Engineers like me may be happy with that but I suspect that physicists will want to sort it out properly, which will be to our advantage.

[Indrek]

FYI. I once calculated the force on a coil at:
http://www.mare.ee/indrek/ephi/force/

I revisited this idea now, added the image coils, and cast it into a simple octave script. This can be found in:
http://www.mare.ee/indrek/octave/ov-0.4.zip

Another two hours wasted The numbers agree with my old C++ code. For example:

Radius: 0.15 m, spacing 0.08 m, current 100000.00 A
Wiffleball radius: 0.15 m
NO WB: Force pushing the top coil away is 9246.98 N
W/ WB: Force pushing the top coil away is 9391.75 N

See polywell_force.m. It appears the force added by the image coils is quite small. At least for the smaller radius WB.

- Indrek

[icarus]

Indrek wrote:

Another two hours wasted Smile The numbers agree with my old C++ code. For example:

Radius: 0.15 m, spacing 0.08 m, current 100000.00 A
Wiffleball radius: 0.15 m
NO WB: Force pushing the top coil away is 9246.98 N
W/ WB: Force pushing the top coil away is 9391.75 N

See polywell_force.m. It appears the force added by the image coils is quite small. At least for the smaller radius WB.

Hello Indrek, good to read your posts again.

I'm going to take a look at that Octave script a little closer shortly but these numbers look good I think. The change in coil force due to presence of WB is positive as I suspected. A quick back-of-envelope says that total normal outward force on all coils is then 868.62 [N] (6 times additional force due to WB on one coil) and since WB radius, R=0.15 [m] for this example then we can calculate an average electron WB plasma pressure, P_wb = 3 072 [Pa] at the WB surface (using surface area=4*pi*R^2).

Anybody have a feel for if this WB pressure number, ~3kPa, is in the ballpark for a WB of this diameter (0.3 [m]))? (Rick Nebel might know I suppose.)

(Now I also wonder how this compares with average mag. field pressure at the same surface? as should be identical)