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XYLENE POWER LTD.

ELECTROMAGNETIC SPHEROMAK

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

ELECTROMAGNETIC SPHEROMAK:
Spheromak mathematics explains the existence of both stable charged particles and semi-stable toroidal plasmas.

In order to exist an electromagnetic spheromak should approximately satisfy the existence condition:
[Pi / 2]
= [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

A common boundary condition which shows the relationship between So^2 and the ratio of the numbers of spheromak current flow path poloidal turns Np to the number of current flow path toroidal turns Nt is given by:
Nr = (Np / Nt)
where:
Nr^2 = {(8 / Pi^2) - [(So^2 - 1) / (So^2 + 1)]^2} / {1 - (16 / [Pi (So^2 - 1)]^2)}

To find the spheromak operating point choose So^2, calculate Nr^2, and then calculate RHS of existence condition. Plot the RHS value versus So^2. Repeat this process to find the So^2 value that gives RHS the closest approach to LHS = (Pi / 2). Use the corresponding value of Nr to find integers Np and Nt.

Typical experimental plasma spheromaks have shape factors of about:
So^2 = 4.2
 

SPHEROMAK GEOMETRY:
The geometry of a spheromak can be characterized by the following parameters:
R = radius of a general point from the spheromak's axis of cylindrical symmetry;
Rc = spheromak's minimum core radius;
Rs = spheromak's maximum equatorial radius;
Rf = (Rs + Rc) / 2 = spheromak's top and bottom radius;
Rw = radius of co-axial cylindrical enclosure;
H = distance of a general point from the spheromak's equatorial plane;
Hs = height of a point on the spheromak wall above the spheromak's equatorial plane;
(2 |Hf|) = spheromak's overall length measured at R = Rf;
 

SPHEROMAK STRUCTURAL ASSUMPTIONS:
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.
3) At a particular spheromak energy the charge hose current Ih is constant;
4) The charge hose current causes a toroidal magnetic field inside the hose spiral and a poloidal magnetic field outside the hose spiral;
5) The net charge causes a cylindrically radial electric field inside the charge hose spiral and a spherically radial electric field outside the charge hose spiral;
6) At the center of the spheromak at (R= 0, H = 0) the electric field is zero;
7) Inside the spheromak wall where:
Rc < R < Rs and |H| < |Hs|
the total field energy density U takes the form:
Ut = Uto (Ro / R)^2
8) Outside the spheromak wall the total field energy density takes the form:
Up = Uo [Ro^2 / (K^2 Ro^2 + R^2 + H^2)]^2;
9) Everywhere at the spheromak surface:
Up = Ut and Up < Upo
 

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

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

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

In an experimental apparatus at R = Rs, the field energy density U must meet both the 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 H = 0 in a cylindrical metal enclosure the electric field Ero is cylindrically radial;
For Rs < R < Rw at H = 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 |H| < |Hs| the internal magnetic field Bti is toroidal;
For Rc < R < Rs and |H| < |Hs| the magnetic field Bti is proportional to (1 / R).

For Rc < R < Rf and H^2 > Hs^2 the external magnetic field is poloidal;
For Rf < R < Rs and H^2 > Hs^2 the poloidal magnetic field energy density is small compared to the spherically radial electric field energy density;
For Rc < R < Rs and H^2 < Hs^2 the electic field component parallel to the main axis of symmetry is zero;
For Rc < R < Rf and H^2 > Hf^2 the electric field is spherically radial;
For Rf < R the spherical electric field is proportional to (R^2 + H^2)^-1;
For Rc < R < Rf the spheromak end funnel area equals the end area of a spherical section with the same solid angle. This condition gives spherical radial electric field continuity outside the spheromak ends.
At R = Rf, H = Hf the spherical radial electric field Ef = Es, where Es is the cylindrically radial electric field at the spheromak equator where R = Rf, H = 0
 

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

DEFINITIONS:
Epsilon = permittivity of free space
Mu = permiability of free space
Bxyz and Exyz by:
B = magnetic field
E = electric field
x = p or t orr indicating a poloidal or toroidal or radial field
y = i or o indicating an inside region or outside region
z = c or s indicating core boundary or outside surface boundary
 

SPHEROMAK CHARGE HOSE PARAMETERS
Define:
Ih = charge hose current;
Lh = axial length of closed spiral of charge motion path;
Nt = Integer number of complete toroidal charge hose turns;
Np = Integer number of complete poloidal charge hose turns;
Lt = length of one purely toroidal charge motion turn
Lp = length of one purely poloidal plasma motion turn at R = Rf;
As = outside surface area of spheromak wall
Vi = ion velocity along charge motion path
Ve = electron velocity along charge motion path
Q = proton charge
Qs = net spheromak charge
Rhoh = charge per unit length along the charge motion path
C = speed of light
 

SPHEROMAK WALL THEORY REVIEW:
On the web page titled PLASMA SHEET 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.
 

ROLES OF SPHEROMAK ELECTRIC AND MAGNETIC FIELDS:
The electric and magnetic fields of a spheromak store energy and act in combination to position and stabilize the spheromak wall.
 

SPHEROMAK WALL:
Under the circumstances of plasma spheromak generation the electrons and ions form a PLASMA SHEET. The plasma sheet is the wall of a hollow torus. Inside the wall the magnetic field is toroidal. Outside the wall the magnetic field is poloidal.

The trapped electrons and ions circulate in the plane of the spheromak wall. The wall is a constant total energy closed path for the trapped charge(s).

The positive ions move opposite to the electrons to approximately balance charge and momentum within the plasma sheet. The electric field component normal to the plasma sheet slightly separates the electron and ion streams, which prevents the energetic electrons being scattered by collisions with the spheromak ions. Hence the electrons and ions follow similar but opposite flowing spiral paths within the plasma sheet.
 

ENERGY DENSITY BALANCE:
For a plasma sheet position to be stable the total field energy density must be the same on both sides of the plasma sheet. This requirement leads to boundary condition equations that determine the shape of spheromaks.
 

PLASMA SHEET POSITION:
A plasma spheromak is a semi-stable energy state. The plasma sheet forming the spheromak positions itself to achieve a total energy relative minimum. At every point on the plasma sheet the sum of the electric and magnetic field energy densities on one side of the plasma sheet equals the sum of the electric and magnetic field energy densities on the other side of the plasma sheet. This general statement resolves into different boundary conditions for different points on the plasma sheet. These boundary conditions establish the spheromak core radius Rc on the equatorial plane, the spheromak outside radius Rs on the equatorial plane and the spheromak length 2 Hf.
 

On the web page titled THEORETICAL SPHEROMAK it is shown that a theoretical spheromak is a stable structure. The question addressed on this web page is under what circumstances does an electromagnetic system form a stable spheromaks?
 

ENERGY DENSITY MATCHING:
Let Qa be the apparent net charge on a spheromak as indicated by an electric field measurement at:
(R^2 + H^2) >> Ro^2.
Note that most field measurements on electrons and protons are far field measurements.

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

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

In the far field:
(R^2 + H^2) >> K^2 Ro^2:

Hence for an electromagnetic spheromak in the far field:
U = Uo [Ro^2 / (K^2 Ro^2 + R^2 + H^2)]^2 ~ [Epsilon / 2][Qa / (4 Pi Epsilon (R^2 + H^2)]^2
or
Uo Ro^4 = [Epsilon / 2][Qa / (4 Pi Epsilon)]^2
or
Uo Ro^4 = [1 / 2 Epsilon][Qa / (4 Pi)]^2

For an electromagnetic spheromak:
Uo / K^4 = (Bpo^2 / 2 Mu)
giving:
(Bpo^2 / 2 Mu) K^4 Ro^4 = [1 / 2 Epsilon][Qa / (4 Pi)]^2
or
(Bpo^2 / 2 Mu) = [1 / 2 Epsilon][Qa / (4 Pi K^2 Ro^2)]^2
or
Bpo^2 = [Mu / Epsilon][Qa / (4 Pi K^2 Ro^2)]^2
= [Mu^2 C^2][Qa / (4 Pi K^2 Ro^2)]^2

Hence:
Bpo = [(Mu C Qa) / (4 Pi K^2 Ro^2)]

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

If the spheromak field energy distribution assumption is correct 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.

BIOT AND SAVART:
The law of Biot and Savart gives an element of axial magnetic field d(Bpo) at (0, 0) due to a current ring with radius R located at height H and having current dI as:
d(Bpo) = [(Mu / 4 Pi) dI 2 Pi R / (R^2 + H^2)] [R / (R^2 + H^2)^0.5]
= (Mu dI R^2 / (2 (R^2 + H^2)^1.5

 

POLOIDAL MAGNETIC FIELD AT THE CENTER OF A SPHEROMAK:
At every point on the charge hose:
Ih^2 = Ip^2 + It^2

dL^2 = [R d(Theta)]^2 + [(Rs - Rc) d(Phi)/ 2]^2

Ip / Ih = R d(Theta) / dL
= R d(Theta) / {[R d(Theta)]^2 + [(Rs - Rc) d(Phi)/ 2]^2}^0.5
= R d(Theta) / {[R d(Theta)]^2 + [d(Theta)(Rs - Rc) d(Phi)/ 2 d(Theta)]^2}^0.5
= R / {[R]^2 + [(Rs - Rc) Nt / 2 Np]^2}^0.5
= Np R / {[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5

Thus:
Ip = Ih Np R / {[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5

The length of a poloidal current hose element is:
R d(Theta)

In a 360 degree current ring there are Nt such poloidal current elements.

Allowing for poloidal magnetic field contributions from both the upper and lower halves of the spheromak in effect there are (2 Nt) such current hose elements.

The distance from the poloidal current element to the center of the spheromak is:
[Hs^2 + R^2]^0.5

cos(A) = [R / (Hs^2 + R^2)^0.5]

The contribution to the poloidal magnetic field at the center of the spheromak due to the (2 Nt) such elements in the matching rings in the upper and lower halves of the spheromak is:
dBpo = (2 Nt)(Mu / 4 Pi) Ih [Ip / Ih][R d(Theta)] [1 / (Hs^2 + R^2)] cos(A)
= (2 Nt)(Mu / 4 Pi) Ih [Np R / {[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5][R d(Theta)][1 / (Hs^2 + R^2)][R / (Hs^2 + R^2)^0.5]
= (2 Nt)(Mu / 4 Pi) (Ih Np R^3 d(Theta)) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Hs^2 + R^2]^1.5)

Hence:
dBpo / d(Phi) = [dBpo / d(Theta)] [d(Theta) / d(Phi)]
= [dBpo / d(Theta)] [Np / Nt]
= (2 Nt)(Mu / 4 Pi) (Ih Np R^3)(Np / Nt) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Hs^2 + R^2]^1.5)
= (2 Np)(Mu / 4 Pi) (Ih Np R^3) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Hs^2 + R^2]^1.5)

Thus:
Bpo = integral from Phi = 0 to Phi = Pi of:
(2 Np)(Mu / 4 Pi) (Ih Np R^3) d(Phi) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Hs^2 + R^2]^1.5)

Recall that:
Hs^2 = (Rs - R)(R - Rc)
= (Rs R - Rs Rc - R^2 + R Rc)

Hence:
Bpo = Integral from Phi = 0 to Phi = Pi of:
(Mu / 2 Pi) Ih R^3 Np^2 d(Phi) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Rs R - Rs Rc + R Rc]^1.5)

Now find d(Phi) in terms of R:
R = [(Rs + Rc) / 2] - [(Rs - Rc) / 2] cos (Phi)
giving:
dR = [(Rs - Rc) / 2] sin (Phi) d(Phi)

Thus:
dR = [(Rs - Rc) / 2] sin (Phi) d(Phi)

= [(Rs - Rc) / 2] {[Hs / [(Rs - Rc) / 2]} d(Phi)
= Hs d(Phi)

Hence:
d(Phi) = dR / Hs
= {dR / [(Rs - R)(R - Rc)]^0.5}

Thus:
Bpo = Integral from Rc to Rs of:
(Mu / 2 Pi) Ih R^3 Np^2 d(Phi) / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Rs R - Rs Rc + R Rc]^1.5)
 
= Integral from Rc to Rs of:
(Mu / 2 Pi) Ih R^3 Np^2 {dR / [(Rs - R)(R - Rc)]^0.5} / ({[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Rs R - Rs Rc + R Rc]^1.5)
 
= Integral from Rc to Rs of:
(Mu / 2 Pi) Ih R^3 Np^2 dR / ([(Rs - R)(R - Rc)]^0.5 {[Np R]^2 + [(Rs - Rc) Nt / 2]^2}^0.5 [Rs R - Rs Rc + R Rc]^1.5)

 
= Integral from Rc to Rs of:
(Mu / 2 Pi) (1 / Rc) Ih (R / Rc)^3 d(R / Rc) Np (Np / Nt)
/ ([((Rs / Rc) - (R / Rc))((R / Rc) - 1)]^0.5 {[(Np / Nt) (R / Rc)]^2 + [((Rs / Rc) - 1) / 2]^2}^0.5
/ [(Rs R / Rc^2) - (Rs / Rc) + (R / Rc)]^1.5)

We can make this integral appear simpler using the substituions:
Z = (R / Rc)
Zo = So^2 = (Rs / Rc)
Nr = (Np / Nt)

Then:
Bpo = Integral from Z = 1 to Z = Zo = Rs / Rc = So^2 of:
(Mu / 2 Pi Rc) Ih Z^3 dZ Np Nr
/ ([(Zo - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(Zo - 1) / 2]^2}^0.5 [Zo Z - Zo + Z]^1.5)

or
Bpo = (Mu Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

which gives the central magnetic field strength for any spheromak.
 

SPECIAL CASE OF So^2 ~ 1:
The special case of So^2 ~ 1 should simplify to the case of a simple current ring. For this special case:
Bpo = (Mu Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 
= (Mu Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 [Nr] )
 
(Mu Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
dZ / ([-Z^2 + (So^2 + 1) Z - So^2]^0.5)
which from 380.001 becomes:
= - 1 arc sin[(-2 Z + (So^2 + 1)) / ((So^2 + 1)^2 - 4 So^2)^0.5|Z = So^2
+ 1 arc sin[(-2 Z + (So^2 + 1)) / ((So^2 + 1)^2 - 4 So^2)^0.5|Z = 1
 
= - arc sin[(-2 So^2 + (So^2 + 1)) / ((So^2 + 1)^2 - 4 So^2)^0.5
+ arc sin[(-2 + (So^2 + 1)) / ((So^2 + 1)^2 - 4 So^2)^0.5
 
= - arc sin[(1 - So^2) / ((So^2 - 1)^2)^0.5
+ arc sin[((So^2 - 1)) / ((So^2 - 1)^2)^0.5
 
= Pi

Hence for this special case:
Bpo = (Mu Np / 2 Pi) (Ih / Rc) Pi
(Mu Np / 2) (Ih / Rc)

From Biot and Savart the magnetic field at the center of a current ring with Np turns is:
(Mu / 4 Pi) Ih Np 2 Pi Rc / Rc^2 = (Mu / 2) Ih Np / Rc

Hence the integral simplifies as expected for this special case.
 

FIND Ih:
Ih = Qs C / Lh = Qs C / [(2 Pi Np (Rs + Rc) / 2)^2 + (2 Pi Nt (Rs - Rc) / 2)^2]^0.5
= Qs C / {Pi [(Np (Rs + Rc))^2 + (Nt (Rs - Rc))^2]^0.5}
= Qs C / {Pi Rc [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2]^0.5}
= [Qs C / {Pi Rc Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}]
= [Qs C Ro / {Pi Rc Ro Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}]
= [Qs C So / {K Pi Ro Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}]
= [(Qs C) / (K Pi Ro Nt)] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]

Hence:
(Ih / Rc) = (1 / Rc) [(Qs C) / (K Pi Ro Nt)] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]
= [(Qs C) / (K^2 Pi Ro^2 Nt)] [So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]
 

Hence:
Bpo = (Mu Np / 2 Pi) (Ih / Rc) Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 
= (Mu Np / 2 Pi) [(Qs C) / (K^2 Pi Ro^2 Nt)] [So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 
= (Mu / 2 Pi) [(Qs C) / (K^2 Pi Ro^2)] [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

This value of Bpo should equal the value of Bpo calculated from the far field condition which gives:
Bpo = [(Mu C Qa) / (4 Pi K^2 Ro^2)]
 

ELECTROMAGNETIC SPHEROMAK EXISTENCE CONDITION:
Hence:
[(Mu C Qa) / (4 Pi K^2 Ro^2)]
= (Mu / 2 Pi) [(Qs C) / (K^2 Pi Ro^2)] [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
and cancelling terms an LHS and RHS gives:
[1 / (4)]
= (1 / 2 Pi) [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
or
[Pi / 2]
= [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

which is referred to herein as the electromagnetic spheromak existence condition. This equation imposes an important constraint on any electromagnetic assembly forming a spheromak.

A critical part of spheromak analysis is finding the So^2 and Nr^2 values that simultaneously meet both the above spheromak existence condition and also meet the common boundary condition that connects Nr and So.
 

ANGULAR PROGRESSION:
Assume that the cross section of a spheromak in free space is round. This assumption is justified at THEORETICAL SPHEROMAK.
Let Theta = angle about the spheromak major axis
Let Phi = angle about spheromak minor axis.

As the charge moves over length Lh Theta increments by (2 Pi Np)

As the charge moves over length Lh Phi increments by (2 Pi Nt)

At every point on the spheromak surface:
d(Theta) /d(Phi)
= (2 Pi Np) / 2 Pi Nt)
= (Np / Nt)
 

CHARGE PER UNIT AREA AT R= Rc, H = 0:
On the equatorial plane at the core at R = Rc, H = 0:
Hose rise = (Rs - Rc) d(Phi) / 2
Hose run = Rc d(Theta)

Hose slope = (Rise) / (Run)
= [(Rs - Rc) d(Phi)] / [2 Rc d(Theta)]
= [(Rs - Rc) Nt] / [2 Rc Np]

In the core the distance between adjacent hoses measured along the inner equator is (2 Pi Rc / Nt).

Geometry shows that in the inner core the distance Dhc between hoses perpendicular to the hose axis is:
Dhc = (2 Pi Rc / Nt) [(Rs - Rc) d(Phi) / 2] / {[(Rs - Rc) d(Phi) / 2]^2 + [Rc d(Theta)]^2}^0.5
= (2 Pi Rc / Nt) [(Rs - Rc) d(Phi) / 2] / [d(Theta){[(Rs - Rc) d(Phi) / 2d(Theta)]^2 + [Rc]^2}^0.5]
= (2 Pi Rc / Nt) [(Rs - Rc) Nt / 2] / [Np {[(Rs - Rc) Nt / 2 Np]^2 + [Rc]^2}^0.5]
= (Pi Rc) [(Rs - Rc)] / {[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5

Hence the charge per unit area Sac at R = Rc, H = 0 is:
Sac = Qs / (Lh Dhc)
= (Qs / Lh){[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5 / [(Pi Rc) (Rs - Rc)]}

Recall that:
Lh = {[2 Pi Np (Rs + Rc) / 2]^2 + [2 Pi Nt (Rs - Rc) / 2]^2}^0.5
= Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5

Hence the charge per unit area at R = Rc, H = 0 is:
Sac = Qs / (Lh Dhc)
= (Qs / Lh){[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5 / [(Pi Rc) (Rs - Rc)]}
= (Qs / Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5) {[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5 / [(Pi Rc) (Rs - Rc)]}
= {Qs / [(Pi^2 Rc) (Rs - Rc)]} {[(Rs - Rc) Nt / 2]^2 + [Np Rc]^2}^0.5 / {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5
= {Qs / [(Pi^2 Rc^2) (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5

Hence:
Sac^2 = {Qs So^2 / [Pi^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 

CHARGE PER UNIT AREA AT R = Rs, H = 0:
On the equatorial plane at the outer perimeter where R = Rs, H = 0:
Hose rise = (Rs - Rc) d(Phi) / 2
Hose run = Rs d(Theta)

Hose slope = (Rise) / (Run)
= [(Rs - Rc) d(Phi)] / [2 Rs d(Theta)]
= [(Rs - Rc) Nt] / [2 Rs Np]

On the outer rim the distance between adjacent hoses measured along the outer equator is (2 Pi Rs / Nt).

Geometry shows that at the outer rim the distance Dhs between hoses perpendicular to the hose axis is:
Dhs = (2 Pi Rs / Nt) [(Rs - Rc) d(Phi) / 2] / {[(Rs - Rc) d(Phi) / 2]^2 + [Rs d(Theta)]^2}^0.5
= (2 Pi Rs / Nt) [(Rs - Rc) d(Phi) / 2] / {[d(Theta)]{[(Rs - Rc) d(Phi) / 2 d(Theta)]^2 + [Rs]^2}^0.5}
= (2 Pi Rs / Nt) [(Rs - Rc) Nt / 2] / {[Np]{[(Rs - Rc) Nt / 2 Np]^2 + [Rs]^2}^0.5}
= (Pi Rs) [(Rs - Rc)] / {Np {[(Rs - Rc) Nt / 2 Np]^2 + [Rs]^2}^0.5}
= (Pi Rs) [(Rs - Rc)] / {[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5

Hence the charge per unit area Sas at R = Rs, H = 0 is:
Sas = Qs / (Lh Dhs)
= (Qs / Lh){[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5 / [(Pi Rs) (Rs - Rc)]

Recall that:
Lh = {[2 Pi Np (Rs + Rc) / 2]^2 + [2 Pi Nt (Rs - Rc) / 2]^2}^0.5
= [Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5]

Hence:
Sas = Qs / (Lh Dhs)
= (Qs / Lh){[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5 / [(Pi Rs) (Rs - Rc)]
= (Qs {[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5 / {[Pi Rs (Rs - Rc)] Pi {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5}
= {Qs / [Pi^2 Rs (Rs - Rc)]} {[(Rs - Rc) Nt / 2]^2 + [Np Rs]^2}^0.5 / {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5
= {Qs / [Pi^2 Rs Rc (So^2 - 1)]} {[(So^2 - 1) Nt / 2]^2 + [Np So^2]^2}^0.5 / {[Np (So^2 + 1)]^2 + [Nt (So^2 - 1)]^2}^0.5
= {Qs / [Pi^2 Ro^2 (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5

Hence:
Sas^2 = {Qs / [Pi^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr So^2]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
 

SURFACE CHARGE RATIO:
A quantity of interest is the surface charge ratio (Sas / Sac). From above this ratio is given by:
(Sas / Sac) = {Qs / [Pi^2 Ro^2 (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5 / {Qs So^2 / [(Pi^2 Ro^2) (So^2 - 1)]} {[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}^0.5
 
= {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {So^2} {[(So^2 - 1) / 2]^2 + [Nr]^2}^0.5
 
= {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[So^2 (So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5

Note that for large Nr this ratio simplifies to:
(Sas / Sac) = {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[So^2 (So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5
= {[Nr So^2]^2}^0.5 / {[Nr So^2]^2}^0.5
= 1

Note that for small Nr this ratio simplifies to:
(Sas / Sac) = {[(So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5 / {[So^2 (So^2 - 1) / 2]^2 + [Nr So^2]^2}^0.5
= {[(So^2 - 1) / 2]^2}^0.5 / {[So^2 (So^2 - 1) / 2]^2}^0.5
= (1 / So)
 

Recall that:
Sac^2 = {Qs / [Pi^2 Rc (Rs - Rc)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

Hence:
[Epsilon Eric^2] = (1 / Epsilon) Sac^2
 

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

FIELD ENERGY DENSITY BALANCE:

In general, neglecting gravitation and kinetic energy:
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 outside the spheromak wall:
Bt = 0
Hence at a spheromak wall:
{[[Bt^2 / 2 Mu] + [(Epsilon / 2) Er^2]}inside
= {[Bp^2 / 2 Mu] + [(Epsilon / 2) Er^2]}outside

This energy density balance (force balance) condition is valid for every element of area dA on the spheromak wall.
 

BOUNDARY CONDITION AT R= Rc, H = 0:
On the equatorial plane at the junction between the core region and the toroidal region field energy density balance gives:
[(Bpoc^2 + Btoc^2) / Mu] + [Epsilon Eroc^2] = [Bpic^2 + Btic^2] / Mu + [Epsilon Eric^2]

In the central core at the spheromak equator the magnetic field is poloidal and the electric field is zero. On the other side of the core wall the magnetic field is purely toroidal and the electric field is cylindrically radial. The field energy density on both sides of this wall must be equal.

However, for a spheromak at R = Rc, H = 0 there are the following important conditions:
Eroc = 0 or no electric field in core
and
Btoc = 0 or no toroidal magnetic field in core
and
Bpic = 0 or no poloidal magnetic field in toroidal region
giving the simplified boundary condition at R = Rc, H = 0 as:
[Bpoc^2 / Mu] = [Btic^2 / Mu] + [Epsilon Eric^2]
or
(Bpoc^2 / 2 Mu) = (Btic^2 / 2 Mu) + (Sac^2 / 2 Epsilon)
or
(Epsilon Bpoc^2 / Mu) = (Epsilon Btic^2 / Mu) + (Sac^2)
or
(Bpoc^2 / Mu^2 C^2) = (Btic^2 / Mu^2 C^2) + (Sac^2)
 

FIND Bpoc:
In the core of the spheromak the electric field is zero. Hence Uo is entirely due to the poloidal magnetic field.
Recall that:
Rs Rc = K^2 Ro^2

At R = 0, H = 0:
Bpo^2 / 2 Mu = Uo / K^4

(Bpoc^2 / 2 Mu) = Uo [Ro^2 / (K^2 Ro^2 + Rc^2)]^2
= [K^4 Bpo^2 / 2 Mu][Ro^2 / (K^2 Ro^2 + Rc^2)]^2
= [Bpo^2 / 2 Mu][K^2 Ro^2 / (K^2 Ro^2 + Rc^2)]^2
= [Bpo^2 / 2 Mu][Rs Rc / (Rs Rc + Rc^2)]^2
= [Bpo^2 / 2 Mu][Rs / (Rs + Rc)]^2

Thus:
Bpoc^2 = Bpo^2 [Rs / (Rs + Rc)]^2
 

FIND Bttc:
From basic electromagnetic theory:
Btic = Mu Nt Ih / 2 Pi Rc
or
Btic^2 / Mu = Mu [Nt Ih / 2 Pi Rc]^2
 

FIND Eric:
From basic electromagnetic theory:
Eric = Sac / Epsilon

Thus substitution of the above partial results into the boundary condition at R = Rc, H = 0 gives:
[Bpoc^2 / Mu] = [Btic^2 / Mu] + [Epsilon Eric^2]
or
[Bpo^2 / Mu] [Rs / (Rs + Rc)]^2
= Mu [Nt Ih / 2 Pi Rc]^2
+ (1 / Epsilon){Sac^2)
or
[Bpo^2 / Mu] Pi^2 [Rs / (Rs + Rc)]^2
= Mu [Nt Ih / 2 Rc]^2
+ (1 / Epsilon){Pi Sac)^2
or
Epsilon [Bpo^2 / Mu] Pi^2 [Rs / (Rs + Rc)]^2
= Epsilon Mu [Nt Ih / 2 Rc]^2
+ [Pi^2 Sac^2]

Recall that:
C^2 = 1 / (Mu Epsilon)
giving:
[Bpo^2 / (Mu^2 C^2)] Pi^2 [Rs / (Rs + Rc)]^2
= [Nt Ih / (2 Rc C)]^2
+ {Pi^2 Sac^2]
 

FIND Ih:
From charge hose theory:
Ih = (Qs / Lh) C
= Qs C / [(2 Pi Np (Rs + Rc) / 2)^2 + (2 Pi Nt (Rs - Rc) / 2)^2]^0.5
= Qs C / {Pi [(Np (Rs + Rc))^2 + (Nt (Rs - Rc))^2]^0.5}
= Qs C / {Pi Rc [(Np (So^2 + 1))^2 + (Nt (So^2 - 1))^2]^0.5}
= [Qs C / {Pi Rc Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}]
 

SUBSTITUTION FOR Ih
Thus the boundary condition at R = Rc, H = 0 becomes:
[Bpo Rs / (Rs + Rc)]^2 / (Mu^2 C^2) = [(Ih Nt / 2 Pi Rc C)^2] + (Sac^2)
or
[Bpo Rs / (Rs + Rc)]^2 / (Mu^2 C^2) = [Ih^2][(Nt / 2 Pi Rc C)^2] + (Sac^2)
or
[Bpo Rs / ((Rs + Rc) Mu C)]^2 = {Qs C / (Pi Rc Nt [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5)}^2 [(Nt / 2 Pi Rc C)^2] + (Sac^2)
or
[Bpo Rs / ((Rs + Rc) Mu C)]^2 = [(Qs C)^2 / ((Pi Rc Nt)^2 [(Nr (So^2 + 1))^2 + (So^2 - 1)^2])] [(Nt / 2 Pi Rc C)^2] + (Sac^2)
or
(Bpo (Rs / (Rs + Rc) Mu C)^2 = [Qs^2 / (4 Pi^4 Rc^4 [(Nr (So^2 + 1))^2 + (So^2 - 1)^2])] + (Sac^2)
or
Bpo^2 = (1 + (Rc / Rs))^2 Qs^2 Mu^2 C^2 / (4 Pi^4 Rc^4 [(Nr (So^2 + 1))^2 + (So^2 - 1)^2])
+ Mu^2 C^2 (1 + (Rc / Rs))^2 (Sac^2)

Recall that:
Sac^2 = {Qs / [Pi^2 Rc^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

Hence the boundary condition at R = Rc, H = 0 gives:
[Bpo^2 / (Mu^2 C^2)] Pi^2 [Rs / (Rs + Rc)]^2
= [Nt Ih / (2 Rc C)]^2
+ {Pi^2 Sac^2]
or
[Bpo^2 / (Mu^2 C^2)] Pi^2 [Rs / (Rs + Rc)]^2
= [Nt Ih / (2 Rc C)]^2
+ {Qs / [Pi Rc (Rs - Rc)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
or
[Bpo^2 / (Mu^2 C^2)] Pi^2 [Rs / (Rs + Rc)]^2
= [Nt Qs C / (Lh 2 Rc C)]^2
+ {Qs / [Pi Rc (Rs - Rc)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

Recall that from far field energy density function matching:
Uo Ro^4 / (R^2 + H^2)^2 = (Epsilon / 2) [Qa / 4 Pi Epsilon)^2] [(1 / (R^2 + H^2)]^2
or
Uo Ro^4 = (Epsilon / 2) [Qa / 4 Pi Epsilon)^2]
or
Uo = (1 / 2 Epsilon)[Qa / 4 Pi Ro^2)]^2
= (Mu C^2 / 2)[Qa / 4 Pi Ro^2)]^2

Recall that:
Uo / K^4 = Bpo^2 / 2 Mu

Thus:
Bpo^2 = 2 Mu (Uo / K^4)
= (2 Mu / K^4) (Mu C^2 / 2)[Qa / 4 Pi Ro^2)]^2
= (1 / K^4) (Mu^2 C^2)[Qa / 4 Pi Ro^2)]^2

Hence:
[Bpo^2 / (Mu^2 C^2)] Pi^2 [Rs / (Rs + Rc)]^2
= [Nt Qs C / (Lh 2 Rc C)]^2
+ {Qs / [Pi Rc (Rs - Rc)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
becomes:
[Mu^2 C^2][Qa / (4 Pi K^2 Ro^2)]^2 / (Mu^2 C^2)] Pi^2 [Rs / (Rs + Rc)]^2
= [Nt Qs / (Lh 2 Rc)]^2
+ {Qs / [Pi Rc (Rs - Rc)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
or
{[Qa / (4 K^2 Ro^2)]^2} [So^2 / (So^2 + 1)]^2
= [Nt Qs C So / (Lh 2 K Ro C)]^2
+ {Qs So^2 / [Pi K^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
or
{[Qa / (4 K^2 Ro^2)]^2} [So^2]^2 [1 / (So^2 + 1)]^2
= [Nt Qs So / (Lh 2 K Ro)]^2
+ {Qs So^2 / [Pi K^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

or
{[Qa / (4 K Ro^2)]^2} [So^2] [1 / (So^2 + 1)]^2
= [Nt Qs / (Lh 2 Ro)]^2
+ {Qs So / [Pi K Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

Recall that:
Lh^2 = [(2 Pi Np (Rs + Rc) / 2)^2 + (2 Pi Nt (Rs - Rc) / 2)^2]
= [(Pi Np (Rs + Rc))^2 + (Pi Nt (Rs - Rc))^2]
= Pi^2 [(Np (Rs + Rc))^2 + (Nt (Rs - Rc))^2]
= Pi^2 Nt^2 [(Nr (Rs + Rc))^2 + ((Rs - Rc))^2]
= Pi^2 Nt^2 Rc^2 [(Nr (So^2 + 1))^2 + ((So^2 - 1))^2]

Hence:
[1 / Lh^2] = 1 / {Pi^2 Nt^2 Rc^2 [(Nr (So^2 + 1))^2 + ((So^2 - 1))^2]}

{[Qa / (4 K Ro^2)]^2} [So^2] [1 / (So^2 + 1)]^2
= [Nt Qs / (Lh 2 Ro)]^2
+ {Qs So / [Pi K Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
becomes:
{[Qa / (4 K Ro^2)]^2} [So^2] [1 / (So^2 + 1)]^2
= [Nt Qs / (2 Ro)]^2 [1 / {Pi^2 Nt^2 Rc^2 [(Nr (So^2 + 1))^2 + ((So^2 - 1))^2]}]
+ {Qs So / [Pi K Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
or
{[Qa / (4 K Ro^2)]^2} [So^2] [1 / (So^2 + 1)]^2 [(Nr (So^2 + 1))^2 + ((So^2 - 1))^2] = [Nt Qs / (2 Ro)]^2 [1 / {Pi^2 Nt^2 Rc^2}]
+ {Qs So / [Pi K Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
or
{[Qa / (4)]^2} [1 / (So^2 + 1)]^2 [(Nr (So^2 + 1))^2 + ((So^2 - 1))^2] = [Nt Qs / (2)]^2 [1 / {Pi^2 Nt^2}]
+ {Qs / [Pi (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
or
[Qa / (4)]^2 Nr^2 + [Qa / (4)]^2} [(So^2 - 1) / (So^2 + 1)]^2
= [Qs]^2 [1 / {4 Pi^2}]
+ {Qs / [2 Pi]}^2 + Qs^2 Nr^2 / [Pi (So^2 - 1)]^2
or
[Qa / (4)]^2 Nr^2 - Qs^2 Nr^2 / [Pi (So^2 - 1)]^2
= [Qs]^2 [2 / {4 Pi^2}] - [Qa / (4)]^2} [(So^2 - 1) / (So^2 + 1)]^2
or
Nr^2 = {[Qs]^2 [2 / {4 Pi^2}] - [Qa / (4)]^2} [(So^2 - 1) / (So^2 + 1)]^2}
/ {[Qa / (4)]^2 - Qs^2 / [Pi (So^2 - 1)]^2}
 
= {[2 / {4 Pi^2}] - [1 / (4)]^2} [(So^2 - 1) / (So^2 + 1)]^2}
/ {[1 / (4)]^2 - 1 / [Pi (So^2 - 1)]}^2}
= [8 / Pi^2 - [(So^2 - 1) / (So^2 + 1)]^2} / {1 - (16 / [Pi (So^2 - 1)]^2)}

Thus the boundary condition at R = Rc, H = 0 is:
Nr^2 = [8 / Pi^2 - [(So^2 - 1) / (So^2 + 1)]^2} / {1 - (16 / [Pi (So^2 - 1)]^2)}

This expression, which is the boundary condition at R = Rc, H = 0 relates the spheromak shape factor So^2 to Nr^2.
 

BOUNDARY CONDITION AT R = Rs, H = 0:
(Epsilon / 2) (Fec Rc / Rs)^2 + (1 / 2 Mu) (Btic Rc / Rs)^2 = (Epsilon / 2) Eros^2 + (Bpos^2 / 2 Mu)
where:
Eric = Sac / Epsilon
and
Eros = (Sas / Epsilon) + (Fec Rc / Rs)

where from the web page titled: PLASMA SHEET PROPERTIES Sas is given by:
Sas^2 = {Qs / [Pi^2 Ro^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr So^2]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

This boundary condition can be simplified by noting that:
[(Epsilon / 2) Eros^2 + (Bpos^2 / 2 Mu)]
is the total field energy density at R = Rs, H = 0 outside the spheromak wall. Outside the spheromak wall the total field energy density is given by:
U = Uo [Ro^2 / (K^2 Ro^2 + R^2 + H^2)]^2.

At R = Rs, H = 0 the field energy density outside the spheromak wall is:
U = Uo [Ro^2 / (K^2 Ro^2 + Rs^2)]^2
= K^4 (Bpo^2 / 2 Mu) [Ro^2 / (K^2 Ro^2 + Rs^2)]^2
= (Bpo^2 / 2 Mu)[K^2 Ro^2 / (K^2 Ro^2 + Rs^2)]^2
= (Bpo^2 / 2 Mu)[Rs Rc / (Rs Rc + Rs^2)]^2
= (Bpo^2 / 2 Mu)[Rc / (Rc + Rs)]^2

Thus the boundary condition at R = Rs, H = 0 becomes:
(Epsilon / 2) (Fec Rc / Rs)^2 + (1 / 2 Mu) (Btic Rc / Rs)^2 = (Bpo^2 / 2 Mu) [Rc / (Rc + Rs)]^2
or
(Epsilon / 2) (Sac Rc / Epsilon Rs)^2 + (1 / 2 Mu) (Btic Rc / Rs)^2 = (Bpo^2 / 2 Mu) [Rc / (Rc + Rs)]^2
or
(1 / 2 Epsilon) (Sac Rc / Rs)^2 + (1 / 2 Mu) (Btic Rc / Rs)^2 = (Bpo^2 / 2 Mu) [Rc / (Rc + Rs)]^2
or
(Sac Rc / Rs)^2 + (Epsilon / Mu) (Btic Rc / Rs)^2 = (Epsilon) (Bpo^2 / Mu) [Rc / (Rc + Rs)]^2
or
(Sac Rc / Rs)^2 + (Btic^2 / C^2 Mu^2) (Rc / Rs)^2 = (Bpo^2 / C^2 Mu^2) [Rc / (Rc + Rs)]^2
or
(Btic^2 / C^2 Mu^2) (Rc / Rs)^2 = (Bpo^2 / C^2 Mu^2) [Rc / (Rc + Rs)]^2 - (Sac Rc / Rs)^2
or
(Btic^2 / C^2 Mu^2) = (Bpo^2 / C^2 Mu^2) [Rs / (Rc + Rs)]^2 - (Sac Rs / Rs)^2

= (Bpo^2 / C^2 Mu^2) [So^2 / (So^2 + 1)]^2 - (Sac)^2

Recall that:
Btic^2 = (Mu Nt Ih / 2 Pi Rc)^2
= (Mu Nt Qs C / Lh 2 Pi Rc)^2
= (Mu Nt Qs C / 2 Pi Rc)^2 (1 / Pi^2)[1 / {[Np (Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}]
= (Mu Nt Qs C / 2 Pi Rc)^2 [1 / (Pi^2 Nt^2 Rc^2)] [1 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}]
= (Mu Qs C / 2 Pi^2 Rc^2)^2 [1 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}]

Thus the boundary condition at R = Rs, H = 0 becomes:
(Mu Qs C / 2 Pi^2 Rc^2 C Mu)^2 [1 / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}]
= (Bpo^2 / C^2 Mu^2) [So^2 / (So^2 + 1)]^2 - (Sac)^2
= (Bpo^2 / C^2 Mu^2) [So^2 / (So^2 + 1)]^2
- {Qs / [Pi^2 Rc^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2} / {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}

 

Multiply through by (Pi^2 / Qs)^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}. Then this outer rim boundary condition becomes:
(1 / 2 Rc^2)^2 = (Bpo^2 / C^2 Mu^2) [So^2 / (So^2 + 1)]^2 (Pi^2 / Qs)^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
- {1 / [Rc^2 (So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
or
(1 / 2)^2 = [(Bpo^2 Rc^4 Pi^4 / Qs^2 C^2 Mu^2)] [So^2 / (So^2 + 1)]^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
- {1 / [(So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}

Recall that from energy density function matching:
Bpo^2 = [Mu^2 C^2][Qa / (4 Pi K^2 Ro^2)]^2

Hence:BR> (Bpo^2 Rc^4 Pi^4 / Qs^2 C^2 Mu^2) = [Mu^2 C^2][Qa / (4 Pi K^2 Ro^2)]^2 (Rc^4 Pi^4 / Qs^2 C^2 Mu^2)
= Pi^2 [Rc^2 / (4 K^2 Ro^2)]^2 [Qa / Qs]^2
= Pi^2 [1 / (4 So^2)]^2 [Qa / Qs]^2

Hence the boundary condition at R = Rs, H = 0 becomes:
(1 / 2)^2 = [Pi / (4 So^2)]^2 [Qa / Qs]^2 [So^2 / (So^2 + 1)]^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
- {1 / [(So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
or
(1 / 4) = [Pi^2 / 16] [Qa / Qs]^2 [1 / (So^2 + 1)]^2 {[Nr (So^2 + 1)]^2 + [(So^2 - 1)]^2}
- {1 / [(So^2 - 1)]}^2 {[(So^2 - 1) / 2]^2 + [Nr]^2}
or
(1 / 4) = [Nr^2 Pi^2 / 16][Qa / Qs]^2 + [Pi^2 / 16][Qa / Qs]^2 {[(So^2 - 1) / (So^2 + 1)]^2} - [1 / 4] - {Nr / [(So^2 - 1)]}^2

Collecting Nr^2 terms gives:
[Nr^2 Pi^2 / 16][Qa / Qs]^2 - {Nr / [(So^2 - 1)]}^2 = (1 / 4) - [Pi^2 / 16][Qa / Qs]^2 {[(So^2 - 1) /(So^2 + 1)]^2} + [1 / 4]
or
Nr^2 [Pi^2[Qa / Qs]^2 - Nr^2 (16 / (So^2 - 1)^2)] = 4 (2) - {Pi^2 [Qa / Qs]^2 [(So^2 - 1) /(So^2 + 1)]^2}
or
Nr^2 = (4(2) - {Pi^2 [Qa / Qs]^2 [(So^2 - 1) / (So^2 + 1)]^2})
/ [Pi^2 [Qa / Qs]^2 - (16 / (So^2 - 1)^2)]

This is the boundary condition at R = Rs, H = 0.

For Qa = Qs:
Nr^2 = (8 - {Pi^2 [(So^2 - 1) / (So^2 + 1)]^2})
/ [Pi^2 - (16 / (So^2 - 1)^2)]
or
Nr^2 = ((8 / Pi^2) - {[(So^2 - 1) / (So^2 + 1)]^2})
/ [1 - (4 / Pi (So^2 - 1))^2]

By comparison the boundary condition at R= Rc, H = 0 gave:
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}]
/ {1 - [4 / (Pi (So^2 - 1))]^2}

These two boundary condition formulae are identical.
 

SUMMARY:
An electromagnetic spheromak is governed by an existence condition and a common boundary condition. There may also be a low energy condition.

The existence condition is:
[Pi / 2]
= [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
X Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

= F I
where:
F = [Nr So^2 / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5
and
I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)

The common boundary condition is:
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}]
/ {1 - [4 / (Pi (So^2 - 1))]^2}
 

RANGE RESTRICTION ON So:
The common boundary condition denominator requires that to keep Nr^2 finite:
1 - [4 / (Pi (So^2 - 1))]^2 > 0
or
1 > [4 / (Pi (So^2 - 1))]
or
(So^2 - 1) > (4 / Pi)
or
So^2 > (4 / Pi) + 1

The common boundary condition numerator requires that in order to keep Nr^2 > 0:
[+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] > 0
Compliance with this condition requires that So^2 < 19
 

SOLUTION FOR SPHEOMAK OPERATING POINT:
To find the spheromak operating point choose So^2, calculate Nr^2, calculate Nr and then calculate RHS of existence condition. Plot the RHS value versus So^2. Repeat this process numerous times to accurateoly determine the So^2 value that gives RHS the closest approach to LHS = (Pi / 2). Use the corresponding value of Nr to find integers Np and Nt that satisfy:
.

Nr = Np / Nt.
 

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


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

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

Nr = 4.0106

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

I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr]^2 [Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 

I = Integral from Z = 1 to Z = 2.3 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 1.01 to Z = 2.29 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.29 to Z = 2.3 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 4.0106 / ([(1.3)(Z - 1)]^0.5 {[16.08516] + 0.4225}^0.5)
 
+ Integral from Z = 1.01 to Z = 2.29 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.29 to Z = 2.3 of:
(2.3)^3 dZ 4.0106 / ([(2.3 - Z)(1.3)]^0.5 {[16.08516] [2.3]^2 + 0.4225}^0.5 [2.3 (1.5) + 2.3]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 4.0106 / 4.6324((Z - 1)]^0.5 )
 
+ Integral from Z = 1.01 to Z = 2.29 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.29 to Z = 2.3 of:
dZ 4.0106 / ([(2.3 - Z)]^0.5 {10.51749})
 
= 0.17315  
+ Integral from Z = 1.01 to Z = 2.29 of:
Z^3 dZ 4.0106 / ([(2.3 - Z)(Z - 1)]^0.5 {[16.08516] [Z]^2 + 0.4225}^0.5 [2.3 (Z - 1) + Z]^1.5)
 
+ 0.07626
= 0.24941 + 1.5667
= 1.8161

(F I) = 0.6936265 (1.8161)
= 1.2597
 
 

TRY So^2 = 2.5
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] / {1 - [4 / (Pi (So^2 - 1))]^2}
= [+ {0.81057} - {0.18367}] / {1 - 0.720507}
= 0.62690 / 0.27949
= 2.243

Nr = 1.49766

F = [Nr So^2 / [(Nr^2 (So^2 + 1)^2) + (So^2 - 1)^2]^0.5
= [(1.49766)(2.5) / [(2.243 )(12.25)) + (2.25)]^0.5
= 3.74415 / 5.45222
= 0.68672

I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr]^2 [Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 

I = Integral from Z = 1 to Z = 2.5 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 1.01 to Z = 2.49 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
Integral from Z = 2.49 to Z = 2.5 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 1.49766 / ([(1.5)(Z - 1)]^0.5 {[2.243] + 0.5625}^0.5)
 
+ Integral from Z = 1.01 to Z = 2.49 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.49 to Z = 2.5 of:
(2.5)^3 dZ 1.49766 / ([(2.5 - Z)(1.5)]^0.5 {[2.243] [2.5]^2 + 0.5625}^0.5 [2.5 (1.5) + 2.5]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 1.49766 / 2.0514((Z - 1)]^0.5 )
 
+ Integral from Z = 1.01 to Z = 2.49 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 2.49 to Z = 2.5 of:
dZ 1.49766 / ([(2.5 - Z)]^0.5 {4.6767})
 
= 0.14601  
+ Integral from Z = 1.01 to Z = 2.49 of:
Z^3 dZ 1.49766 / ([(2.5 - Z)(Z - 1)]^0.5 {[2.243] [Z]^2 + 0.5625}^0.5 [2.5 (Z - 1) + Z]^1.5)
 
+ 0.064047
= 0.21005 + 1.3987
= 1.6087

(F I) = 0.6867 (1.6087)
= 1.1047
 
 

TRY So^2 = 4.0
Nr^2 = [+ {8 / Pi^2} - {[(So^2 - 1) / (So^2 + 1)]^2}] / {1 - [4 / (Pi (So^2 - 1))]^2}
= [+ {0.81057} - {0.3600}] / {1 - 0.180126}
= 0.45057 / 0.819874
= .54956

Nr = 0.7413

F = [Nr So^2 / [(Nr^2 (So^2 + 1)^2) + (So^2 - 1)^2]^0.5
= [(.7413)(4.0) / [(.54956)(25)) + (9)]^0.5
= 2.96529 / 4.7685
= 0.62185

I = Integral from Z = 1 to Z = So^2 of:
Z^3 dZ Nr / ([(So^2 - Z)(Z - 1)]^0.5 {[Nr]^2 [Z]^2 + [(So^2 - 1) / 2]^2}^0.5 [So^2 (Z - 1) + Z]^1.5)
 

= Integral from Z = 1 to Z = 4.0 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 1.01 to Z = 3.99 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 3.99 to Z = 4.0 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 0.7413 / ([(3.0)(Z - 1)]^0.5 {[0.54956] + 2.25}^0.5)
 
+ Integral from Z = 1.01 to Z = 3.99 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 3.99 to Z = 4.0 of:
(4.0)^3 dZ 0.7413 / ([(4.0 - Z)(3.0)]^0.5 {[0.54956] [4.0]^2 + 2.25}^0.5 [4.0 (3.0) + 4.0]^1.5)
 
= Integral from Z = 1 to Z = 1.01 of:
dZ 0.7413 / 2.898047 ((Z - 1)]^0.5 )
 
+ Integral from Z = 1.01 to Z = 3.99 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ Integral from Z = 3.99 to Z = 4.0 of:
dZ 0.7413 / ([(4.0 - Z)]^0.5 {5.7557})
 
= 0.05115  
+ Integral from Z = 1.01 to Z = 3.99 of:
Z^3 dZ 0.7413 / ([(4.0 - Z)(Z - 1)]^0.5 {[0.54956] [Z]^2 + 2.25}^0.5 [4.0 (Z - 1) + Z]^1.5)
 
+ 0.025758
= 0.07690 + 0.6605
= 0.7374

(F I) = 0.62185 (0.7374)
= 0.45856
 
 

SPHEROMAK EVOLUTION:
The above equations indicate that when a spheromak is initially formed the spheromak will operate at large Nr corresponding to So^2 = 2.27. However, over time the spheromak will quickly lose energy by radiation until its energy falls to a stable minimum energy point at So^2 = 3.765 where the spheromak will no longer spontaneously radiate. As the spheromak energy decreases, Nr^2 decreases, So^2 increases and the volume of the toroidal region increases.
 

The value of:
So^2 = 3.765
is comparable to the ratio:
So^2 = (Rs / Rc) = 4.2
obtained from a General Fusion plasma spheromak photograph. However, a plasma spheromak may be affected by inertial forces and other issues that do not affect a charged particle spheromak.
 

PLASMA SPHEROMAKS:
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 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

and that this energy density in the far field must equal
Uo [Ro^2 / (K^2 Ro^2 + R^2 + H^2)]^2
where:
Uo / K^4 = (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) [K^2 Ro^2 / (K^2 Ro^2 + R^2 + H^2)]^2
or
[Qs^2 / (32 Pi^2 Epsilon)] = (Bpo^2 / 2 Mu) [K^2 Ro^2]^2
or
Qs^2 = (Bpo^2 / 2 Mu) [K^2 Ro^2]^2 (32 Pi^2 Epsilon)
= (Bpo^2 / Mu) [K^2 Ro^2]^2 (16 Pi^2 / C^2 Mu)
= (Bpo^2 / Mu^2) K^22 [Ro^2]^2 (16 Pi^2 / C^2)

Thus:
Qs = (Bpo / Mu) K [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 compression.
 

This web page last updated August 31, 2016.

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