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**SPHEROMAK APPROXIMATION:**

This web page uses relatively simple approximations to show that net charge circulating around a closed path forms the field energy density distribution required for the existence of a spheromak. A more precise 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 hold a stable amount of energy.

A spheromak quasi-toroid has a major axis and a minor axis. The spheromak's circulting current follows a closed spiral path which is characterized by **Np** turns around the quasi-toroid's major axis and by **Nt** turns around the quasi-toroid's minor axis. This current path has the quasi-toroidal shape of the glaze on a 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 for an isolated spheromak is zero. Outside the spheromak wall the electric field for an isolated spheromak is normal to the spheromak surface. 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 spacial distribution of the circulating charge is constant over time. Hence the electric and magnetic field geometry is constant over time.

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

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

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

**SPHEROMAK CONCEPT:**

Conceptually a spheromak wall is a quasi-toroidal surface formed from the closed current path of a spheromak. The current path within the spheromak wall conforms to the quasi-toroid surface curvature.

The magnetic field of a spheromak has both toroidal and poloidal components. The current path gradually changes direction over the surface of the quasi-toroid while following a closed spiral.

In a spheromak quantized positive and/or negative charge moves along a closed spiral path embedded in the spheromak wall. 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 current path 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 forms the interface between the toroidal and poloidal magnetic fields.

**SPHEROMAK GEOMETRY APPROXIMATION:**

Assume that th spheromak shape is a round toroid.

Rs = maximum spheromak wall radius in the equatorial plane;

Rc = minimum spheromak wall radius in the equatorial plane:

Ro = (Rs + Rc) / 2 = Spheromak characteristic radius;

H = Z value of spheromak wall;

Outside zone electric field set by an assumed charged ring of radius Ro with charge Qs located at (R = Ro, Z = 0);

Outside zone magnetic field set by an assumed ring of radius Ro carring current Np I located at (R = Ro, Z = 0);

Inside zone magnetic field B = Muo Nt I / 2 Pi R

Lh^2 = Np^2 Lp^2 + Nt^2 Lt^2

= Np^2 [2 Pi (Rs + Rc) / 2]^2 + Nt^2 [2 Pi (Rs - Rc) / 2)]^2

= Np^2 Pi^2 [(Rs + Rc)]^2 + Nt^2 Pi^2 [(Rs - Rc))]^2

Spheromak existence requirement:

I = Qs C / Lh

**FIND ELECTRIC FIELD CONTRIBUTION TO THE PLANCK CONSTANT**:

In the outside region where R > Rs:

Electric Field = Qs / (4 Pi Epsilono R^2)

Field Energy Density = (Epsilono / 2)(Electric Field)^2

= (Epsilon / 2)[Qs / (4 Pi Epsilono R^2)]^2

Field Energy

= Integral from R = Rs to R = infinity of:

(Field Energy Density) 4 Pi R^2 dR

= Integral from R = Rs to R = infinity of:

(Epsilon / 2)[Qs / (4 Pi Epsilono R^2)]^2 [4 Pi R^2 dR]

= Integral from R = Rs to R = infinity of:

(Qs^2 / 8 Pi Epsilono) dR / R^2

= (Qs^2 / 8 Pi Epsilono Rs)

Recall that:

C^2 = 1 / (Muo Epsilono)

which gives the electric field energy for R > Rs as:

Muo C^2 Qs^2 / 8 Pi Rs

Recall that:

Lh^2 = Np^2 Pi^2 [(Rs + Rc)]^2 + Nt^2 Pi^2 [(Rs - Rc))]^2

If Rs >> Rc then:

Lh^2 ~ Pi^2 Rs^2 (Np^2 + Nt^2)

or

Lh ~ Pi Rs (Np^2 + Nt^2)^0.5

or

(1 / Rs) = Pi (Np^2 + Nt^2) / Lh

giving:

Electric field energy for R > Rs

= Muo C^2 Qs^2 / 8 Pi Rs

= [Muo C^2 Qs^2 / 8 Pi] [Pi (Np^2 + Nt^2) / Lh]

= **[Muo C^2 Qs^2 / 8] [(Np^2 + Nt^2) / Lh]**

Recall that:

F = C / Lh

giving:

Electric field energy for R > Rs

= [Muo C^2 Qs^2 / 8] [(Np^2 + Nt^2) / Lh]

=** [Muo C Qs^2 / 8] [(Np^2 + Nt^2)] F**

Hence the electric field contribution to the Planck constant is about:

**[Muo C Qs^2 / 8] [(Np^2 + Nt^2)]**

**FIND THE CONTRIBUTION TO h FROM THE TOROIDAL REGION:**

Approximate a spheromak by a round cross section toroid.

Inside the spheromak wall:

Bt(R> = Muo Nt I / 2 Pi R

Toroidal magnetic field energy density:

[Bt(R)]^2 / 2 Muo = (1 / 2 Muo)[Muo Nt I / 2 Pi R]^2

= (Muo / 2)[Nt I / 2 Pi R]^2

The magnetic energy in the toroidal region is:

Integral from R = Rc to R = Rs of:

(Muo / 2)[Nt I / 2 Pi R]^2 2 Pi R dR 2 H

= Integral from R = Rc to R = Rs of:

[(Muo / 2)(Nt I)^2 dR / 2 Pi R] 2 H

Recall that:

I = Q C / Lh

giving:

magnetic fied energy intoroidal region

= Integral from R = Rc to R = Rs of:

[(Muo / 2)(Nt I)^2 dR / 2 Pi R] 2 H

= Integral from R = Rc to R = Rs of:

[(Muo / 2)(Nt Q C / Lh)^2 dR / 2 Pi R] 2 H

=

**INNER WALL BOUNDARY CONDITION AT R = Rc, Z = 0:**

At R = Rc, Z = 0:

Btc = Muo Nt I / 2 Pi Rc

Bpc = Muo I Np 2 Pi Ro / 4 Pi Ro^2

= Muo I Np / 2 Ro

However, due to energy density balance at R = Rc, Z = 0:

Btc = Bpc

or

Muo Nt I / 2 Pi Rc = Muo I Np / 2 Ro

or

Nt / Pi Rc = Np / Ro

or

**Np / Nt = Ro / Pi Rc**

**Et** ~ [Bta^2 / 2 Muo] Vt

= (Muo / 2)[Nt I / Pi (Rs + Rc)]^2 Pi^2 [(Rs - Rc) / 2]^2 [(Rs + Rc)]

= (Muo / 2)[(Nt I)^2 / Pi^2 (Rs + Rc)^2] Pi^2 [(Rs - Rc) / 2]^2 [(Rs + Rc)]

= **(Muo / 8)[(Nt I)^2 / (Rs + Rc)] [(Rs - Rc)]^2**

which is anapproximation of the energy contained in the toroidal magnetic field inside the spheromak wall.

**Approximation of the energy contained in the poloidal magnetic field:**

Bp = Muo Np I / 2 Pi ((Rs + Rc) / 2)

= Muo Np I / (Pi (Rs + Rc))

Bp^2 / 2 Muo = (Muo /2) [Np I / (Pi (Rs + Rc))]^2

Vp ~ Pi [Rc]^2 2 (Rs + Rc) / 2

= Pi Rc^2 (Rs + Rc)

Ep = [Bp^2 / 2 Muo] Vp

= (Muo /2) [Np I / (Pi (Rs + Rc))]^2 [Pi Rc^2 (Rs + Rc)]

= (Muo / 2) [(Np I)^2 / Pi] [Rc^2 / (Rs + Rc)]

**E** = Et + Ep

= **(Muo / 8)[(Nt I)^2 / (Rs + Rc)] [(Rs - Rc)]^2
+ (Muo / 2) [(Np I)^2 / Pi] [Rc^2 / (Rs + Rc)]**

At the spheromak's stable operating point:

dE = dEt + dEp = 0

**CURRENT PATH CONDITION:**

Spheromak winding analysis shows that:

(Np / Nt) = Np / [P - 2 Np]

or

Nt = P - 2 Np

or

dNt = - 2 dNp

dE = (Muo / 8)[(2 Nt dNt I^2) / (Rs + Rc)] [(Rs - Rc)]^2

+ (Muo / 2) [(2 Np dNp I^2) / Pi] [Rc^2 / (Rs + Rc)]

= 0

or

(Muo / 8)[(2 Nt (- 2 dNp) I^2) / (Rs + Rc)] [(Rs - Rc)]^2

+ (Muo / 2) [(2 Np dNp I^2) / Pi] [Rc^2 / (Rs + Rc)]

= 0

or

(Muo / 2) [(2 Np dNp I^2) / Pi] [Rc^2 / (Rs + Rc)]

=(Muo / 8)[(2 Nt (2 dNp) I^2) / (Rs + Rc)] [(Rs - Rc)]^2

or

[(Np) / Pi] [Rc^2 ]

= [(Nt / 2)] / [(Rs - Rc)]^2

or

**[Np / Nt] = [(Rs - Rc)]^2 [Pi / 2 Rc^2]**

**INNER RIM BOUNDARY CONDITION:**

For an isolated spheromak the electric field on the equatorial plane in the spheromak core and inside the spheromak wall is zero. Hence at the inner rim the toroidal magnetic field intensity inside the spheromak wall equals the poloidal magnetic field intensity outside the spheromak wall.

Inside the spheromak wall:

Bt = Muo Nt I /2 Pi Rc

Outside the spheromak wall to a first approximation:

Bp 2 Pi [(Rs - Rc) / 2} = Muo Np I

or

Bp = Muo Np I / Pi (Rs - Rc)

Now apply the inner rim boundary condition:

Bp = Bt

or

Muo Np I / Pi (Rs - Rc) = Muo Nt I / 2 Pi Rc

or

Np / (Rs - Rc) = Nt / (2 Rc)

or

**Np / Nt = (Rs - Rc) / 2 Rc**

which is the inner rim boundary condition.

Recall that the winding condition resulted in the formula:

[Np / Nt] = [(Rs - Rc)]^2 [Pi / 2 Rc^2]

Note that Pi appears in one (Np / Nt) expression but not the other. In order for the equation to be valid Np and Nt must have nontrivial values.

Equate the two expressions for Np / Nt to get:

(Rs - Rc) / 2 Rc = [(Rs - Rc)]^2 Pi / 2 Rc^2

or

[1] = (Rs - Rc) Pi / Rc

or

(Rs - Rc) / Rc = 1 / Pi

or

(Rs / Rc) = (Pi + 1) / Pi

Hence:

Np / Nt = 1 / 2 Pi

However, Np and Nt are integers.

Hence we are looking for integer values of Np and Nt with a ratio that is close to:Np / Nt = (1 / 2 Pi)

Recall that:

(Np / Nt) = Np / [P - 2 Np]

Hence:

1 / 2 Pi = Np / [P - 2 Np]
or

[P - 2 Np] = 2 Pi Np

or

P = [2 Pi + 2] Np

= 8.2831853 Np

However, Np is an integer. Thus in principle we can find an approximate solution to this equation.

Try Np = 7:
Then:

P = 57.9822971.
However, 58 is not a prime number. However, if it was prime then:

Nt = P - 2 Np

= 58 - 14

= 44

In fact this solution is not correst due to the simpifying approximations used, but this web page shows the mathematical sequence for finding a solution to the spheromak problem.

I = Q C / L
or

(Nt I)^2 = (Q C)^2 Nt^2 / L^2

and

(Np I)^2 = (Q C)^2 Np^2 / L^2

Recall that:

E = **(Muo / 8)[(Nt I)^2 / (Rs + Rc)] [(Rs - Rc)]^2
+ (Muo / 2) [(Np I)^2 / Pi] [Rc^2 / (Rs + Rc)]**

= (Muo / 8)[(Q C)^2 Nt^2 / L^2 (Rs + Rc)] [(Rs - Rc)]^2

+ (Muo / 2) [(Q C)^2 Np^2 / L^2 Pi] [Rc^2 / (Rs + Rc)]

recall that:

F = C / L

or

E = (Muo F)[(Q^2 C) Nt^2 / L 8 (Rs + Rc)] [(Rs - Rc)]^2

+ (Muo F) [(Q^2 C) Np^2 / L Pi] [Rc^2 / 2 (Rs + Rc)]

Thus:

**h = [E / F]** = (Muo)[(Q^2 C) Nt^2 / L 8 (Rs + Rc)] [(Rs - Rc)]^2

+ (Muo) [(Q^2 C) Np^2 / L Pi] [Rc^2 / 2 (Rs + Rc)]

= [Muo Q^2 C / L]

{[Nt^2 / 8 (Rs + Rc)] [(Rs - Rc)]^2 + [Np^2 / Pi] [Rc^2 / 2 (Rs + Rc)]}

Recall that:

(Rs - Rc) / Rc = 1 / Pi

and

Np / Nt = (1 / 2 Pi)

which gives:

**h** = [Muo Q^2 C / L]

{[Nt^2 / 8 (Rs + Rc)] [(Rs - Rc)]^2 + [Np^2 / Pi] [Rc^2 / 2 (Rs + Rc)]}

= [Muo Q^2 C / L]

{[Nt^2 / 8 (Rs + Rc)] [(Rc / Pi)]^2 + [Nt^2 / 4 Pi^3] [Rc^2 / 2 (Rs + Rc)]}

= [Muo Q^2 C / L][Nt^2 / 8 (Rs + Rc)] {[(Rc / Pi)]^2 + [1 / Pi^3] [Rc^2]}

= [Muo Q^2 C / L][Nt^2 Rc^2 / 8 Pi^2 (Rs + Rc)] {(Pi + 1) / Pi}

= [Muo Q^2 C / L][Nt^2 Rc^2 / 8 Pi^2 Rc (2 + (1 / Pi))] {(Pi + 1) / Pi}

= **[Muo Q^2 C / L][Nt^2 Rc / 8 Pi^2 (2 + (1 / Pi))] {(Pi + 1) / Pi}**

Recall that:

L^2 = [Np (2 Pi)(Rs + Rc)/2]^2 + [Nt (2 Pi) (Rs - Rc)/ 2]^2

= [Np (Pi)(Rs + Rc)]^2 + [Nt (Pi) (Rs - Rc)]^2

= Pi^2 { [Np(Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}

or

L = Pi {[Np(Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5

Recall that:

Np = Nt / 2 Pi

and

(Rs - Rc) = Rc / Pi

or

Rs = Rc (1 + (1 / Pi))

giving:

L = Pi {[Np(Rs + Rc)]^2 + [Nt (Rs - Rc)]^2}^0.5

= Pi {[(Nt / 2 Pi)(Rs + Rc)]^2 + [Nt (Rc / Pi)]^2}^0.5

= Pi {[(Nt^2 / 4 Pi^2)(Rs + Rc)^2] + Nt^2 Rc^2 / Pi^2}^0.5

= Nt {[(Rs + Rc) / 2]^2 + [Rc]^2}^0.5

= Nt {[Rc(2 + (1 / Pi))/2]^2 + [Rc]^2}^0.5

= Nt Rc {1 + (1 / Pi) + (1 / 4 Pi^2) + (1)^2}^0.5

= Nt Rc {2 + (1 / Pi) + (1 / 4 Pi^2)}^0.5

Hence:

**h** = [Muo Q^2 C / L][Nt^2 Rc / 8 Pi^2 (2 + (1 / Pi))] {(Pi + 1) / Pi}

= [Muo Q^2 C][Nt^2 Rc / 8 Pi^2 (2 + (1 / Pi))] {(Pi + 1) / Pi}

/ Nt Rc {2 + (1 / Pi) + (1 / 4 Pi^2)}^0.5

= [Muo Q^2 C][Nt / 8 Pi^3 (2 + (1 / Pi))] {(Pi + 1)}

/ {2 + (1 / Pi) + (1 / 4 Pi^2)}^0.5

= **[Muo Q^2 C][Nt / 4 Pi (2 Pi + 1)] {(Pi + 1)}
/ {8 Pi^2 + 4 Pi + 1}^0.5**

Thus we have developed a crude expression for the Planck constant:

h = (E / F)

However, to get the lead coeeficient right we cannot use the approximations used on this web page. Instead exact expressions must be used. The exact expressions are mathematically much more complex.

**SOLUTION APPROXIMATION:**

Note that a neutral spheromak cannot exist because a spheromak relies on the distributed charge on the current filament to balance the attractive magnetic foces between adjacent current filaments. However, to obtain an approximate spheromak solution it may be practical to assume that the field energy contained in the spheromak magnetic fields is much greater than the energy contained in the electric field. Then for spheromak wall geometric stability the poloidal magnetic energy density outside the spheromak wall equals the toroidal magnetic energy density inside the spheromak wall at every point on the spheromak wall. This requirement leads to the approximate spheromak geometry and the ratio of number of poloidal turns Np to the number of toroidal turns Nt. These values in turn lead to the spheromak frequency and the static magnetic energy of the spheromak. The ratio:

(static magnetic energy) / (frequency) = Planck constant.

Note that Np and Nt are both integers which leads to a unique solution. Solve the problem at the inside wall on the equatorial plane. Find that the spheromak cross section is ellipsoidal.

A real charged spheromak in a vacuum has an external radial electric field which may change the field energy distribution. However, it is instructive to examine the case where the magnetic field energy density is dominant.

A real spheromak case may be an electron spheromak around a positive nucleus. At large distances the electric fields cancel. Inside the spheromak the electric field is zero. At the spheromak inside wall on the equatorial plane the poloidal magnetic energy density outside the wall plus the external electric field energy density equals the toroidal magnetic field energy density inside the wall.

In a plasma or a crystal the electric fields almost cancel so spheromak energy is largely magnetic. Hence it is informative to calculate the theoretical Planck constant for the purely magnetic case.

This web page last updated April 21, 2021.

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