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XYLENE POWER LTD.

MAGNETIC FLUX QUANTUM

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

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]

or
(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]
 

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:>BR> 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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