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On this web page we find the poloidal magnetic flux of a charged particle spheromak.

**PARAMETER DEFINITIONS:**

The Planck Constant h is a function of:

**Mu** = permiability of free space;

**C** = speed of light in a vacuum;

**Q** = proton charge;

**Pi** = (circumference / diameter) of a circle

= 3.141592653589793

**Pi^2** = 9.869604401

Define for a spheromak in free space:

**Rc** = minimum radius of inner spheromak wall;

**Rs** = maximum radius of outer spheromak wall;

**(K Ro) = (Rs Rc)^0.5** = radius of atomic particle where the potential energy well is deepest;

**So = [Rs / K Ro] = [K Ro / Rc] ** = spheromak shape parameter;

**So^2 = (Rs / Rc)**;

**Hs** = distance of spheromak wall from the equatoral plane;

**Hf** = maximum value of **|Hs|**

**2 Hf** = spheromak overall length;

**Lh** = charge hose length;

**Np** = number of poloidal charge motion path turns contained in Lh;

**Nt** = number of toroidal charge motion path turns contained in Lh;

**Nr = Np / Nt**;

**Rf** = wall radius at **H = Hf** and **H = -Hf**;

**Lp = Pi (Rs + Rc)**;

**Lt = Pi (Rs - Rc)**;

**Bpo** = poloidal magnetic field strength at the center of the spheromak;

**Upo = (Bpo^2 / 2 Mu)** = maximum field energy density at the center of the spheromak;

**Upo = (Uo / K^4)** = maximum field energy density at the center of the spheromak;

From the definitionof the Planck Constant:

**Ett = h Fh**

and

**dEtt = h dFh**

Recall that:

**Fh** = (C / Lh)

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

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

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

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

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

Hence:

dFh = [C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5] d(1 / (K Ro))

which gives:

dEtt = h dFh

= [h C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5] d(1 / (K Ro))

Recall that:

Efs = (Bpo^2 / 2 Mu) (K Ro)^3 Pi^2

Ett = (Ett / Efs) Efs

or

Efs = Ett / (Ett / Efs)

or

Efs = h Fh / (Ett / Efs)

or

Efs = [h / (Ett / Efs)] [C / (Pi Nt K Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Recall that:

Efs = (Bpo^2 / 2 Mu) (K Ro)^3 Pi^2

Equating the two expressions for Efs gives:

(Bpo^2 / 2 Mu) (K Ro)^3 Pi^2 = [h / (Ett / Efs)] [C / (Pi Nt K Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

**FIND POLOIDAL MAGNETIC FLUX Phip:**

As shown on the web page titled ELECTROMAGNETIC SPHEROMAK the peak magnetic field strength Bpo at the center of a spheromak can be expressed as:

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

In the core of the spheromak on the equatorial plane the energy density U is given by:

U = Uo [Ro^2 / ((K Ro)^2 + R^2 + H^2)]^2

= [Bpo^2 / 2 Mu][K^2 Ro^2 / ((K Ro)^2 + R^2 + H^2)]^2

Within the core of the spheromak on the equatorial plane:

U = Bp^2 / 2 Mu

which gives:
Bp^2 / 2 Mu = [Bpo^2 / 2 Mu][K^2 Ro^2 / ((K Ro)^2 + R^2)]^2

or

Bp = Bpo [K^2 Ro^2 / ((K Ro)^2 + R^2)]

The poloidal magnetic flux Phip through the spheromak core is:

**Phip** = Integral from R = 0 to R = Rc of:

Bp 2 Pi R dR

= Integral from R = 0 to R = Rc of:

Bpo [K^2 Ro^2 / ((K Ro)^2 + R^2)] 2 Pi R dR

= Bpo [K^2 Ro^2 2 Pi Integral from R = 0 to R = Rc of:

R dR / ((K Ro)^2 + R^2)]

= Bpo K^2 Ro^2 2 Pi {(1 / 2)Ln((K Ro)^2 + Rc^2) - (1 / 2)Ln((K Ro)^2)}

= Bpo K^2 Ro^2 Pi Ln{[(K Ro)^2 + Rc^2] / [(K Ro)^2]}

= Bpo K^2 Ro^2 Pi Ln{[1 + (Rc^2 / (K Ro)^2)]}

= Bpo K^2 Ro^2 Pi Ln{[1 + (1 / So^2)]}

Thus:

**Bpo K^2 Ro^2 Pi = Phip / Ln{[1 + (1 / So^2)]**

**ELIMINATE Bpo (K Ro)^2:**

From the expressions for Efs above:

**(Bpo^2) (K Ro)^4 Pi^2 = [h 2 Mu/ (Ett / Efs)] [C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]**

Substituting in the expression for Phip gives:

{Phip / Ln[1 + (1 / So^2)]}^2 = [h 2 Mu/ (Ett / Efs)] [C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

or

**(Phip)^2 = [(h 2 Mu) / (Ett / Efs)] [C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5] {Ln[1 + (1 / So^2)]}^2**

**PLANCK CONSTANT:**

On the web page titled: PLANCK CONSTANT it is shown that before adjustment for recoil kinetic energy the Planck Constant h is given by:

**h = {1 - [(So -1)^2 / (So^2 + 1)]^2}
[(Mu C Qa^2) / (4 Pi)] [Pi^2 / 8] Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5 / [So]**

where:

(Ett / Efs) = {1 - [(So -1)^2 / (So^2 + 1)]^2}

Rearranging the expression for h gives:

[So] / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5

= {1 - [(So -1)^2 / (So^2 + 1)]^2} {1 / h} [(Mu C Qa^2) / (4 Pi)] [Pi^2 / 8] Nt

= {Ett / Efs} {1 / h} [(Mu C Qa^2) / (4 Pi)] [Pi^2 / 8] Nt

Substituting this expression into the equation for (Phip)^2 gives:

**(Phip)^2**

= [(h 2 Mu) / (Ett / Efs)] [C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5] {Ln[1 + (1 / So^2)]}^2

= [(h 2 Mu) / (Ett / Efs)] [C / (Pi Nt)] {Ett / Efs} {1 / h} [(Mu C Qa^2) / (4 Pi)] [Pi^2 / 8] Nt {Ln[1 + (1 / So^2)]}^2

= [2] [(Mu^2 C^2 Qa^2) / (4)] [1 / 8] {Ln[1 + (1 / So^2)]}^2

= ** [Mu^2 C^2 Qa^2] [1 / 16] {Ln[1 + (1 / So^2)]}^2 **

Hence the poloidal flux quantum is:

**(Phip) = [Mu C Qa] [1 / 4] {Ln[1 + (1 / So^2)]} **

This is the poloidal magnetic flux for an atomic particle spheromak.

Note that for an atomic particle spheromak with a single charge Phip is independent of the particle size. In an atomic nucleus containing multiple nucleons the nucleons tend to align to minimize the net external magnetic flux.

**NUMERICAL EVALUATION OF Phip:**

Pi = 3.14159265359

Qa = 1.60217662 X 10^-19 coulombs

C = 2.99792458 X 10^8 m / s

Mu = 4 Pi X 10^-7 T^2 m^3 / J

So = 2.025950275

**Phip** = [Mu C Qa] [1 / 4] {Ln[1 + (1 / So^2)]}

= [(Mu C Qa) / (4 Pi)] Pi {Ln[1 + (1 / So^2)]}

= [(10^-7 T^2 m^3 / J) (2.99792458 X 10^8 m / s) (1.60217662 X 10^-19 coulombs) (3.141592654) (Ln(1 + (1 / (2.025950275)^2))

= [(10^-7 T^2 m^3 / J) (2.99792458 X 10^8 m / s) (1.60217662 X 10^-19 coulombs) (3.141592654) (Ln(1.2436365473)

= 3.290157699 X 10^-18 T^2 m^4 coul / J-s

= 3.290157699 X 10^-18 T m^2 (T m^2 coul / J-s)

= **3.290157699 X 10^-18 T m^2**

**Phip** is the theoretical poloidal magnetic flux associated with a stationary free electron or free proton.

Units check:

F = Q V B

F D = Q V B D

J = coul (m / s) T m

Note that this flux quantum is three orders of magnitude smaller than the flux quantum:

h / Q

obtained from Josephson junction super conductivity measurements.

This web page last updated October 11, 2016.

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