Home Energy Nuclear Electricity Climate Change Lighting Control Contacts Links


XYLENE POWER LTD.

SPHEROMAK MAGNETIC MOMENT

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

MAGNETIC MOMENT:
Each charged particle type has a characteristic magnetic moment that can be accurately measured. This web page focuses on measurement of a particle's magnetic moment and calculation of particle spheromak parameters Ro and Bpo that can be calculated from the magnetic moment.

Let an externally applied magnetic field be dBx. A spheromak can be viewed as a constant current loop that interacts with the externally applied magnetic field. Assume for a moment that the symmetry axis of the spheromak is perpendicular to the applied magnetic field. For ease of calculation assume that the current loop is rectangular with sides of length La and Lb.

The energy difference between a spheromak position aligned to the applied magnetic field and a spheromak position directly opposed to the applied magnetic field can be calculated by integrating the (torque X change in angle) necessary to change the spheromak from an opposed to an aligned state.

Consider a thin rectangular current loop with sides of length La and Lb. Let the charge on this loop be Qs. The charge per unit length on this loop is:
Qs / [2(La + Lb)]

An element of charge is:
dQ = Qs dL / [2(La + Lb)]
where dL is an element of loop length, giving:
dQ / dL = Qs / [2 (La + Lb)]

The current I due to charge flowing at velocity V around this loop is:
I = Qs V / [2(La + Lb)]

The peak force acting on a segment of the loop of length La is:
Fp = (dQ / dL) V dBx La
= {Qs / [2 (La + Lb)]} V dBx La
= I dBx La,
where I is the loop current.

The peak torque resisting twisting the loop is:
T = 2 Fp (Lb / 2)
= Fp Lb
= I dBx La Lb

The torque as a function of the angle = Angle between the external magnetic field axis and the spheromak axis of symmetry is:
T = I dBx La Lb sin(Angle)

An elemental change in energy dE is:
dE = T d(Angle)

The change in spheromak potential energy when the spheromak changes from an aligned to an opposed orientaion is:
E2 - E1 = Integral from Angle = 0 to Angle = Pi of:
I Bx La Lb sin(Angle) d(Angle)
= 2 I dBx La Lb
= 2 (I A) dBx
where:
A = loop cross sectional area.

The product:
M = (I A) for a thin current ring is known as the ring's magnetic moment.

For a the more complex geometry of a spheromak the magnetic moments add.

However, I and A both vary with position on the spheromak wall.

The distance from a point on the spheromak wall to the axis of symmetry of the spheromak is Rw.

Let Phi be the angle between the spheromak equatorial plane and a vector from the toroidal region center line to a point (Rw, Hw) on the spheromak wall.

To evaluate the spheromak magnetic moment we need to find for points on the spheromak wall expressions for Rw as a function of Phi and (poloidal surface current density) as a function of Phi.

From the web page titled: ELECTROMAGNETIC SPHEROMAK:
Rw = [(Rs + Rc) / 2] - [(Rs - Rc) / 2] cos (Phi)
or
[(Rs + Rc) / 2] - Rw = [(Rs - Rc) / 2] cos (Phi}
or
cos(Phi) = {[(Rs + Rc) / 2] - Rw} {2 / (Rs - Rc)
= (Rs + Rc - 2 Rw) / (Rs - Rc)

Hence:
- sin(Phi) dPhi = - 2 dRw / (Rs - Rc)
or
dPhi = 2 dRw / [(Rs - Rc) sin(Phi)]
= 2 dRw / [(Rs - Rc) (1 - cos^2(Phi))^0.5]
= 2 dRw / [(Rs - Rc) (1 - [(Rs + Rc - 2 Rw) / (Rs - Rc)]^2)^0.5]
= 2 dRw / [((Rs - Rc)^2 - [(Rs + Rc - 2 Rw)]^2)^0.5]

From the web page titled: ELECTROMAGNETIC SPHEROMAK:
At every point on the charge hose:
Ih^2 = Ip^2 + It^2

Define:
dL = element of length along the charge hose;
d(Theta) = element of angle around the spheromak axis of symetry;
d(Phi) = element of angle around the torus center line with respect to the equatorial plane;

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

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

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

There are Np charge hoses at equally spaced angular intervals of dPhi = (2 Pi / Np) radians. Hence the poloidal surface current density is:
(Poloidal current per hose)(Hoses per radian)(dPhi)
Ip (Np / 2 Pi) dPhi.

The corresponding element of magnetic moment is:
Ip (Np / 2 Pi) dPhi (Pi Rw^2)

The total spheromak magnetic moment is:
M = Integral from Phi = 0 to Phi = 2 Pi of:
Ip (Np / 2 Pi) d(Phi)(Pi Rw^2).  
= Integral from Phi = 0 to Phi = Pi of:
2{Ih Np Rw / {[Np Rw]^2 + [(Rs - Rc) Nt / 2]^2}^0.5} {(Np / 2 Pi) d(Phi)}(Pi Rw^2)
 
= Integral from Phi = 0 to Phi = Pi of:
{2 Ih Np^2 Rw^3 / {[Np Rw]^2 + [(Rs - Rc) Nt / 2]^2}^0.5} {(1 / 2 ) d(Phi)}
 
= Integral from Rw = Rc to Rw = Rs of:
{Ih Np^2 Rw^3 / {[Np Rw]^2 + [(Rs - Rc) Nt / 2]^2}^0.5}
{2 dRw / [((Rs - Rc)^2 - [(Rs + Rc - 2 Rw)]^2)^0.5]}
 
= Integral from Rw = Rc to Rw = Rs of:
(Q Np Fh) {Np Rw^3 / Rc {[Np Rw / Rc]^2 + [((Rs / Rc) - 1) Nt / 2]^2}^0.5}
{2 dRw / [Rc ([(Rs / Rc) - 1]^2 - [(Rs / Rc) + 1 - 2 (Rw / Rc)]^2)^0.5]}
 
M = Integral from Rw = Rc to Rw = Rs of:
(Q Np Fh S^2) {Np Rw^3 / Ro {[Np S Rw / Ro]^2 + [(S^2 - 1) Nt / 2]^2}^0.5}
{2 dRw / [Ro ([S^2 - 1]^2 - [S^2 + 1 - 2 (S Rw / Ro)]^2)^0.5]}
 
M = Integral from Rw / Ro = 1 / S to Rw / Ro = S of:
(Q Np Fh S^2 Ro^2) {Np Rw^3 / Ro^3 {[Np S Rw / Ro]^2 + [(S^2 - 1) Nt / 2]^2}^0.5}
{2 (dRw / Ro) / [([S^2 - 1]^2 - [S^2 + 1 - 2 (S Rw / Ro)]^2)^0.5]}
 
M = Integral from Rw / Ro = 1 / S to Rw / Ro = S of:
(Q Np Fh S^2 Ro^2) {Np Rw^3 / Ro^3 {[Np S Rw / Ro]^2 + [(S^2 - 1) Nt / 2]^2}^0.5}
{2 (dRw / Ro) / [([S^4 - 2 S^2 + 1] - [(S^2 + 1)^2 + 4 (S Rw / Ro)^2 - 4 (S^2 + 1)(S Rw / Ro)])^0.5]}
 
M = Integral from Rw / Ro = 1 / S to Rw / Ro = S of:
(Q Np Fh S^2 Ro^2) {Np Rw^3 / Ro^3 {[Np S Rw / Ro]^2 + [(S^2 - 1) Nt / 2]^2}^0.5}
{2 (dRw / Ro) / [( - [ 4S^2 + 4 (S Rw / Ro)^2 - 4 (S^2 + 1)(S Rw / Ro)])^0.5]}
 
M = Integral from Rw / Ro = 1 / So to Rw / Ro = So of:
(Q Np Fh So^2 Ro^2) {Np Rw^3 / Ro^3 {[Np So Rw / Ro]^2 + [(So^2 - 1) Nt / 2]^2}^0.5}
{(dRw / Ro) / [([ - So^2 - (So Rw / Ro)^2 + (So^2)(S Rw / Ro) + (So Rw / Ro)])^0.5]}

Let X = So Rw / Ro  
M = Integral from X = 1 to X = So^2 of:
[(Q Np Fh Ro^2)/ So^2] {Np X^3 / {[Np X]^2 + [(So^2 - 1) Nt / 2]^2}^0.5}
{dX / [([ - So^2 - (X)^2 + (So^2 + 1)(X)])^0.5]}
 
M = Integral from X = 1 to X = So^2 of:
[(Q Np Fh Ro^2)][1 / So^2] {X^3 / {[X]^2 + [(So^2 - 1) / (2 Nr)]^2}^0.5}
{dX / [[(X - 1)(So^2 - X)]^0.5]}
 

APPROXIMATE SOLUTION:
For a charged particle spheromak the total circulating current is:
Q Np Fh
and the approximate area enclosed by the circulating current is:
Pi Ro^2

Hence the approximate magnetic moment is:
M ~ Q Np Fh Pi Ro^2
 

PRECISE SOLUTION:
Hence the exact magnetic moment is given by:
M = Integral from X = 1 to X = So^2 of:
[(Q Np Fh Pi Ro^2)][1 / (Pi So^2)] {X^3 / {[X]^2 + [(So^2 - 1) / (2 Nr)]^2}^0.5}
{dX / [[(X - 1)(So^2 - X)]^0.5]}
or
M = [(Q Np Fh Pi Ro^2)] Im
where:
Im = Integral from X = 1 to X = So^2 of:
[1 / (Pi So^2)] {X^3 / {[X]^2 + [(So^2 - 1) / (2 Nr)]^2}^0.5}
{dX / [[(X - 1)(So^2 - X)]^0.5]}

Note that the denominator of this integral blows up at X = 1 and at X = So^2.

Note that Im is a scaling factor of the order of unity. The integration for Im needs precise numerical evaluation. This integration provides a precise relationship between magnetic moment M and the product (Fh Ro^2).
 

NUMERICAL EVALUATION OF Im:
So = 2.026
So^2 = 4.104676
[1 / (Pi So^2)] = 0.0775481804
Nr = (222 / 305) = 0.7278688525
[(So^2 - 1) / (2 Nr)]^2 = [3.104676 / 1.455737705]^2 = 4.548480189

Thus:
Im = Integral from X = 1 to X = So^2 of:
[1 / (Pi So^2)] {X^3 / {[X]^2 + [(So^2 - 1) / (2 Nr)]^2}^0.5}
{dX / [[(X - 1)(So^2 - X)]^0.5]
 
Im = Integral from X = 1 to X = 4.104676 of:
[0.0775481804] {X^3 / {[X]^2 + 4.548480189}^0.5}
{dX / [[(X - 1)(4.104676 - X)]^0.5]}
or
Im = Integral from X = 1 to X = 4.104676 of:
[0.0775481804 X^3 dX] / [[(X^2 + 4.548480189)(X - 1)(4.104676 - X)]^0.5]

Numerical evaluation of this integral using the Graph program gives:
Im = 1.544

From the web page titled: PLANCK CONSTANT:
Fh = [C / (Pi Nt Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
or
Fh Ro = [C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Hence:
M = [(Q Np Fh Pi Ro^2)] Im
= [(Q Np Pi Ro)] Im [C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
or
[1 / Ro] = [(Q Np Pi)] Im [C / (M Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
  = [Q C / M] Im [So Nr / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

This equation allows precise determination of Ro from an experimental measurement of magnetic moment M.
 

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

The total electromagnetic field energy content of a spheromak is given by:
Ett = h Fh = [h C / (Pi Nt Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
 

OBTAINING SPHEROMAK ENERGY FROM MEASUREMENT OF MAGNETIC MOMENT:
Ett = [h C / (Pi Nt Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Recall that:
[1 / Ro] = [Q C / M] Im [So Nr / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Hence:
Ett = [h C / (Pi Nt Ro)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
 
= [h C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5][Q C / M]
Im [So Nr / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5
 
= [h C / (Pi Nt)] [So^2 Nr / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}][Q C / M] Im
 
= [1 / M] [Q C^2 Im h / Pi Nt][So^2 Nr / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}]

This equation gives the spheromak energy in terms of the spheromak magnetic moment M.
 

NUMERICAL EVALUATION OF Ett:
Nr = 0.7278688525
Nr^2 = 0.5297930664
So = 2.025950275
So^2 = 4.104474517
Nt = 305
h = 6.62606597 X 10^-34 m^2 kg / s
Im = 1.544
Q = 1.60217656 X 10^-19 coul
C = 2.99792458 X 10^8 m / s

[So^2 Nr / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}]
= [4.104474517 (0.7278688525) / {[ 0.5297930664 (5.104474517)^2] + [(3.104474517)]^2}]
= [4.104474517 (0.7278688525) / {13.80410806 + 9.637762027}
= [4.104474517 (0.7278688525) / {23.44187009}]
=0.1274437212

[Q C^2 Im h / Pi Nt]
= [1.60217656 X 10^-19 coul X (2.99792458 X 10^8 m / s)^2 X 1.544 X 6.62606597 X 10^-34 m^2 kg / s] / [3.14159 X 305]
= 0.1537465877 X 10-37 coul m^2 m^2 kg / s^3
= 0.1537465877 X 10-37 coul m^2 J / s

Thus the spheromak energy is given by:
Ett = [1 / M] [0.1274437212] [0.1537465877 X 10-37 coul m^2 J / s]
= [1 / M] [0.0195940373 X 10^-37 coul m^2 J / s]
= [1 / M] [1.95940373 X 10^-39 coul m^2 J / s]

Eg For a proton M = 14.10626497 X 10^-27 J / T

Hence for a proton:
Ett = [1.95940373 X 10^-39 coul m^2 J / s] / [14.10626497 X 10^-27 J / T]
= 0.138903 X 10^-12 coul (m^2 J T) / (s-J)
= 0.138903 X 10^-12 coul (m^2 kg) / (s^2 coul)
= 0.138903 X 10^-12 J
= 0.138903 X 10^-12 J X 1 eV / 1.60217656 X 10^-19 J
= 0.0866964 X 10^7 eV
= 0.866964 MeV

The spheromak field energy calculared from the proton magnetic moment is only a small fraction of the proton rest mass. This calculation of spheromak field energy indicates that energy released by hydrogen to form helium-4 is more than just spheromak electromagnetic field energy. There is not enough positive spheromak electromagnetic field energy available to account for the total release of energy (17.6 MeV) that occurs when tritium and deuterium combine to form He-4. The spheromak field energy Ett calculated from the measured proton magnetic moment is only a small fraction of the proton's rest mass energy.

An important conclusion is that most of a proton's inertial rest mass is not directly due to its spheromak electric and magnetic fields. When nucleons get sufficiently close they form a deep mutual potential energy well. This deep nuclear potential energy well is not accounted for by the electromagnetic spheromak model. There must be another short range inter-nucleon interaction that produces negative energy to account for nuclear binding energy.

The result is analogous for electrons. The spheromak field energy calculated from the electron magnetic moment only accounts for a small fraction of the electron rest mass.

One possible explanation of this data is that somehow the formulae connecting the spheromak field energy and the spheromak magnetic moment are incorrect. Eg If the field energy is systematically two orders of magnitude larger for a particular measured magnetic moment then the spheromak field energies would account for much if not all of the proton and electron rest masses. This issue needs further investigation. Another possible route is to recalculate the magnetic moment from the change in energy due to a change in the magnetic flux quantum in the presence of an externally applied magnetic field.
 

MEASUREMENT OF MAGNETIC MOMENT OF ATOMIC PARTICLES:
If the orientation of a magnetic moment transitions in an external magnetic field dBx the amount of energy (E2 - E1) absorbed during the transition is:
E2 - E1 = Integral from Angle = 0 to Angle = Pi of:
I A Bx sin(Angle)d(Angle)
= I A dBx [- Cos(Pi) + Cos(0)]
= 2 I A dBx
= 2 M dBx

However in terms of photon emission or absorption:
E2 - E1 = h dFh

Thus: h dFh = 2 M dBx
or
dFh / dBx = 2 M / h

Hence:
M = (h / 2) (dFh / dBx)

This formula is used to measure the magnetic moments of various atomic particles by measuring the frequency of the photons emitted or absorbed from a sample in a strong magnetic field.
 

NUMERICAL DATA:
Mu = 4 Pi X 10^-7 T^2 m^3 / J
Qa = 1.602 X 10^-19 coul
C = 2.99792458 X 10^8 m / s
h = 6.62606597 X 10^-34 m^2 kg / s
Me = 9.109 10-31 kg
Pi = 3.14159
Np = 222
Nt = 305
Nr = 0.7278688525
Nr^2 = 0.5297930664
So = 2.025950275
So^2 = 4.104474517

Units:
coul (m / s) T = kg m / s^2
or
coul / kg = (1 / T-s)
or
T = kg / coul s
and
J = kg m^2 / s^2
giving:
J / T = kg m^2 coul s/ s^2 kg = m^2 coul / s

Thus:
J / T^2 m^2 coul = (m^2 coul) / (s T m^2 coul) = 1 / s T
 

EXPERIMENTAL DATA:
For a proton experimental measurements give:
(dFh / dBx) = (42.5781 X 10^6 Hz / T)

Hence the magnetic moment M of a proton is:
M = (h / 2) (dFh / dBx)
= (6.62606597 X 10^-34 m^2 kg / s)(1 / 2)(42.5781 X 10^6 Hz / T)
= 141.0626497 X 10^-28 J / T
= 14.10626497 X 10^-27 J / T
 

PHYSICAL SIZE OF A PROTON SPHEROMAK:
Find the nominal radius Ro of a proton spheromak:
[1 / Ro] = [Q C / M] Im [So Nr / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Numerical Term Evaluation:
[So^2 / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}]
= [4.104474517 / {[0.5297930664 (5.104474517)^2] + [(3.104474517)]^2}]
= [4.104474517 / {[13.80410806] + [9.637762027]}]
= [4.104474517 / {23.44187009}]
= 0.175091599

[So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
= 0.418439

Ro = [M / (Q C Im Nr)] [{[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5] / So]
= [M / (Q C Im Nr)] [{[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5] / So]
= [(14.10626497 X 10^-27 J / T) / (1.602 X 10^-19 coul X 2.99 X 10^8 m / s X Im X (222 / 305))][1 / 0.418439]
= ( 1 / Im)(305 / 222)[(14.10626497 X 10^-27 J / T) / (1.602 X 10^-19 coul X 2.9979 X 10^8 m / s X 0.418439)]
= ( 1 / Im)(305 / 222)[(14.10626497 X 10^-27 J / T) / (1.602 X 10^-19 coul X 2.9979 X 10^8 m / s X 0.418439)]
= (1 / Im) 9.6437 X 10^-16 J s / T coul m
= (1 / Im) 9.6437 X 10^-16 J s coul s / kg coul m
= (1 / Im) 9.6437 X 10^-16 m
= (1 / Im) 0.96437 fermi

= (1 / 1.544) 0.96437 fermi
Ro = 0.62459 fermi

As shown on the web page titled: SPHEROMAK ENERGY the relationship between the classical proton charge radius Re and Ro is:
Re = (4 / Pi) Ro

Hence we calculate the classical proton charge radius Re to be:
Re = (4 / Pi) Ro
= (4 / Pi) (0.62459 fermi)
= 0.795 fermi

We calculate Rs to be:
Rs = So Ro = 2.026 (0.62459 fermi)
= 1.265 fermi

These figures compares to a published "proton charge radius" in the range .84 fermi to 0.87 fermi.
 

QUANTIFICATION OF Bpo FOR A PROTON:
Recall from web page titled: MAGNETIC FLUX QUANTUM that:
Bpo Ro^2 Pi = Phip / {Ln[1 + (1 / So^2)]}
or
Bpo = [Phip / {Ro^2 Pi Ln[1 + (1 / So^2)]}
= [3.290157699 X 10^-18 T m^2 / {(0.93001 X 10^-30 m^2 / Im^2) Pi Ln[1 + (1 /4.104474517)]}
= Im^2 [3.290157699 X 10^-18 T m^2 / {(0.93001 X 10^-30 m^2) Pi [0.2180397]}
= Im^2 X 4.6331 X 10^12 T
= 11.045 X 10^12 T

Thus Bpo for a proton is about 11.045 X 10^12 T

Hence the externally applied magnetic field is tiny compared to the spheromak core magnetic field strength.
 

This web page last updated February 18, 2018.

Home Energy Nuclear Electricity Climate Change Lighting Control Contacts Links