Home Energy Nuclear Electricity Climate Change Lighting Control Contacts Links


XYLENE POWER LTD.

MAGNETIC FLUX QUANTUM

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

It is shown herein that atomic particle spheromaks with quantum charges have both poloidal and toroidal magnetic flux quanta. The poloidal magnetic flux can be calculated by integrating across the core at Z = 0. Finding the toroidal magnetic flux requires first developing an equation for H(R) where H is the distance of the spheromak wall above the plane Z = 0. That requires a 3 dimensional solution to the field energy distribution.

FIX On this web page we find the poloidal magnetic flux quantum Phip and the toroidal magnetic flux quantum Phit. These flux quanta depend on the atomic particle charge but are independent of the physical size of the atomic particle.
Phip = [(Muo C Qs) (0.0020787035)]
Phit = [(Muo C Qs) / (4 Pi)][0.4347969536]
Phit = __________ Phip  

FIX

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;
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 filament turns contained in Lh;
Nt = number of toroidal charge filament turns contained in Lh;
Rf = wall radius at H = Hf and H = -Hf;
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;
Phip = poloidal magnetic flux;
Phit = toroidal magnetic flux;
 

From the definition of the Planck Constant:
Ett = h Fh
and
dEtt = h dFh

From the web page titled ELECTROMAGNETIC SPHEROMAK:
Fh = (C / Lh)
Np = 363
Nt = 362
P = Np + 2 Nt = 1087
Lh / 2 Pi Ro = [Np P]^0.5 = 628.156

Bpo = Muo Np I / 2 Ro

Bto = Bt|(R = K Ro)
= Muo Nt I / (2 Pi K Ro)

[Bpo / Bto] = [Muo Np I / 2 Ro] / [Muo Nt I / (2 Pi K Ro)]
= (Np Pi K) / Nt

Uo / K^4 = Bpo^2 / 2 Muo
= [1 / 2 Muo]{Muo Q C Np / 4 (Np + Nt) Pi Ro^2}^2 ??? = Muo {Q C / Pi Ro^2}^2 (1 / 2) (Np / 4 (Np+ Nt))^2 Thus: [1 / K^4] = [Muo {Q C / Pi Ro^2}^2 (1 / 2) (Np / 4 (Np+ Nt))^2] / Upo However: Uo = [(Muo C^2 Q^2 / 32 Pi^2 Ro^4) giving: [1 / K^4] = (1 / 2) (Np / 4 (Np+ Nt))^2] / [1 / 32] = Np^2 / (Np + Nt)^2 or K^2 = (Np + Nt) / Np = 625 / 312 = 2 + (1 / 312) Somewhere there is a confusion between K^2 and K^3 Pi. FIX

From the web page titled: SPHEROMAK ENERGY:
Efs = (Bpo^2 / 2 Mu) (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 Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Recall that:
Efs = (Bpo^2 / 2 Mu) (Ro)^3 Pi^2

 

POLOIDAL MAGNETIC FLUX Phip:
The field energy density in the core of a spheromak is:
U = Uo {Ro^2 / [(K Ro - R)^2 + Z^2]}^2

At Z = 0 this field is entirely poloidal magnetic, where: U = Bp^2 / 2 Muo

Hence for 0 < R < Rc, at Z = 0:
Bp^2 / 2 Muo = Uo {Ro^2 / [(K Ro - R)^2]}^2

Thus:
Bp^2 = 2 Muo Uo {Ro^2 / [(K Ro - R)^2]}^2
or
Bp = [2 Muo Uo]^0.5 {Ro^2 / [(K Ro - R)^2]}

Then the poloidal magnetic flux Phip through the center of the spheromak is:
Phip = Integral from R = 0 to R = Rc of:
Bp 2 Pi R dR
= Integral from R = 0 to R = Rc of:
[2 Muo Uo]^0.5 {Ro^2 / [(K Ro - R)^2]} 2 Pi R dR
= [2 Muo Uo]^0.5 [Ro^2] 2 Pi {R dR /(K Ro - R)^2}
which from Dwight 91.2 is:
a = (K Ro)
b = (- 1)
giving:
Phip = [2 Muo Uo]^0.5 [Ro^2] 2 Pi {Ln|(K Ro - R)| + [(K Ro) / (K Ro - R)]}|(R = Rc)
- [2 Muo Uo]^0.5 [Ro^2] 2 Pi {Ln|(K Ro - R)| + [(K Ro) / (K Ro - R)]}|(R = 0)
 
= [2 Muo Uo]^0.5 [Ro^2] 2 Pi {Ln|(K Ro - Rc)| + [(K Ro) / (K Ro - Rc)] - Ln|(K Ro| - 1}
= [2 Muo Uo]^0.5 [Ro^2] 2 Pi {Ln[(K Ro - Rc) / (K Ro)] + [Rc / (K Ro - Rc)]}

As shown on the web page titled ELECTROMAGNETIC SPHEROMAK:
Uo = [(Muo C^2 Q^2) / (32 Pi^2 Ro^4)]

Hence:
Phip = [(Muo C Q) / (4 Pi Ro^2)][Ro^2] 2 Pi {Ln[(K Ro - Rc) / (K Ro)] + [Rc / (K Ro - Rc)]}
= [(Muo C Q) / 2] {Ln[1 - (Rc / (K Ro))] + [Rc / (K Ro - Rc)]}
= [(Muo C Q) / 2] {Ln[(1 - (Rc / K Ro)] + [(Rc / K Ro) / (1 - (Rc / K Ro))]}

We must confirm that: (Rc / Ro) < K

The web page titled: Theoretical Spheromak gives:
Rc = K^2 Ro + (Ho^2 / 2 Ro) - (Ho^2 / 2 Ro)[(4 K^2 Ro^2 / Ho^2) + 1]^0.5
or
[Rc / Ro] = K^2 + [1 / 2][Ho^2 / Ro^2] - [1 / 2][Ho^2 / Ro^2][[(4 K^2 Ro^2 / Ho^2) + 1]^0.5

For:
K^2 = [1 / 4][Ho^2 / Ro^2]
= [1 / 4][Np / Nt]
Np / Nt = 363 / 362
= 1.002762431
= [Ho^2 / Ro^2]
= 4 K^2
then:
K^2 = 0.2506906077
and
K = 0.5006901314

[Rc / Ro] = K^2 + [1 / 2][Ho^2 / Ro^2] - [1 / 2][Ho^2 / Ro^2][[(4 K^2 Ro^2 / Ho^2) + 1]^0.5
= 0.2506906077 + 0.5 [1.002762431] - 0.5 [1.002762431][2]^0.5
= 0.7520718233 - 0.7090601149
= 0.0430117084

[Rc / K Ro] = 0.0430117084 / 0.5006901314
= 0.0859048455

Ln[(1 - (Rc / K Ro)
= Ln(1 - 0.0859048455)
= Ln(.9140951545)
= -.0898206052

[(Rc / K Ro) / (1 - (Rc / K Ro))] = 0.0859048455 / (1 - 0.0859048455)
= 0.0859048455 / (0.9140951545)
= 0.0939780121

Hence:
Phip = [(Muo C Q) / 2] {Ln[(1 - (Rc / K Ro)] + [(Rc / K Ro) / (1 - (Rc / K Ro))]}
= [(Muo C Q) / 2]{-.0898206052 + 0.0939780121}
= [(Muo C Q) / 2][.0041574069]
= [(Muo C Q)][0.0020787035]
 

NUMERICAL EVALUATION OF Phip:
Phip = [(Mu C Qs)][0.0020787035]
= [(4 Pi X 10^-7 T^2 m^3 / J) (2.99792458 X 10^8 m / s) (1.60217662 X 10^-19 coulombs) [0.0020787035]
= [(10^-7 T^2 m^3 / J) (2.99792458 X 10^8 m / s) (1.60217662 X 10^-19 coulombs) 4 (3.141592654) [0.0020787035]
= 0.12546815 X 10^-18 T^2 m^4 coul / J-s
= 0.12546815 X 10^-18 T m^2 (T m^2 coul / J-s)
= 0.12546815 X 10^-18 T m^2

Phip is the poloidal magnetic flux through the core of a single free quantum charged particle which enables the existence of free electrons and free protons. Note that Phip is dependant on the particle's quantum charge Qs but is independent of the particle's rest energy. Thus Phip is a poloidal magnetic flux quantum for an atomic particle spheromak.

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

TOROIDAL MAGNETIC FLUX Phit:
We need to also calculate the toroidal magnetic flux quantum for a charged particle and the ratio of the poloidal magnetic flux to the toroidal magnetic flux.

Inside the spheromak wall the toroidal magnetic field is given by:
Bt = Bto (K Ro / R)

Inside the spheromak wall:
dA = 2 Zw dR

Hence an element of magnetic flux is:
dPhit = Bt dA
= Bto (K Ro / R)[2 Zw dR]

The web page titled Theoretical Spheromak gives the spheromak wal equation as:
[Uo / Uto]^0.5 = [K / (Rw Ro)][(K Ro - Rw)^2 + Zw^2]
or
Zw^2 = [(Rw Ro) / K][Uo / Uto]^0.5 - (K Ro - Rw)^2
or
Zw = {[(Rw Ro) / K][Uo / Uto]^0.5 - (K Ro - Rw)^2}^0.5

Phit = Integral from R = Rc to R = Rs of:
Bto (K Ro / R)[2 Zw dR]
= Integral from R = Rc to R = Rs of:
Bto (K Ro / R)[2 dR]{[(R Ro) / K][Uo / Uto]^0.5 - (K Ro - R)^2}^0.5
= Integral from R = Rc to R = Rs of:
Bto (K Ro)[2 dR]{[(Ro) / R K][Uo / Uto]^0.5 - ((K Ro / R) - 1)^2}^0.5
= Integral from R = Rc to R = Rs of:
(2 Bto Ro) dR]{[(Ro K) / R][Uo / Uto]^0.5 - K^2 ((K Ro / R) - 1)^2}^0.5

Recall that:
[Uo / Uto]^0.5 = [Ho^2 / Ro^2] = 4 K^2

Hence:

Phit = Integral from R = Rc to R = Rs of:
(2 Bto Ro) dR]{[(Ro K) / R][Uo / Uto]^0.5 - K^2 ((K Ro / R) - 1)^2}^0.5
= Integral from R = Rc to R = Rs of:
(2 Bto Ro) dR]{[(Ro K) / R][4 K^2] - K^2 ((K Ro / R) - 1)^2}^0.5
= Integral from R = Rc to R = Rs of:
(2 Bto Ro K) dR]{[(4 Ro K) / R] - ((K Ro / R) - 1)^2}^0.5
= Integral from R = Rc to R = Rs of:
(2 Bto Ro K) dR]{[(4 Ro K) / R] - [(K Ro / R)^2 - 2 (K Ro / R) + 1]}^0.5
= Integral from R = Rc to R = Rs of:
(2 Bto Ro K) dR]{[(6 Ro K) / R] - (K Ro / R)^2 - 1]}^0.5
Try X = (1 / R) dX = - dR / R^2 = - dR / X^2 dR = - X^2 dX Phit = integral from X = Xc to X = Xs of:
(2 Bto Ro K) (- X^2 dX)]{[(6 Ro K X)] - (K Ro X)^2 - 1]}^0.5
Phit = integral from X = Xc to X = Xs of:
(- 2 Bto Ro K) (X^2 dX)]{[- (K Ro X)^2 + (6 Ro K X) - 1]}^0.5

a = - (K Ro)^2
b = 6 Ro K
c = -1
Dwight 380.021 gives:
Hs^2 = [(Rs - R)(R - Rc)]

Hence:
dPhit = [(Mu Nt Ih) / (Pi R)][Hs dR]
= [(Mu Nt Ih) / (Pi R)][(Rs - R)(R - Rc)]^0.5 dR
= [(Mu Nt Ih) / Pi][(Rs - R)(R - Rc)]^0.5 [dR / R]
= [(Mu Nt Ih Ro) / Pi][((Rs / Ro) - (R / Ro))((R / Ro) - (Rc / Ro))]^0.5 [dR / R]

Let Z = (R / Ro)
giving:
dZ = dR / Ro

Then:
dPhit = [(Mu Nt Ih Ro) / Pi][((Rs / Ro) - (R / Ro))((R / Ro) - (Rc / Ro))]^0.5 [dR / R]
= [(Mu Nt Ih Ro) / Pi][(So - Z)(Z - (1 / So))]^0.5 [dZ / Z]

Phit = [(Mu Nt Ih Ro) / Pi] Integral from Z = (1 / So) to Z = So of:
[(So - Z)(Z - (1 / So))]^0.5 [dZ / Z]
 
= [(Mu Nt Ih Ro) / Pi] Integral from Z = (1 / So) to Z = So of:
[- Z^2 + Z (So + (1 / So)) - 1]^0.5 [dZ / Z]
 

Dwight 380.311 gives:
Integral from Z = (1 / So) to Z = So of:
[- Z^2 + Z (So + (1 / So)) - 1]^0.5 [dZ / Z]
 
= [- Z^2 + Z (So + (1 / So)) - 1]^0.5|Z = So
- [- Z^2 + Z (So + (1 / So)) - 1]^0.5|Z = (1 / So)
+ [(So + (1 / So)) / 2] Integral from Z = (1 / So) to Z = So of:
dZ / [- Z^2 + Z (So + (1 / So)) - 1]^0.5
- Integral from Z = (1 / So) to Z = So of:
dZ / {Z [- Z^2 + Z (So + (1 / So)) - 1]^0.5}
 
= + [(So + (1 / So)) / 2] Integral from Z = (1 / So) to Z = So of:
dZ / [- Z^2 + Z (So + (1 / So)) - 1]^0.5
- Integral from Z = (1 / So) to Z = So of:
dZ / {Z [- Z^2 + Z (So + (1 / So)) - 1]^0.5}
which from Dwight 380.001 becomes:
 
= + [(So + (1 / So)) / 2] {- arc sin[(-2 Z + (So + (1 / So))) / ((So + (1 / So))^2 - 4)^0.5]}|Z = So
- [(So + (1 / So)) / 2] {- arc sin[(-2 Z + (So + (1 / So))) / ((So + (1 / So))^2 - 4)^0.5]}|Z = (1 / So)
- Integral from Z = (1 / So) to Z = So of:
dZ / {Z [- Z^2 + Z (So + (1 / So)) - 1]^0.5}
 
= + [(So + (1 / So)) / 2] {- arc sin[(-So + (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
- [(So + (1 / So)) / 2] {- arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
- Integral from Z = (1 / So) to Z = So of:
dZ / {Z [- Z^2 + Z (So + (1 / So)) - 1]^0.5}
which from Dwight 380.111 becomes:
 
= + [(So + (1 / So)) / 2] {- arc sin[(-So + (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
- [(So + (1 / So)) / 2] {- arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
- {arc sin{[(So + (1 / So)) Z - 2] / [|Z|((So + (1 / So))^2 - 4)^0.5]}|Z = So
+ {arc sin{[(So + (1 / So)) Z - 2] / [|Z|((So + (1 / So))^2 - 4)^0.5]}|Z = (1 / So)
%nbsp;
= + [(So + (1 / So)) / 2] {- arc sin[(-So + (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
- [(So + (1 / So)) / 2] {- arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
- {arc sin{[So^2 - 1] / [|So|((So + (1 / So))^2 - 4)^0.5]}
+ {arc sin{[(1 / So)^2 - 1] / [|(1 / So)|((So + (1 / So))^2 - 4)^0.5]}
%nbsp;
= - [(So + (1 / So)) / 2] {arc sin[(-So + (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
+ [(So + (1 / So)) / 2] {arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
- {arc sin{[So - (1 / So)] / [((So + (1 / So))^2 - 4)^0.5]}
+ {arc sin{[(1 / So) - So] / [((So + (1 / So))^2 - 4)^0.5]}
%nbsp;
= {1 - [(So + (1 / So)) / 2]} {arc sin[(-So + (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
+ {[(So + (1 / So)) / 2] - 1} {arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
 
= {1 - (So / 2) - (1 /2 So))]} {arc sin[(-So + (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
+ {(So / 2) + (1 / 2 So) - 1} {arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
 
= {- 1 + (So / 2) + (1 /2 So))]} {arc sin[(+So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
- {-(So / 2) - (1 / 2 So) + 1} {arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
 
= {So + (1 / So) - 2}{arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}

 

Thus:
Phit = [(Mu Nt Ih Ro) / Pi] Integral from Z = (1 / So) to Z = So of:
[- Z^2 + Z (So + (1 / So)) - 1]^0.5 [dZ / Z]
 
=[(Mu Nt Ih Ro) / Pi]{So + (1 / So) - 2}{arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}
= [(Mu Qs Nt Fh Ro) / Pi]{So + (1 / So) - 2}{arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]}

Recall that:
Fh = [C / (Pi Nt Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
or
Nt Ro Fh = [C / Pi] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Thus:
Phic = [(Mu Qs Nt Fh Ro) / Pi]{So + (1 / So) - 2}{arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5] = [C / Pi] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5][(Mu Qs) / Pi]
{So + (1 / So) - 2}{arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]
= [(Mu Qs C) / (4 Pi)][4 / Pi][So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]{So + (1 / So) - 2}
{arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]

 

NUMERICAL EVALUATION OF Phit:
From the web page titled: PLANCK CONSTANT Nr^2 = (0.7278688525)^2
= 0.5297930664
and
So = 2.025950275
and
So^2 = 4.104474517

[So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
= [2.025950275 / {[0.5297930664 (5.104474517)^2] + [(3.104474517)]^2}^0.5]
= [2.025950275 / {[13.80410806] + [9.637762027]}^0.5]
= [2.025950275 / {4.841680503}]
= 0.4184394806

{So + (1 / So) - 2} = 2.025950275 + (1 / 2.025950275) - 2
= .025950275 + 0.4935955301
= 0.5195458051

[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]
= [(2.025950275 - (1 / 2.025950275)) / ((2.025950275 + (1 / 2.025950275))^2 - 4)^0.5]
= [(1.532354745) / (2.348111064)^0.5]
= [(1.532354745) / 1.532354745]
= 1.00000000

Hence:
arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5] = (Pi / 2)

Thus:
Phit = [(Mu Qs C) / 4 Pi][4 / Pi][So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]{So + (1 / So) - 2}
{arc sin[(So - (1 / So)) / ((So + (1 / So))^2 - 4)^0.5]
 
= [(Mu Qs C) / (4 Pi)][4 / Pi] [0.4184394806] [0.5195458051][Pi / 2]
= [(Mu Qs C) / (4 Pi)] [0.4184394806] [0.5195458051][2]
= [(Mu Qs C) / (4 Pi)][0.4347969537]

Thus:
Phit / Phip = [0.4347969537] / [0.684991614]
= 0.6347478493
where Phip is the poloidal magnetic flux quantum and Phic is the toroidal magnetic flux quantum.

It is informative to compare (Phip) to the apparent magnetic flux quantum (h / Q) measured using Josephson junction super conductivity techniques.

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

UNITS CHECK:
h / (Q Phip) = J s / (coul T m^2)
coul (m /s) T = kg m / s^2
1 / T = coul (m / s) s^2 / kg m
h / (Q Phip) = [J s / (coul T m^2)][coul (m / s) s^2 / kg m]
= [kg m^2 s / (s^2 coul m^2)][coul (m / s) s^2 / kg m]
= 1
 

NUMERICAL EVALUATION OF [h / Q Phip]:
Pi = 3.14159265359
Qs = 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
Nt = 305
Nr = (Np / Nt) = 222 / 305 = 0.7278688525

[h / Q Phip] = {1 - [(So -1)^2 / (So^2 + 1)]^2}
[Pi / 8] Nt {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5
/ {[So] {Ln[1 + (1 / So^2)]}}
 
= {1 - [(1.025950275)^2 / (2.025950275^2 + 1)]^2}
[Pi / 8] Nt {[Nr^2 (2.025950275^2 + 1)^2] + [(2.025950275^2 - 1)]^2}^0.5
/ {[2.025950275] {Ln[1 + (1 / 2.025950275^2)]}}
 
= {1 - [(1.059875912 / (5.104474517)]^2}
[Pi / 8] Nt {[0.7278688525^2 (5.104474517)^2] + [(3.104474517)]^2}^0.5
/ {[2.025950275] {Ln[1.243636547]}}
 
= {1 - 0.0431129722}
[Pi / 8] Nt {[13.80410806] + 9.637762027}^0.5
/ {0.4417377661}
 
= {0.9568870278}
[Pi / 8] (305) 4.841680503
/ {0.4417377661}
 
= 1256.17927

Hence a magnetic flux quantum (h / Q) as measured with a Josephson junction is:
1256.17927 X Phip
 

Note that for a particular Phip direction Phit has two possible directions. In quantum mechanics these two Phit values are referred to as spin states.

Units check:
F = Q V B
F D = Q V B D
J = coul (m / s) T m
1 / T = coul m^2 / s J

This web page last updated February 16, 2018.

Home Energy Nuclear Electricity Climate Change Lighting Control Contacts Links