XYLENE POWER LTD.

SPHEROMAK STRUCTURE:

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

SPHEROMAK STRUCTURE:
This web page shows that net charge circulating around a closed path forms the external field energy density distribution required for the existence of a spheromak. The more elaborate mathematical proof of spheromak wall position stability is developed on the web page titled:THEORETICAL SPHEROMAK. The electric and magnetic fields of a spheromak maintain its quasi-toroidal geometry and to contain a stable amount of energy.

A spheromak has a major axis and a minor axis. The circulating current I follows a closed spiral path which is characterized by Np poloidal turns around the spheromak's major axis and by Nt toroidal turns around the spheromak's minor axis. The current filament path has the shape of the glaze on a distorted doughnut and is referred to as the spheromak wall. Inside the spheromak wall the magnetic field is toroidal. Outside the spheromak wall the magnetic field is poloidal. Inside the spheromak wall the electric field is zero. Outside the spheromak wall the local electric field is normal to the spheromak wall. Within the core of the spheromak the electric field components parallel to the equitorial plane cancel. In the far field the electric field is nearly spherically radial. The relative geometry of the spheromak wall is constant over time independent of the spheromak contained energy.

The spheromak field structure allows semi-stable plasma spheromaks and discrete stable atomic charged particles to exist and act as stores of energy.
 

SPHEROMAK CONCEPT:
Conceptually a spheromak wall is a quasi-toroidal surface formed by the closed filament of a spheromak. This filament path conforms to the spheromak wall surface curvature.

The current filament axis gradually changes direction over the surface of the quasi-toroid following a closed spiral.

A spheromak is cylindrically symmetric about the spheromak major axis and is mirror symmetric about the spheromak's equatorial plane. The net charge Qs is uniformly distributed over the filament length Lh.

For an isolated spheromak in a vacuum, at the center of the spheromak the net electric field is zero. Inside the spheromak wall the magnetic field is purely toroidal and the electric field is zero. In the region outside the spheromak wall the magnetic field is purely poloidal. Outside the spheromak wall the electric field is normal to the spheromak wall and is radial in the far field. The current circulates within the spheromak wall which separates the regions with toroidal and poloidal magnetic fields.
 

SPHEROMAK WALL:
Outside the spheromak wall the field energy density U has electric and poloidal magnetic field components.
Thus outside the spheromak wall:
U = Up + Ue
= ([Bp(R, Z)]^2 / 2 Muo) + ([Epsilono / 2] [E(R, Z)]^2)

Inside the spheromak wall:
U = Ut(R)
= (Muo / 2) [Nt I / 2 Pi R]^2
= (Muo / 2) [Nt I / 2 Pi Ro]^2 [Ro / R]^2

At the spheromak wall:
Ut(R) = Up(R, Z) + Ue(R, Z)
or
([Bp(R, Z)]^2 / 2 Muo) + ([Epsilono / 2] [E(R, Z)]^2) = (Muo / 2) [Nt I / 2 Pi Ro]^2 [Ro / R]^2

The functions [Bp(R, Z)]^2 and [Ep(R, Z)]^2 are developed in terms of Ro on the web page titled Theoretical spheromak.

This equation must be simplified to find the equation that defines the spheromak wall.

At R = Rc, Z = 0:
Utc = Btc^2 / 2
= (Bpc^2 / 2) + (Epsilono / 2) Ec^2]
and at R = Rs, Z = 0
Uts = Bts^2 / 2
= (Bps^2 / 2) + (Epsilono / 2) Es^2]

Bto = Muo Nt / 2 Pi Ro
or
Uto = Muo Nt^2 / 8 Pi^2 Ro^2
and
Utc = [Muo Nt^2 / 8 Pi^2 Ro^2] [Ro^2 / Rc^2]
and
Uts = [Muo Nt^2 / 8 Pi^2 Ro^2] [Ro^2 / Rs^2]

ELEMENTARY SPHEROMAK:
An approximate plan view of the current path of a theoretical elementary spheromak with Np = 3 and Nt = 4 is shown below. The blue lines show the current path on the upper face of the spheromak. The red lines show current path on the lower face of the spheromak. Note that the current path never intersects itself except at the point where the current starts to retrace its previous path.

In the diagram yellow shows the region of toroidal magnetic field. Outside the yellow region is a poloidal magnetic field and a spherically radial electric field.

This elementary spheromak winding pattern was generated using a polar graph and formulae of the form:
R = Rc + K [t - t(2N)] where t(2N) = (2 N) (3 Pi / 4) and t(2N) < t < t(2N+1)
and
R = Rs - K [t - t(2N + 1)]
where:
t(2N + 1) = (2N + 1)(3 Pi / 4)
and
t(2N + 1) < t < t(2N + 2)
where:
N = 0, 1, 2, 3. Use:
Rc = 1000,
Rs = 4105,
K = (4140 / Pi)
Top to bottom connection points were depicted by adjusting the torus Rs to 4045 and Rc to 1060.
 

ATOMIC PARTICLE SPHEROMAKS:
Atomic particle spheromaks have a quantized charge that superficially appears to be at rest with respect to an inertial observer. Isolated stable atomic particles such as electrons and protons hold specific amounts of energy (rest mass). When these particles aggregate with opposite charged particles the assembly emits photons. This photon emission decreases the total amount of energy in the assembly, creating a mutual potential energy well.

In an atomic particle spheromak net charge moves uniformly at the speed of light around a closed spiral path. The spheromak net charge is uniformly distributed along this current path. The uniform charge distribution along the current path and the uniform current cause constant electric and magnetic fields. The time until an element of net charge retraces its previous path is (1 / Fh) where Fh is the characteristic frequency of the spheromak.

An isolated spheromak in free space has a distorted toroid cross section. However, the fields of an atomic particle spheromak may be further distorted by external electric and magnetic fields.
 

LOCATION IN A SPHEROMAK:
A spheromak has both cylindrical symmetry about its main axis of symmetry and has mirror symmetry about its equatorial plane. A position in a spheromak can be defined by:
(R, Z)
where:
R = radius from the main axis of cylindrical symmetry;
and
Z = height above (or below) the spheromak equatorial plane.
 

SPHEROMAK CROSS SECTIONAL DIAGRAM:
The following diagram shows the approximate cross sectional shape of a real spheromak.

Note that the cross secion of a real spheromak may not be round.

In this diagram the axis of symmetry is R = 0

At R = 0, Z = 0 the field energy density is maximum and is entirely due to the poloidal magnetic field.

Note that a spheromak in free space is quasi-toroidal.
 

GEOMETRICAL FEATURES OF A SPHEROMAK:
Important geometrical features of a spheromak include:
Rc = the spheromak wall inside radius on the equatorial plane;
Rs = the spheromak wall outside radius on the equatorial plane;
Ro - the imaginary ring radius on the equatorial plane;
Ho - thespheromak wall height above the equatorial plane at R = Ro;
Rm = the value of R at the spheromak end where: Z = Hm;
(2 |Hm|) = the overall spheromak length;
Np = number of poloidal charge filament turns about the major axis of symmetry;
Nt = number of toroidal charge filament turns about the minor axis in the spheromak;

The subscript c refers to spheromak wall inside radius (core) on the equatorial plane;
s refers to the spheromak wall outside radius on the equatorial plane.

In order to understand the material on this web page it is essential for the reader to study the spheromak cross sectional diagram and to identify the above mentioned parameters.

When an element of net charge moving along the spheromak filament has passed through the spheromak core hole Nt times it has also circled around the main axis of spheromak symmetry Np times, after which it reaches the point in its closed path where it originally started.

Define:
Lt = 2 Pi K Ro = one purely toroidal filament turn length;
Lp = 2 Pi Ro = one purely poloidal filament turn of length;
 

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 inner wall minimum radius is Rc and the spheromak's outer wall maximum radius is Rs.

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

The spheromak wall contains Nt quasi-toroidal turns equally spaced around 2 Pi radians about the main spheromak axis of symmetry.
Each purely toroidal winding turn has length:
Lt = 2 Pi K Ro
so the purely toroidal spheromak winding length is:
(Nt Lt)

The spheromak wall contains Np poloidal turns which are spaced around the spheromak wall perimeter. The spheromak field calculations are based on the poloidal turns being concentrated at R = Ro. The purely poloidal turn length is:
Lp = 2 Pi Ro
and the purely poloidal winding length is:
(Np Lp)

The total spheromak winding length Lh is given by:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2

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 on average:
(poloidal angle advance) / (toroidal angle advance) = Np / Nt

Note that Np and Nt cannot be equal except at Nt = Np = 1.
 

SPHEROMAK SHAPE:
The shape of the spheromak is in part defined by the ratio:
Nr = (Np / Nt).

The energy and frequency of a spheromak also involve Np and Nt.

There are Nt filament paths that go through the equatorial plane at the inside diameter of the spheromak and hence form the spheromak inner wall. There are Nt filament paths that go through the equitorial plane at the outside diameter of the spheromak wall and hence form the outer wall.
 

SPHEROMAK PARAMETER DEFINITIONS:
The remainder of this web page is primarily devoted to spheromak parameter definitions.

Define:
R = radial distance of a point from the major axis of symmetry of spheromak;
Z = axial distance of a point above the spheromak equatorial plane (Z is negative for points below the equatorial plane);
H = axial distance of a point on the spheromak wall above the spheromak equatorial plane (H is negative for points below the equatorial plane);
Ro = characteristic spheromak radius
Ho = H|(R = Ro)
(Rx, Zx) = a point located at R = Rx, Z = Zx
Hm = maximum value of |Z| on the spheromak wall
2 Hm = spheromak overall length
Ue = electric field energy density as a function of position outside the spheromak wall;
Up = magnetic field energy density as a function of position outside the spheromak wall;
U = Ue + Up + Ut = total field energy density as a function of position;
Upo = (Bpo^2 / 2 Muo) = maximum magnetic field energy density at R = 0, Z = 0;
Lp = 2 Pi Ro
Lt = 2 Pi K Ro = single turn toroidal winding length
Utc = toroidal field energy density at R = Rc, Z = 0
Zw = spheromak wall height at radius R = Rw
Ut = Uto (Ro / R)^2
= toroidal magnetic energy density function inside the spheromak wall
Uts = Uto (Ro / Rs)^2 = toroidal magnetic energy density at R = Rs
Utc = Uto (Ro / Rc)^2 = toroidal magnetic energy density at R = Rc
Ue|(R = 0) = radial electric field energy density on the spheromak axis as calculated herein;Up|(R = 0) = _________
= poloidal magnetic field energy density on the spheromak axis as calculated herein.
 

SPHEROMAK FILAMENT PARAMETERS
Define:
Ih = filament current;
Lh = overall length of closed filament loop;
Dh = center to center distance between adjacent filament paths
As = outside surface area of spheromak wall
Q = proton net charge
Qs = net charge on spheromak
Nnh = integer number of negative charge quanta
Nph = integer number of positive charge quanta
Vn = velocity of negative charge quanta along charge hose
Vp = velocity of positive charge quanta along charge hose
C = speed of light
Nr = Np / Nt
= ratio of two integers which have no common factors. This ratio must be inherently stable.
 

SPHEROMAK CHARGE DISTRIBUTION ASSUMPTION:
Assume that the spheromak charge is uniformly distributed over the filament length.
 

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 net electric field is zero;
For R < Rc the toroidal magnetic field is zero;
For R < Rc the magnetic field Bp is purely poloidal;
For R = 0 the magnetic field is parallel to the axis of cylindrical symmetry;

For Rc < R < Rs the electric field is zero;
For Rc < R < Rs the poloidal magnetic field is zero;
For Rc < R < Rs the toroidal magnetic field Bt is proportional to (1 / R).

For Rs < R in free space the electric field Ero is spherically radial;
For Rs < R in the far field the electric field Ero is proportional to (1 / R^2);
For Rs < R in free space the toroidal magnetic field is zero;
For Rs < R in the far field the poloidal magnetic field Bp is proportional to (1 / R^3);
 

FILAMENT CURRENT:
Ih = [Qp Np Vp + (- Q Ne Ve)] / Lh
 

SPHEROMAK FILAMENT WINDING GEOMETRY:
A very important issue in understanding natural spheromaks is grasping that:
Np cannot equal Nt and that Np and Nt can have no common factors other than one. Otherwise the windings would fall on top of one another or the spheromak would collapse.
 

SPHEROMAK CHARACTERISTIC FREQUENCY Fh:
A spheromak's characteristic frequency Fh is given by:
Fh = C / Lh
where:
C = speed of light.

This equation relates the change in spheromak frequency to a change in Lh assuming that the spheromak geometry is stable so that the parameters Np, and Nt remain constant. To find the Planck constant we need to find the change in spheromak energy with respect to a change in Ro.

The spheromak energy will have to be expressed in terms of the toroidal magnetic energy density:
Bt = Muo Nt I / (2 Pi R)
= Muo Nt Qs C / (2 Pi R Lh)
or
Ut = Bt^2 / 2 Muo
= [Muo Nt Qs C / (2 Pi R Lh)]^2 / 2 Muo
= [Nt Qs C / Pi R Lh]^2 / 8 Muo
which is the energy density as a function of R inside the spheromak wall.

To proceed we need to quantify the total spheromak energy and find how it changes with Rc.
 

SPHEROMAK REGIONS:
A spheromak has two main regions. The outside region in which field energy is calculated using vertical cylindrical elements. The toroidal region in which field energy is calculated using vertical cylinder elements.

In the core region the energy density is approximately set by the energy density at spheromak axis at (R = 0, Z). At (R = 0, Z = 0) the energy density is Upor. Hence an element of core energy is:
U(R, Z) 2 Pi R dR dZ.
In the adjacent toroidal region:
Bt = Muo Nt I / (2 Pi R)
so the field energy density at the inside wall is
Bt^2/2 Muo = (1 / 2 Muo)[Muo Nt I / 2 Pi R]^2

The energy density at the inside wall is:
U(R = 0, Z) = Uo (Rc / Rw)^2

An element of core energy is:
dE = U(R = 0, Z) Pi Rw^2 dZ
= Uo (Rc / Rw)^2 Pi Rw^2 dZ
= Uo Rc^2 Pi dZ

Thus:
Core field energy = 2 Uo Rc^2 Pi Hm

Bo = (Muo Np I 2 Pi Ro( / 4 Pi Ro^2
= Muo Np I / 2 Ro

Uo = (Bo^2 / 2 Muo)
= [ Muo Np I / 2 Ro]^2 [1 / 2 Muo]
= Muo Np^2 I^2 / 8 Ro^2

Thus core field energy = 2 Uo Rc^2 Pi Hm
= 2 [Muo Np^2 I^2 / 8 Ro^2] Rc^2 Pi Hm

Recall that:
I = Qs C / Lh
and
F = C / Lh
which gives:
Core field energy = 2 [Muo Np^2 I^2 / 8 Ro^2] Rc^2 Pi Hm
= 2 [Muo Np^2 / 8 Ro^2] Rc^2 Pi Hm Qs^2 C^2 / Lh^2

= 2 [Muo Np^2 / 8 Ro^2] Rc^2 Pi Hm Qs^2 C F/ Lh

= [Muo Qs^2 C F] [Rc^2 Pi Np^2 Hm / 4 Ro^2 Lh]

OUTER WALL BOUNDARY CONDITION AT R = Rs, Z = 0:
At R = Rs, Z = 0
Electric field energy density + Poloidal magnetic field energy density ~ toroidal energy density (Epsilono / 2)[Qs / 4 Pi Epsilono Rs^2]^2 + Bps^2 / 2 Muo = (Muo Nt I / 2 Pi Rs)^2 (1 / 2 Muo) I = Qs C / Lh (Epsilono / 2)[Qs / 4 Pi Epsilono Rs^2]^2 + Bps^2 / 2 Muo = (Muo Nt / 2 Pi Rs)^2 (1 / 2 Muo)[Qs C / Lh]^2
or
(1 / Epsilono)[Qs / 4 Pi Rs^2]^2 + Bps^2 / Muo = Muo (Nt / 2 Pi Rs)^2 [Qs C / Lh]^2
or
[Qs / 4 Pi Rs^2]^2 + Bps^2 Epsilono / Muo = (1 / C^2) (Nt / 2 Pi Rs)^2 [Qs C / Lh]^2
or
[1 / 4 Rs^2]^2 + [Bps^2 Epsilomo Pi^2 / Muo Qs^2] = (Nt / 2 Rs)^2 [1 / Lh]^2
 

Recall that:
Lh^2 = (Np 2 Pi Ro)^2 + (Nt 2 Pi K Ro)^2
giving:
Lh^2 = (2 Pi Ro)^2 [Np^2 + (K Nt)^2]

Finding K requires doing a surface integral.

OUTER WALL SHAPE:
Outside energy density is proportional to [1 / (R^2 + Z^2)^2]
Toroidal region energy density proportional to (1 / R^2)
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX On outside wall:
(K / Rw^2) = 1 / (Rw^2 + Zw^2)^2
At Rw = Rs, Zw = 0
K / Rs^2 = 1 / Rs^4
Hence:
K = (1 / Rs^2)

Thus on the outside wall:
1 / Rs^2 Rw^2 = 1 / (Rw^2 + Zw^2)^2
or
(Rw^2 + Zw^2)^2 = Rs^2 Rw^2
or
Rw^2 + Zw^2 = Rs Rw
or
Zw^2 = Rw (Rs - Rw)
or
Zw = [Rw (Rs-Rw)]^0.5
defines the shape of the outer wall.

By comparison, for a circle:
Z^2 + R^2 = Rs^2
or
Z^2 = Rs^2 - R^2
= Rs^2 - R Rs + R Rs -R^2
= Rs(Rs - R) + R (Rs - R)
since Rs > R the first term holds a significant volume an hence a significant amount of electricfield energy.

Find the peak in Z at Hm:
d[R(Rs-R)] / dR = -R + (Rs - R) = 0
Functional peak in Z at R = Rs / 2 = Rm
The corresponding Hm value is given by:
Hm^2 = (Rs / 2)(Rs - (Rs/ 2)) = (Rs / 2)^2
Hence:
Hm = (Rs / 2)
Check this value against the position of the inner wall. Hence the spheromak core energy for Z values up to Hm is:
Uo Pi Rc^2 Rs
 

We still have unquantified toroidal energy and end cap energy.

FIELD ENERGY IN THE CORE:
The field energy in the core is:
2 Uo Rc^2 Pi Hm
= Muo C^2 Qs^2 / (Ro^2)] {Nr^2 / (32 Rs^2)} {Rc^2 Pi Hm)
= Muo C^2 Qs^2 / Lh (Ro^2)] {Nr^2 / (32 Rs^2)} {Lh Rc^2 Pi Hm)
= Muo C Qs^2 F /(Ro^2)] {Nr^2 / (32 Rs^2)} {Lh Rc^2 Pi Hm)

= Muo C Qs^2 F] {Nr^2 Pi / (64 Rs Ro^2)} {Lh Rc^2)
 

From inner wall boundary condition at R = Rc, Z = 0:
Rc = (Nt / Np) Ro / Pi
= Ro / Pi Nr

Hence core energy becomes:
Muo C Qs^2 F] {Nr^2 Pi / (64 Rs Ro^2)} {Lh Rc^2)

Muo C Qs^2 F] {Nr^2 Pi / (64 Rs Ro^2)} {Lh Ro^2 / Nr^2 Pi^2)
= Muo C Qs^2 F] { 1 / (64 Rs)} {Lh / Pi)

Now apply the outer wall boundary condition at R = Rs, Z = 0 which is:
Lh = Nt 2 Rs
so that the spheromak core energy becomes:
Muo C Qs^2 F] { 1 / (64 Rs)} {Lh / Pi)

= Muo C Qs^2 F] { 1 / (64 Rs)} {Nt 2 Rs / Pi)

= [Muo C Qs^2] {Nt / (32 Pi)} F

Thus the core for -(Rs / 2) < Z < +(Rs / 2) contribution to h is:
[Muo C Qs^2] {Nt / (32 Pi)}
which suggests that Nt ~ 1000

CALCULATE THE ENERGY CONTRIBUTION OF THE OUTER PART OF THE TOROIDAL REGION

Calculate the cap energy which to a first approximation is the electric field energy trapped between a sphere of radius Rs and the surface at:
Z = Hm = Rs / 2
with fringes for R > Rs / 2. Use vertical cylinder elements. In most of the cap zone we can use U from the axial calculation. In the fringes we can use U from the radial calculation.
 

OUTER WALL BOUNDARY CONDITION AT R = Rs, Z = 0:
At R = Rs, Z = 0
Ignore poloidal magnetic field. Electric field energy density ~ toroidal energy density (Epsilono / 2)[Qs / 4 Pi Epsilono Rs^2]^2 = (Muo Nt I/ 2 Pi Rs)^2 (1 / 2 Muo) I = Qs C / Lh (Epsilono / 2)[Qs / 4 Pi Epsilono Rs^2]^2 = (Muo Nt / 2 Pi Rs)^2 (1 / 2 Muo)[Qs C / Lh]^2 or
(1 / Epsilono)[Qs / 4 Pi Rs^2]^2 = Muo (Nt / 2 Pi Rs)^2 [Qs C / Lh]^2 or
[Qs / 4 Pi Rs^2]^2 = (1 / C^2) (Nt / 2 Pi Rs)^2 [Qs C / Lh]^2 or
[1 / 4 Rs^2]^2 = (Nt / 2 Rs)^2 [1 / Lh]^2 or
Lh^2 = (Nt / 2 Rs)^2 [4 Rs^2]^2 or
Lh = (Nt / 2 Rs) 4 Rs^2
= Nt 2 Rs

Lh = Nt 2 Rs
 

INNER WALL BOUNDARY CONDITION AT (R = Rc, Z = 0):
Btc = Bpc
or
Muo Nt I / 2 Pi Rc = Muo Np I 2 Pi Ro / 4 Pi Ro^2
= Muo Np I / 2 Ro
or
Nt / Pi Rc = Np / Ro
or
Np / Nt = Ro / Pi Rc
 

WINDING CONDITION:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2
= (Np 2 Pi Ro)^2 + (Nt 2 Pi K Ro)^2
 

INNER WALL SHAPE:
For the inner wall:
At Z = 0 then R = Rc

It is shown herein that:
U|(R = 0, Z) = [Mu C^2 Qs^2 / (Ro^2 + Z^2)^3] {(Z^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}
giving:
Rw = Rc [Uo / U|(R = 0, Zw)]^0.5
= Rc [(Mu C^2 Qs^2 / Ro^6)(Ro^4 Np^2 /8 Lh^2)]^0.5 / [[Mu C^2 Qs^2 / (Ro^2 + Zw^2)^3] {(Zw^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}]^0.5

= Rc [Np^2 / 8 Ro^2 Lh^2]^0.5 / [1 / (Ro^2 + Zw^2)^3] {(Zw^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}]^0.5

= Rc [Np^2 / 8 Ro^2 Lh^2]^0.5 [(Ro^2 + Zw^2)^3]^0.5 / {(Zw^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}]^0.5

Check this function at Zw = 0:
Rw = Rc[Np^2 / 8 Ro^2 Lh^2]^0.5 [(Ro^2)^3]^0.5 / { (Ro^4 Np^2 / [8 Lh^2])}^0.5

= Rc[1 / Ro^2]^0.5 [(Ro^2)^3]^0.5 / { (Ro^4 )}^0.5
= Rc
as expected.

However, inverting the function Rw(Z)to find Z(Rw) is extremely difficult. What can be done is to find Rw at Zw:
Rw = Rc [Np^2 / 8 Ro^2 Lh^2]^0.5 [(Ro^2 + Zw^2)^3]^0.5 / {(Zw^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}]^0.5

or
Rw^2 = Rc^2 [Np^2 / Ro^2 Lh^2] [(Ro^2 + Zw^2)^3] / {(Zw^2 / 4 Pi^2) + (Ro^4 Np^2 / [Lh^2])}]

This expression needs computer evaluation.
 

SPHEROMAK SURFACE CHARGE DENSITY:
Charge filament current continuity means that Ih is everywhere constant for a particular spheromak. Force balance between adjacent charge filament turns causes the charge filament linear charge density:
Rhoh = [(Qp Nph + Qn Nnh) / Lh]
to be uniform everywhere on that spheromak.

The charge per unit area Sa on the spheromak surface is:
Sa = Rhoh / Dh
where Dh is the distance between adjacent plasma hoses. Note that the toroidal winding component causes Dh to vary over the spheromak surface.

The spheromak wall charge per unit area Sa is inversely proportional to Dh.
 

TOTAL SURFACE CHARGE:
The net charge per unit area Sa at any point on spheromak wall is:
Sa = (Rhoh / Dh)
where Dh is position dependent.

Recall that:
(Ih / C)^2 = Rhoh^2
Hence the local spheromak wall surface charge per unit area Sa is given by:
Sa = (Rhoh / Dh)
= [Ih / (Dh C)]
In this formula Ih is constant for a particular charge hose and hence for a particular spheromak wall formed from that charge hose. Hence:
Sa is proportional to (1 / Dh)

CHECK FROM HERE ON

Recall that:
Ih = [Qp Nph Vp + (Qn) Nnh Vn] [ 1 / Lh]
Hence at any particular point on a spheromak wall:
Sa^2 = [Ih / (Dh C)]^2
= [Qp Nph Vp + (Qn) Nnh Vn]^2 [ 1 / Lh]^2 / (Dh C)^2
= [Qp Nph Vp + (Qn) Nnh Vn]^2 [1 / (Lh Dh C)^2]
or
Sa^2 Dh^2 Lh^2 = [Qp Nph Vp + (Qn) Nnh Vn]^2 / C^2

Note that because Sa is proportional to (1 / Dh) both the left hand side and the right hand side of this equation are constant independent of position on the spheromak wall.

The total charge Qs on the spheromak wall is:
Integral from X = 0 to X = Lh of:
Sa(X) Dh(X) dX.
The product:
[Sa(X) Dh(X)]
is constant.

Hence:
Qs^2 = [Qp Nph Vp + (Qn) Nnh Vn]^2 / C^2
 

SPECIAL CASES:
For a plasma with Ve >> Vi and Nnh ~ Nph the equation for Qs simplifies as follows:
Qs^2 = [Q Ni Vi + (-Q) Ne Ve]^2 / C^2
~ [Q Ne Ve]^2 / C^2

For an atomic particle:
Qs^2 C^2 = [Qp Nph Vp + (Qn) Nnh Vn]^2
This equation indicates that a free atomic charged particle is simply a spheromak with charge filament current:
Ih = Qs C / Lh
 

NUCLEAR PARTICLE SPHEROMAKS:
Quark theory indicates that for the special case of a proton:
Nph = 2;
Nnh = 1;
Qp = (2 Q / 3);
Qn = (- Q / 3);
Qs = 2 Qp + Qn = Q;

This web page last updated June 17, 2024.