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ELECTROMAGNETIC SPHEROMAK

By Charles Rhodes, P.Eng., Ph.D.

ELECTROMAGNETIC SPHEROMAK:
On the web page titled THEORETICAL SPHEROMAK it is shown that a theoretical spheromak is a stable structure. This web page identifies the circumstances that enable an electromagnetic system to form a stable spheromak.

In order for a spheromak to represent a real physical entity it is necessary to show that a spheromak is an energy stable configuration and that the energy density functions that cause spheromak formation arise from the electric and magnetic field energy densities due to net electric charge and electric charge flow around a closed path that is everywhere tangent to the spheromak wall.

Then spheromak mathematics explains the existence and behavior of both stable charged atomic particles and semi-stable toroidal plasmas.
 

EXPERIMENTAL VALUE OF So:
Plasma spheromaks have been experimentally observed and photographed. Typical plasma spheromaks produced in a laboratory have shape parameters So of about:
So^2 = (Rs / Rc) ~ 4.0
 

SPHEROMAK MODEL:
An electromagnetic spheromak consists of a single layer closed toroidal winding forming a current path with Np poloidal turns and Nt toroidal turns. The dimensions and distributed charge are such that there is no net force on the winding anywhere along the current path.

A spheromak is governed by two energy density equations and various boundary conditions. There is the equation for spheromak energy content which is derived on the web page titled: SPHEROMAK ENERGY and there is the equation for energy density balance at the inner spheromak wall, referred to herein as the "boundary condition". These two equations together with the far field condition lead to the Planck Constant and the Fine Structure Constant which are the subjects of the web page titled: PLANCK CONSTANT.
 

SPHEROMAK WALL:
Under the circumstances of plasma spheromak generation the electrons and ions follow a closed spiral path. This path traces out a three dimensional surface in the shape of a hollow torus, like the icing on a doughnut, known as the spheromak wall. The formation of this wall is discussed on the web page titled CHARGE HOSE SHEET. Inside the spheromak wall the magnetic field is purely toroidal and the net electric field is zero. Outside the spheromak wall the magnetic field is purely poloidal and the net electric field is mainly radial.

The trapped electrons and ions circulate on closed spiral paths within the thin surface known as the spheromak wall. The spheromak wall position corresponds to a system total energy minimum.

In a plasma spheromak the electrons and ions follow similar but opposite spiral paths within the spheromak wall. The positive ions move opposite to the electrons to approximately balance both charge and momentum within the spheromak wall. Within the spheromak wall there is sufficient separation between the opposite flowing electron and ion streams to prevent the energetic electrons being scattered by collisions with the spheromak ions.

A fundamental difference between a plasma spheromak and a quantum particle spheromak is that in a plasma the electrons and ions are of two types and are subject to inertial forces whereas in a quantum charged particle the circulating charge may also be composed of two types but has no inertial mass. High energy plasma spheromaks are also subject to relativistic effects.
 

ROLES OF SPHEROMAK ELECTRIC AND MAGNETIC FIELDS:
The electric and magnetic fields of a spheromak store energy and act in combination to position and stabilize the spheromak wall. It is not sufficient to simply position the spheromak wall. The spheromak wall position must be at a relative energy minimum so that if the spheromak is moderately disturbed it spontaneously returns to its stable equilibrium geometry.
 

DEFINITIONS:
U = with no subscripts indicates total field energy density at any point in space

Subscripts for Uxyz are defined as follows:
x = o implies center of spheromak at R = 0, Z = 0;
x = e implies that Ue is electric field portion of the field energy density;
x = m implies that Um is the magnetic field portion of the energy density;
y = o implies that the expression for U, Ue or Um is only valid outside the spheromak wall;
y = i implies that the expression for U, Ue or Um is only valid inside the spheromak wall;
z = c implies that the value for U, Ue or Um is only valid at R = Rc, Z = 0;
z = s implies that the value for U, Ue or Um is only valid at R = Rs, Z = 0; Uo = total field energy density at R = 0, Z = 0

Uc = value of U at R = Rc, Z = 0;
Us = value of U at R = Rs, Z = 0;
Ue = electric field energy density;
Uei = electric field energy density inside the spheromak wall;
Ueis = electric field energy density inside the spheromak wall at R = Rs, Z = 0;
Ueo = electric field energy density outside spheromak wall;
Ueoc = Ueo evaluated at R = Rc, Z = 0;
> Ueos = Ueo evaluated at R = Rs, Z = 0;
Um = magnetic field energy density;
Umi = magnetic field energy density inside the spheromak wall;
Umis = magnetic field energy density inside the spheromak wall at R = Rs, Z = 0;
Umo = magnetic field energy density outside spheromak wall;
Umoc = Umo evaluated at R = Rc, Z = 0;
Umos = Umo evaluated at R = Rs, Z = 0.

Epsilon = permittivity of free space
Muo = permeability of free space
Bxyz and Exyz by:
B = magnetic field
E = electric field
x = p or t or r subscripts where p indicates a poloidal poloidal magnetic field, t indicates a toroidal magnetic field, r indicates a radial electric field
y = i or o subscripts indicating an inside spheromak wall or outside spheromak wall
z = c or s subscripts indicating core wall or outside wall intersection with the equatorial plane;
 

FIELD DEFINITIONS:
Eroc = electric field outside the wall at R = Rc, Z = 0;
Bpoc = poloidal magnetic field outside the wall at R = Rc, Z = 0;
Btoc = toroidal magnetic field outside the wall at R = Rc, Z = 0;
Bpic = poloidal magnetic field inside the wall at R = Rc, Z = 0;
Btic = toroidal magnetic field inside the wall at R = Rc, Z = 0;
 

SPHEROMAK CHARGE HOSE PARAMETERS
Define:
Ih = charge hose current;
Lh = axial length of closed spiral of charge motion path;
Nt = Integer number of complete toroidal charge hose turns contained in Lh;
Np = Integer number of complete poloidal charge hose turns contained in Lh;
Lt = length of one purely toroidal charge motion turn
Lp = length of one purely poloidal charge motion turn at R = Rf;
As = outside surface area of spheromak wall
Vi = ion velocity along charge motion path
Ve = electron velocity along charge motion path
Q = proton charge
Qs = net spheromak charge
Rhoh = charge per unit length along the charge motion path
C = speed of light
Theta = angle around the main spheromak axis of symmetry
Phi = angle around the toroidal axis of symmetry measured with respect to the spheromak equatorial plane
 

SPHEROMAK GEOMETRY:
The geometry of a spheromak can be characterized by the following parameters:
R = radial distance from the spheromak's axis of cylindrical symmetry to a general point (R, Z);
Rc = spheromak's minimum core radius;
Rs = spheromak's maximum equatorial radius;
Rf = [(Rs + Rc) / 2] = spheromak's top and bottom radius;
Rw = radius of co-axial cylindrical enclosure such as a vacuum chamber;
Z = distance of a general point from the spheromak's equatorial plane;
Zs = distance of a point on the spheromak wall from the spheromak's equatorial plane;
(2 |Zf|) = spheromak's overall length measured at R = Rf;
 

EQUATORIAL PLANE:
On the spheromak's equatorial plane:
Z = 0
For points on the spheromak's equatorial plane the following statements can be made:

For R = 0 the radial electric field is zero;
For R = 0 the axial electric field is zero;
For R < Rc the toroidal magnetic field Btoc = 0
For R < Rc the magnetic field Bp is purely poloidal;
For R = 0 the magnetic field Bpo is along the spheromak axis of cylindrical symmetry;
Hence, for R = 0 the magnetic field Bpo > Btic;

For Rc < R < Rs the electric field Eri = 0;
For Rc < R < Rs the poloidal magnetic field Bpi = 0;
For Rc < R < Rs the toroidal magnetic field Bti is proportional to (1 / R);
For R = Rc the toroidal magnetic field Bti < Bpo

BASIC ELECTROMAGNETIC THEORY:
FIND AN EXPRESSION FOR U(R, Z):
At any point in space the electromagnetic field energy density U is given by: U = (Epsilono / 2)[Ex^2 + Ey^2 + Ez^2] + (1 / 2 Muo)[Bx^2 + By^2 + Bz^2]

The electric field energy densities are continuous functions except at a charge where there is a step change. The magnetic field energy densities are continuous functions except at a current where there is a step change.

In cylindrical co-ordiantes:
Ex^2 + Ey^2 = Er^2 + Et^2
and
Bx^2 + By^2 = Br^2 + Bt^2

Hence:
U = (Epsilono / 2)[Er^2 + Et^2 + Ez^2] + (1 / 2 Muo)[Br^2 + Bt^2 + Bz^2]

The spheromak symetrical structure gives:
Et(R, Z) = Et(R, - Z) = 0
Er(R, Z) = Er(R, - Z)
Ez(R, Z) = - Ez (R, - Z)
Br(R, Z) = - Br(R, - Z)
Bz(R, Z) = Br(R, - Z)
Bt(R, Z) = Bt)R, - Z)
Et(R, 0) = 0
Er(R, 0) = Function;
Ez(R, 0) = 0
Br(R, 0) = 0
Bz(R, 0) = Function
Bt(R, 0) = Function
Bz(0, 0) = [2 Muo Uo]^0.5
 

Since Et(R, Z) = 0:
U = (Epsilono / 2)[Er^2 + Ez^2] + (1 / 2 Muo)[Br^2 + Bt^2 + Bz^2]

Outside the spheromak wall:
Bt = 0
which gives:
U = (Epsilono / 2)[Er^2 + Ez^2] + (1 / 2 Muo)[Br^2 + Bz^2]

Inside the spheromak wall:
Br = 0 and Bz = 0 giving:
U = (Epsilono / 2)[Er^2 + Ez^2] + (1 / 2 Muo)[Bt^2]

Inside the spheromak wall of an isolated spheromak:
Er = 0 and Ez = 0 giving:
U = + (1 / 2 Muo)[Bt^2]

All fields decrease with increasing distance from R = 0, Z = 0.

Separation of variables:
In general the energy density is of the form:
U = Ur(R)Uz(Z)
where:
Uz(Z) = Uz(-Z)

U(0, 0) = Uo

Uz = A / (1 + C Z^2 + D Z^4 + E Z^6)

Ur = G / (1 + H R + I R^2 + J R^3 + K R^4)

U = (A G) /(1 + C Z^2 + H R + H R C Z^2 + D Z^4 + I R^2 + D Z^4 H R + I R^2 C Z^2)
= (A G) /(1 + H R + C Z^2 + I R^2 + H R C Z^2 + D Z^4 + I R^2 C Z^2 + D Z^4 H R + ..)

We can evaluiate U along the Z axis to find C, D, E

We can evaluate U along Z = 0 to find H, I, J

Combine these two to find an expression for U (R, Z)

Note that if H = 0 the above expression simplifies to the form of our existing approximation for U. We need to focus on the value of H.

CONTINUE FROM HERE

In an experimental apparatus at R = Rs, the field energy density U must meet both the nearly spherically radial requirements of free space and the cylindrically radial requirement imposed by the proximity of a cylindrical metal enclosure wall.

For Rs << R in free space the electric field Ero is spherically radial;
For Rs << R in free space the electric field Ero is proportional to (1 / R^2);
For Rs < R in free space the toroidal magnetic field Bto = 0;
For Rs << R in free space the poloidal magnetic field Bpo is approximately proportional to (1 / R^3);

For Rs < R < Rw at Z = 0 in a cylindrical metal enclosure the electric field Ero is cylindrically radial;
For Rs < R < Rw at Z = 0 in a cylindrical metal enclosure the electric field Ero is proportional to (1 / R);
For Rs < R < Rw the toroidal magnetic field Bto = 0;
 

SPHEROMAK END CONDITIONS:
The spheromak ends are mirror images of each other. Let Rf be the radius of the spheromak end funnel face at the spheromak's longest point. Then the following boundary conditions apply outside the spheromak end face:
For R < Rc the spheromak has no physical end and the magnetic field is entirely poloidal;

For Rc < R < Rs and |Z| < |Zs| the internal magnetic field Bti is toroidal;
For Rc < R < Rs and |Z| < |Zs| the magnetic field Bti is proportional to (1 / R).

Outside the spheromak wall the magnetic field is purely poloidal;
For Rc < R < Rs and Z^2 < Zs^2 the electic field component parallel to the main axis of symmetry is zero;
For (A^2 R^2 + B^2 Z^2) >> Zf^2 the electric field is spherically radial;
For (A^2 R^2 + B^2 Z^2) >> Zf^2 spherical electric field is proportional to (A^2 R^2 + B^2 Z^2)^-1;
 

When these constraints are properly applied the quantitative agreement between the engineering model and published spheromak photographs is remarkable.
 

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ELECTROMAGNETIC SPHEROMAK STRUCTURE:
1) A spheromak wall is composed of a closed spiral of charge hose or plasma hose of overall length Lh;

2) Spheromak net charge Qs is uniformly distributed over charge hose or plasma hose length Lh resulting in a net charge per unit length:
[Qs / Lh];

3) The spheromak net charge circulates at the speed of light C (constant velocity) along the charge hose path, which gives the spheromak a natural frequency:
Fh = C / Lh

4) The flow of net charge Ih has two orthogonal charge circulation velocity components, a component which contributes to the external poloidal magnetic field and a component which contributes to the internal toroidal magnetic field. The orthogonal current flows can each be either positive or negative, so a spheromak has 4 possible quantum states relate to an observer. Hence each spheromak has two orthogonal magnetic vectors (poloidal and toroidal) each of which has two possible orientations. There is poloidal up and poloidal down magnetic vector. For each of the two possible poloidal magnetic vector directions there is toroidal clockwise (CW) and toroidal counter clockwise (CCW) magnetic vector.

5)Define:
Lp = length of one average charge path poloidal turn
Np = number of charge path poloidal turns
Lt = length of one charge path toroidal turn
Nt = number of charge path toroidal turns.
Rs = toroid outside radius on the equitorial plane
Rc = toroid inside radius on the equitorial plane.

(A / B) = elliptical toroidal path [(major radius) / (minor radius)]
6) Lp = Pi (Rs + Rc)
7)Kc = Lt / [Pi (Rs - Rc)]
8) Lt = Pi (Rs - Rc) Kc
 

SPHEROMAK CURRENT PATH LENGTH Lh:
Electromagnetic spheromaks arise from the electric current formed by distributed net charge Qs circulating at the speed of light C around the closed spiral path of length Lh which defines the spheromak wall. On the equatorial plane measured from the main axis of symmetry the spheromak inside radius is Rc and the spheromak outside radius is Rs.

Let Np be the integer number of poloidal currrent path turns in Lh and let Nt be the integer number of toroidal current path turns in Lh.

The spheromak wall contains Nt quasi-toroidal turns equally spaced around 2 Pi radians in angle Theta about the main spheromak axis of symmetry.
Each purely toroidal winding turn has length:
2 Pi (Rs - Rc) Kc / 2 = Pi (Rs - Rc) Kc
so the purely toroidal spheromak winding length is:
Nt Pi (Rs - Rc) Kc

Note that for a round spheromak cross section toroid Kc = 1. If for an elliptical cross section spheromak:
[A / B] > 1
then:
Kc > 1

The spheromak wall contains Np poloidal turns which are equally spaced around the ellipse perimeter. The average purely poloidal turn length is:
2 Pi (Rs + Rc) / 2 = Pi (Rs + Rc)
and the purely poloidal winding length is:
Np Pi (Rs + Rc)

In one spheromak cycle period the poloidal angle advances Np (2 Pi) radians. In the same spheromak cycle period the toroidal angle advances Nt (2 Pi) radians.
Hence:
(poloidal angle advance) / (toroidal angle advance) = Np / Nt

While a current point moves radially outward from Rc to Rs the toroidal angle advance is Pi radians and the toroidal travel is Lt / 2. The corresponding distance along the equatorial outer circumference is:
Pi (Np / Nt) Rs.
Thus Pythagoras theorm gives the current point travel distance along the winding for the toroidal half turn as:
[(Lt / 2)^2 + ( Pi Np Rs / Nt)^2]^0.5

While the current point moves radially inward from Rs to Rc the toroidal angle advance is Pi radians and the toroidal point travel is Lt / 2. The corresponding distance along the equatorial inner circumference is:
Pi (Np / Nt) Rc.
Thus Pythagoras theorm gives the current point travel distance along this winding toroidal half turn as:
[(Lt / 2)^2 + ( Pi Np Rc / Nt)^2]^0.5

Thus the total winding length Lh is:
Lh = Nt [(Lt / 2)^2 + (Pi Np Rs / Nt)^2]^0.5
+ Nt [(Lt / 2)^2 + ( Pi Np Rc / Nt)^2]^0.5
 
= [(Nt Lt / 2)^2 + (Pi Np Rs)^2]^0.5
+ [(Nt Lt / 2)^2 + (Pi Np Rc)^2]^0.5

Recall that:
Lt = [Kc Pi (Rs - Rc)]
and from spheromak geometry:
Rs = Ro So / A
and
Rc = Ro / A So

Thus:
Lh = {(Nt Lt / 2)^2 + (Pi Np Rs)^2}^0.5
+ {(Nt Lt / 2)^2 + (Pi Np Rc)^2}^0.5
 
= {(Nt [Kc Pi (Rs - Rc)] / 2)^2 + (Pi Np Rs)^2}^0.5
+ {(Nt [Kc Pi (Rs - Rc)] / 2)^2 + (Pi Np Rc)^2}^0.5
 
= {[Nt Kc Pi (Rs - Rc) / 2]^2 + (Pi Np Rs)^2}^0.5
+ {[Nt Kc Pi (Rs - Rc) / 2]^2 + (Pi Np Rc)^2}^0.5
 
= {[Nt Kc Pi ((Ro So / 2 A) - (Ro / 2 A So))]^2 + (Pi Np (Ro So / A))^2}^0.5
+ {[Nt Kc Pi ((Ro So / 2 A) - (Ro / 2 A So))]^2 + (Pi Np (Ro / A So))^2}^0.5
 
= [Pi Ro / A]{[Nt Kc ((So / 2) - (1 / 2 So))]^2 + [Np So]^2}^0.5
+[Pi Ro / A] {[Nt Kc ((So / 2) - (1 / 2 So))]^2 + [(Np / So)]^2}^0.5
 
= [Pi Ro / 2 So A] {[Nt Kc ((So^2) - (1))]^2 + [Np (2 So^2)]^2}^0.5
+[Pi Ro / 2 So A] {[Nt Kc ((So^2) - (1))]^2 + [2 Np]^2}^0.5
 
= [Pi Ro / 2 So A] {[Nt Kc (So^2 - 1)]^2 + [Np (2 So^2)]^2}^0.5
+[Pi Ro / 2 So A] {[Nt Kc (So^2 - 1)]^2 + [2 Np]^2}^0.5
 
= [Pi Ro Nt / 2 So A] {[Kc (So^2 - 1)]^2 + [(Np / Nt) (2 So^2)]^2}^0.5
+[Pi Ro Nt / 2 So A] {[Kc (So^2 - 1)]^2 + ([Np / Nt][2])^2}^0.5

Thus:
[Lh A / 2 Pi Ro]
= [Nt / 4 So] {[Kc (So^2 - 1)]^2 + (Np / Nt)^2 [2 So^2]^2}^0.5
+ [Nt / 4 So]{[Kc (So^2 - 1)]^2 + (Np / Nt)^2 [2]^2}^0.5

 
= Nt {[Kc (So^2 - 1) / 4 So]^2 + (Np / Nt)^2 [2 So^2 / 4 So]^2}^0.5
+ Nt {[Kc (So^2 - 1)/ 4 So]^2 + (Np / Nt)^2 [(2 / 4 So)]^2}^0.5
 
= + Nt {[Kc (So^2 - 1) / 4 So]^2 + (Np / Nt)^2 [So / 2]^2}^0.5
+ Nt {[Kc (So^2 - 1)/ 4 So]^2 + (Np / Nt)^2 [1 / (2 So)]^2}^0.5

= [Lh A / 2 Pi Ro]
=La + Lb
where:
La = Nt {[Kc (So^2 - 1) / 4 So]^2 + (Np / Nt)^2 [So / 2]^2}^0.5
and
Lb = Nt {[Kc (So^2 - 1)/ 4 So]^2 + (Np / Nt)^2 [1 / (2 So)]^2}^0.5

This equation is the result of spheromak geometric analysis.
 

10) Fh = C / Lh
11) Define the spheromak nominal radius Ro and shape parameter So by:
(A Rs / Ro) = (Ro / A Rc) = So
12) So^2 = Rs / Rc
13) So = A Rs / Ro
14) So = Ro / A Rc
15) Ro^2 = A^2 Rs Rc

As discussed on the web page: SPHEROMAK STRUCTURE:
Np and Nt share no common factors other than unity.

Hence:
[Lh A / 2 Pi Ro]
is a constant common geometric ratio for all spheromaks.

16) In a stable spheromak of a particular size the circulating current:
Ih = Qs Fh = C / Lh
is constant;

17) The circulating current Ih causes a purely toroidal magnetic field inside the spheromak wall and a purely poloidal magnetic field outside the spheromak wall;

18) The net charge Qs causes an electric field outside the spheromak wall which is normal to the spheromak wall in the near field and is spherically radial in the far field;

19) At the center of the spheromak at (R = 0, Z = 0) the net electric field is zero;

20) In the region enclosed by the spheromak wall where:
Rc < R < Rs and |Z| < |Zs|
the total field energy density U takes the form:
Uim = Uio (Ro / R)^2
where:
Uim = toroidal magnetic field energy density inside the spheromak wall;

21) Outside the spheromak wall the total field energy density takes the form:
U = Ue + Um
= Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2

where:
Uo = [(Muo Qs^2 C^2) / (32 Pi^2 Ro^4)];

22) Note that this value of Uo comes from integration over a sphere, where:
A > 1
and
B < 1 .

23) Everywhere on the thin spheromak wall:
U = Ueo + Umo = Uei + Umi
where:

Uei = 0

24) In a electromagnetic spheromak the static electric and magnetic field energy density functions are a result of distributed circulating charge that causes the static electric field and that circulates within the spheromak wall with a characteristic frequency:
Fh = C / Lh
causing the static magnetic fields. The charge circulation pattern is described by five parameters: Np, Nt, So, A, B and Ro where:
Np = number of poloidal turns per charge circulation cycle;
Nt = number of toroidal turns per charge circulation cycle;
So^2 = (Rs / Rc) = spheromak shape parameter
Rs = maximum radius from spheromak symmetry axis to spheromak wall;
Rc = minimum radius from spheromak symmetry axis to spheromak wall;
Ro^2 = (A^2 Rs Rc) where Ro is the nominal spheromak radius.
A = [2 Zf / (Rs - Rc)]
B = 1.000
 

25) In order for a spheromak to exist Np and Nt are integers that do not have any common factors. Let P be a prime number. Then functions which produces integer Np, Nt pair values without common factors are:
SOLUTION FAMILY A: [Np / Nt] = (P - 2 Nt) / Nt
or
P = 2 Nt + Np
which for:
Np = (Nt + 1)
gives:
P = 2 Nt + (Nt + 1)
= 3 Nt + 1

For this case:
2 dNt + dNp = 0
Note that in FAMILY A Np double increments for each increment in Nt. This does not result in a prime number compliant solution because the calculated value of P becomes even which is impossible for a prime number.
 

SOLUTION FAMILY B:
[Nt / Np] = (P - 2 Np) / Np or
P = 2 Np + Nt
which for:
Np = (Nt + 1)
P = 2(Nt + 1) + Nt
= 3 Nt + 2
For this case:
2 dNp + dNt = 0
Note that in FAMILY B Nt is odd so that P can be odd. Np changes in single steps while Np changes in double steps. Hence only FAMILY B gives a real solution.
Np = Nt + 1  

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SPHEROMAK WALL POSITION:
A spheromak is a stable energy state. The spheromak wall positions itself to achieve a total energy relative minimum consistent with the spheromaks natural frequency Fh. At every point on the spheromak wall the sum of the electric and magnetic field energy densities on the outside side of the spheromak wall equals the toroidal magnetic field energy density on the inside of the spheromak wall. This general statement resolves into different detailed boundary conditions at different points on the spheromak wall. This general boundary condition establishes the spheromak core radius Rc on the equatorial plane, the spheromak outside radius Rs on the equatorial plane and the spheromak length 2 Zf.
 

ENERGY DENSITY BALANCE AT THE SPHEROMAK WALL
Inside the spheromak wall the magnetic field energy density is given by:
Umi = Umic [(Rc / R)^2]

At all points on the spheromak wall the total inside energy density equals the total outside energy density. Hence at all points on the spheromak wall:
Umo + Ueo = Umi

For a spheromak wall position to be stable the total field energy density must be the same on both sides of a thin spheromak wall. This requirement leads to boundary condition equations that determine the shape of a spheromak.

In general the total field energy density U at any point in a spheromak is given by:
U = [Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]
where:
Bp = poloidal magnetic field strength;
Bt = toroidal magnetic field strength;
Er = radial electric field strength.

Thus at a spheromak wall:
{[Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}inside
= {[Bp^2 / 2 Mu] + [Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}outside

 

However, inside the spheromak wall:
Bp = 0
and
Er = 0
and outside the spheromak wall:
Bt = 0.

Hence at every point on a spheromak wall:
{[[Bt^2 / 2 Mu]}inside
= {[Bp^2 / 2 Mu] + [(Epsilon / 2) Er^2]}outside

At R = Rc and Z = 0:
Symmetry gives:
Eroc = 0
so that:
Umoc = (Umic).
 

APPLICATION OF BOUNDARY CONDITIONS:
1) The energy density in the far field points to Uo in terms of Epsilono.

2) Knowledge of Uo enables calculation of Bpo.

3 Knowledge of Bpo enables calculation of an estimate of Np by assuming that the poloidal current is concentrated in a ring at Z = 0, R = Ro.

4) The boundary condition at the spheromak inner wall enables calculation of Btc in terms of So and hence Bto in terms of So.

5) Bto enables calculation of Nt as a function of So.

6) For a particular So value we can integrate to find the Np value that will yield the required Bpo value.

7) Thus we can find (Np).

8) FIX

9) Knowing the values of Np and (Np / Nt) we can find a consistent prime number Nt.
 

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THEORETICAL SPHEROMAK Ett = [Uo Pi^2 Ro^3 / (A^2 B)] {4 So (So^2 - So + 1) / (So^2 + 1)^2}
 

Note that Ett is a function of Uo, Ro, A, B and So. Hence in electromagnetic analysis of a spheromak we seek to find expressions for these parameters.

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CHARGE CIRCULATION AROUND PATH Lh:
The spheromak wall contains Nt toroidal turns equally spaced around 2 Pi radians in angle Theta about the main spheromak axis of symmetry.
Each purely toroidal turn has length:
2 Pi (Rs - Rc) Kc / 2 = Pi (Rs - Rc) Kc
so the purely toroidal winding length is:
Nt Pi (Rs - Rc) Kc

Note that for a round spheromak cross section toroid Kc = 1

The spheromak wall contains Np poloidal turns which are equally spaced around the ellipse perimeter. The average poloidal turn length is:
2 Pi (Rs + Rc) / 2 = Pi (Rs + Rc)
and the purely poloidal winding length is:
Np Pi (Rs + Rc)

Assume that the spheromak wall is composed of a current path of length Lh containing uniformly distributed charge Qs that is circulating around a closed spiral path at the speed of light C. The spheromak inside radius measured from the axis of symmetry is Rc and the spheromak outside radius measured from the axis of symmetry is Rs.

In one spheromak cycle the poloidal angle advances Np (2 Pi) radians. In the same cycle the toroidal angle advances Nt (2 Pi) radians. Hence:
(poloidal angle advance) / (toroidal angle advance) = Np / Nt While a current point moves from Rc to Rs the toroidal angle advance is Pi radians and the toroidal travel is Lt / 2. The corresponding distance along the equatorial outer circumference is:
Pi (Np / Nt) Rs.
Thus Pythagoras theorm gives the current point travel distance along the winding for the half toroidal turn as:
[(Lt / 2)^2 + ( Pi Np Rs / Nt)^2]^0.5

While the current point moves from Rs to Rc the toroidal angle advance is Pi radians and the toroidal point travel is Lt / 2. The corresponding distance along the equatorial inner circumference is:
Pi (Np / Nt) Rc.
Thus Pythagoras theorm gives the current point travel distance along this winding half turn as:
[(Lt / 2)^2 + ( Pi Np Rc / Nt)^2]^0.5

Thus the total winding length Lh is:
Lh = Nt [(Lt / 2)^2 + (Pi Np Rs / Nt)^2]^0.5
+ Nt [(Lt / 2)^2 + ( Pi Np Rc / Nt)^2]^0.5
 
= [(Nt Lt / 2)^2 + (Pi Np Rs)^2]^0.5
+ [(Nt Lt / 2)^2 + (Pi Np Rc)^2]^0.5

Recall that:
Lt = [Kc Pi (Rs - Rc)]
and
Rs = Ro So / A
and
Rc = Ro / A So

Thus:
Lh = {(Nt Lt / 2)^2 + (Pi Np Rs)^2}^0.5
+ {(Nt Lt / 2)^2 + (Pi Np Rc)^2}^0.5
 
= {(Nt [Kc Pi (Rs - Rc)] / 2)^2 + (Pi Np Rs)^2}^0.5
+ {(Nt [Kc Pi (Rs - Rc)] / 2)^2 + (Pi Np Rc)^2}^0.5
 
= {[Nt Kc Pi (Rs - Rc) / 2]^2 + (Pi Np Rs)^2}^0.5
+ {[Nt Kc Pi (Rs - Rc) / 2]^2 + (Pi Np Rc)^2}^0.5
 
= {[Nt Kc Pi ((Ro So / 2 A) - (Ro / 2 A So))]^2 + (Pi Np (Ro So / A))^2}^0.5
+ {[Nt Kc Pi ((Ro So / 2 A) - (Ro / 2 A So))]^2 + (Pi Np (Ro / A So))^2}^0.5
 
= [Pi Ro / A]{[Nt Kc ((So / 2) - (1 / 2 So))]^2 + [Np So]^2}^0.5
+[Pi Ro / A] {[Nt Kc ((So / 2) - (1 / 2 So))]^2 + [(Np / So)]^2}^0.5
 
= [Pi Ro / 2 So A] {[Nt Kc ((So^2) - (1))]^2 + [Np (2 So^2)]^2}^0.5
+[Pi Ro / 2 So A] {[Nt Kc ((So^2) - (1))]^2 + [2 Np]^2}^0.5
 
= [Pi Ro / 2 So A] {[Nt Kc (So^2 - 1)]^2 + [Np (2 So^2)]^2}^0.5
+[Pi Ro / 2 So A] {[Nt Kc (So^2 - 1)]^2 + [2 Np]^2}^0.5
 
= [Pi Ro Nt / 2 So A] {[Kc (So^2 - 1)]^2 + [(Np / Nt) (2 So^2)]^2}^0.5
+[Pi Ro Nt / 2 So A] {[Kc (So^2 - 1)]^2 + ([Np / Nt][2])^2}^0.5

Thus:
[Lh A / 2 Pi Ro] = [Nt / 4 So] {[Kc (So^2 - 1)]^2 + (Np / Nt)^2 [2 So^2]^2}^0.5
+ [Nt / 4 So]{[Kc (So^2 - 1)]^2 + (Np / Nt)^2 [2]^2}^0.5

or
[Lh A / 2 Pi Ro] = Nt {[Kc (So^2 - 1) / 4 So]^2 + (Np / Nt)^2 [So / 2]^2}^0.5
+ Nt {[Kc (So^2 - 1)/ 4 So]^2 + (Np / Nt)^2 [(1 / 2 So)]^2}^0.5

This equation is the result of spheromak geometric analysis.

...................................................................................................................................

Use this equation with dLh = 0 to relate dNp to dNt. Note that the spheromak stability analysis should not involve A.

[Lh A 2 So / Pi Ro] = {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2) [2 So^2]^2)}^0.5
+ {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)[2]^2}^0.5

= La + Lb
where:
La = {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)[2 So^2]^2}^0.5
and
Lb = {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)[2]^2}^0.5

d[Lh A 2 So / Pi Ro] = 0
Implies that:
dLa + dLb = 0
or
{1 / 2 La} {2 Nt dNt [Kc (So^2 - 1)]^2 + 2 Np dNp [2 So^2]^2)}
+ {1 / 2 Lb} {2 Nt dNt [Kc (So^2 - 1)]^2 + 2 Np dNp [2]^2}
= 0

Make substitution:
dNt = - 2 dNp
to get:
{1 / 2 La} {2 Nt (- 2 dNp) [Kc (So^2 - 1)]^2 + 2 Np dNp (4 So^4)}
+ {1 / 2 Lb} {2 Nt (-2 dNp) [Kc (So^2 - 1)]^2 + 2 Np dNp 4}
= 0

or
{1 / 2 La} {- 4 Nt [Kc (So^2 - 1)]^2 + 8 Np (So^4)}
+ {1 / 2 Lb} {-4 Nt [Kc (So^2 - 1)]^2 + 8 Np}
= 0

which sets the relationship between Np and Nt at the spheromak operating point.

Bring terms to a common denominator to get:
Lb {- 4 Nt [Kc (So^2 - 1)]^2 + 8 Np (So^4)}
+ La {-4 Nt [Kc (So^2 - 1)]^2 + 8 Np}
= 0

Thus:
Lb [8 Np So^4 - {4 Nt [Kc (So^2 - 1)]^2}]
= La [{4 Nt [Kc (So^2 - 1)]^2} - 8 Np]
or
Lb^2 [8 Np So^4 - {4 Nt [Kc (So^2 - 1)]^2}]^2
= La^2 [{4 Nt [Kc (So^2 - 1)]^2} - 8 Np]^2<

Rearrangiing gives:
La^2 / Lb^2
= [8 Np So^4 - {4 Nt [Kc (So^2 - 1)]^2}]^2
/ [{4 Nt [Kc (So^2 - 1)]^2} - 8 Np]^2

However, from the geometric expression:
La = {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4 So^4)}^0.5
and
Lb = {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4)}^0.5

Hence:
La^2 / Lb^2
= {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4 So^4)}
/ {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4)}

Equating the two expressions for (La^2 / Lb^2} gives:
[8 Np So^4 - {4 Nt [Kc (So^2 - 1)]^2}]^2
/ [{4 Nt [Kc (So^2 - 1)]^2} - 8 Np]^2
= {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4 So^4)}
/ {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4)}

Rearrange to eliminate divides to get:
[8 Np So^4 - {4 Nt [Kc (So^2 - 1)]^2}]^2 {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4)}
= {[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4 So^4)}[{4 Nt [Kc (So^2 - 1)]^2} - 8 Np]^2

Expand to get:
{[64 Np^2 So^8 - 64 Np So^4 Nt [Kc (So^2 - 1)]^2 + 16 Nt^2 [Kc (So^2 - 1)]^4}
{Nt^2 [Kc (So^2 - 1)]^2 + 4 Np^2}
={16 Nt^2 [Kc (So^2 - 1)]^4 - 64 Nt[Kc (So^2 - 1)]^2 Np + 64 Np^2}
{[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4 So^4)}

Divide both sides by 16 to get:
{[4 Np^2 So^8 - 4 Np So^4 Nt [Kc (So^2 - 1)]^2 + Nt^2 [Kc (So^2 - 1)]^4}
{Nt^2 [Kc (So^2 - 1)]^2 + 4 Np^2}
 
= {Nt^2 [Kc (So^2 - 1)]^4 - 4 Nt[Kc (So^2 - 1)]^2 Np + 4 Np^2}
{[Nt^2][Kc (So^2 - 1)]^2 + (Np^2)(4 So^4)}

 

Further expand both sides to get:
{4 Np^2 So^8 Nt^2 [Kc (So^2 - 1)]^2
- 4 Np So^4 Nt [Kc (So^2 - 1)]^2 [Nt^2][Kc (So^2 - 1)]^2
+ Nt^2 [Kc (So^2 - 1)]^4 Nt^2 [Kc (So^2 - 1)]^2
+ 4 Np^2 So^8 4 Np^2
- 4 Np So^4 Nt [Kc (So^2 - 1)]^2 4 Np^2
+ Nt^2 [Kc (So^2 - 1)]^4 4 Np^2
= Nt^2 [Kc (So^2 - 1)]^4 [Nt^2][Kc (So^2 - 1)]^2
- 4 Nt [Kc (So^2 - 1)]^2 Np [Nt^2][Kc (So^2 - 1)]^2
+ 4 Np^2 [Nt^2] [Kc (So^2 - 1)]^2
+ Nt^2 [Kc (So^2 - 1)]^4 (Np^2)(4 So^4)
- 4 Nt [Kc (So^2 - 1)]^2 Np (Np^2)(4 So^4)
+ 4 Np^2 (Np^2) (4 So^4)

OK

Cancelling equal Nt^4 terms in gives:
4 Np^2 So^8 Nt^2 [Kc (So^2 - 1)]^2
- 4 Np So^4 Nt [Kc (So^2 - 1)]^2 [Nt^2] [Kc (So^2 - 1)]^2
+ 4 Np^2 So^8 4 Np^2
- 4 Np So^4 Nt [Kc (So^2 - 1)]^2 4 Np^2
+ Nt^2 [Kc (So^2 - 1)]^4 4 Np^2
= - 4 Nt [Kc (So^2 - 1)]^2 Np [Nt^2][Kc (So^2 - 1)]^2
+ 4 Np^2 [Nt^2][Kc (So^2 - 1)]^2
+ Nt^2 [Kc (So^2 - 1)]^4 (Np^2) (4 So^4)
- 4 Nt [Kc (So^2 - 1)]^2 Np (Np^2) (4 So^4)
+ 4 Np^2 (Np^2) (4 So^4)

Collecting terms gives: Nt^3 { - 4 Np So^4 [Kc (So^2 - 1)]^2 [Kc (So^2 - 1)]^2
+ 4 [Kc (So^2 - 1)]^2 Np [Kc (So^2 - 1)]^2}
+ Nt^2 {4 Np^2 So^8 [Kc (So^2 - 1)]^2
+ [Kc (So^2 - 1)]^4 4 Np^2
- 4 Np^2 [Kc (So^2 - 1)]^2
- [Kc (So^2 - 1)]^4 (Np^2)(4 So^4)
} + Nt{- 4 Np So^4 [Kc (So^2 - 1)]^2 4 Np^2
+ 4 [Kc (So^2 - 1)]^2 Np (Np^2)(4 So^4)}
+ 1 {+ 4 Np^2 So^8 4 Np^2 - 4 Np^2 (Np^2)(4 So^4)}
= 0

Simplifying terms gives: Nt^3 {4 Np [Kc (So^2 - 1)]^4 (1 - So^4)}
+ Nt^2 {4 Np^2 [Kc (So^2 - 1)]^2 [So^8 + [Kc (So^2 - 1)]^2 -1 + So^4[Kc (So^2 - 1)]^2 ] +1{ 16 Np^4 So^4 (So^4 - 1)}
= 0

or
Nt^3 {4 Np [Kc (So^2 - 1)]^4 (So^4 - 1)}
- Nt^2 {4 Np^2 [Kc (So^2 - 1)]^2 [So^8 + [Kc (So^2 - 1)]^2 -1 + So^4[Kc (So^2 - 1)]^2 ] - {16 Np^4 So^4 (So^4 - 1)}
= 0

Note that:
[So^8 + [Kc (So^2 - 1)]^2 -1 + So^4[Kc (So^2 - 1)]^2 ]
= [So^8 - 1] + [So^4 + 1][Kc (So^2 - 1)]^2
= [So^4 - 1] [So^4 + 1] + [So^4 + 1][Kc (So^2 - 1)]^2
= [So^2 + 1] [So^4 - 1 + [Kc (So^2 - 1)]^2]

Thus:
Nt^3 {4 Np [Kc (So^2 - 1)]^4 (So^4 - 1)}
- Nt^2 {4 Np^2 [Kc (So^2 - 1)]^2 [So^2 + 1] [So^4 - 1 + [Kc (So^2 - 1)]^2]
- {16 Np^4 So^4 (So^4 - 1)}
= 0

or
Nt^3 {4 Np [Kc (So^2 - 1)]^4 (So^4 - 1)}
- Nt^2 {4 Np^2 [Kc (So^2 - 1)]^4 [So^2 + 1]
- Nt^2 {4 Np^2 [Kc (So^2 - 1)]^2 [So^2 + 1] [So^4 - 1]
- {16 Np^4 So^4 (So^4 - 1)}
= 0

Divide through by 4 Np^4 to get:
[Nt / Np]^3 {[Kc (So^2 - 1)]^4 (So^4 - 1)}
- [Nt / Np]^2 {[Kc (So^2 - 1)]^4 [So^2 + 1]
- [Nt / Np]^2 {[Kc (So^2 - 1)]^2 [So^2 + 1] [So^4 - 1]
- {4 So^4 (So^4 - 1)}
= 0

Divide through by [Kc (So^2 - 1)]^4 to get:
[Nt / Np]^3 (So^4 - 1)
- [Nt / Np]^2 [So^2 + 1]
- [Nt / Np]^2 [So^2 + 1] {[So^4 - 1] / [Kc (So^2 - 1)]^2}
- 4 {So^4 /[Kc (So^2 - 1)]^2} {(So^4 - 1) / [Kc (So^2 - 1)]^2}
= 0

Factor out (So^2 + 1) to get:
[Nt / Np]^3 (So^2 - 1)
- [Nt / Np]^2
- [Nt / Np]^2 {[So^4 - 1] / [Kc (So^2 - 1)]^2}
- 4 {So^4 /[Kc (So^2 - 1)]^2} {(So^2 - 1) / [Kc (So^2 - 1)]^2}
= 0

Further simplify: [Nt / Np]^3 (So^2 - 1)
- [Nt / Np]^2
- [Nt / Np]^2 {[So^2 + 1] / [Kc^2 (So^2 - 1)]}
- 4 {So^4 /[Kc^4 (So^2 - 1)^3]}
= 0

Thus for every So^2 and Kc value there is a corresponding (Nt / Np) value.
Nt = P - 2 Np

Thus for each value of So^2 and Kc and trial P increment through Np to find an approximate zero solution.

The correct value of P will give an exact zero solution.

This is the expression for Lh obtained from Lh stability. Recall that from spheromak geometry we have the expressions:
or
(La / Nt) = {[Kc (So^2 - 1)]^2 + (Np / Nt)^2 (4) (So^4)}^0.5
and
(Lb / Nt) = {[Kc (So^2 - 1)]^2 + (Np / Nt)^2 (4)}^0.5

We also have the expression from spheromak geometry:
[Lh A So / 2 Pi Ro] = (La + Lb) / 4
= Nt [(1 / 4)[(La / Nt) + (Lb / Nt)]]
or
A = [2 Pi Ro (La + Lb) / 4 Lh So]
= [2 Pi Ro Nt / Lh] [(La / Nt) + (Lb / Nt)] / 4 So]

The solution procedure is to substitute the stable value of (Ro / Lh) into the geometric formula.

Thus:
(La + Lb) / 4 = {Np [Lb(So^4) + La]} / {2 Nt [Kc (So^2 - 1)]^2} or
(La + Lb) = La (4 Np / {2 Nt [Kc (So^2 - 1)]^2})
+ Lb (4 Np So^4 / {2 Nt [Kc (So^2 - 1)]^2})
which indicates that:
(4 Np / {2 Nt [Kc (So^2 - 1)]^2}) = 1
and
(4 Np So^4 / {2 Nt [Kc (So^2 - 1)]^2}) = 1 dLh = (1 / 2){[(Nt Lt / 2)^2 + (Pi Np Rs)^2]^-0.5} {2 (Nt Lt / 2)(Lt / 2) dNt + 2 Pi Np Rs Pi Rs}
+ (1 / 2){[(Nt Lt / 2)^2 + (Pi Np Rc)^2]^-0.5}{{2 (Nt Lt / 2)(Lt / 2) dNt + 2 Pi Np Rc Pi Rc}dNp
= 0

dNt = - 2 dNp dLh = {[(Nt Lt / 2)^2 + (Pi Np Rs)^2]^-0.5} {2 (Nt Lt / 2)(Lt / 2) (-2 dNp) + 2 Pi Np Rs Pi Rs}dNp
+ {[(Nt Lt / 2)^2 + (Pi Np Rc)^2]^-0.5}{{2 (Nt Lt / 2)(Lt / 2)(- 2 dNp) + 2 Pi Np Rc Pi Rc}dNp
= 0

Thus:
Lh^2 = [(Nt Lt / 2)^2 + (Pi Np Rs)^2]
+ [(Nt Lt / 2)^2 + ( Pi Np Rc)^2]
+ 2 {[(Nt Lt / 2)^2 + ( Pi Np Rs)^2]^0.5}{[(Nt Lt / 2)^2 + ( Pi Np Rc)^2]^0.5}
 
= {[(Nt Lt / 2)^2 + (Pi Np Rs)^2]}
+ {[(Nt Lt / 2)^2 + (Pi Np Rc)^2]}
+ 2 {{[(Nt Lt / 2)^2 + (Pi Np Rs)^2]}{[(Nt Lt / 2)^2 + (Pi Np Rc)^2]}}^0.5
 
= {[(Nt Lt / 2)^2 + (Pi Np Rs)^2]}
+ {[(Nt Lt / 2)^2 + (Pi Np Rc)^2]}
+ 2 {[(Nt Lt /2)^2]^2 + [(Pi Np)^4 Rs^2 Rc^2]
+ [(Nt Lt /2)^2 (Pi Np)^2 (Rs^2 + Rc^2)]}^0.5
 
Lt = Pi Kc (Rs - Rc)

Lh^2 = 2 (Nt Lt / 2)^2 + (Pi Np)^2 (Rs^2 + Rc^2) + root term CONTINUE,SIMPLIFY

Note that Np and Nt are positive integers. The quantity [(Lh A / (2 Pi Ro)] is believed to be a geometric constant for a stable spheromak. The stability of this quantity relies on the stabilities of Np, Nt and So.

Note that:
Np = Nt + 1

Note that with increasing spheromak energy Ro and Lh both decrease so that the ratio:
(Lh A / Ro)
remains constant. As Lh decreases the spheromak frequency:
Fh = C / Lh
increases.

The circulating current Ih is given by:
Ih = Qs Fh
= Qs C / Lh where:
Ih = Qs Fh = Qs C / Lh
This equation imposes an important relationship between Nr and So in a spheromak.
 

NATURAL FREQUENCY:
The natural frequency Fh of a spheromak is:
Fh = C / Lh

Rearranging this equation gives:
dits axis of symmetry.

Note that this formula applies to all spheromaks

*****************************************************************

**************************************************************************

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DEVELOP AN EXPRESSION FOR THE ENERGY DENSITY AS A FUNCTION OF POSITION OUTSIDE THE SPHEROMAK WALL:

AXIAL ELECTRIC FIELD ENERGY DENSITY DUE TO A RING OF CHARGE:
Assume that a thin ring of radius "Ro" has net charge Qs. Then the linear charge density along the thin ring is:
Qs / (2 Pi Ro)
and an element of charge is:
dQs = [Qs / (2 Pi Ro)] dL
where dL is an element of the ring's circumferential length.

Consider a point at distance "Z" along the ring axis, where Z = 0 on the ring plane.

The electric field along distance (Ro^2 + Z^2)^0.5 due to charge dQs is:
dE =(1 / 4 Pi Epsilon) [dQs / (Ro^2 + Z^2)]
where:
Epsilon = permittivity of free space

The component of this electric field along the ring axis is:
dE = (1 / 4 Pi Epsilon) [dQ / (Ro^2 + Z^2)] cos(Theta)
where:
cos(Theta) = Z / (Ra^2 + Z^2)^0.5

The net electric field E at distance Z along the ring axis is:
E = (1 / 4 Pi Epsilon) [Qs / (Ro^2 + Z^2)] cos(Theta)
= (1 / 4 Pi Epsilon) [Qs / (Ro^2 + Z^2)] [Z / (Ro^2 + Z^2)^0.5]
= (1 / 4 Pi Epsilon) [Qs Z / (Ro^2 + Z^2)^1.5]

The electric field energy density Ue at Z = Z, R = 0 is given by:
Ue = (Epsilon / 2) E^2
= (Epsilon / 2){(1 / 4 Pi Epsilon) [Qs Z / (Ro^2 + Z^2)^1.5]}^2
= [Qs^2 / (32 Pi^2 Epsilon)] [Z^2 / (Ro^2 + Z^2)^3]
= [Mu C^2 Qs^2 / (32 Pi^2)] [Z^2 / (Ro^2 + Z^2)^3]

Note that at Z = 0 the net electric field is zero and the electric field energy density is zero.

In the far field where:
Z >> Ro
then along the ring axis in the far field:
Ue = [Qs^2 / (32 Pi^2 Epsilon)] [1 / Z^4]

Thus for Z >> Ro along the ring axis the electric field energy density Ue is proportional to (1 / Z)^4.
 

AXIAL MAGNETIC FIELD ENERGY DENSITY DUE TO A RING OF CURRENT:
The law of Biot and Savart gives an element of magnetic field dB at a measurement point on the axis of a ring due to an electric current I is:
dB = (Mu / 4 Pi) Ip dL X R / |R|^3
where:
Ip = poloidal current around the ring
|R| = distance from current element Ip dL to the measurement point
Mu = magnetic permeability of free space
dL = an element of length along the direction of electric current flow around the ring
[R / |R|] = unit vector along the direction of R

Consider a ring of radius "Ro" and an axial measurement point at distance "Z" along the ring axis from the plane of the current ring.

Then along the ring axis the net magnetic field B at the measurement point is axial and is given by:
B = [(Mu / 4 Pi) Ip 2 Pi Ro / (Ro^2 + Z^2)] sin(Theta)
where:
(Theta) = angle between unit vector (R / |R|) and the ring axis.

However:
sin(Theta) = Ro / (Ro^2 + Z^2)^0.5

Thus the net magnetic field B along the current ring axis due to ring current I is given by::
B = [(Mu / 4 Pi) Ip 2 Pi Ro / (Ro^2 + Z^2)] sin(Theta)
= [(Mu / 4 Pi) Ip 2 Pi Ro / (Ro^2 + Z^2)][ Ro / (Ro^2 + Z^2)^0.5]
= [(Mu Ip) / 2] [Ro^2 / (Ro^2 + Z^2)^1.5]

The magnetic field energy density along the ring axis at Z = Z is:
Um = B^2 / 2 Mu
= [(Mu Ip) / 2]^2 [Ro^2 / (Ro^2 + Z^2)^1.5]^2 / 2 Mu
= [Mu Ip^2 / 8] [Ro^4 / (Ro^2 + Z^2)^3]

Thus for Z >> Ro the magnetic field energy density Um is proportional to (1 / Z)^6. Hence at large distances the magnetic field energy density becomes negligibly small as compared to the electric field energy density.

At the center of the ring where Z = 0 the magnetic field is:
Bpo = [(Mu Ip) / (2 Ro)

The corresponding magnetic field energy density Umo at the center of the ring is:
Umo = Bpo^2 / 2 Mu
= [Mu Ip^2 / 8 Ro^2]

Note that Umo is the magnetic field energy density at R = 0, Z = 0 corresponding to current Ip circulating at R = Ro.

TOTAL AXIAL ENERGY DENSITY DUE TO A RING OF CIRCULATING CHARGE:

Along the ring axis the total electromagnetic energy density U is given by:
U = Ue + Um
= [Mu C^2 Qs^2 / (32 Pi^2)] [Z^2 / (Ro^2 + Z^2)^3] + [Mu Ip^2 / 8] [Ro^4 / (Ro^2 + Z^2)^3]
= [Mu C^2 Qs^2 / (32 Pi^2)] {[Z^2 / (Ro^2 + Z^2)^3] + {[Mu Ip^2 Ro^2 / 8] /[Mu C^2 Qa^2 / (32 Pi^2)]} [Ro^2 / (Ro^2 + Z^2)^3]

If:
{[Mu Ip^2 Ro^2 / 8] /[Mu C^2 Qs^2 / (32 Pi^2)]} = 1
or if:
{[ Ip^2 Ro^2 ] / [ C^2 Qs^2 / (4 Pi^2)]} = 1
or if
{[ Ip^2 Ro^2 4 Pi^2 ] / [ C^2 Qs^2]} = 1
or if:
{[ Ip Ro 2 Pi ] / [ C Qs]} = 1
or if
Ip = C Qs / (2 Pi Ro)
meaning that the charge circulates at the speed of light then the total energy density U along the ring axis is given by:
U = [Mu C^2 Qs^2 / (32 Pi^2)] [1 / (Ro^2 + Z^2)^2]
and
Uo = [Mu C^2 Qs^2 / (32 Pi^2 Ro^4)]
 

FIND EXPRESSIONS FOR THE ELECTRIC AND MAGNETIC FIELD ENERGY DENSITIES AS A FUNCTION OF POSITION OUTSIDE THE SPHEROMAK WALL:
For R > Rc the magnetic field energy density outside the spheromak wall will likely be:
Um = Umo [(Ro^2 Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
and the electric field energy density outside the spheromak wall will likely be:
Ue = Ueo [(((A R)^2 + Z^2) Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
where:
Umo = Ueo = Uo

For R < Rc the magnetic field energy density outside the spheromak wall will likely be:
Um = Umo [(Ro^2 + (A R)^2) Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
and the electric field energy density outside the spheromak wall will likely be:
Ue = Ueo [((B Z)^2 Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
where:
Umo = Ueo = Uo

Note that both Ue and Um have step changes at R = Rc but the steps are equal and opposite so that U is a smoothly continuous function.

Note that for R > Rc:
U = Um + Ue
= Umo [(Ro^2 Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
+ Ueo [(((A R)^2 + (B Z)^2) Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
= Uo [((Ro^2 Ro^4) + ((A R)^2 + (B Z)^2) Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
= Uo [Ro^4 / (Ro^2 + (A R)^2 + (B Z)^2)^2]
= Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2
as expected.

Note that for R < Rc:
U = Um + Ue
= Umo [(Ro^2 + (A R)^2) Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
+ Ueo [(Z^2 Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
where:
Umo = Ueo = Uo

Hence:
U = Uo [(Ro^2 + (A R)^2 + (B Z)^2) Ro^4) / (Ro^2 + (A R)^2 + (B Z)^2)^3]
= Uo [Ro^4 / (Ro^2 + (A R)^2 + (B Z)^2)^2]
= Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2
as expected
 

Note that both Ue and Um have step changes at R = Rc but the steps are equal and opposite in magnitude so that the total energy density U remains a smoothly continuous function.

Note that in the near field A and B are both of the order of unity and in the far field:
A = B = 1.000

Note that there is a geometric relationship between A and Kc. The parameter Kc involves the perimeter length of an ellipse whereas the parameter A involves the ratio of the major and minor axis of that same ellipse.

Thus we have shown the origin of the expression for the total energy density outside the spheromak wall. This expression may not accurately reflect experimental reality, but it is at least mathematically tractable.
 

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FIND Uo WITH RESPECT TO THE ELECTRIC FIELD FAR FIELD DISTRIBUTION:
Let Qs be the net charge of a spheromak as indicated by an electric field measurement at:
(A R)^2 + (B Z)^2 >> Ro^2.
Note that most field measurements on electrons and protons are far field measurements.

In the far field the energy field density of an electromagnetic system is almost purely electric. The electric field energy density in the far field is:
[Epsilono / 2] E^2
~ [Epsilono / 2][Qs / (4 Pi Epsilono (R^2 + Z^2)]^2

The theoretical spheromak electric energy density in the far field is:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2

The electric field due to a single particle is:
Q / [4 Pi Epsilno [(Ro^2 + (A R)^2 + (B Z)^2)]]
= Q / [4 Pi Epsilono [(Ro^2 + (A X)^2 + (A Y)^2 + (B Z)^2)]]

For a collection of particles the average electric field at three dimensional radial vector Rx is: Q / [4 Pi Epsilono [(Ro^2 + (A Rx)^2 / 3 + (A Rx)^2 / 3 + (B Rx / 3)^2)]]
= Q / [4 Pi Epsilono [(Ro^2 + Rx^2 {(A)^2 / 3 + (A)^2 / 3 + (B / 3)^2}]]
= Q / [4 Pi Epsilono [(Ro^2 + Rx^2 {(2 A^2 + B^2)/ 3}]]

To match standard electrostatics:
(2 A^2 + B^2)/ 3 = 1

Spheromak mathematics set the ratio (A / B). B = (B / A) A
or
B^2 = (B / A)^2 A^2
Hence:
(2 A^2 + B^2)/ 3 = 1
becomes:
(2 A^2 + (B / A)^2 A^2)/ 3 = 1
or
A^2 (2 + (B / A)^2)/ 3 = 1
or
A^2 = 3 / (2 + (B / A)^2) or
A = {3 / (2 + (B / A)^2)}^0.5
and B = (B / A) A
= (B / A){3 / (2 + (B / A)^2)}^0.5
= {3 (B / A)^2 / (2 + (B / A)^2)}^0.5 = {3 / (2 (A / B)^2 + 1)}^0.5

Inspection shows that if (A / B) > 1 then A > 1 amd B < 1.
 

For the classical case of randomly oriented electrons:
U = (Epsilono / 2)[Qs / (4 Pi Epsilono Rx^2]^2 as compared to: U = Uo [Ro^2 / (Ro^2 + Rx^2)]^2

Comparing equations for Rx^2 >> Ro^2 gives:
Uo Ro^4 = (Epsilono / 2)[Qs / (4 Pi Epsilono]^2 or Uo = [Qs^2 / (32 Epsilono Pi^2 Ro^4)]

Recall that:
C^2 = 1 / (Muo Epsilono)

Uo = (1 / Ro)^4 [Muo C^2 / 2][Qs / 4 Pi]^2

= (Muo / 32)[Qs C / (Pi Ro^2)]^2
Note that in general A > 1 and B < 1 in order to maintain Uo.
 

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FIND THE MAGNETIC FIELD Bt INSIDE THE SPHEROMAK WALL AND HENCE Nt:
The boundary condition at Z = 0 at the inner spheromak wall gives:
Uo {Ro^2 / [Ro^2 + (A Rc)^2]}^2 = Btc^2 / 2 Muo

Recall that:
Uo = [Muo C^2 Qs^2 / (32 Pi^2 Ro^4)]
and
Btc = Muo Nt Ih / 2 Pi Rc
= Muo Nt Qs C / (Lh 2 Pi Rc)

Combining these equations gives:
[Muo C^2 Qs^2 / (32 Pi^2 Ro^4)]{Ro^2 / [Ro^2 + (A Rc)^2]}^2
= [Muo Nt Qs C / (Lh 2 Pi Rc)]^2 / 2 Muo
or
[1 / (16)]{1 / [Ro^2 + (A Rc)^2]}^2
= [Nt / (Lh 2 Rc)]^2
or
[1 / 4]{1 / [Ro^2 + (A Rc)^2]}
= [Nt / (Lh 2 Rc)]
or
{1 / [Ro^2 + (A Rc)^2]} = 2 Nt / (Lh Rc)
or
Nt = [Lh Rc / 2]{1 / [Ro^2 + (A Rc)^2]}
= [Lh / 2 Ro]{(Rc Ro) / [Ro^2 + (A Rc)^2]}
= [Lh / 2 Ro]{(Rc / Ro) / [(Ro^2 / Ro^2) + (A Rc / Ro)^2]}
= {(Lh / 2 Ro) (Rc / Ro){1 / [1 + (A Rc / Ro)^2]}

Recall that:
(A Rc / Ro) = (1 / So)

Hence:
Nt = {(Lh / 2 Ro) (1 / A So){1 / [1 + (1 / So)^2]}
= {(Lh / 2 Ro A) (1 / So){So^2 / [So^2 + 1]}
= [Lh / (2 Ro A)][So / (So^2 + 1)]

Thus we have an explicit equation for Nt.
 

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ESTIMATE OF Np = Npe
Assume that Uo arises from a current ring at Z = 0, R = Ro

For such a current ring:
Bpo = (Muo Ih Npe) / (2 Ro)

Uo = (Bpo^2 / 2 Muo)
= (1 / 2 Muo) [(Muo Ih Npe) / (2 Ro)]^2
= (Muo / 8)[(Ih Npe) / Ro]^2
= (Muo / 8)[(Qs C Npe) / Lh Ro]^2

Recall that:
Uo = (Muo / 32)[Qs C / (Pi Ro^2)]^2

Equate the to expressions for Uo to get:
(Muo / 8)[(Qs C Npe) / Lh Ro]^2 = (Muo / 32)[Qs C / (Pi Ro^2)]^2
or
[(Npe) / Lh]^2 = (1 / 4)[1 / (Pi Ro)]^2
or
[Npe / Lh] = (1 / 2)[1 / Pi Ro]
or
Npe = (Lh / 2 Pi Ro)

Recall that:
Nt = [Lh / (2 Ro A)][So / (So^2 + 1)]
Hence:
(Npe / Nt) = (Lh / 2 Pi Ro) / {[Lh / (2 Ro A)][So / (So^2 + 1)]}
= (1 / 2 Pi) / {[1 / (2 A)][So / (So^2 + 1)]}
= (A / Pi) [(So^2 + 1) / So]

An important issue is whether the axial poloidal magnetic field produced by Npe is sufficient to provide the required size of Uo. If Npe = Nt and if So = 2 then:
A > (Pi / 2.5)

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FIND EXACT EXPRESSION FOR (Np / Nt) IN TERMS OF So:
Recall that:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2
where
Lp = Pi (Rs + Rc)
and
Lt = Pi (Rs - Rc) Kc

Hence:
(Lh / Ro)^2 = (Np Lp / Ro)^2 + (Nt Lt / Ro)^2
= [Np Pi (Rs + Rc) / Ro]^2 + [Nt Pi (Rs - Rc) Kc / Ro]^2
= [Np Pi ((So / A) + (1 / A So))]^2 + [Nt Pi Kc ((So / A) - (1 / So A))]^2
= [Pi Nt / A]^2 {[(Np / Nt) ((So) + (1 / So))]^2 + [Kc ((So) - (1 / So))]^2}

Recall that:
Nt = [Lh / (2 Ro A)][So / (So^2 + 1)]
= [Lh / Ro] (1 / 2 A)[So / (So^2 + 1)]

Hence:
Nt^2 = [Lh / Ro]^2 (1 / 2 A)^2 [So / (So^2 + 1)]^2

Thus:
(Lh / Ro)^2 = [Lh / Ro]^2 (1 / 2 A)^2 [So / (So^2 + 1)]^2 [Pi / A]^2
{[(Np / Nt) ((So) + (1 / So))]^2 + [Kc ((So) - (1 / So))]^2}
 
or
1 = [(1 / 2 A)^2 [So / (So^2 + 1)]^2 [Pi / A]^2
{[(Np / Nt) ((So) + (1 / So))]^2 + [Kc ((So) - (1 / So))]^2}
or
(2 A)^2 [(So^2 + 1) / So]^2 [A / Pi]^2
= {[(Np / Nt) ((So) + (1 / So))]^2 + [Kc ((So) - (1 / So))]^2}
or
(2 A)^2 [(So^2 + 1) / So]^2 [A / Pi]^2 - [Kc ((So) - (1 / So))]^2
= [(Np / Nt) ((So) + (1 / So))]^2
or
[Np / Nt]^2 = {(2 A)^2 [(So^2 + 1) / So]^2 [A / Pi]^2 - [Kc ((So) - (1 / So))]^2}
/ [So + (1 / So)]^2
or
[Np / Nt]^2 = {(2 A^2 / Pi)^2 [(So^2 + 1)]^2 - [Kc ((So^2) - 1)]^2}
/ [So^2 + 1)]^2
or
[Np / Nt]^2 = {(2 A^2 / Pi)^2 - [Kc ((So^2) - 1) / (So^2 + 1)]^2}
where:
Kc = [1 + (A / B)] [Kh / 2]

To a first approximation:
Kh = 1.00

Thus:
[Np / Nt]^2 = {(2 A^2 / Pi)^2
- [(1 + (A / B)) / 2]^2 Kh^2 [((So^2) - 1) / (So^2 + 1)]^2}

or
(2 A^2 / Pi)^2
= [Np / Nt]^2 + [(1 + (A / B)) / 2]^2 Kh^2 [((So^2) - 1) / (So^2 + 1)]^2

This equation provides an approximate solution for A as a function of So.

Once the A versus So curve is known We can then move along this curve the until the magnetic integration:
(Nt / Np) = [1 / A][So / (So^2 + 1)][INTEGRAL]

which forces Bpo^2 / 2 Muo to equal Uo is satisfied. That satisfaction point gives us both A and So values that are required to evaluate the fine structure constant.

Remember that Np / Nt is a ratio of integers derived from a prime number P.

Clearly it is absolutely necessary to take into account the elliptical cross section of a spheromak. A circular approximation does not yield a real (Np / Nt) value.

It appears that the So versus A curve is only weakly dependent on the choice of prime number. Hence there may be different quantum states corresponding to nearby prime numbers. It appears that the Fine Structure Constant coefficient is also only weakly dependent on the choice of prime number. The important parameter that really depends on the prime number is (Lh / Ro).
 

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AXIAL MAGNETIC FIELD AT CENTER OF A SPHEROMAK:
The poloidal magnetic field Bpo at the center of a spheromak is key to electromagnetic spheromak calculations. It is precisely derived below for both an ellipsoidal cross section spheromak and a round cross section spheromak.

FIND MAGNETIC FIELD Bpo ALONG THE MAIN AXIS OF SYMMETRY At R = 0, Z = 0 FOR AN ELLIPSOIDAL CROSS SECTION SPHEROMAK:
The law of Biot and Savart gives an element of axial magnetic field:
d(Bpo) at R = 0, Z = 0 due to a current ring winding with radius R located at height Zs and having current Ih dNp as:
d(Bpo) = [(Muo / 4 Pi) Ih dNp 2 Pi R / (R^2 + Zs^2)] [R / (R^2 + Zs^2)^0.5]
= (Muo Ih dNp R^2) / [2 (R^2 + Zs^2)^1.5]
 

Assume that the poloidal windings are equally spaced around the ellipse. Let dS = an element of distance tangent to the ellipse.
(dS)^2 = (dR)^2 + (dZs)^2
Then:
dNp = [dS / (ellipse perimeter)] Np
= Np [(dR)^2 + (dZs)^2]^0.5 / (ellipse perimeter)

From the web page titled: THEORETICAL SPHEROMAK:
(ellipse perimeter) = Pi (a + b) Kh

Hence:
dNp = Np [(dR)^2 + (dZs)^2]^0.5 / Pi (a + b) Kh

a = (Rs - Rc) / 2
b = Zf

From the web page titled: THEORETICAL SPHEROMAK:
Zs = (A / B) [(Rs - R) (R - Rc)]^0.5

Hence:
Zf = (A / B) [(Rs - Rf) (Rf - Rc)]^0.5
where:
Rf = [(Rs + Rc) / 2]

Thus:
Zf = (A / B) [(Rs - Rf) (Rf - Rc)]^0.5
= (A / B) [(Rs - [(Rs + Rc) / 2]) ([(Rs + Rc) / 2] - Rc)]^0.5
= (A / B) [(Rs - Rc) / 2]) ([(Rs - Rc) / 2])]^0.5
= (A / B) (Rs - Rc) / 2
= b

Hence:
dNp = Np [(dR)^2 + (dZs)^2]^0.5 / Pi (a + b) Kh
= Np [(dR)^2 + (dZs)^2]^0.5 / Pi ([(Rs - Rc) / 2] + [(A / B) (Rs - Rc) / 2]) Kh
= Np [(dR)^2 + (dZs)^2]^0.5 / [Pi Kh (1 + (A / B))(Rs - Rc) / 2]

Recall that:
Zs = (A / B) [(Rs - R) (R - Rc)]^0.5
or
dZs = (A / 2 B)[(Rs - R) (R - Rc)]^-0.5 [(Rs - R) dR - dR (R - Rc)]
or
(dZs)^2 = (A / 2 B)^2 [(Rs + Rc - 2 R) dR]^2 / [(Rs - R) (R - Rc)]

(dS)^2 = (dR)^2 + (dZs)^2
= {(dR)^2 [(Rs - R) (R - Rc)] + (A / 2 B)^2 [(Rs + Rc - 2 R) dR]^2} / [(Rs - R) (R - Rc)]
= (dR)^2 {[(Rs - R) (R - Rc)] + (A / 2 B)^2 [(Rs + Rc - 2 R)]^2} / [(Rs - R) (R - Rc)]

Thus:
dS = dR {[(Rs - R) (R - Rc)] + (A / 2 B)^2 [(Rs + Rc - 2 R)]^2}^0.5 / [(Rs - R) (R - Rc)]^0.5
giving:
dNp = Np dS / (elipse perimeter length)
= Np dS / [Pi Kc (Rs - Rc)]
= {Np / [Pi Kc (Rs - Rc)]}
dR {[(Rs - R) (R - Rc)] + (A / 2 B)^2 [(Rs + Rc - 2 R)]^2}^0.5 / [(Rs - R) (R - Rc)]^0.5

Recall that:
d(Bpo) = [(Muo / 4 Pi) Ih dNp 2 Pi R / (R^2 + Zs^2)] [R / (R^2 + Zs^2)^0.5]
= (Muo Ih dNp R^2 / (2 (R^2 + Zs^2)^1.5

The term:
{R^2 + Zs^2}^1.5 = {R^2 + (A / B)^2 [(Rs - R) (R - Rc)]}^1.5

Both the top and bottom halves of the spheromak contribute equally to the magnetic field at R = 0, Z = 0. Hence:
Bpo = 2 X Integral from R = Rc to R = Rs of:
(Muo Ih dNp R^2 / [2 (R^2 + Zs^2)^1.5]
 
= Integral from R = Rc to R = Rs of:
(Muo Ih dNp R^2 / [(R^2 + Zs^2)^1.5]
 
= Integral from R = Rc to R = Rs of:
{Muo Ih R^2 / {R^2 + (A / B)^2 [(Rs - R) (R - Rc)]}^1.5}
{Np / [Pi Kc (Rs - Rc)]}
dR {[(Rs - R) (R - Rc)] + (A / 2 B)^2 [(Rs + Rc - 2 R)]^2}^0.5 / [(Rs - R) (R - Rc)]^0.5
 
= Integral from R = Rc to R = Rs of:
{Muo Ih R^2 / {R^2 + (A / B)^2 [(Rs - R) (R - Rc)]}^1.5}
{Np / [Pi Kc (Rs - Rc)]}
dR {[(Rs - R) (R - Rc)] + (A / 2 B)^2 [(Rs + Rc - 2 R)]^2}^0.5 / [(Rs - R) (R - Rc)]^0.5
 
OK to here!
Bpo = Integral from (R / Ro) = (Rc / Ro) to (R / Ro) = (Rs / Ro) of:
(1 / Ro){(Muo Ih R^2 / Ro^2) / {(R / Ro)^2 + (A / B)^2 [((Rs / Ro) - (R / Ro)) ((R / Ro) - (Rc / Ro)]}^1.5}
(1 / Ro){2 Np / [Pi Kc ((Rs / Ro) - (Rc / Ro))]}
(Ro) d(R / Ro) {[((Rs / Ro) - (R / Ro)) ((R / Ro) - (Rc / Ro))] + (A / 2 B)^2 [((Rs / Ro) + (Rc / Ro) - (2 R / Ro)]^2}^0.5
/ [((Rs / Ro) - (R / Ro)) ((R / Ro) - (Rc / Ro))]^0.5
 

Define:
X = R / Ro
Xs = Rs / Ro = (So / A)
Xc = Rc / Ro = (1 / (A So))

For each specified A and So pair the integral finds the corresponding value of Bpo.

Then:
Bpo = Integral from X = Xc to X = Xs of:
(1 / Ro){(Muo Ih X^2) / {X^2 + (A / B)^2 [Xs - X) (X - Xc)]}^1.5}
{Np / [Pi Kc (Xs - Xc)]} dX
{[(Xs - X) (X - Xc)] + (A / 2 B)^2 [Xs + Xc - 2 X]^2}^0.5 / [(Xs - X)(X - Xc)]^0.5

 
= [Muo Ih Np / 2 Pi Ro] Integral from X = Xc to X = Xs of:
{(X^2) / {X^2 + (A / B)^2 [Xs - X) (X - Xc)]}^1.5}
{2 / [Kc (Xs - Xc)]} dX
{[(Xs - X) (X - Xc)] + (A / 2 B)^2 [Xs + Xc - 2 X]^2}^0.5 / [(Xs - X)(X - Xc )]^0.5

 

Hence:
Bpo = [Muo Ih Np / 2 Pi Ro] [INTEGRAL]
where:
[INTEGRAL] = Integral from X = Xc to X = Xs of:
{(X^2) / {X^2 + (A / B)^2 [Xs - X) (X - Xc)]}^1.5}
{2 / [Kc (Xs - Xc)]} dX
{[(Xs - X) (X - Xc)] + (A / 2 B)^2 [Xs + Xc - 2 X]^2}^0.5
/ [(Xs - X)(X - Xc )]^0.5

 

where:
X = R / Ro
Xs = Rs / Ro = (So / A)
Xc = Rc / Ro = (1 / (A So))

This is the center field boundary condition.

This equation is exact for an elliptical cross section spheromak.

Recall that:
Bpo = [Muo Ih Np / 2 Pi Ro] [INTEGRAL] Ih = Q C / Lh Bpo = [Muo Q C Np / 2 Pi Lh Ro] [INTEGRAL]

Recall that:
Uo = Bpo^2 / 2 Muo = {[Muo Q C Np / 2 Pi Lh Ro] [INTEGRAL]}^2 / 2 Muo = (Muo / 32)[Qs C / (Pi Ro^2)]^2
or
{[Np] [INTEGRAL]}^2 = (1 / 4)[Lh / (Ro)]^2
or
Np [INTEGRAL] = (1 / 2)[Lh / Ro]
or
Np = (1 / 2) [Lh / Ro] / [INTEGRAL]

Recall that:
Bto = (Muo Nt Ih) / (2 Pi Ro)

Then:
[Bpo / Bto] = [Muo Ih Np / 2 Pi Ro] [INTEGRAL] / (Muo Nt Ih) / (2 Pi Ro)
= [Np] [INTEGRAL] / Nt

On the web page titled: THEORETICAL SPHEROMAK it is shown that:
Uto = [Uo / A^2] [So^2 /(So^2 + 1)^2]

Hence:
Uto / Uo = (Bto / Bpo)^2
= [1 / A^2] [So^2 /(So^2 + 1)^2]
or
(Bto / Bpo) = [1 / A] [So /(So^2 + 1)]
or
(Bpo / Bto) = A [(So^2 + 1) / So]

Thus, equating the two expressions for (Bpo / Bto) gives:
[Np] [INTEGRAL] / Nt = A (So^2 + 1) / So
or
[Np / Nt] = A (So^2 + 1) / {So [INTEGRAL]}
Which expresses [Np / Nt] as a function of A and So.

where:
[INTEGRAL] = Integral from X = Xc to X = Xs of:
{(X^2) / {X^2 + (A / B)^2 [Xs - X) (X - Xc)]}^1.5}
{2 / [Kc (Xs - Xc)]}
dX {[(Xs - X) (X - Xc)] + (A / 2 B)^2 [Xs + Xc - 2 X]^2}^0.5
/ [(Xs - X)(X - Xc )]^0.5

where:
X = R / Ro
Xs = Rs / Ro = (So / A)
Xc = Rc / Ro = (1 / (A So))

Thus if we specify A and So we can calculate a real number value of:
[Np / Nt].
 

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DELETE THIS SECTION AS TO BEING NON-ELLIPSE COMPLIANT

POLOIDAL MAGNETIC FIELD AT THE CENTER OF A ROUND CROSS SECTION SPHEROMAK:
Earlier on this web page it is shown that in order to provide the poloidal magnetic field density required for a spheromak to exist the poloidal magnetic field at R = Rc, Z = 0 must satisfy:
Bpc = Btc
= Muo Nt Q Fh / 2 Pi Rc
= Muo Nt Q Fh So / 2 Pi Ro
in order to balance the toroidal magnetic field at R = Rc, Z = 0.

Recall that:
Fh = C / Lh

Thus:
Bpc = Muo Nt Q Fh So / 2 Pi Ro
= Muo Nt Q C So / 2 Pi Ro Lh

To achieve this magnetic field in the core of the spheromak there must be the required number of poloidal turns in the spheromak. However, in general:
Bpo > Bpc.

Hence in general Bpo must satisfy:
Bpo > Muo Nt Q C So / 2 Pi Ro Lh

Now let us attempt a precise calculation of Bpo.

The distance from the spheromak symmetry axis to the toroidal centerline is:
(Rs + Rc) / 2

The distance from the toroidal centerline to the spheromak wall is:
(Rs - Rc) / 2

The radial distance R from the spheromak symmetry axis to a point on the spheromak wall is:
R = [(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)
where:
Phi = angle about spheromak minor axis between the spheromak equatorial plane and a point on the spheromak wall, as measured a the toroidal center line.

Hs = height of a point on the spheromak wall above the spheromak equatorial plane, given by:
Hs = [(Rs - Rc) / 2] Sin(Phi)

D = distance from a point on the spheromak wall to the center of the spheromak given by:
D^2 = (R^2 + Hs^2)
= {[(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)}^2 + {[(Rs - Rc) / 2] Sin(Phi)}^2
= [(Rs + Rc) / 2]^2 + [(Rs - Rc) / 2]^2 [Cos(Phi)]^2
- 2 [(Rs + Rc) / 2][(Rs - Rc) / 2] Cos(Phi) + {[(Rs - Rc) / 2] Sin(Phi)}^2
 
= [(Rs + Rc) / 2]^2 + [(Rs - Rc) / 2]^2 - [(Rs^2 - Rc^2) / 2] Cos(Phi)
 
= [(Rs^2 + Rc^2) / 2] - [(Rs^2 - Rc^2) / 2] Cos(Phi)

Recall that:
d(Phi) / d(Theta) = (Nt / Np)

Consider two adjacent spheromak windings. Let dTheta and dPhi be the winding spacings in the poloidal and toroidal directions. At any point on a winding the slope of the winding is:
{[(Rs - Rc) / 2] d(Phi)} / {R d(Theta)}

In an elemental slope box with side lengths:
{[(Rs - Rc) / 2] d(Phi)}
and
{R d(Theta)}
contains the fraction [d(Theta) / 2 Pi] of one poloidal winding.

The poloidal windings are equally spaced. Hence the number of poloidal windings per radian in Phi is:
Np / 2 Pi

Hence:
dNp = Np d(Phi) / (2 Pi)

dBpo = (Muo Qs Fh dNp / 4 Pi) 2 Pi R [R / (R^2 + Hs^2)^1.5
= (Muo Qs Fh [Np d(Phi) / (2 Pi)] / 4 Pi) 2 Pi R [R / (R^2 + Hs^2)^1.5]
 
= [Muo Qs Fh Np d(Phi) / 4 Pi] [R^2 / (R^2 + Hs^2)^1.5]
 
= {(Muo Qs Fh Np d(Phi) / 4 Pi}
{[(Rs + Rc) / 2] - [(Rs - Rc) / 2] Cos(Phi)}^2
/ {[(Rs^2 + Rc^2) / 2] - [(Rs^2 - Rc^2) / 2] Cos(Phi)}^1.5
 
= {(Muo Qs Fh Np d(Phi) / 4 Pi}
Ro^2 (1 / 2)^2 {[(So + (1 / So))] - [(So - (1 / So))] Cos(Phi)}^2
/ Ro^3 (1 / 2)^1.5 {So^2 + (1 / So)^2 - [(So^2 - (1 / So)^2)] Cos(Phi)}^1.5
 
= (Muo Qs Fh Np d(Phi) / 4 Pi
(1 / 2)^0.5 {[(So + (1 / So))] - [(So - (1 / So))] Cos(Phi)}^2
/ Ro {So^2 + (1 / So)^2 - [(So^2 - (1 / So)^2)] Cos(Phi)}^1.5
 
= (Muo Qs Fh Np So / 4 Pi Ro)(1 / 2)^0.5 d(Phi)
{[(So^2 + 1)] - [(So^2 - 1)] Cos(Phi)}^2
/ {(So^4 + 1) - [(So^4 - 1)] Cos(Phi)}^1.5

Recall that:
Fh = C / Lh

Bpo = 2 X Integral from Phi = 0 to Phi = Pi of:
(Muo Qs C Np So / 4 Pi Lh Ro)(1 / 2)^0.5 d(Phi)
{[So^2 + 1] - [(So^2 - 1) Cos(Phi)]}^2
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5
 
= Integral from Phi = 0 to Phi = Pi of:
(Muo Qs C Np So / 2 Pi Lh Ro)(1 / 2)^0.5 d(Phi)
{[So^2 + 1] - [(So^2 - 1) Cos(Phi)]}^2
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

 

Recall that:
Bto = Muo Nt Ih / 2 Pi Ro
= Muo Nt Qs C / (2 Pi Ro Lh)

Hence for a spheromak:
(Bpo / Bto) = Integral from Phi = 0 to Phi = Pi of:
(Np / Nt)(1 / 2)^0.5 d(Phi)
{[So^2 + 1] - [(So^2 - 1) Cos(Phi)]}^2
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

 

The issue that remains to be determined is what is the value of So?

****************************************************************

**********************************************************************

MAGNETIC FLUX THROUGH THE CORE OF A SPHEROMAK:
On the Z = 0 plane for 0 < R < Rc:
U = Uo {Ro^2 / [Ro^2 + (A R)^2]}^2
= Uo {A^2 Rs Rc / [A^2 Rs Rc + (A R)^2]}^2
= Uo {Rs Rc / [Rs Rc + (R)^2]}^2

Uo = Bpo^2 / (2 Muo)
U = Bp^2 /(2 Muo)

Thus:
[Bp^2 / 2 Muo] = [Bpo^2 / 2 Muo]{Rs Rc / [Rs Rc + (R)^2]}^2
or
Bp = Bpo {Rs Rc / [Rs Rc + (R)^2]}

Thus the magnetic flux through the spheromak core is:
Integral from R = 0 to R = Rc of:
Bp 2 Pi R dR
 
= Integral from R = 0 to R = Rc of:
Bpo {Rs Rc / [Rs Rc + (R)^2]} 2 Pi R dR

Let X = [Rs Rc + R^2]
dX = 2 R dR
Then the magnetic flux through spheromak core is:
Integral from X = Rs Rc to X = Rs Rc + Rc^2 of:
Bpo Pi dX / X
 
= Bpo Rs Rc Pi Ln[(Rs Rc + Rc^2) / Rs Rc]
 
= Bpo Rs Rc Pi Ln[(Rs + Rc) / Rs]
 
= Bpo (Pi Ro^2 / A^2) Ln[1 + (1 / So^2)]
 

*********************************************************************

*********************************************************************

*********************************************************

Problem here???? FIX THIS SECTION

EQUATE THE TWO EXPRESSIONS FOR Bpo TO SHOW CONSISTENCY:
For an ideal spheromak with an outside energy density defined by:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2
the magnetic field at R = 0, Z = 0 is given by:
Bpo = [Muo C Qs / (4 Pi Ro^2)]

FIX

Equating this value to our integral gives:
Bpo = [Muo C Qs / (4 Pi Ro^2)]
FIX = [2^0.5 Muo Qs C / (4 Pi^2 Ro^2)]
{[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
Integral from Phi = 0 to Phi = Pi of:
[So^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

Cancel out common terms to get:
[1]
= [2^0.5 / Pi]
{[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
Integral from Phi = 0 to Phi = Pi of:
[So^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5
or
[Pi / 2^0.5] = 2.22144
= {[Np^2 (Rs + Rc)^2] / [Np^2 (Rs + Rc)^2 + Nt^2 (Rs - Rc)^2]}
Integral from Phi = 0 to Phi = Pi of:
[So^2] {[So^2 + 1] - [(So^2 - 1) Cos(Phi)]} dPhi
/ {[So^4 + 1] - [(So^4 - 1) Cos(Phi)]}^1.5

Numerical Evaluation of Integral
So    Integral
1.3xxxx
1.52.1164
1.72.055
1.91.9939
2.01.9638
2.51.8249

Note that the spheromak tends to operate with a central magnetic field strength slightly less than the electric field seems to indicate.

Note that So < 2 and Nt < Np.

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********************************************************************

The characteristic frequency of a spheromak is:
F = C / Lh
= C / (Pi {[Nt (Rs - Rc) Kc]^2 + [Np (Rs + Rc)]^2}^0.5)
= C / (Pi Ro {[Nt (So - (1 / So)) Kc]^2 + [Np (So + (1 / So))]^2}^0.5)
= C So / (Pi Ro {[Nt (So^2 - 1) Kc]^2 + [Np (So^2 + 1)]^2}^0.5)

Efs = upper limit on spheromak field energy

From the web page titled: SPHEROMAK ENERGY: For electromagnetic spheromaks at R = 0, Z = 0 the electric field is zero. Hence:
Uo = (Bpo^2 / 2 Muo)
which gives:
Efs = [Uo Ro^3 Pi^2 / (A^2 B)]
= [(Bpo^2 / 2 Muo) Ro^3 Pi^2 / (A^2 B)]
= [Qs^2 / (32 Pi^2 Epsilon Ro^4)][Ro^3 Pi^2 / (A^2 B)]
= [Qs^2 / (32 Epsilon Ro A^2 B)]
= [Qs^2 Muo C^2 / (32 Ro A^2 B)]
 

FIELD ENERGY DENSITY INSIDE THE SPHEROMAK WALL:
In the toroidal region inside the spheromak wall where Rc < R < Rs and |Z| < |Zs| the total field energy density is given by:
Ut = Uc (Rc / R)^2

This energy density function arises purely from a toroidal magnetic field.
 

CLASSICAL CHARGED PARTICLE RADIUS Re:(this section is not part of the Planck constant derivation)
The classical expression for particle field only considers the electric field energy and assumes a particle radius of Re. The above expression can be compared to classical electric field energy for an electron given by:
Classical electic field = (1 / 4 Pi Epsilon) Q / R^2
El;ectric field Energy density = (Epsilon / 2)(Electric Field)^2
= (Epsilon / 2) [(1 / 4 Pi Epsilon) Q / R^2]^2
and
Electric Field energy = Integral from R = Re to R = infinity of:
(Epsilon / 2) [(1 / 4 Pi Epsilon) Qs / R^2]^2 4 Pi R^2 dR
= Integral from R = Re to R = infinity of:
[(Qs^2 / 8 Pi Epsilon) / R^2] dR
= [(Qs^2 / 8 Pi Epsilon) / Re]
= [(Qs^2 Mu C^2 / 8 Pi) / Re]
where Re is the classical electron radius.

Equating the two expressions for field energy gives:
[(Qs^2 Muo C^2 / 8 Pi) / Re] = (Muo C^2 / 2)[Qs / 4]^2 [1 / Rs Rc]^0.5
or
[(1 / Pi) / Re] = [1 / 4] [1 / Rs Rc]^0.5
or
Re = (4 / Pi) [Rs Rc]^0.5
= (4 / Pi) Ro

It is shown herein that the ratio of Rs to Rc corresponds to a spheromak energy minimum that sets the Planck constant but the precise reasons why electrons and protons have different mass (rest energy) are uncertain.
 

Inside the spheromak wall the energy density function arises from a toroidal magnetic field and a cylindrically radial electric field.

TOROIDAL MAGNETIC FIELD ENERGY:
From Gauss Law the toroidal magnetic field is given by:
Bt = [(Muo Ih Nt) / (2 Pi R)]

The toroidal magnetic field energy density is given by:
Bt^2 / 2 Muo = [(Mu Ih Nt) / (2 Pi R)]^2 / (2 Muo)
= (Muo / 8)[(Ih Nt) / (Pi R)]^2

Ih = (Qs C) / Lh
and
Lh^2 = {[Nt 2 Pi (Rs - Rc) Kc / 2)]^2 + [Np 2 Pi (Rs + Rc) / 2]^2}^0.5
= Pi^2 {[Nt (Rs - Rc) Kc)]^2 + [Np (Rs + Rc)]^2}
or
(Ih Nt) = (Qs C Nt) / Lh

Thus:
(Bt^2 / 2 Muo) = (Mu / 8)[(Ih Nt) / (Pi R)]^2
= (Muo / 8)[(Qs C Nt) / (Lh Pi R)]^2

On the spheromak wall:
Zs^2 = (A / B)^2 (Rs - R) (R - Rc)

An element of toroidal magnetic volume is:
dV = 2 Zs 2 Pi R dR
= 4 Pi R dR (A / B) [(Rs - R) (R - Rc)]^0.5

An element of toroidal magnetic energy dEtm is:
dEtm = (Bt^2 / 2 Muo) dV
= (Muo / 8)[(Qs C Nt) / (Lh Pi R)]^2 4 Pi R dR (A / B) [(Rs - R) (R - Rc)]^0.5 = (A / B) (Muo / 2)[(Qs C Nt)^2 / (Lh^2 Pi] [(Rs - R) (R - Rc)]^0.5 (dR / R)

Toroidal magnetic energy Etm =
Integral from R = Rc tpo R = Rs of:
(A / B) (Muo / 2)[(Qs C Nt)^2 / (Lh^2 Pi] [(Rs - R) (R - Rc)]^0.5 (dR / R)

The characteristic frequency of this particle is:
F = C / Lh
= C / (Pi {[Nt (Rs - Rc) Kc]^2 + [Np (Rs + Rc)]^2}^0.5)

Let X = (R / Rs)

Thus the toroidal magnetic energy Et is given by:
Et = Integral from X = Xa = (Rc / Rs) to X = Xb = 1 of:
(A / B) (Muo / 2)[(Qs C Nt)^2 Rs / (Lh^2 Pi)] [(1 - X)(X - Xa)]^0.5 dX / X)
= Integral from X = Xa = (Rc / Rs) to X = Xb = 1 of:
(A / B) (Muo / 2)[(Qs^2 C Nt^2) Rs / (Lh Pi)] Fh [(1 - X)(X - Xa)]^0.5 (dX / X)
= Integral from X = Xa = (Rc / Rs) to X = Xb = 1 of:
(A / B)[(Muo Qs^2 C) / 4 Pi)] Fh Nt (2 Rs Nt / Lh) [(1 - X)(X - Xa)]^0.5 (dX / X)

This is an accurate expression for the toroidal magnetic energy.


 

CONTINUING WITH THE DEVELOPMENT OF Efs:
Recall that from THEORETICAL SPHEROMAK:
Lp = Pi (Rs + Rc) Lt = Pi (Rs - Rc) Kc Lh = {[Pi (Rs + Rc)]^2 + [Pi (Rs - Rc)]^2 Kc^2}^0.5

Thus:
Fh = C / Lh
= C / {[Pi Np (Rs + Rc)]^2 + [Pi Nt (Rs - Rc)]^2 Kc^2}^0.5
= C / Rc {[Pi Np (So^2 + 1)]^2 + [Pi Nt (So^2 - 1)]^2 Kc^2}^0.5
= (Ro / Rc) [C / Ro Nt](1 / {[Pi Nr (So^2 + 1)]^2 + [Pi (So^2 - 1)]^2 Kc^2}^0.5})
= [So C / Ro Nt Pi] [1 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2 Kc^2}^0.5}]

Hence:
(1 / Ro) = [Fh Nt Pi {[Nr (So^2 + 1)]^2 + [(So^2 - 1) Kc^2]^2}^0.5}] / So C]

Thus:
Efs = [Muo C^2 Qs^2 / 32 Ro A^2 B]
= [ Muo C^2 Qs^2 / 32 A^2 B] [Fh Nt Pi {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2 Kc^2}^0.5}] / So C]
= [ Muo C Qs^2 Pi / 32 A^2 B] [Fh Nt {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2 Kc^2}^0.5}] / So]
= [Muo C Qs^2 / 4 Pi A^2 B] [Pi^2 / 8] Fh Nt ({[(Nr^2) (So^2 + 1)^2] + [(So^2 - 1)]^2 Kc^2} / So^2)^0.5

Note that Efs is dependent on Nr^2 and So^2. Since Nr^2 and So^2 are related by the common boundary condition there is a low energy operating value of So^2 and there is a corresponding operating value of Nr^2.

***************************************************************

SPHEROMAK WALL BOUNDARY CONDITION AT R = Rc. Z = 0:

Outside the spheromak wall the total field energy density is given by:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2
At R = Rc, Z = 0 the total field energy density is given by:
Upc = Uo [Ro^2 / (Ro^2 + (A Rc)^2)]^2
= Uo [(A^2 Rs Rc) / ((A^2 Rs Rc) + (A Rc)^2)]^2
= Uo [Rs / (Rs + Rc)]^2

Inside the spheromak walls where the static field energy is entirely toroidal magnetic, at R = Rc, Z = 0:
Btc = (Muo Nt Ih / 2 Pi Rc)

The corresponding inside field energy density at R = Rc, Z = 0 is:
Utc = Btc^2 / 2 Muo
= (1 / 2 Muo)[(Muo Nt Ih) / (2 Pi Rc)]^2
= (Muo / 2) [(Nt Ih) / (2 Pi Rc)]^2

Hence the boundary condition at R = Rc, Z = 0 is:
Upc = Utc
or
Uo [Rs / (Rs + Rc)]^2 = (Muo / 2) [(Nt I) / (2 Pi Rc)]^2
or
Uo [Rs / (Rs + Rc)]^2 = (Muo / 2) [(Nt I) / (2 Pi Ro)]^2 [Ro / Rc]^2
or
Uo [So^2 / (So^2 + 1)]^2 = (Muo / 2) [(Nt I) / (2 Pi Ro)]^2 [A So]^2
or
Uo = (Muo / 2) [(Nt I) / (2 Pi Ro)]^2 [A]^2 [(So^2+ 1) / (So)]^2
which is the wall boundary condition at R = Rc, Z = 0.
 

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**********************************************************

Note that Nr has a lower limit set by the magnetic field configuration. For A = 1 the highest possible value of Nr is (2 / Pi) at which point So = 1. For So = 2 likely A > 1. If the presence of enclosure walls increases A the probability of spheromak formation increases.

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SPHEROMAK WALL BOUNDARY CONDITION AT R = Rc, Z = 0 LIMIT ON THE RANGE OF So^2:

Note that:
0 < Nr^2 < (4 A^4 / Pi^2)
which restricts the range of [((Rs - Rc)) / (Rs + Rc)]^2 to:
0 < [((Rs - Rc)) / (Rs + Rc)]^2 < [4 A^4 / Kc^2 Pi^2]
or
0 < [((Rs - Rc)) / (Rs + Rc)] < [2 A^2 / Kc Pi]
or
0 < [(So^2 - 1) / (So^2 + 1)] < [2 A^2 / Kc Pi]

At:
[(So^2 - 1) / (So^2 + 1)] = [2 A^2 / Kc Pi]
or
(So^2 - 1) = [2 A^2 / Kc Pi](So^2 + 1)
or
So^2 {1 - [2 A^2 / Kc Pi]} = {1 + [2 A^2 / Kc Pi]}
or
So^2 = {1 + [2 A^2 / Kc Pi]} / {1 - [2 A^2 / Kc Pi]}

Thus for a spheromak:
1 < So^2 < {1 + [2 A^2 / Kc Pi]} / {1 - [2 A^2 / Kc Pi]}

For A^2 / Kc = 1 This expression simplifies to:
1 < So^2 < {1 + [2 / Pi]} / {1 - [2 / Pi]}
or
1 < So^2 < {Pi + 2} / {Pi - 2}
or
1 < So^2 < {5.14} / {1.14}
or
1 < So^2 < 4.51
 

 
 

SPHEROMAK WALL BOUNDARY CONDITION AT R = Rf, Z = Zf:
Immediately inside the spheromak wall:
Btf = Muo Nt It / 2 Pi Rf
or
Utf = (1 / 2 Muo)[ Muo Nt It / 2 Pi Rf]^2
= (Muo / 2) [Nt It / Pi (Rs + Rc)]^2

Immediately outside the spheromak wall:
Upf = Uo [Ro^2 / (Ro^2 + (A Rf)^2 + (B Zf)^2]^2
= Uo [A^2 Rs Rc / (A^2 Rs Rc + (A (Rs + Rc) / 2)^2 + (B Zf)^2]^2
= Uo [Rs Rc / (Rs Rc + ((Rs + Rc) / 2)^2 + (B^2 Zf^2 / A^2)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs + Rc) / 2)^2 + (B / A)^2 (Rs - Rf)(Rf - Rc)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs + Rc) / 2)^2 + (B / A)^2 ((Rs - ((Rs + Rc) / 2))(((Rs + Rc / 2) - Rc))]^2
= Uo [Rs Rc / (Rs Rc + ((Rs^2 + Rc^2 + 2 Rs Rc) / 4) + (B / A)^2 ((Rs - Rc) / 2))((Rs - Rc) / 2)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs^2 + Rc^2 + 2 Rs Rc) / 4) + (B / A)^2 ((Rs^2 - 2 Rs Rc + Rc^2) / 4)]^2
= Uo [Rs Rc / (Rs Rc + ((Rs^2 + Rc^2) / 4) + (B / A)^2 ((Rs^2 + Rc^2) / 4)]^2
= Uo [Rs Rc / (Rs Rc + (1 + (B / A)^2)((Rs^2 + Rc^2) / 4)]^2

Utf = Upf
or
(Muo / 2) [Nt I / Pi (Rs + Rc)]^2 = Uo [Rs Rc / (Rs Rc + (1 + (B / A)^2)((Rs^2 + Rc^2) / 4)]^2

Carry along (B / A) to get the correct wall boundary condition at Rf, Zf. **************************************************************************************************************************************

which for A = B gives:
(Muo / 2) [Nt I / Pi (Rs + Rc)]^2 = Uo [Rs Rc / (Rs Rc + 2((Rs^2 + Rc^2) / 4)]^2
or
(Muo / 2) [Nt I / Pi (Rs + Rc)]^2 = Uo [Rs Rc 2 / (Rs + Rc)^2]^2

Recall that:
I = Qs C / Lh
and
Uo = [Muo C^2 Qs^2 / 32 Pi^2 Ro^4]

Hence:
(Muo / 2) [Nt I / Pi]^2 = Uo [Rs Rc 2 / (Rs + Rc)]^2
or
(Muo / 2) [Nt Qs C / Lh Pi]^2 = [Muo C^2 Qs^2 / 32 Pi^2 Ro^4] [Rs Rc 2 /(Rs + Rc)]^2
or cancelling equal terms:
[Nt / Lh]^2 = [1 / 4 Ro^4] [Rs Rc /(Rs + Rc)]^2
or
[Nt / Lh]^2 = [1 / (4 A^4 Rs^2 Rc^2)] [Rs Rc / (Rs + Rc)]^2
or
[Nt / Lh]^2 = [1 / (4 A^4 (Rs + Rc)^2)]
or
[Nt^2 / (Np^2 Lp^2 + Nt^2 Lt^2)] = [1 / (4 A^4 (Rs + Rc)^2)]
This is the result of the boundary condition at Rf, Zf
 

Rearrange this formula to get:
Nt^2 = [1 / 4 A^4 (Rs + Rc)^2](Np^2 Lp^2 + Nt^2 Lt^2)
or
Nt^2 {1 - [(Lt^2) / (4 A^4 (Rs + Rc)^2]} = [(Np^2 Lp^2) / (4 A^4 (Rs + Rc)^2)]

or
Nr^2 = Np^2 / Nt^2
= {1 - [(Lt^2) / (4 A^4 (Rs + Rc)^2)]} / {Lp^2) / (4 A^4 (Rs + Rc)^2)}
= {4 A^4 (Rs + Rc)^2 - Lt^2} / {Lp^2}
= {4 A^4 (Rs + Rc)^2 - Pi^2 (Rs - Rc)^2 Kc^2} / {Pi^2 (Rs + Rc)^2}
= {(4 A^4 / Pi^2) (Rs + Rc)^2 - (Rs - Rc)^2 Kc^2} / {(Rs + Rc)^2}
 

= {(4 A^4 / Pi^2) - [(Rs - Rc) / (Rs + Rc)]^2 Kc^2}
This is the boundary condition at R = Rf, Z = Zf. Note that it is identical to the boundary condition at R = Rc, Z = 0.

 

Thus: Nr^2 = {(4 A^4 / Pi^2) - [((Rs - Rc) Kc) / (Rs + Rc)]^2}
or
Nr^2 + [((Rs - Rc) Kc) / (Rs + Rc)]^2 = (4 A^4 / Pi^2)
is the wall boundary condition for an electromagnetic spheromak. Note that Kc is a function of A via ellipse theory.

Recall that So^2 = (Rs / Rc)
which gives:
Nr^2 + [(So^2 - 1) Kc / (So^2 + 1)]^2 = (4 A^4 / Pi^2)

THIS IS THE BOUNDARY CONDITION EQUATION FOR AN ELECTROMAGNETIC SPHEROMAK
 

SPHEROMAK STATIC FIELD ENERGY:
The web page titled SPHEROMAK ENERGY shows that the energy density functions:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2
outside a spheromak wall and
U = Uto [(Ro / R)^2]
inside a spheromak wall result in spheromaks with total energy given by:
Ett = [Uo Pi^2 Ro^3 / A^2]
X {1 - [(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}
 
= [Uo Pi^2 Ro^3 / A^2]
X {[(So^2 + 1)^2 - (So - 1)^2 (So^2 + 1) + 2 So (So - 1)^2] / [(So^2 + 1)^2]}
 
= [Uo Pi^2 Ro^3 / A^2]
X {[(So^4 + 2 So^2 + 1) - (So^2 -2 So + 1) (So^2) - (So^2 -2 So + 1) + 2 So (So^2 - 2So + 1)] / [(So^2 + 1)^2]}
 
= [Uo Pi^2 Ro^3 / A^2]
X {[2 So (2 So^2 - 2So + 2)] / [(So^2 + 1)^2]}
 
= [Uo Pi^2 Ro^3 / A^2] {4 So (So^2 - So + 1) / (So^2 + 1)^2}
 

Note that Ett is a function of Uo, Ro, A and So. Hence in electromagnetic analysis of a spheromak we seek to find expressions for these parameters.

CHECK THE FOLLOWING SPHEROMAK SHAPE PARAMETER when the magnetic vectors are in balance:
Nr^2 = R^2 = (So^2 - 1)^2 Kc^2 /(So^2 + 1)^2

Hence when the magnetic vectors are in balance:
2 (So^2 - 1)^2 Kc^2 /(So^2 + 1)^2 = [4 A^4 / Pi^2]
or
(So^2 - 1)^2 Kc^2 /(So^2 + 1)^2 = 2 [A^4 / Pi^2]
or
(So^2 - 1) Kc / (So^2 + 1) = 2^0.5 [A^2 / Pi]
or
(So^2 - 1) Kc = 2^0.5 [A^2 / Pi](So^2 + 1)
or
So^2 {Kc - 2^0.5 [A^2 / Pi]} = {Kc + 2^0.5 [A^2 / Pi]}
or
So^2 = {Kc + 2^0.5 [A^2 / Pi]} / {Kc - 2^0.5 [A^2 / Pi]}
or
So^2 = {1 + 2^0.5 [A^2 / Kc Pi]} / {1 - 2^0.5 [A^2 / Kc Pi]}

If (A^2 / Kc) = 1 then:
So^2 = {1 + 0.4501581586} / {1 - 0.4501581586}
= {1.4501581586} / {0.5498418414}
= 2.637409614
and
So = 1.624010349
 

NUMBER THEORY:
The ratio Nr^2 = [(Np / Nt)^2] is locked at an integer ratio which is equal to [2 A^4 / Pi^2]. Hence:
[(So^2 - 1) Kc / (So^2 + 1)]^2 = 2 A^4 / Pi^2
or
[(So^2 - 1) Kc / (So^2 + 1)] = 2^0.5 A^2 / Pi
or
[(So^2 - 1) / (So^2 + 1)] = 2^0.5 A^2 / (Kc Pi)

The relationship between A and Kc is complex and in practise requires a computer program for accurate evaluation.
 

THE NO COMMON FACTOR CONSTRAINT:
An important constraint on the existence of a spheromak is that Np and Nt have no common factors. A mathematical expression of this statement of no common factors is:
[Np / Nt] = [(N + 1) / ((prime) - 2(N + 1))]
 
for
0 <= N <= [(Prime - 1) / 2]

This mathematical function was devised by Heather Rhodes. The quantity (prime) can be any prime number.
 

Prove that the function:
(Np / Nt) = {[(N + 1)] / [prime - 2(N + 1)]}
generates (Np / Nt) values with no common factors.
If (N + 1) = (Fo M)
where Fo is a factor of (N + 1) then the denominator is
(prime - 2 Fo M) which can only be reduced if Fo is a factor of prime, which it is not due to the definition of a prime number.

However, the prime number can in principle adopt a range of values. The value that the prime number adopts will minimize the error in (Np / Nt) where:
Np = N + 1
and
Nt = [prime - 2(N + 1)]

[Np / Nt] = [(N + 1) / [(prime) - 2 (N + 1)]

Thus if A is unique for all spheromaks this formula will pick a particular (prime) that makes (N + 1) an integer. Hence we will have a unique solution.

If we assume that A is unique and is very close to unity (which may or may not be true) then experimental data suggests that:
Np = (N + 1) = 223
and
Nt = 495 and
(prime) = 941

We will test this assumption on the web page titled PLANCK CONSTANT.

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MAGNETIC FLUX THROUGH THE CORE OF A SPHEROMAK:
On the Z = 0 plane for 0 < R < Rc:
U = Uo {Ro^2 / [Ro^2 + (A R)^2]}^2
= Uo {A^2 Rs Rc / [A^2 Rs Rc + (A R)^2]}^2
= Uo {Rs Rc / [Rs Rc + (R)^2]}^2

Uo = Bpo^2 / (2 Muo)
U = Bp^2 /(2 Muo)

Thus:
[Bp^2 / 2 Muo] = [Bpo^2 / 2 Muo]{Rs Rc / [Rs Rc + (R)^2]}^2
or
Bp = Bpo {Rs Rc / [Rs Rc + (R)^2]}

Thus the magnetic flux through the spheromak core is:
Integral from R = 0 to R = Rc of:
Bp 2 Pi R dR
 
= Integral from R = 0 to R = Rc of:
Bpo {Rs Rc / [Rs Rc + (R)^2]} 2 Pi R dR

Let X = [Rs Rc + R^2]
dX = 2 R dR
Then the magnetic flux through spheromak core is:
Integral from X = Rs Rc to X = Rs Rc + Rc^2 of:
Bpo Pi dX / X
 
= Bpo Rs Rc Pi Ln[(Rs Rc + Rc^2) / Rs Rc]
 
= Bpo Rs Rc Pi Ln[(Rs + Rc) / Rs]
 
= Bpo (Pi Ro^2 / A^2) Ln[1 + (1 / So^2)]
 

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CHECK FROM HERE ONWARDS-- SUSPECT ERRORS

Substitution of Nr^2 into the So^2 dependent term of Efs:
{[(Nr^2) (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5
gives:
{[(Nr^2) (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5
= {[{[8](So^2 + 1)^2 - [Pi^2] [(So^2 - 1)]^2} / {[Pi^2] - [4 / (So^2 - 1)]^2}] + [(So^2 - 1)^2]}^0.5
 
= [({[8](So^2 + 1)^2] - [Pi^2] [(So^2 - 1)]^2 + [(So^2 - 1)^2] [[Pi^2] - [16]}
/ {[Pi^2] - [4 / (So^2 - 1)]^2})^0.5]
 
= [({[8 (So^2 + 1)^2] - [16]} / {[Pi^2] - [16 / (So^2 - 1)^2]})^0.5]
 
= [({[8 (So^2 + 1)^2] - [16]} / {[Pi^2] - [4 / (So^2 - 1)]^2})^0.5] / So
 
= [({[8 (So^2 + 1)^2] - [16]} / {So^2 [Pi^2] - [16 So^2 / (So^2 - 1)^2]})^0.5]
 

Let:
X = So^2

Then the So^2 dependent term of Efs becomes:
[({[8(X + 1)^2] - [16]} / {X [Pi^2] - [16 X / (X - 1)^2]})^0.5]
 
=[({[8(X + 1)^2 (X - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^0.5]
 
= [({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^0.5]

Hence:
Efs = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt [({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^0.5]
= [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt [(8 {[(So^4 - 1)^2] - [2 (So^2 - 1)^2]} / {So^2 (So^2 - 1)^2 [Pi^2] - [16 So^2]})^0.5]
 
= [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] Fh Nt [So^2 - 1] (8 {So^4 + 2 So^2 - 1} / {So^2 (So^2 - 1)^2 [Pi^2] - [16 So^2]})^0.5]
 

DIFFERENTIATION OF Efs:
Differentiate Efs with respect to X = So^2 to get:

d(Efs / dX) = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5
{{X (X - 1)^2 [Pi^2] - [16 X]} {16 (X^2 -1)(2 X) - 32 (X - 1)}
- {[8 (X^2 - 1)^2] - [16 (X - 1)^2]} {(X - 1)^2 [Pi^2] + 2 X (X -1)[Pi^2] -16}}

Factor out 8 (X - 1) from both product terms to get:
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{{X (X - 1)^2 [Pi^2] - [16 X]} {2 (X + 1)(2 X) - 4}
- {[(X^2 - 1)(X + 1)] - [2 (X - 1)]} {(X - 1)^2 [Pi^2] + 2 X (X -1)[Pi^2] -16}}
or
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{4 X {(X - 1)^2 [Pi^2] - [16]} {X^2 + X - 1}
- (X - 1) {[(X + 1)(X + 1)] - [2]} {(X - 1)^2 [Pi^2] + 2 X (X -1)[Pi^2] -16}}
or
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{4 X {(X - 1)^2 [Pi^2] - [16]} {X^2 + X - 1}
- (X - 1) {X^2 + 2 X - 1} {(X - 1)^2 [Pi^2] + 2 X (X -1)[Pi^2] -16}}
or
d(Efs / dX) = [Mu C Qa^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
4 X {(X - 1)^2 [Pi^2]} {X^2 + X - 1} - (X - 1) {X^2 + 2 X - 1} {(X - 1)^2 [Pi^2]} - (X - 1) {X^2 + 2 X - 1} {2 X (X - 1) [Pi^2]}
+ 4 X {- [16]}{X^2 + X - 1} - (X - 1) {X^2 + 2 X - 1}{-16}
or
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{(X - 1)^2 [Pi^2]} {(4 X)(X^2 + X - 1) - (X - 1) {X^2 + 2 X - 1} - (X^2 + 2 X - 1) (2 X)}
-64 X^3 - 64 X^2 + 64 X + 16 X^3 + 32 X^2 - 16 X - 16 X^2 - 32 X + 16
or
d(Efs / dX) = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{(X - 1)^2 [Pi^2]} {(4 X)(X^2 + X - 1) - (3 X - 1) (X^2 + 2 X - 1)}
- 48 X^3 - 48 X^2 + 16 X + 16
or
d(Efs / dX) = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{(X - 1)^2 [Pi^2]} {(4 X^3 + 4 X^2 - 4 X - 3 X^3 - 6 X^2 + 3 X + X^2 + 2 X - 1}
- 16 (+ 3 X^3 + 3 X^2 - X - 1)
or
d(Efs / dX) = [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8
{(X - 1)^2 [Pi^2]} {X^3 - X^2 + X - 1} + 16 {(1 - 3 X^2) (X + 1)}
or
d(Efs / dX) = [Mu C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt]
{0.5 ({[8 (X^2 - 1)^2] - [16 (X - 1)^2]} / {X (X - 1)^2 [Pi^2] - [16 X]})^-0.5 (X - 1) 8}
{{(X - 1)^2 [Pi^2]} {(X^2 + 1)(X - 1} - 16 {(3 X^2 - 1) (X + 1)}}

CHECK THIS PLOT
A plot of the third line of this expression shows that its zero value is at approximately:
X = 3.765 = So^2

Note that the second line of this expression equals:
4 (X - 1) [Muo C Qs^2 / 4 Pi] [Pi^2 / 8] [Fh Nt / Efs]

In a real spheromak, although Efs is the largest single energy term the total energy is significantly affected by Eft and Efst. Hence:
d(Ett) / d(So^2) = 0 at an So^2 value that is somewhat removed from So^2 = 3.765. A photograph of a spheromak plasma suggests that So^2 ~ 4.2. Calculating the So^2 value at which Ett is a relative minimum is a major mathematical exercise. The importance of this exercise is that the steady state value of So^2 determines the steady state value of Nr^2 which in turn determines the Planck constant.
 

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ELEMENTAL STRIP ANALYSIS:
Consider an elemental strip of constant radius R. The strip length is:
2 Pi R

The strip width is:
[(Rs - Rc) / 2] dPhi

The strip contains Nt partial windings.

Let L be the length of each partial winding. Then:
L^2 = (R dTheta)^2 + [(Rs - Rc) dPhi / 2]^2

The total charge hose length on the strip is:
Nt L = Nt {(R dTheta)^2 + [(Rs - Rc) dPhi / 2]^2}^0.5
= {(Nt R dTheta)^2 + [(Rs - Rc) Nt dPhi / 2]^2}^0.5
 

SURFACE CHARGE DENSITY:
As shown on the web page titled: THEORETICAL SPHEROMAK the surface charge dQs contained on the elemental strip of constant R is:
dQs = 2 Pi R dLt Sac (Rc / R)
or
Qs = 2 Pi Lt Sac Rc
or
Qs = 2 Pi Lt Sa R
or
Sa = Qs / 2 Pi Lt R

Hence:
dQs = 2 Pi R dLt Sa
= 2 Pi R dLt Qs / 2 Pi Lt R
= dLt Qs / Lt

The corresponding element of area dA is:
dA = 2 Pi R dLt

Hence the charge / unit area Sa on the spheromak wall is:
dQs / dA = (dLt Qs / Lt) / (2 Pi R dLt)
Sa = Qs / {Lt (2 Pi R)]

At R = Rc the charge per unit area Sac is given by:
Sac = Qs / {Lt (2 Pi Rc)]

In general the surface charge density Sa is given by:
Sa = Qs / {Lt (2 Pi R)]

= Sac Ro / R So

This charge distribution equation is required to find the electric field distribution.
 

POLOIDAL TURNS CONTAINED IN AN ELEMENTAL STRIP:
Recall that:
[dTheta / dPhi] = [Np [(Rs + Rc) / 2] / R Nt]

The number of poloidal turns contained in an elemental strip is:
Nt R dTheta / 2 Pi R = [Nt / 2 Pi] [Np (Rs + Rc) / 2 R Nt] dPhi
= [Np (Rs + Rc) / 4 Pi R] dPhi
where:
[Np (Rs + Rc) / 4 Pi R]
is the number of poloidal turns per radian in Phi
 

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Ett = [Uo Pi^2 Ro^3 / A^2] {4 So (So^2 - So + 1) / (So^2 + 1)^2}
 

Note that Ett is a function of Uo, Ro, A and So. Hence in electromagnetic analysis of a spheromak we seek to find expressions for these parameters.

SPHEROMAK STATIC FIELD ENERGY:
The web page titled SPHEROMAK ENERGY shows that the energy density functions:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + (B Z)^2)]^2
outside a spheromak wall and
U = Uoc [(Rc / R)^2]
inside a spheromak wall result in spheromaks with total energy given by:
Ett = Uo Pi^2 Ro^3/ A^2
X {1 - [(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}
 
= Uo Pi^2 Ro^3 {4 So (So^2 - So + 1) / (So^2 + 1)^2}
Note that this formula applies to all spheromaks.
 

The energy density distribution:
U = Uo [Ro^2 / (Ro^2 + (A R)^2 + Z^2)]^2
provides an energy density of:
U = Uo [Ro^2 / (Ro^2 + (A Rc)^2)]^2
at R = Rc, Z = 0.

FIX A^2 FROM HERE ONWARDS

SPHEROMAK ENERGY CONTENT:
Substitution for Uo gives:
Efs = Uo Ro^3 Pi^2 / A^2
= (Muo / 2) Ro^3 Pi^2 [(Qs C) /(4 Pi Ro^2)]^2
= (Muo / 32) (Qs C)^2 /(Ro)
and
Ett = [Muo (Qs C)^2 / (32 Ro)] {1 -[(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}
or
Ett = [(Muo C^2 Qs^2) / (32 Ro)]{4 So (So^2 - So + 1) / (So^2 + 1)^2}

This equation gives the static electromagnetic field energy content of a spheromak in terms of its nominal radius Ro and its shape factor So where:
So = Rs / Ro = Ro / Rc

FIX A ENTRIES

Thus an electromagnetic spheromak has a net charge Qs, a nominal radius Ro and a theoretical peak central poloidal magnetic field strength given by:
Bpo = [(Muo C Qs) / (4 Pi Ro^2)]

This value of Bpo should equal the value of Bpo obtained by applying the law of Biot and Savart to the circulating charge in the spheromak.
 

MAXIMUM ENERGY VALUE OF So:
An important issue is finding the value of So which maximizes a spheromak's energy content at any particular value of Ro. At that energy maximum:
dEtt / dSo = 0

Recall that:
Ett = [(Muo C^2 Qs^2) / (32 Ro)]{4 So (So^2 - So + 1) / (So^2 + 1)^2}

The So dependent portion of the function is:
S(So) = {So (So^2 - So + 1) / (So^2 + 1)^2}

dS(So) / dSo = {(So^2 + 1)^2 [(So^2 - So + 1) + So (2 So - 1)]
- [So (So^2 - So + 1)] 2 (So^2 + 1) 2 So}
/ (So^2 + 1)^4
= 0

Hence:
{(So^2 + 1)^2 [(So^2 - So + 1) + So (2 So - 1)]
- [So (So^2 - So + 1)] 2 (So^2 + 1) 2 So}
= 0

Cancelling (So^2 + 1) terms gives:
{(So^2 + 1) [(So^2 - So + 1) + So (2 So - 1)]
- [4 So^2 (So^2 - So + 1)]}

Hence:
(So^2 + 1 - 4 So^2)(So^2 - So + 1) + (So^2 + 1) So (2 So - 1) = 0
or
(- 3 So^2 + 1)(So^2 - So + 1) + (So^2 + 1)(2 So^2 - So) = 0
or
- 3 So^4 + 3 So^3 - 3 So^2 + So^2 - So + 1 + 2 So^4 - So^3 + 2 So^2 - So = 0
or
- So^4 + 2 So^3 - 2 So + 1 = 0
or
So^4 - 2 So^2 + 2 So - 1 = 0
or
So^4 - So^2 = So^2 - 2 So + 1
or
So^2 (So^2 - 1) = (So - 1)^2
or
So^2 = (So - 1)^2 / [(So - 1) (So + 1)]
= (So - 1) / (So + 1)

This equation has a solution of So = 1 which says that at a particular Ro the spheromak's energy is maximum at So = 1, at which point Rc = Ro = Rs corresponding to no volume inside the spheromak wall.

Thus at the maximum energy state: {So (So^2 - So + 1) / (So^2 + 1)^2}
= {1 (1 - 1 + 1) / (1 + 1)^2}
= 1 / 4

At So = 2:
(spheromak energy)
= {So (So^2 - So + 1) / (So^2 + 1)^2}
= {2 (4 - 2 + 1) / (4 + 1)^2}
= {6 / 25}
= which is only slightly less than the spheromak maximum energy.

Thus for modest So values of the order of So < 2 the static field energy content of a spheromak is only weakly dependent on the spheromak's So value. However, at larger So values the spheromak static field energy is proportional to (1 / So).
 

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CONFINING THE RANGE OF So:
In the toroidal region: Ut = Utc (Rc / R)^2 at R = Rc the total field is (Eric + Btc).
In general inside the spheromak wall the field is:
(Btc)(Rc / R)
and the field energy density is:
+ (Btc^2 / 2 Muo)(Rc / R)^2
 
= + [(Muo / 2) (Nt Q C / 2 Pi R Lh) ^2)]

Recall that for an ideal spheromak the energy density at R = Ro is (1 / 4) the energy density at R = 0. Hence at R = Ro:
(1 / 4)[Muo / 2][(C Qs) / (4 Pi Ro^2)]^2
= [(Epsilono / 2) (Sac / Epsilono)^2 (Rc / Ro)^2]
+ [(Muo / 2) (Nt Qs C / 2 Pi Ro Lh) ^2)]

or
(1 / 4)[Muo / 2][(C Qs) / (4 Pi Ro^2)]^2
= [(1 / 2 Epsilono) (Sac / So)^2]
+ [(Muo / 2) (Nt Qs C / 2 Pi Ro Lh) ^2)]

FIX FROM HERE ONWARD
From the spheromak charge distribution:
Sac = Qs So^2 / 2 Ro^2 Pi^2 (So^2 -1)

Hence:
(1 / 4)[Muo / 2][(C Qs) / (4 Pi Ro^2)]^2
= (1 / 2 Epsilono) [Qs So / 2 Ro^2 Pi^2 (So^2 -1)]^2
+ [(Muo / 2) (Nt Qs C / 2 Pi Ro Lh)^2)]
or (1 / 4)[Muo / 2][(C Qs) / (4 Pi Ro^2)]^2
= (Muo C^2 / 2) [Qs So / 2 Ro^2 Pi^2 (So^2 -1)]^2
+ [(Muo / 2) (Nt Qs C / 2 Pi Ro Lh)^2)]
or cancelling terms:
(1 / 4)[1 / (4 Ro)]^2
= [So / 2 Ro Pi (So^2 -1)]^2
+ [(Nt / 2 Lh)^2)]
or
[Lh / Ro]^2 [(1 / 8)^2]
= [Lh / Ro]^2 [So / 2 Pi (So^2 - 1)]^2 + [Nt / 2]^2
or
[Lh / Ro]^2 = [Nt / 2]^2 / {[(1 / 8)^2] - [So / 2 Pi (So^2 - 1)]^2}
 
= Nt^2 / [(1 / 4)^2 - {So / Pi (So^2 - 1)}^2]
 
= Nt^2 Pi^2 (So^2 - 1)^2 / {[Pi^2 (So^2 - 1)^2 / 4^2] - So^2}
 
= Nt^2 Pi^2 (So^2 - 1)^2 / So^2 {[Pi^2 (So^2 - 1)^2 / (4 So)^2] - 1}
 

Comparison with the previous result greatly limits the range of So.

Recall that:
(Lh / Ro)^2 = (Pi^2 / So^2) [Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2]
= (Pi^2 / So^2) [Nt^2 (So^2 - 1)^2][{Np^2 (So^2 + 1)^2 / Nt^2 (So^2 - 1)^2 + 1]
 
= [Nt^2 Pi^2 (So^2 - 1)^2 / So^2][{Np^2 (So^2 + 1)^2 / Nt^2 (So^2 - 1)^2 + 1]

Comparison of the two expressions for (Lh / Ro)^2 implies that:
[{Np^2 (So^2 + 1)^2 / Nt^2 (So^2 - 1)^2 + 1]
= 1 / {[Pi^2 (So^2 - 1)^2 / (4 So)^2] - 1}
or
[{Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2]
= Nt^2 (So^2 -1)^2 / {[Pi^2 (So^2 - 1)^2 / (4 So)^2] - 1}

Hence:
1 < [Pi^2 (So^2 - 1)^2 / (4 So)^2] < 2
or
1 < [Pi (So^2 - 1) / (4 So)] < 2^0.5

This expression restricts the range of So. We can easily calculate So values at the extreme ends of the possible range of So.

At:
[Pi (So^2 - 1) / (4 So)] = 2^0.5
or
Pi So^2 - 2^0.5(4 So) - Pi = 0
or
So = {2^0.5(4) +/- [32 + 4 Pi^2]^0.5} / 2 Pi
= {5.6568 + [71.4784]^0.5} / 6.2831853
= 2.24588

At:
1 = [Pi (So^2 - 1) / (4 So)]
or
Pi (So^2 - 1) = 4 So
or
Pi So^2 - 4 So - Pi = 0
or
So = {4 +/- [16 + 4 Pi^2]^0.5} / 2 Pi = {4 + [55.47841]^0.5} / 6.2831853
= 1.8220

Thus the possible range of So is restricted to:
1.8220 < So < 2.24588

WRONG - FIX

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MERGING AXIAL ELECTRIC FIELD AND MAGNETIC FIELDS OF A SPHEROMAK

Hence the Fine Structure constant definitely arises from the spheromak relationship and the relative strength of electric and magnetic fields. We need a more exact solution to properly evaluate it.

At R = Rs, Z = 0:
Toroidal magnetic field must approximately match the increase in the electric field. The increase in electric field energy density is given by:
(Epsilono / 2)(Sac Rc / Epsilono Rs)^2

The drop in toroidal magnetic field energy density is:
(1 / 2 Muo)(Muo Nt Q Fh / 2 Pi Rs)^2
= (1 / 2 Muo)(Muo Nt Q C / 2 Pi Rs Lh)^2

Thus, equating these two expressions gives:
(1 / 2 Epsilono)(Sac / So^2)^2 = (Muo / 2)(Nt Q C / 2 Pi Ro So Lh)^2
or
(Sac / So)^2 = (Epsolono Muo) (Nt Q C / 2 Pi Ro Lh)^2
= (Nt Q / 2 Pi Ro Lh)^2
or
(Sac / So) = (Nt Q / 2 Pi Ro Lh)
or
Sac = (Nt Q So / 2 Pi Ro Lh)

Recall that the spheromak surface charge distribution is given by:
Sac = [Qs So^2] / [2 Pi^2 Ro^2 ( So^2 - 1)]WRONG

Equating the two expressions for Sac gives:
(Nt Q So / 2 Pi Ro Lh) = [Qs So^2] / [2 Pi^2 Ro^2 ( So^2 - 1)]
or
(Nt / Lh) = So / [Pi Ro (So^2 - 1)]
or
(Lh / Ro) = Pi (So^2 - 1) Nt / So

This is a tremendously revealing result. It is part of:
(Lh / Ro) = (Pi / So)[Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2]^0.5

Note that the poloidal and toroidal magnetic fields are orthogonal.

Hence at R = Rs the poloidal magnetic field energy density is given by:
(Epsilono / 2)[Sac Rc / Epsilono Rs]^2 = Bps^2 / 2 Muo
FIX where:
Bps = Muo Np Qs Fh / 2 Pi Rx
= (Bpc Rc / Rs)^2 / 2 Muo
due to the change in perimeter lengths.

The change in radial electric field at R= Rc, H = 0 causes an energy density balance at R = Rc of:
[(Bpc)^2 / 2 Muo] - [(Btc^2 / 2 Muo] = (Epsilono / 2)[Sac / Epsilono]^2
or
[(Bpc)^2 / 2 Muo] = [(Btc^2 / 2 Muo] + (Epsilono / 2)[Sac / Epsilono]^2

If the spheromak was a straight core the poloidal magnetic field at the spheromak wall would be:
Muo Np Q Fh / [2 Pi (Rs - Rc) / 2]

At the outer perimeter this field is reduced by a factor of:
Ro / Rs = 1 / So

At the inner perimeter this field is increased by a factor of:
Ro / Rc = So.

The thesis is that the total energy contained in a spheromak is:
Outside Electric field energy + Outside Poloidal Magnetic field energy + Toroidal magnetic field energy.

From the web page titled: SPHEROMAK ENERGY:
Toroidal magnetic field energy + internal electric field energy
= Poloidal magnetic field energy + external electric field energy = ????????????????

THE INDEX N:
On this web page the total field energy density U is expressed as:
U = Ue + Um
where Ue is the electric field energy density component and Um is the magnetic field energy density component.

At large R and/or large Z, Ue is approximately:
Ue ~ U = Uo [Ro^4 / (Ro^2 + (A R)^2 + Z^2)^2}

In general due to spheromak symmetry at R = 0, Z = 0:
Ueo = 0

At R = Rc, Z = 0:
U = Uo [Ro^2 / (Ro^2 + A^2 Rc^2)]^2

Note that Uoc reflects the reality that R = Rc and Z = 0 there is a radial electric field Eroc pointing toward the spheromak axis of symmetry.

Note that at R = 0, Z = 0:
Umo = Uo

Note that in general outside the spheromak wall:
U = Uo [Ro^2 / (Ro^2 + A^2 R^2 + Z^2)]^2 which is the equation for the total field energy density outside a spheromak wall.

The expressions for Ue and Um in the far field match the well known far field dependences of electric and magnetic field energy densities on distance.
 

STRATEGY:
Find the electric and magnetic field energy densities along the spheromak main axis of symmetry. Compare this function to:
U = Uo [Ro^2 / (Ro^2 + Z^2)]^2 and try to determine A via its effect on the electric and magnetic field energy densities along the Z axis.
 

*************************************************************** ***************************************************************

ELECTRIC FIELD ALONG THE Z AXIS:
Consider an elemental ring of charge on the spheromak wall. The circumference of this ring is:
2 Pi R

If the spheromak cross section is circular the width of this ring is:
[(Rs - Rc) / 2] d(Phi)
where Phi is an angle measured at the toroidal axis from the spheromak equatorial plane to the ring.
At R = Rc, Phi = 0.
At R = Rs, Phi = Pi.

Note that:
R = [(Rs + Rc) / 2] - [(Rs - Rc) / 2] cos (Phi)
and
R^2 = [(Rs + Rc) / 2]^2 + [(Rs - Rc) / 2]^2 [cos (Phi)]^2
- [(Rs^2 - Rc^2) / 2] cos(Phi)

These expressions for R and R^2 are valid for both the upper and lower halves of the spheromak.

UPPER HALF OF SPHEROMAK:
In the upper half of the spheromak where d(Phi) is positive the area dA of this elemental ring is:
dA = 2 Pi R (Rs - Rc) d(Phi) / 2
= Pi R (Rs - Rc) d(Phi)

At R = Rc the charge per unit area on the elemental ring is Sac. At other radii the charge per unit area on the elemental ring is: Sac (Rc / R)

Thus for the upper half of the spheromak the charge dQs on the elemental ring is:
dQs = Sac (Rc / R) dA
= Sac (Rc / R) 2 Pi R (Rs - Rc) d(Phi) / 2
= Sac Pi Rc (Rs - Rc) d(Phi)

For the upper half of the spheromak the straight line distance from the elemental ring to the point R= 0, H = Rc on the spheromak symmetry axis is given by Pythagoras theorem as:
{{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^0.5

The vertical distance between Z = Rc and the elemental ring is:
{Rc - [(Rs - Rc) / 2] sin(Phi)}
Note that for the upper half of the spheromak this term is positive if the elemental ring is below H = Rc and is negative if the elemental ring is above H = Rc.

For the upper half of the spheromak the element of axial electric field at R = 0, Z = Rc, due to the elemental charge ring is:
dEo = [dQs / (4 Pi Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^1.5
 
= [Sac Pi Rc (Rs - Rc) d(Phi) / (4 Pi Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^1.5
 
= [Sac Rc (Rs - Rc) d(Phi) / (4 Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + [(Rs + Rc) / 2]^2
+ [(Rs - Rc) / 2]^2 [cos (Phi)]^2 - [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5
 
= [Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs + Rc) / 2]^2
+ [(Rs - Rc) / 2]^2 - [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5
 
= [Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs^2 + Rc^2) / 2]
- [(Rs^2 - Rc^2) / 2] cos(Phi)}^1.5
 

SPHEROMAK LOWER HALF:
For the lower half of the spheromak where d(Phi) is negative the area dA of and elemental ring is:
dA = - 2 Pi R (Rs - Rc) d(Phi) / 2
= - Pi R (Rs - Rc) d(Phi)

Thus for the lower half of the spheromak the charge dQs on the elemental ring is:
dQs = Sac (Rc / R) dA
= - Sac (Rc / R) 2 Pi R (Rs - Rc) d(Phi) / 2
= - Sac Pi Rc (Rs - Rc) d(Phi)

For the lower half of the spheromak the straight line distance from the elemental ring to the point R= 0, H = Rc on the spheromak symmetry axis is given by Pythagoras theorem as:
{{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^0.5
Note that for the lower half of the spheromak sin(Phi) is negative.

For the lower half of the spheromak the vertical distance between H = Rc and the elemental ring is:
{Rc - [(Rs - Rc) / 2] sin(Phi)}
Note that for the lower half of the spheromak where sin(Phi) is always negative this term is always positive.

For the lower half of the spheromak where d(Phi) and sin(Phi) are negative the contribution of an elemental ring to the axial electric field at H = Rc is:
dEo = [dQs / (4 Pi Epsilono)]{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^1.5
 
= [- Sac Pi Rc (Rs - Rc) d(Phi)/ (4 Pi Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + R^2}^1.5
 
= [- Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{[(Rc - [(Rs - Rc) / 2] sin(Phi)}^2 + [(Rs + Rc) / 2]^2
+ [(Rs - Rc) / 2]^2 [cos (Phi)]^2
- [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5
 
= [- Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs + Rc) / 2]^2
+ [(Rs - Rc) / 2]^2
- [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5
 
= [- Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs^2 + Rc^2) / 2]
- [(Rs^2 - Rc^2)/ 2] cos(Phi)}^1.5

 

TOTAL AXIAL ELECTRIC FIELD:
Thus the total electric field Eo at R = 0, Z = Rc is:
Eo = Integral from Phi = 0 to Phi = Pi of:
[Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {{Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs^2 + Rc^2) / 2]
- [(Rs^2 - Rc^2) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- Sac Rc (Rs - Rc) d(Phi)/ (4 Epsilono)]
{Rc - [(Rs - Rc) / 2] sin(Phi)}
/ {Rc^2 - Rc [(Rs - Rc)] sin(Phi) + [(Rs^2 + Rc^2) / 2]
- [(Rs^2 - Rc^2) / 2] cos(Phi)}^1.5

FIX A ENTRIES

Substitute:
Rs = So Ro
and
Rc = Ro / So
to get:
Eo = Integral from Phi = 0 to Phi = Pi of:
[Sac (Ro / So)^2 (So^2 - 1) d(Phi)/ (4 Epsilono)]
(Ro / So){1 - [(So^2 - 1) / 2] sin(Phi)}
/ (Ro / So)^3 {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- Sac (Ro / So)^2 (So^2 - 1) d(Phi)/ (4 Epsilono)]
(Ro / So){1 - [(So^2 - 1) / 2] sin(Phi)}
/(Ro / So)^3 {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
 
= Integral from Phi = 0 to Phi = Pi of:
[Sac (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- Sac (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 

FIX CHARGE DISTRIBUTION

FIX THE CHARGE DISTRIBUTION EQUATION Substitute the charge distribution expression into the integration for Eo:
Eo = Integral from Phi = 0 to Phi = Pi of:
[Sac (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- Sac (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
 
Eo = Integral from Phi = 0 to Phi = Pi of:
[{Qs So^2 / [2 Ro^2 Pi^2 (So^2 - 1)]} (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- {Qs So^2 / [2 Ro^2 Pi^2 (So^2 - 1)]} (So^2 - 1) d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
 
= Integral from Phi = 0 to Phi = Pi of:
[{Qs So^2 / [2 Ro^2 Pi^2]} d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[- {Qs So^2 / [2 Ro^2 Pi^2]} d(Phi)/ (4 Epsilono)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5

Factor out:
[Qs / (4 Pi Epsilono Ro^2)] to get:
Eoc = Integral from Phi = 0 to Phi = Pi of:
[Qs / (4 Pi Epsilono Ro^2)][{ So^2 / [2 Pi]} d(Phi)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5
 
+ Integral from Phi = 0 to Phi = - Pi of:
[Qs / (4 Pi Epsilono Ro^2)][- { So^2 / [2 Pi]} d(Phi)]
{1 - [(So^2 - 1) / 2] sin(Phi)}
/ {1 - [(So^2 - 1)] sin(Phi) + [(So^4 + 1) / 2]
- [(So^4 - 1) / 2] cos(Phi)}^1.5

In order to perform these integrations numerically we must first quantify So^2. Thus the order of calculation is to:
a) choose the Np / Nt pair to be investigated;
b) Calculate So;
c) Calculate (1 / Alpha) and the corresponding F value;
d) If (1 / Alpha) is approximately correct perform the this integration;
e) Use the result of this integration to recalculate F;
f) Compare the two calculated F values. If they are almost the same output all the relevant data for this Np/Nt pair;
g) Investigate the stability of adjacent Np/Nt pairs.

Let us assume that the result of this integration is:
Eoc = [Qs / (4 Pi Epsilono Ro^2)] Ic,
where Ic is a unitless integration result.

Recall that:
Eoc = [Qs / (4 Pi Epsilono Ro^2)] [So^2 /(So^2 + 1)] [1 / (So^2 + 1)]^(N /2)

Equate the two expressions for Eoc to get:
[Qs / (4 Pi Epsilono Ro^2)] Ic
= [Qs / (4 Pi Epsilono Ro^2)] [So^2 /(So^2 + 1)] [1 / (So^2 + 1)]^(N /2)
or
Ic = [So^2 /(So^2 + 1)] [1 / (So^2 + 1)]^(N /2)
 

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SPHEROMAK ENERGY RATIO:
An issue in electromagnetic spheromaks is the ratio of electric field energy to total energy. On the web page titled SPHEROMAK ENERGY it was shown that the total contained electomagnetic energy of a spheromak before adjustment for the toroidal portion is:
Efs = Uo Ro^3 Pi^2
and the spheromak energy Ett after adjustment for the toroidal portion is:
Ett / Efs
= {1 -[(So - 1)^2 / (So^2 + 1)] + [2 So (So - 1)^2 / (So^2 + 1)^2]}

For R > Rc the electric field energy density of a spheromak outside the spheromak wall is given by:
Ueo = Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2 [(R^2 + H^2) / (Ro^2 + R^2 + H^2)]^N

Let Z^2 = R^2 + H^2

FIX N

Then:
Ueo = Uo [Ro^2 / (Ro^2 + Z^2)]^2 [(Z^2) / (Ro^2 + Z^2)]^N

The corresponding total electric field energy is given by:
Integral from Z = 0 to Z = infinity of:
Uo 4 Pi Z^2 dZ Ro^4 (Z^2)^(2 N) / (Ro^2 + Z^2)^(N + 2)
 

From Dwight XXXX the solution to this integral is:
DO INTEGRATION

The corresponding total magnetic field energy is given by:
Integral from Z = 0 to Z = infinity of:
Uo 4 Pi Z^2 dZ Ro^4 [(Ro^2 + Z^2)^N - (Z^2)^N] / (Ro^2 + Z^2)^(N + 2)
 

From Dwight XXXX the solution to this integral is:
DO INTEGRATION

Thus the electric field energy contained in Efs is about ____% of the total field energy of Efs and the magnetic field energy is about ______% of the total field energy of Efs.
 

******************************************************************************* *************************************************************************

SPHEROMAK WALL THEORY REVIEW:
On the web page titled CHARGE HOSE PROPERTIES it was shown that for a charge hose (charge motion path) current Ih is given by:
Ih = (1 / Lh) [Qp Nph Vp + Qn Nnh Vn]
= Qs C / Lh
and
Rhoh = (1 / Lh) (Qp Nph + Qn Nnh)
= Qs / Lh
giving:
(Ih / C)^2 = Rhoh^2
= (Qs / Lh)^2
= (1 / Lh)^2 [Qp Nph + Qn Nnh]^2

and that for practical ionized gas plasmas where Ve^2 >> Vi^2 and Ne ~ Ni:
Ih^2 ~ [Q Ne Ve / Lh]^2
and
[(Ni - Ne) / Ne]^2 ~ (Ve / C)^2
and
Qs^2 ~ [Q Ne Ve / C]^2

These equations allow the development of electromagnetic spheromak theory for atomic particles and plasma.
 

******************************************************************** ********************************************************************************

CONNECTING SPHEROMAK THEORY TO THE FINE STRUCTURE CONSTANT:

The total spheromak energy E is of the form:
E = R(Ro) S(So) where:
R(Ro) is a function proportional to (1 / Ro) and hence to natural frequency Fh and S(So) is a function consisting of two orthogonal terms, one proportional to the number of spiral path poloidal turns Np and the other proportional to the number of spiral path toroidal turns Nt.

Existence and boundary conditions impose a mathematical relationship between these two orthogonal energy terms.
 

To find the spheromak operating point at particular Np, Nt integer values calculate Nr, then calculate So, and then calculate:
Z^2 = {[Np (So^2 + 1)]^2 + [Np (So^2 -1)Kc]^2}.
and then calculate [1 / Alpha]^2.

It appears that (1 / Alpha) is primarily set by the spheromak boundary condition which sets:
Nr^2 + R^2 = 1 / [(Pi / 2 A^2)^2]

Thus we can readily compute the corresponding [Z^2 / Nt^2] value. We can use this methodology to find [Z / Nt] as a function of So and hence (1 / (Alpha Nt) as a function of So.

We can find the So value corresponding to the stable So value in a plot of (1 / Alpha Nt)^2 versus So. The Alpha value at this point is the Fine Structure constant. However, to determine Alpha we must first determine (1 / Alpha Nt) and then determine Nt.

(1 / Alpha)^2
= (Pi / 4)^2 [(So^2 - So + 1) / (So^2 + 1)^2]^2 [Z^2]
or
(1 / Alpha Nt)^2
= (Pi / 4)^2 [(So^2 - So + 1) / (So^2 + 1)^2]^2 [Z^2 / Nt^2]

We know (Z / Nt) from the aforementioned computation process. Hence using a computer we can readily determine the So value of the relative minimum in a plot of:
(1 / Alpha Nt)^2 versus So

Plot:
(Pi / 4)^2 [(So^2 - So + 1) / (So^2 + 1)^2]^2 [Z^2 / Nt^2] versus So
and find the So value of the stable operating point. At that So value we will know the corresponding value of:
(1 / Alpha Nt)
and we can calculate Nr^2 using the formula:
Nr^2 = {Z^2 / [Nt^2 (So^2 + 1)^2]} - {[(So^2 - 1)^2 / (So^2 + 1)^2]}

Hence the mathematical procedure is to slowly vary So to find the spheromak lock point in the function:
(1 / Alpha Nt) vs So
where the term:
[Z / Nt}
in the expression for (1 / Alpha Nt) comes from the aforementioned calculation.

Then we recognize that in a spheromak:
Nr = Np / Nt
where Np and Nt are integers with no common factors.
 

Recall that:
Nr^2 = Np^2 / Nt^2
where Np and Nt are integers with no common factors. Thus to find Np and Nt it is necessary to test both the Np and Nt values for integer and factor compliance. This is not a huge task because we know that:
(1 / Alpha) ~ 137
constrains the maximum size of the Np and Nt integer values to less than about 500.

We know that with the simple boundary condition:
(1 / Alpha Nt)^2 = (Pi / 4)^2 [(So^2 - So + 1)^2 / (So^2 + 1)^8] {2 (So^2 + 1)^2 - 4}
/ {- [4 / [(So^2 - 1)^2]] + [Pi^2 / 4]}

We know that:
(1 / Alpha) ~ 137

Hence we can estimate Nt using the equation:
Nt|estimate = (1 / Alpha) / (1 / (Alpha Nt))
= 137 /(1 / (Alpha Nt))

Then we can estimate Np using the equation:
Np|estimate = Nr Nt|estimate
where to calculate Nr we first calculate:
[Z^2 / Nt^2] = {2 - [4 / (So^2 + 1)^2]}
/ {+ [Pi^2 / 4][1 / (So^2 + 1)^2] - [4 / [(So^2 + 1)^2 (So^2 - 1)^2]]} and then calculate Nr^2 using the equation:
Nr^2 = {Z^2 / [Nt^2 (So^2 + 1)^2]} - {[(So^2 - 1)^2 / (So^2 + 1)^2]}
and the calculate Nr using the equation: Nr = [Nr^2]^0.5

These estimates in combination with a list of prime numbers lead to only a few Np, Nt combinations that need to be fully tested.

Nr = Np / Nt
where Np and Nt are both integers with no common factors and usually either Np or Nt is prime.
 

value of (1 / Alpha) is given by:
(1 / Alpha) = 137.035999

Based on experience with plasmas I expected that the operating So^2 value would be:
So^2 ~ 4.1.

Moving the operating So value requires a change in the electric field energy distribution function. This issue needs more study.

 

Calculate the corresponding [Z^2 / Nt^2] value.

Calculate the corresponding value of Nr^2 using the formula:
Nr^2 = {Z^2 / [Nt^2 (So^2 + 1)^2]} - {[(So^2 - 1)^2 / (So^2 + 1)^2]}

Calculate the exact value of Nr using the formula:
Nr = [Nr^2]^0.5

Find Np and Nt which are the smallest integers with no common factors that precisely satisfy the equation:
Nr = (Np / Nt)

Usually Np and Nt have no common factors because one of them is prime.

Calculate the error in Nr base on Np = 223 and Nt = 303.

It is the precision of this ratio of integers coincident with real numbers which are a function of Pi that causes spheromak stability and hence quantization of energy.

Once Nt is precisely determined use the formula:
[1 / Alpha] = [Nt] [1 / (Alpha Nt)]
to determine the calculated value of:
[1 / Alpha].

Check if the calculated value of [1 / Alpha] is close to the value:
(1 / Alpha) = 137.035999
which is published at:
Fine Structure Constant


 

***********************************************************

SUMMARY:
An electromagnetic spheromak is governed by an existence condition and a common boundary condition. After an electromagnetic spheromak forms it spontaneously emits photons until it reaches an energy minimum also known as a ground state .

SPHEROMAK SHAPE PARAMETER:
See the new web page titled: SPHEROMAK SHAPE PARAMETER.
 

CHECK TABLE
So^2Nr^2NrFI(F I)
2.3 16.08516 4.0106 0.69362651.81611.2597
2.5 2.243 1.4976 0.68672 1.6087 1.10476
3.0
3.5
4.0 .549560.7413 0.62185 0.7374 0.45856
4.5
5.0
5.5
6.0

These equations seem to indicate that when a spheromak is initially formed the spheromak will try to operate at So^2 = 2.27, Nr large rather than at its low energy point. This issue needs further investigation. The toroidal region will reduce the spheromak energy. We must figure out how energy minimization affects the existence and/or boundary condition.

NUMERICAL CALCULATIONS:
TRY: So^2 = 2.3
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] / {1 - [4 / (Pi (So^2 - 1))]^2}
= [+ {0.81057} - {0.155188}] / {1 - 0.959255}
= 0.655382 / 0.0407445
= 16.08516

Nr = 4.0106

F = [Nr So^2 / [(Nr^2 (So^2 + 1)^2) + (So^2 - 1)^2]^0.5
= [(4.0106)(2.3) / [(16.08516 )(10.89)) + (1.69)]^0.5
= 9.22438 / 13.29877
= 0.6936265

SPHEROMAK EVOLUTION:
The above equations indicate that when a spheromak is initially formed the spheromak will operate at large Nr corresponding to So^2 = 2.27. However, over time the spheromak will lose energy by radiation until its energy falls to a stable minimum energy point at So^2 = 4.0________ where the spheromak can no longer spontaneously radiate. As the spheromak energy decreases, Nr^2 decreases, So^2 increases and the volume of the toroidal region increases.
 

The value of:
So^2 = 4.1
is comparable to the ratio:
So^2 = (Rs / Rc) = 4.2
obtained from a General Fusion plasma spheromak photograph. However, a plasma spheromak may be affected by inertial forces and other issues that do not affect a charged particle spheromak.
 

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PLASMA SPHEROMAKS:

SPHEROMAK WALL THEORY REVIEW:
On the web page titled CHARGE HOSE PROPERTIES it was shown that for a charge hose (charge motion path) current Ih is given by:
Ih = (1 / Lh) [Qp Nph Vp + Qn Nnh Vn]
= Qs C / Lh
and
Rhoh = (1 / Lh) (Qp Nph + Qn Nnh)
= Qs / Lh
giving:
(Ih / C)^2 = Rhoh^2
= (Qs / Lh)^2
= (1 / Lh)^2 [Qp Nph + Qn Nnh]^2

and that for practical ionized gas plasmas where Ve^2 >> Vi^2 and Ne ~ Ni:
Ih^2 ~ [Q Ne Ve / Lh]^2
and
[(Ni - Ne) / Ne]^2 ~ (Ve / C)^2
and
Qs^2 ~ [Q Ne Ve / C]^2

These equations allow the development of electromagnetic spheromak theory for atomic particles and plasma.
 

Define:
Ih = plasma hose current
C = speed of light
Rhoh = (Qs / Lh)

Recall from PLASMA HOSE THEORY that:
(Ih / C) = Rhoh
= (Qs / Lh)
= Qs / [(Np Lp)^2 + (Nt Lt)^2]^0.5
or
Ih = C Qs / [(Np Lpf)^2 +(Nt Lt)^2]^0.5
= Qs C / {Pi Ro [[(Np (Rs + Rc))^2 / (Ro)^2] + [(Nt (Rs - Rc))^2 / (Ro)^2]]^0.5}
Thus if the charge Q on an atomic particle spheromak is replaced by the net charge Qs on a plasma spheromak the form of the spheromak equations is identical.

However, in a plasma spheromak:
Qs = Q (Ni - Ne)
where (Ni - Ne) is positive.

The web page PLASMA HOSE THEORY shows that for a plasma spheromak:
(Ni - Ne)^2 C^2 = (Ne Ve)^2
where Ve = electron velocity.

The kinetic energy Eke of a free electron with mass Me is given by:
Eke = (Me / 2) Ve^2

Hence:
(Ni - Ne)^2 C^2 = Ne^2 (2 Eke / Me)
or
Qs = Q (Ni - Ne) = Q (Ne / C)[2 Eke / Me]^0.5

Thus in a plasma spheromak Qs can potentially be obtained via measurements of Ne and Eke. However, due to the free electrons being confined to the spheromak wall Ne is not easy to accurately directly measure.
 

CALCULATION OF PLASMA SPHEROMAK Ne FROM THE FAR FIELD:
An approximate expression for the distant radial electric field is:
[Qs / 4 Pi Epsilon] [1 / (R^2 + H^2)]

The corresponding far field energy density is:
Ue = (Epsilon / 2)[Qs / (4 Pi Epsilon)]^2 [1 / (R^2 + H^2)]^2
= [Qs^2 / (32 Pi^2 Epsilon)][1 / (R^2 + H^2)]^2

***********************************************************

FIX THE A and Z TERMS FROM HERE ONWARDS:
and that this energy density in the far field must equal
Uo [Ro^2 / (Ro^2 + R^2 + H^2)]^2
where:
Uo = (Bpo^2 / 2 Mu)
Equating the two energy density expressions in the far field gives:
[Qs^2 / (32 Pi^2 Epsilon)] [1 / (R^2 + H^2)]^2
= (Bpo^2 / 2 Mu) [Ro^2 / (Ro^2 + R^2 + H^2)]^2
or
[Qs^2 / (32 Pi^2 Epsilon)] = (Bpo^2 / 2 Mu) [Ro^2]^2
or
Qs^2 = (Bpo^2 / 2 Mu) [Ro^2]^2 (32 Pi^2 Epsilon)
= (Bpo^2 / Mu) [Ro^2]^2 (16 Pi^2 / C^2 Mu)
= (Bpo^2 / Mu^2) [Ro^2]^2 (16 Pi^2 / C^2)

Thus:
Qs = (Bpo / Mu) [Ro^2] (4 Pi / C)

Recall that the formula for a plasma spheromak gave:
Qs = Q (Ne / C)[2 Eke / Me]^0.5
or
Ne = Qs C /(Q [2 Eke / Me]^0.5)
= (Bpo / Mu) [Ro^2] (4 Pi / C) C /(Q [2 Eke / Me]^0.5)
= (Bpo / Mu) [Ro^2] (4 Pi /(Q [2 Eke / Me]^0.5)
= (Bpo / Mu) [Rs Rc] (4 Pi /(Q [2 Eke / Me]^0.5)

This equation can be used to estimate Ne in experimental plasma spheromaks.
 

For a spheromak compressed from state a to state b this equation can be written in ratio form as:
(Neb / Nea) = (Bpob / Bpoa)(Rsb Rcb / Rsa Rca) (Ekea / Ekeb)^0.5
or
(Neb / Nea)^2 = (Bpob / Bpoa)^2(Rsb Rcb / Rsa Rca)^2 (Ekea / Ekeb)
or
(Neb / Nea)^2(Roa^6 / Rob^6)
= (Bpob / Bpoa)^2(Rsb Rcb / Rsa Rca)^2 (Roa^6 / Rob^6)(Ekea / Ekeb)
= (Bpob / Bpoa)^2(Rsb Rcb / Rsa Rca)^2 [(Rsa Rca)^3 / (Rsb Rcb)^3] (Ekea / Ekeb)
= (Bpob / Bpoa)^2 [(Rsa Rca) / (Rsb Rcb)] (Ekea / Ekeb)
 

EXPERIMENTAL PLASMA SPHEROMAK DATA:
General Fusion has reported spheromak free electron kinetic energies ranging from 20 eV - 25 eV for low energy density spheromaks at the spheromak generator to 400 ev - 500 eV for higher energy density spheromaks at the downstream end of the conical plasma injector. General Fusion reports a spheromak linear size reduction between these two positions of between 4X and 5X. The corresponding observed apparent electron densities rise from 2 X 10^14 cm^-3 to 2 X 10^16 cm^-3. The corresponding observed magnetic field increases from .12 T to 2.4 T to 3 T. At this time this author does not know for certain: where on the spheromak the electron kinetic energy was measured, where on the spheromak the apparent electron density was measured, where on the spheromak the magnetic field was measured or the absolute dimensions of the measured spheromaks and their enclosure.

Hence:
16 < [Ekeb / Ekea] < 25
20 < (Bpob / Bpoa) < 25
400 < (Bpob / Bpoa)^2 < 625<
4 < (Rca / Rcb) < 5
16 < (Rca / Rcb)^2 < 25
64 < (Rca / Rcb)^3 < 125
[(Nea / Rca^3) / (Neb / Rcb^3)]^2 = 10^-2

It appears that during the plasma spheromak compression Nea decreases to Neb while Qs remains constant. This effect might be due to electron-ion recombination during spheromak compresion.
 

This web page last updated September 26, 2020.

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