XYLENE POWER LTD.

ELECTROMAGNETIC SPHEROMAK

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

INTRODUCTION:
It is implicity assumed herein that spheromaks are the result of well known electromagnetic field interactions.

This web page is concerned with finding mathematical expressions for the spheromak geometry and spheromak electric and magnetic field energy densities for use by other web pages on this web site. It is assumed herein that a spheromak is cylindrically symmetric with any point on the spheromak being defined by its radius R from the main axis of symmetry and its distance Z above the equatorial plane. It is shown that that outside the spheromak wall the electromagnetic field energy density is given by:
Up(R, Z) = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2
and inside the spheromak wall the electromagnetic field energy density is given by:
Ut(R, Z) = Uto (Ro / R)^2.

The position of the spheromak wall is defined by the locus of points where:
Up(R, Z) = Ut(R, Z).

It is assumed that inside the spheromak wall there is no net poloidal magnetic field and there is no net radial electric field.

It is further assumed that outside the spheromak wall there is no net toroidal magnetic field.

The general approach is to find the axial electric and magnetic fields associated with a thin circular closed current ring of radius Ro located at R = Ro, Z = 0 containing uniformly distributed net charge Q and having Np poloidal turns carrying current I, and then to modify those expressions to describe a spheromak.

The modified expressions are tested at both large and small R and Z values.

The modified expressions are then used to find the spheromak geometry in terms of well known physical parameters.

An important parameter is [Ho / Ro] which indicates the exact shape of the spheromak. This parameter affects the calculated values of the Planck Constant and the Fine Structure Constant.

It is assumed that inside the spheromak wall the only non-zero net field is the toroidal magnetic field given by:
Bt = Muo Nt I / 2 Pi R
so that inside the spheromak wall the toroidal magnetic field energy density is:
Ut = Bt^2 / 2 Muo
= (Muo / 8) [Nt I / Pi R]^2

Inside the spheromak wall all the other field energy density contributions are assumed to be zero.
 

STRATEGY:
This problem is attacked by first finding the field energy densities along the Z axis, then finding the field energy densities at selected points on the equatorial plane and then finally finding the field energy densities elsewhere inside and outside the spheromak wall.

The expressions calculated on this web page are used on the web page titled: SPHEROMAK ENERGY to find the total field energy content of a spheromak. That expression for total field enegy is then used on the web age titled: PLANCK CONSTANT to find both the Planck Constant and the Fine Structure Constant in terms of other physical parameters.

This work provides the fundamental connection between units of energy and units of time which currently are connected only by experimental measurements of the Planck Constant.
 

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

Inside the spheromak wall the magnetic field is assumed to be purely toroidal and given by:
Bt = Muo Nt I / (2 Pi R)
where:
R = radius from main axis of symmetry
Muo = permiability of free space
I = circulating current
Pi = length of the perimeter of a circle / circle radius
= 3.14159265

Outside the spheromak wall the magnetic field is assumed to be purely poloidal and similar to the magnetic field produced by a thin ring of radius Ro, in the Z = 0 plane, centered at the origin and carrying current (Np I). However, due to spacial distribution of the poloidal turns there is no certainty that the magnetic field at the center of the spheromak is the same as would be the case if the all the poloidal turns were concentrated in this thin ring.

Inside the spheromak wall the net electric field is assumed to be zero. Outside the spheromak wall the electric field is assumed to be similar to the electric field produced by a thin ring of radius Ro centered at the origin with uniformly distributed charge Q.

At the spheromak wall the field energy density inside the wall equals the field energy density outside the wall. This issue determines the exact position of the spheromak wall.

The spheromak wall intersects the equatorial plane at two different radii, R = Rc and R = Rs. At the inner spheromak wall on the equatorial plane, where R = Rc, the toroidal magnetic field energy density just inside the spheromak wall balances the poloidal magnetic field energy density just outside the spheromak wall. The net electric fiel for R < Rc and Z = 0 is assumed to be zero.

The exact spheromak cross sectional shape is determined by compliance with the assumed energy and vector field distributions. Note that the filament turns are more closely spaced at smaller R values than at larger R values.

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 will spontaneously return to its stable equilibrium geometry. The spheromak wall position corresponds to a system total energy minimum.
 

PLASMA 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 distorted hollow pseudo ellipsoid, known as the spheromak wall. The mechanism of 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 electric field is mainly normal to the spheromak suface.

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 be composed of one or two types but has no inertial mass. High energy plasma spheromaks are also subject to relativistic effects.
 

PARAMETER DEFINITIONS:
R = radius from the spheromak main axis of symmetry to point (R, Z);
Ro = characteristic radius
K Ro = an adjusted value of Ro necesssary to compy with value of Bpo
Rw = value of R of a point on the spheromak wall;
Z = height of point (R, Z) above or below the equatorial plane;
Zw = value of Z of a point on the spheromak wall;

Subscripts:
c = inner wall on equatorial plane;
s = outer wall on equatorial plane;
p = poloidal;
t = toroidal;
z = parallel to main axis of symmetry;
Lp = poloidal path length around the major axis of symmetry;
Lpc = 2 Pi Rc;
Lpo = 2 Pi Ro;
Lps = 2 Pi Rs;
Lt = spheromak perimeter length around the spheromak minor axis of symmetry;
Rc = spheromak wall minimum radius where Z = 0;
Rs = spheromak wall maximum radius where Z = 0;
Ro = radius of the thin charged current ring used to calculate the electric and magnetic fields outside the spheromak wall;
U = with no subscripts indicates total field energy density at any point (R, Z) in space
Uo = field energy density at R = 0, Z = 0 as projected from the electric far field;
Upo = Uo / K^4 = poloidal magnetic field energy density at R = 0, Z = 0;
(0 < K < 1) Uto = toroidal magnetic field energy density at R = Ro, Z = 0;
Ue = electric field portion of the field energy density at (R, Z);
Um = the magnetic field portion of the energy density at (R, Z);
U = Ue + Um;
Uc = U|(R = Rc, Z = 0);
Us = U|(R = Rs, Z = 0);
Ut = toroidal magnetic field energy density;
B = magnetic field;
Bt = toroidal magnetic field;
Bto = Bt|(R = Ro, Z = 0);
Btc = Bt|(R = Rc, Z = 0;
Bts = Bt|(R = Rs, Z = 0);
Bp = poloidal magnetic field:
Bpo = Bp|(R = 0, Z = 0);
Bpc = Bp|(R = Rc, Z = 0);
Bps = Bp|(R = Rs, Z = 0);
E = electric field;
Ep = 0 = poloidal electric field;
Epc = Ep|(R = Rc, Z = 0) = 0
Er = radial electric field;
Erc = Er|(R = Rc, Z = 0)= 0 by cancellation;
Ers = Er|(R = Rs, Z = 0) = [2 Q / (4 Pi Epsilono Rs^2)]
Et = toroidal electric field;
Etc = Et|(R = Rc, Z = 0) = 0
Ets = Et(R = Rs, Z = 0) = 0
Ez = electric field parallel to the Z axis;
Ezc = Ez|(R = Rc, Z = 0) = 0
Ezs = Ez|(R = Rs, Z = 0) = 0
Epsilono = permittivity of free space
Muo = permeability of free space
 

FIELD RELATIONSHIPS:
Upc = Utc
Ups = Uts
Utc = Uto (Ro / Rc)^2
Uts = Uto (Ro / Rs)^2
Bp = poloidal magnetic field;
Bpo = Bp|(R= 0, Z = 0) = Muo Np I / 2 Ro
Btc = Bt|(R = Rc, Z = 0) = Muo Nt I / 2 Pi Rc
Bpo / Btc = (Np / Nt)(Pi Rc / Ro)
Btc = Bto (Ro / Rc)
Bpo / Bto = (Bpo / Btc)(Btc / Bto) = (Np Pi / Nt)
Bts = Bto (Ro / Rs)
 

SPHEROMAK FILAMENT PARAMETERS
Define:
I = charge hose current;
Lh = length of closed filament defining charge motion path;
Nt = Integer number of toroidal filament turns contained in Lh;
Np = Integer number of poloidal filament turns contained in Lh;
Lto = length of one purely toroidal filament turn;
Lpo = 2 Pi Ro = length of one poloidal filament turn at R = Ro;
Lh^2 = (Np Lpo)^2 + (Nt Lto)^2
= (Np 2 Pi Ro)^2 + (Nt Lto)^2

SPHEROMAK GEOMETRY:
The geometry of a spheromak can be further characterized by the following parameters:
Rw = radius of apoint on the spheromak wall from the main axis of symmetry;
Zw = distance of a point on the spheromak wall from the spheromak's equatorial plane;
At Rw = K Ro, Zw = +/- Ho;
Note that:
Ho = Zw|(R = K Ro), not at R = Ro and not at R = Rm, where:
dZw / dRw = 0
 

SPHEROMAL SYMMETRY:
At any point in space the electromagnetic field energy density U is given by: U = (Epsilono / 2)[Er^2 + Ez^2 + Et^2] + (1 / 2 Muo)[Br^2 + Bz^2 + Bt^2]

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

The spheromak cylindrical symmetry 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) = Bz(R, - Z)
Bt(R, Z) = Bt(R, - Z)
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:
Er = 0 and Ez = 0 giving:
U = + (1 / 2 Muo)[Bt^2]

All total field energy densities decrease with increasing distance from R = 0, Z = 0.

Separation of variables:
In general the field energy density is of the form:
U = Ur(R)Uz(Z)
where:
Uz(Z) = Uz(-Z)
and
Ur|(R = 0) = 1

Uz|(Z = 0) = Uo

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

ELECTRIC FIELD ENERGY DENSITY ALONG Z AXIS DUE TO A THIN RING OF CHARGE:
Assume that a thin ring of radius "Ro" has a net charge Q. Then the linear charge density along the thin ring is:
Q / (2 Pi Ro)
and an element of charge is:
dQ = [Q / (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 dQ is:
dE =(1 / 4 Pi Epsilono) [dQ / (Ro^2 + Z^2)]
where:
Epsilono = permittivity of free space

The component of this electric field along the ring axis at point (R = 0, Z = Z) is:
dE = (1 / 4 Pi Epsilon) [dQ / (Ro^2 + Z^2)] cos(Gamma)
where:
cos(Gamma) = Z / (Ro^2 + Z^2)^0.5

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

The electric field energy density Ue at Z = Z, R = 0 due to a thin ring charge is given by:
Ue|(R = 0, Z = Z) = (Epsilon / 2) E^2
= (Epsilon / 2){(1 / 4 Pi Epsilon) [Q Z / (Ro^2 + Z^2)^1.5]}^2
= [Q^2 / (32 Pi^2 Epsilon)] [Z^2 / (Ro^2 + Z^2)^3]
= [Muo C^2 Q^2 / (32 Pi^2)] [Z^2 / (Ro^2 + Z^2)^3]
= [Muo C^2 Q^2 / (32 Pi^2 Ro^4)] [Z^2 Ro^4 / (Ro^2 + Z^2)^3]

If charge Q is concentrated at a point the electric field sensed at a remote point is:
Q / (4 Pi Epsilono R^2)

The related electric field energy distribution is:
U = (Epsilono / 2)[Q / 4 Pi Epsilono Z^2]^2
= (1 / 2 Epsilono)[Q^2 / 16 Pi^2 Z^4]
= [(Muo C^2 Q^2 / 32 Pi^2) / Z^4]
= [(Muo C^2 Q^2 / 32 Pi^2 Ro^4) (Ro^4 / Z^4)]

Note that at R = 0, Z = 0 the net electric field is zero and the electric field energy density is:
Ue|(R = 0, Z = 0) = 0.

In the far field where:
R = 0, Z >> Ro
Ue|(R = 0, Z = Z) = [Q^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.

Hence for a thin charged ring the electric field energy density on the Z axis is:
Ue|(Z = 0) = [Muo C^2 Q^2 / (32 Pi^2)] [Z^2 / (Ro^2 + Z^2)^3]
 

MAGNETIC FIELD ENERGY DENSITY ALONG THE Z AXIS DUE TO A CURRENT RING AT Z = 0:
The law of Biot and Savart gives the net magnetic field B at a measurement point Z on the axis of a ring with Np turns and circulating electric current I as:
B = [Muo Np I / 2][Ro^2 / (Ro^2 + Z^2)^1.5]
where:
Ro = ring radius
Muo = magnetic permeability of free space

The magnetic field energy density along the Z axis is:
Um|(R = 0, Z = Z) = B^2 / 2 Mu
= [(Muo Np I) / 2]^2 [Ro^2 / (Ro^2 + Z^2)^1.5]^2 / 2 Mu
= [Muo Np^2 I^2 / 8] [Ro^4 / (Ro^2 + Z^2)^3]

Thus for Z >> Ro the magnetic field energy density Um|(R = 0, Z = Z) 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 thin ring where Z = 0 and R = 0 the magnetic field is:
Bpo = [(Muo Np I) / (2 Ro)

The corresponding magnetic field energy density Umo at the center of the thin ring is:
Umo = Upm|(R = 0, Z = 0)
= Bpo^2 / 2 Muo
= [Muo Np^2 I^2 / 8 Ro^2]

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

TOTAL FIELD ENERGY DENSITY ALONG THE Z AXIS DUE TO A THIN CURRENT RING AT R = Ro, Z = 0:
Along the Z axis the total electromagnetic energy density U is given by:
Up|(R = 0, Z = Z) = Upe|(R = 0, Z = Z) + Upm|(R = 0, Z = Z)
= [Muo C^2 Q^2 / (32 Pi^2)][K^2] [Z^2 / (Ro^2 + Z^2)^3]
+ [Muo Np^2 I^2 / 8] [Ro^4 / (Ro^2 + Z^2)^3]
= {[Muo C^2 Q^2 / (32 Pi^2)] [Z^2] + [Muo Np^2 I^2 Ro^4 / 8]} /[Ro^2 + Z^2]^3

COMBINED CASE:
Recall that:
Up|(R = 0, Z = Z) = {[(Muo C^2 Q^2) / (32 Pi^2)] [Z^2] + [Muo Np^2 I^2 Ro^4 / 8]} /[Ro^2 + Z^2]^3
= {[(Muo C^2 Q^2) / (32 Pi^2)] [Z^2] + [4 Muo Pi^2 Np^2 I^2 Ro^4 / 32 Pi^2]} /[Ro^2 + Z^2]^3

However, for a spheromak:
I = Q C / Lh
which gives:
Up|(R = 0, Z = Z) = {[(Muo C^2 Q^2) / (32 Pi^2)][Z^2] + [4 Muo Pi^2 Np^2 Q^2 C^2 Ro^4 / 32 Pi^2 Lh^2]} /[Ro^2 + Z^2]^3
= [(Muo C^2 Q^2) / (32 Pi^2)] {[Z^2] + [4 Pi^2 Np^2 Ro^4 / Lh^2]}{Ro^2 /[Ro^2 + Z^2]}^3

Now make the key assumption that for the hypothetical current ring:
= [4 Pi^2 Np^2 Ro^2 / Lh^2] = 1
or
2 Pi Np Ro / Lh = 1
or
Lh / 2 Pi Ro = Np = integer
so that:
Up|(R = 0, Z = Z) = [(Muo C^2 Q^2) / (32 Pi^2)] {[Z^2] + [Ro^2]} /[Ro^2 + Z^2]^3
= [(Muo C^2 Q^2) / (32 Pi^2)]{1 / [Ro^2 + Z^2]}^2
= [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)]{Ro^2 / [Ro^2 + Z^2]}^2

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

SPHEROMAK GUESS:
Based on the form of the equations relevant to a simple current ring, we guess that the external energy density function of a spheromak is:
U = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2
where:
Uo = [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)].

Then in the far field where:
R >> K Ro
and
Z >> K Ro
U = Uo {Ro^2 / [R^2 + Z^2]}^2
while in the near field:
U = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2
where for R = 0, Z = 0:
Upo = Uo / K^4
where Upo is the poloidal magnetic field energy density at R = 0, Z = 0 and K is a constant slightly less than unity
(0 < K < 1)
that is used to raise the magnetic field energy density at the origin with respect to the projected electric field energy density at the origin so that Upo is larger than Uo.

Note that in the far field the effect of K disappears. However, that far field expression is not valid in the near field and cannot be integrated over all space to find the total field energy whereas the near field expression, which is correct, can be so integrated to find total field energy.

The question that immediately arises is "Does this guessed external energy density function accurately represent the energy density at general R and Z values at places where that energy density can be calculated by alternative means?"

The validity of this guessed external energy density function can be quantitatively checked for both small and large R and Z values and can be qualitatively checked at certain midrange R and Z values.
 

DATA SUPPORTING THE GUESSED SPHEROMAK EXTERNAL FIELD ENERGY DENSITY FUNCTION:
At R = 0, Z >> K Ro the guessed external energy density function simplifies to:
U = [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)] [Ro^2 / Z^2]^2.
This expression is identical to the corresponding far electric field energy density from a point charge.

At R >> (K Ro), Z = 0 this guessed external energy density function simplifies to:
U = [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)] [Ro^2 / (K Ro -R)^2]^2
~ [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)] [Ro^2 / R^2]^2
which expression is identical to the corresponding far electric field energy density from a point charge.

At R >> K Ro, Z = >> Ro the guessed external energy density function:
U = [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)] {Ro^2 / [(K Ro - R)^2 + Z^2]}^2.
~ [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)] {Ro^2 / [(R)^2 + Z^2]}^2.
which expression is identical to the far electric field energy density from a point charge.

At R = 0 , Z = 0 the guessed external energy density function becomes:
U = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2
= Uo / K^4
= Upo
 

SPHEROMAK INSIDE WALL BOUNDARY CONDITION:
The web page titled: THEORETICAL SPHEROMAK gives the interior wall boundary condition as:
[Uto / Uo]^0.5 = {Ro^2 / [(K Ro - Rc)^2]} [Rc / K Ro]
= {Ro Rc / K) / [(K Ro) - Rc)]^2}

Also: (K^2 Ro - Rc) / Ro = = (Ho^2 / 2 Ro^2) {-1 + [(4 K^2 Ro^2 / Ho^2) + 1]^0.5}

Basic physics gives:
Bpo = Muo Np I / 2 Ro
where Ro is the radius of the poloidal current path and Btc = Muo Nt I / 2 Pi Rc. Thus:
Btc / Bpo = [Muo Nt I / 2 Pi Rc] / [Muo Np I / 2 Ro]
= [Nt Ro / Np Pi Rc]

Thus:
[Uto / Uo]^0.5 = [Nt / Np] Pi Ro^2 / K Rc^2

The web page titled: THEORETICAL SPHEROMAK gives the interior wall boundary condition as:
[Uto / Uo]^0.5 = {Ro^2 / [(K Ro - Rc)^2]} [Rc / K Ro]
= {(Ro Rc / K) / [(K Ro) - Rc)]^2}

Equating he two expressions for [Uto / Uo]^0.5 gives:
[Nt / Np] Pi Ro^2 / K Rc^2 = (Ro Rc / K) / [(K Ro) - Rc]^2}
or
[Nt / Np] Pi Ro / Rc^2 = {Rc / [(K Ro) - Rc)]^2}
or
[Nt / Np] Pi Ro = Rc^3 / [(K Ro) - Rc]^2}
or
[(K Ro) - Rc]^2 = Np Rc^3 / [Nt Pi Ro]
or
K^2 Ro^2 - 2 K Ro Rc + Rc^2 = Np Rc^3 / [Nt Pi Ro]
or K^2 (Ro^3 / Rc^3) - 2 K (Ro^2 / Rc^2) + (Ro / Rc) = [Np / Nt Pi] [K Ro / Rc]^2 (Ro / Rc) - 2 [K Ro / Rc] + (Ro / Rc) = [Np / Nt Pi] {[K Ro / Rc] - 1}^2 = [Rc / Ro][Np / Nt Pi]

We need another near field data point to resolve K from Rc.

For R < Rc, Z = 0 the energy field is entirely poloidal magnetic. Hence:
Bpc^2 / 2 Muo = Uo {Ro^2 / [(K Ro - Rc)^2]}^2
and
Bpo^2 / 2 Muo = Uo {Ro^2 / [(K Ro)^2]}^2
= Uo / K^4
giving:
(Bpc / Bpo)^2 = Uo {Ro^2 / [(K Ro - Rc)^2]}^2 / Uo {1 / K^4}
= (K Ro)^4 / (K Ro - Rc)^4
or
(Bpc / Bpo) = (K Ro)^2 / (K Ro - Rc)^2 or
Bpc = Bpo {(K Ro)^2 / (K Ro - Rc)^2}

Recall that inside the spheromak wall:
Ut = Uto [K Ro / R]^2
where Uto is the toroidal magnetic energy field density at:
R = K Ro

Inside the spheromak wall the entire energy field is toroidal magnetic. Hence:
Ut / Uto = (Bt / Bto)^2 = [K Ro / R]^2

Hence:
Btc = Bto [K Ro / Rc]

Now apply the inside wall boundary condition at R = Rc, Z = 0 which is:
Bpc = Btc
or
Bpo {(K Ro)^2 / (K Ro - Rc)^2} = Bto [K Ro / Rc]

Rearrange this equation to get:
Bpo / Bto = [K Ro / Rc] / {(K Ro)^2 / (K Ro - Rc)^2}
= (K Ro - Rc)^2 / (K Rc Ro)

However:
Bpo / Bto = [Upo / Uto]^0.5

Hence equating expressions for [Bpo / Bto] gives:
[Upo / Uto]^0.5 = (K Ro - Rc)^2 / (K Rc Ro)

Recall that:
Bpo^2 / 2 Muo = Upo = Uo / K^4
or
K^4 Bpo^2 / 2 Muo = K^4 Upo = Uo
and
Bto^2 / 2 Muo = Uto
giving:
Upo / Uto = Bpo^2 / Bto^2
or
[Upo / Uto]^0.5 = [Bpo / Bto]

Thus equating expressions for [Upo / Uto]^0.5 gives:
[Bpo / Bto] = (K Ro - Rc)^2 / (K Rc Ro)
 

DETERMINATION OF Bpo
Recall that:
Upo = Bpo^2 / (2 Muo)
= Uo / K^4 = [(Muo C^2 Q^2) / (32 Pi^2 Ro^4 K^4)]
or
Bpo^2 = [(Muo^2 C^2 Q^2) / (16 Pi^2 Ro^4 K^4)
or
Bpo = (Muo C Q) / (4 Pi Ro^2 K^2)

NEAR FIELD:

Bpo = Muo Np I / 2 Ro

Bto = Muo Nt I / (2 Pi K Ro)
where Bto is the toroidal magnetic field at R = K Ro

[Bpo / Bto] = [Muo Np I / 2 Ro] / [Muo Nt I / (2 Pi K Ro)]
= (Np Pi K) / Nt

Recall that:
Bpo / Bto = [Upo / Uto]^0.5.

Hence equating expressions for Bpo / Bto gives:
(Np Pi K) / Nt = [Upo / Uto]^0.5
or
[(Np Pi K) / Nt]^2 = [Upo / Uto]
= [Uo / K^4 Uto]

Hence:
K^4 [Np Pi K / Nt]^2 = [Uo / Uto]
or
[Np Pi K^3 / Nt]^2 = [Uo / Uto]

The spheromak wall occurs at:
Up = Ut
where:
Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2 = Uto [K Ro / R]^2

At R = K Ro:
Uo {Ro^2 / Ho^2}^2 = Uto
or
[Ho / Ro]^4 = [Uo / Uto]

Equating expressions for Uo / Uto gives:
[Np Pi K^3 / Nt]^2 = [Ho / Ro]^4
or
[Np Pi K^3 / Nt] = [Ho / Ro]^2
 

At R = Rs, Z = 0:
Uo {Ro^2 / [(K Ro - Rs)^2]}^2 = Uto [K Ro / Rs]^2
or
[Uto / Uo] = [Rs / K Ro]^2 {Ro^2 / [(K Ro - Rs)^2]}^2
or
[Uto / Uo]^0.5 = [Rs / K Ro] {Ro^2 / [(K Ro - Rs)^2]}
= [Rs Ro / K] / [(K Ro - Rs)^2]

In summary:
[Uto / Uo]^0.5
= [Rs Ro / K] / [(K Ro - Rs)^2]
= [Ro / Ho]^2
= [Nt / Np Pi K^3]
= [Rc Ro / K] / [(K Ro - Rc)^2]
 

EXPERIMENTAL EVALUATION OF K:
From above:
[Rs Ro / K] / [(K Ro - Rs)^2] = [Rc Ro / K] / [(K Ro - Rc)^2]
or
[Rs] / [(K Ro - Rs)^2] = [Rc] / [(K Ro - Rc)^2]
or
Rs [(K Ro - Rc)^2] = Rc [(K Ro - Rs)^2] or
Rs [(K Ro)^2 - 2 K Ro Rc + Rc^2] = Rc [(K Ro)^2 - 2 K Ro Rs + Rs^2]
or
Rs [(K Ro)^2 + Rc^2] = Rc [(K Ro)^2 + Rs^2]
or
(Rs - Rc)(K Ro^2) = Rc Rs^2 -Rs Rc^2 = Rc Rs (Rs - Rc)
or
K Ro^2 = Rc Rs
or
K = Rc Rs / Ro^2

In order to make use of this equation we need Ro as a function of Rm, the radius at which the spheromak length is maximum.

The web page titled: Theoretical Spheromak gives:
[Rm / Ro] = [1 / 2][Ho / Ro]^2 + K

The web page titled: Theoretical Spheromak also gives:
[Hm / Ho]^2 = [(Ho^2 / 4 Ro^2) + K]

FIND EXPRESSION FOR [Lh / (2 Pi Ro)]^2:
Recall that:
[Lh / (2 Pi Ro)]^2 = [Np Lp / (2 Pi Ro)]^2 + [Nt Lt / (2 Pi Ro)]^2
where:
[Lp / 2 Pi Ro]^2 = 1
and from the web page titled THEORETICAL SPHEROMAK:
[Lt / (2 Pi Ro)]^2 = [Ho / Ro]^2 {[(Ho^2 / 4 K^2 Ro^2) + 1]}
giving:
[Lh / (2 Pi Ro)]^2 = [Np Lp / (2 Pi Ro)]^2 + [Nt Lt / (2 Pi Ro)]^2
= Np^2 + Nt^2 [Ho / Ro]^2 {[(Ho^2 / 4 K^2 Ro^2) + 1]}
= Np^2 + Nt^2 [Ho / Ro]^4 [1 / 4 K^2] + Nt^2 [Ho / Ro]^2
 

WINDING STABILITY:
When the spheromak filament length is stable implying that the spheromak contained energy is stable:
d{[Lh / (2 Pi Ro)]^2 = 0
implies that:
2 Np dNp + 2 Nt dNt [Ho / Ro]^4 [1 / 4 K^2] + 2 Nt dNt [Ho / Ro]^2 = 0
or
Np dNp = - {Nt dNt [Ho / Ro]^4 [1 / 4 K^2] + Nt dNt [Ho / Ro]^2
or
- dNp / dNt = (Nt / Np) {[Ho / Ro]^4 [1 / 4 K^2] + [Ho / Ro]^2}

However, Np, Nt, dNp and dNt are all integers. For a stable winding prime number theory indicates that:
dNp / dNt = - 2:
P = prime number which is odd
P = (Np + 2 Nt), Np = odd, 2 Nt = even

With this winding at a nearly constant energy the spheromak can trade off Np and Nt values in whole number increments with certainty of no collapse due to a common factor in Np and Nt.

Hence:
2 = [Nt / Np]{[Ho^4 / Ro^4][1 / 4 K^2] + [Ho^2 / Ro^2]}
or
2 Np / Nt = [Ho^2 / Ro^2] {[Ho^2 / Ro^2][1 / 4 K^2] + 1}

Note that Np and Nt are positive integers where Np is not equal to Nt.

An approximate solution is:
Np ~ Nt
Ho ~ Ro
K ~ (1 / 2)
 

[Ho^2 / Ro^2] {[Ho^2 / Ro^2][1 / 4 K^2] + [Ho^2 / Ro^2] - 2 Np / Nt = 0
or
[Ho^2 / Ro^2] {[Ho^2 / Ro^2] + [4 K^2 Ho^2 / Ro^2] - 8 K^2 Np / Nt = 0

The solution to this quadratic equation is:
[Ho^2 / Ro^2] = {- 4 K^2 +/- [16 K^4 + 4(1) 8 K^2 Np / Nt)]^0.5} / 2
= {- 4 K^2 + 4 K^2 [1 + (2 Np / K^2 Nt)]^0.5 } / 2
= 2 K^2 {- 1 + [1 + (2 Np / K^2 Nt)]^0.5 }

The term:
[1 + (2 Np / K^2 Nt)] must be a perfect square.

FIRST SOLUTION ATTEMPT:
Try ( 2 Np / K^2 Nt) = 3:
Then:
Ho^2 / Ro^2 = 2 K^2
Then:
2 Np / Nt = 3 K^2 = (3 / 2) [Ho^2 / Ro^2]
or
Np / Nt = (3 / 4) [Ho^2 / Ro^2]
Problem because Np / Nt is a rational number.
 

EXACT SOLUTION:
[Ho^2 / Ro^2] = 2 K^2 {- 1 + [1 + (2 Np / K^2 Nt)]^0.5 }

Try:
(2 Np / K^2 Nt) = 8

Then:
Ho^2 / Ro^2 = 4 K^2
Then:
2 Np / Nt = 8 K^2 = 2 Ho^2 / Ro^2
or
Np / Nt = [Ho^2 / Ro^2]
is an exact solution.
 

EXACT SOLUTION SUMMARY:
4 K^2 = [Ho / Ro]^2 = [Np / Nt]
K ~ (1 / 2)

Recall that:
P = Np + 2 Nt
or
(P - 2 Nt) / Nt = 4 K^2
or
P = 4 K^2 Nt + 2 Nt
= Np + 2 Nt Thus: Np = 4 K^2 Nt = Nt + 1 or (4 K^2 - 1) Nt = 1 or Nt = 1 / (4 K^2 - 1)

FIND EXPRESSION FOR Nt:
From above:
K^2 = [1 / 4][Ho^2 / Ro^2]
= [1 / 4][Np / Nt]
or
Np / Nt = 4 K^2

If Np ~ Nt then K ~ (1 / 2)

If K = (1 / 2) + D
then
K^2 = (1/ 2)^2 + D + D^2
then
4 K^2 = 1 + 4 D + 4 D^2

Thus:
Np / Nt = (1 + 4 D + 4 D^2)
= (1 + 2 D)^2

Now assume that:
Np = Nt + 1,
then:
4 D^2 + 4 D = (1 / Nt)
or
D^2 + D - (1 / 4 Nt) = 0
or
D = {- 1 +/- [1 + 4(1)(1 / 4 Nt)]^0.5} / 2
or
= {-1 +/- [1 + (1 / Nt)]^0.5 } / 2
~ (1 / 4 Nt)

Thus:
[K - (1 / 2)] = D ~ (1 / 4 Nt)
or
4 Nt = 1 / [K - (1 / 2)]
or
Nt = [1 / (4 K - 2)]

We need to further investigate K to determine Nt.
 

FIND EXPRESSION FOR [Lh / 2 Pi Ro]:
Recall that:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Uo / Uto][1 / (4 K^2)] + Nt^2 [Uo / Uto]^0.5
 

Recall that:
[Uo / Uto]^0.5 = [Ho^2 / Ro^2]

Hence:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Ho / Ro]^4 [1 / (4 K^2)] + Nt^2 [Ho / Ro]^2
 

Recall that:
Ho^2 / Ro^2 = 4 K^2
?????????????? giving:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Ho / Ro]^4 [1 / (4 K^2)] + Nt^2 [Ho / Ro]^2
= Np^2 + Nt^2 [Ho / Ro]^2 + Nt^2 [Ho / Ro]^4
= Np^2 + 2 Nt^2 [Ho / Ro]^2

Recall that:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Lt / (2 Pi Ro)]^2
which gives:
[Lt / (2 Pi Ro)]^2 = 2 [Ho / Ro]^2

Thus:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 [Lt / (2 Pi Ro)]^2
= Np^2 + Nt^2 2 [Ho / Ro]^2

Recall that:
[Ho / Ro]^2 = Np / Nt

Hence:
[Lh / 2 Pi Ro]^2 = Np^2 + Nt^2 2 [Ho / Ro]^2
= Np^2 + Nt^2 2 (Np / Nt)
= Np^2 + 2 Np Nt
= Np (Np + 2 Nt)

Note that for this stable solution:
Np + 2 Nt = P = prime number

[Lh / 2 Pi Ro] = {Np (Np + 2 Nt)}^0.5 = {Np P}^0.5
 

SOLUTION SUMMARY:
[Lh / 2 Pi Ro]^2 = Np P = Np (Np + 2 Nt)

Np / Nt = 4 K^2 = [Ho^2 / Ro^2]

P = Np + 2 Nt, Np = odd, 2 Nt = even

Np ~ Nt

Ho ~ Ro

K = (1 / 2) + D:
where:
D << (1 / 2)
and
Nt = [1 / (4 K - 2)]
and
D ~ (1 / 4 Nt)
= 1 / {4[1 / (4 K - 2)]}
= (4 K - 2) / 4
= K - (1 / 2)
as expected.
 

We need to further investigate the size of K which is set by electromagnetic phenomena .

This expression should be used on the web page titled: PLANCK CONSTANT to determine the Planck Constant and the Fine Structure Constant.

Hence the key issue in computation of [Lh / 2 Pi Ro] is precise evaluation of:
D ~ [1 / (4 Nt)]
 

The value of [Ho^2 / Ro^2] can be used to solve the space energy integral.
 

DETERMINATION OF Np AND Nt:
Np / Nt = N + F
where:
Np = unknown integer
Nt = Unknown integer
N = known integer
F = known decimal fraction

Thus:
Np = Nt N + Nt F
= Nt N + I
where:
I = Nt F = whole integer.

Hence:
Nt = (1 / F) I

This problem is easier to solve because we can start with Ii = 10,000,000 calculate Nti and then simplify.

Then:
Np = Nt N + I
 

If the guessed external energy density function is incorrect close to R = 0, Z = 0 the effect is to cause a delta function add-on to U(R,Z) at R = 0, Z = 0. This delta fuction adds an offset onto the values of Efs = energy calculated by integrating the field energy density over all space.

SUMMARY:
The spheromak energy density function outside the spheromak wall is:
Up|(R, Z)
= [(Muo C^2 Q^2) / (32 Pi^2)] / [(K Ro - R)^2 + Z^2]^2
= [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)]{Ro^2 / [(K Ro - R)^2 + Z^2]}^2
= Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2

The corresponding energy density function inside the spheromak wall is:
Ut = Uto (K Ro / R)^2

At the spheromak wall:
Ut = Up
or
Uto (K Ro / R)^2 = Upo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2
or
[Uto / Upo] = (R / K Ro)^2 {Ro^2 / [(K Ro - R)^2 + Z^2]}^2
or
[Uto / Upo]^0.5 = (R / K Ro){Ro^2 / [(K Ro - R)^2 + Z^2]}
= [R Ro / K] / [(K Ro - R)^2 + Z^2]
or
[(K Ro - R)^2 + Z^2] = [R Ro / K] [Uo / Uto]^0.5

On the spheromak wall at R = K Ro:
[Zw|(R = K Ro)]^2 = Ro^2 [Uo / Uto]^0.5
= Ho^2
where:
Ho^2 = Z^2|(R = K Ro)
giving:
[Uo / Uto]^0.5 = (Ho / Ro)^2

On the spheromak wall at Z = 0:
(K Ro - R)^2 [K / R Ro] = [Uo / Uto]^0.5
= (Ho / Ro)^2

This equation is quadratic with two real solutions, R = Rc and R = Rs:
Expanding the equation gives:
Ro (K^2 Ro^2 - 2 K Ro R + R^2) = R Ho^2
or
R^2 + (-2 Ro - (Ho^2 / Ro)) R + Ro^2 = 0
or
R = {(2 Ro + (Ho^2 / Ro)) +/- [(2 Ro + (Ho^2 / Ro))^2 - 4(1)Ro^2]^0.5} / 2
= {(2 Ro + (Ho^2 / Ro)) +/- [(4 Ro(Ho^2 / Ro) + (Ho^2 / Ro)^2]^0.5} / 2
= {(2 Ro + (Ho^2 / Ro)) +/- Ho [(4) + (Ho^2 / Ro^2)]^0.5} / 2

Thus:
Rc = {(2 Ro + (Ho^2 / Ro)) - Ho [(4) + (Ho^2 / Ro^2)]^0.5} / 2
and
Rs = {(2 Ro + (Ho^2 / Ro)) + Ho [(4) + (Ho^2 / Ro^2)]^0.5} / 2

These expressions can be rewritten as:

2 (Rs - Ro) = {(Ho^2 / Ro) + Ho [(4) + (Ho^2 / Ro^2)]^0.5}

2 (Ro - Rc) = - (Ho^2 / Ro)) + Ho [(4) + (Ho^2 / Ro^2)]^0.5}

Note that:
(Rs - Ro) > (Ro - Rc)

The general equation for the spheromak wall is:
[(K Ro - R)^2 + Z^2] = R Ro [Uo / Uto]^0.5

At R = K Ro, Z = +/- Ho

This equation describes the spheromak wall which is a toroid with a distorted cross section.
This distorted toroid has a height Z = Ho above the equatorial plane at R = Ro. The spheromak wall passes through the equatorial plane at R = Rc and at R = Rs where Z = 0. Note that Ho is not the maximum spheromak height. This maximum height is at the R value where:
dZ / dR = 0.

There is a family of possible spheromaks that have different (Ho / Ro) ratios. One of these ratios should optimize the spheromak structural stability.
 

Thus the spheromak relative geometry can be chacterized by the ratio:
(Ho / Ro) where:
the spheromak wall is defined by:
[(1 - (R / Ro))^2 + (Z / Ro)^2] = (Ho / Ro)^2 (R / Ro)
or
(Z / Ro) = + / - [(Ho / Ro)^2 (R / Ro) - (1 - (R / Ro))^2]^0.5
Which equation is only valid for:
(Rc / K Ro) < (R / K Ro) < (Rs / K Ro)

Recall that:
[Upo / Uto]^0.5 = Ho^2 / Ro^2
and that:
[Uo / Uto]^0.5 = [Lh / 2 Ro Nt]
which gives:
[(1 - (R / Ro))^2 + (Z / Ro)^2] = [Lh / 2 Ro Nt] (R / Ro)
where:

Lh = [(Np Lp)^2 + (Nt Lt)^2]^0.5
and

Lp = 2 Pi Ro

(Ho / Ro)^2
Check = 4 K^2
= [Uo / Uto]^0.5
= [Lh / 2 Ro Nt]
= [(Np Lp)^2 + (Nt Lt)^2]^0.5 / 2 Ro Nt
= [((Np / Nt) Nt Lp)^2 + (Nt Lt)^2]^0.5 / 2 Ro Nt
= [((Np / Nt) Lp)^2 + (Lt)^2]^0.5 / 2 Ro
= [((Np / Nt) 2 Pi Ro)^2 + (Lt)^2]^0.5 / 2 Ro
= [((Np / Nt) Pi)^2 + (Lt / 2 Ro)^2]^0.5
 

TOROIDAL MAGNETIC FIELD ENERGY:
The toroidal magnetic field energy is:
Integral from R = Rc to R = Rs of:
Uto (Ro / R)^2 2 Pi R dR (2 Zs)
= Integral from R = Rc to R = Rs of:
Uto (Ro / R)^2 2 Pi R dR (2 [Ho^2 (R / Ro) - (Ro - R)^2]^0.5)
= Integral from R = Rc to R = Rs of:
Uto (Ro^2 / R) 4 Pi dR ([Ho^2 (R / Ro) - (Ro - R)^2]^0.5)
which may need numerical integration.

Note that 113 is prime and 355 = 5(71). Ideally for highly stable spheromaks both Np and Nt should be prime. However, if 113 is disturbed to either 112 = (7 X 2 X 2 X 2 X 2) or to 114 = (107 X 2) it still shares no common factors with 355, which potentially provides the required spheromak stability.

This expression is used on the web page titled: PLANCK CONSTANT to determine the Planck Constant and the Fine Structure Constant.

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Define:
F = C / Lh

Hence:
Upo = [Muo / 2] [F Qs Np / 2 Ro]^2
 

MAGNETIC FIELD ENERGY DENSITY ON THE EQUATORIAL PLANE:

Assume that outside the spheromak wall the poloidal magnetic field is the same as the magnetic field of a thin ring having Np turns located at R = Ro, Z = 0.

The ring has Np turns at R = Ro, Z = 0

Find the poloidal magnetic field at R = Rc where Rc < Ro.

Let Theta = angle at centre of ring with respect to the +ve X axis integrated over.

Let Phic = Cross product angle between a normal to the element of current and a vector to R = Rc on the +ve X axis.

Let Dc = distance from current element (Ro dTheta) at Z = 0 to sense position at R = Rc, Theta = 0, Z = 0

dBpc = (Uo / 4 Pi) I Np Ro dTheta cos(Phic) / Dc^2

Bpc = 2 (Uo / 4 Pi) I Np Ro Integral from Theta = 0 to Theta = Pi of
dTheta cos(Phic) / Dc^2

Now find cos(Phic) and Dc in terms of Ro , Rc and Theta:
Geometry gives:
Ro = Rc cos(Theta) + Dc cos(Phic)
or
cos(Phic) = [Ro – Rc cos(Theta)] / Dc

The current path element Ro d(Theta) is located at point (X, Y) on the equatorial plane where Z = 0.

On the current ring:
X^2 + Y^2 = Ro^2

For point (X, Y) on the current ring geometry gives:
Dc^2 = Y^2 + (X – Rc)^2
= Y^2 + X^2 – 2 Rc X + Rc^2
= Ro^2 – 2 Rc X + Rc^2

From geometry:
X = Ro cos(Theta)

Hence:
Dc^2 = Ro^2 – 2 Rc Ro cos(Theta) + Rc^2

dBp = (Uo / 4 Pi) I Np Ro dTheta cos(Phic) / Dc^2
= (Uo / 4 Pi) I Np Ro dTheta [Ro – Rc cos(Theta)] / Dc^3
 
= (Uo / 4 Pi) I Np Ro dTheta [Ro – Rc cos(Theta)] / [Ro^2 – 2 Rc Ro cos(Theta) + Rc^2]^1.5
 
= (Uo / 4 Pi) I Np Ro dTheta Ro [1 – (Rc / Ro)cos(Theta)] / Ro^3 [1 – 2 (Rc / Ro) cos(Theta) + (Rc / Ro)^2]^1.5
 
= (Uo / 4 Pi) I Np dTheta [1 – (Rc / Ro) cos(Theta)] / Ro [1 – 2 (Rc / Ro) cos(Theta) + (Rc / Ro)^2]^1.5

For the special case of (Rc / Ro) = 0 this equation simplifies to:
Bpo = 2 (Uo / 4 Pi) I Np Integral from Theta = 0 to Theta = Pi of dTheta / Ro
= (Uo I Np) / (2 Ro)
as expected.

(Bpc / Bpo) = [2 (Uo / 4 Pi) I Np / Bpo ]Integral from Theta = 0 to Theta = Pi of : dTheta [1 – (Rc / Ro) cos(Theta)]
/ Ro [1 – 2 (Rc / Ro) cos(Theta) + (Rc / Ro)^2]^1.5
 
= [2 (Uo / 4 Pi) I Np / [(Uo I Np) / (2 Ro)] ] Integral from Theta = 0 to Theta = Pi of:
dTheta [1 – (Rc / Ro) cos(Theta)]
/ Ro [1 – 2 (Rc / Ro) cos(Theta) + (Rc / Ro)^2]^1.5
or
Bpc / Bpo = [(1 / Pi) ] Integral from Theta = 0 to Theta = Pi of:
dTheta [1 – (Rc / Ro) cos(Theta)]
/ [1 – 2 (Rc / Ro) cos(Theta) + (Rc / Ro)^2]^1.5
 

PERFORM THIS INTEGRATION NUMERICALLY:

Thus the magnetic field in the poloidal region (spheromak core) is least at the origin and increases slowly out to (R / Ro) = (1 / 2) and then increases rapidly as Rc approaches Ro.

By following an analogous procedure we can plot [Bps / Bpo] vs [Rs / Ro].

There is a family of toroidal region curves of the form:
Bt / Bto = Ro / R

Recall that:
Bto / Bpo = [Nt / Pi Np]

Hence within the toroidal region: Bt / Bpo = (Bt / Bto)(Bto / Bpo)
= (Ro / R)[Nt / Pi Np]

This equation describes a family of toroidal region curves that intersect the Bp(R,Z)curves in two places. The particular curve is set by the ratio [Nt / Np].

At R = Rc:
Btc / Bpo = (Ro / Rc)[Nt / Pi Np]

Generally Btc = Bpc and Bpc > Bpo

Generally (Ro / Rc) > 1

Hence:[Nt /Pi Np] ~ 1
or
Nt / Np ~ Pi

The spheromak will tend to adopt lower energy curves. However, Nt and Np must be suitable integers.

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.

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 spheromak 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

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
 

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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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

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.

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

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)

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

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

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

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)

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

Note that this formula applies to all spheromaks

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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.

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
&n

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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?

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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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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

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

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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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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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)]

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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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.

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

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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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]

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.

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.
 

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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

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.
 

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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.
 

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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.
 

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) an

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.
 

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:

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

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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.
 

SPHEROMAK EVOLUTION:
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.