# XYLENE POWER LTD.

## SPHEROMAK MAGNETIC MOMENT

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

MAGNETIC MOMENT:
Each charged particle type has a characteristic parameter known as its magnetic moment that can be accurately measured. This web page focuses on definition and measurement of a particle's magnetic moment and calculation of a spheromak's parameters that can be determined from the magnetic moment.

QUALITATIVE DESCRIPTION:
When a spheromak is immersed in a uniform external magnetic field Bz the system has two energy extremes. At the higher energy extreme the spheromak poloidal core magnetic field is aligned parallel with the external magnetic field so that the two magnetic field vectors add in the spheromak core and cancel at the spheromak perimeter.

The lower energy extreme occurs when the spheromak poloidal core magnetic field is aligned anti-parallel to the external magnetic field so the two magnetic field vectors cancel in the core and add at the spheromak perimeter.

Thus changing a spheromak from an aligned parallel to an aligned antiparallel state in a external magnetic field causes a change in total system energy.

DEFINITION OF MAGNETIC MOMENT M:
Let an externally applied uniform magnetic field be Bz.

The magnetic field configuration B(R, H) around a spheromak can be expressed as:
B(R, H) = Br(R, H) + Bh(R, H)
where Br is the cylindrically radial magnetic field vector component and Bh is the axial magnetic field vector component.

Note that Bh switches sign at:
R = (Rs + Rc) / 2
and Br switches sign at
H = 0

If in state "a" a small uniform external magnetic field Bz is applied parallel to the spheromak axis the magnetic field configuration becomes:
Ba(R, H) = Br(R, H) + Bh(R, H) + Bz
and the magnetic field energy density distribution becomes:
[Ba(R, H)]^2 / 2 Mu = [Br(R, H)]^2 / 2 Mu + [Bh(R, H) + Bz]^2 / 2 Mu

If in state "b" the magnetic field Bz is reversed in direction the magnetic field energy distribution becomes:
[Bb(R, H)]^2 / 2 Mu = [Br(R, H)]^2 / 2 Mu + [Bh(R, H) - Bz]^2 / 2 Mu

The difference in the magnetic field energy distributions between state "b" and state "a" is:
[Bb(R, H)]^2 / 2 Mu - [Ba(R, H)]^2 / 2 Mu
= [Br(R, H)]^2 / 2 Mu + [Bh(R, H) - Bz]^2 / 2 Mu - [Br(R, H)]^2 / 2 Mu + [Bh(R, H) + Bz]^2 / 2 Mu
= - 4 Bh(R, H) Bz / 2 Mu

Hence the change in energy DeltaE between state "b" and state "a" is given by:
DeltaE = Integral over volume V of [- 4 Bh(R, H) Bz dV / 2 Mu]

Thus:
DeltaE / Bz = [- 2 / Mu] Integral {Bh(R, H) 2 Pi R dR dH}
= - 2 M
where:
M = [1 / Mu] Integral {Bh(R, H) 2 Pi R dR dH}
is known as the magnetic moment

The parameter M can in principle be precisely computed for a spheromak of nominal radius Ro. However, that integration is difficult.

IDEAL CURRENT LOOP METHOD:
A spheromak can be viewed as an ideal 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 ideal 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 ideal 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 Bz La
= {Qs / [2 (La + Lb)]} V Bz La
= I Bz La,
where I is the loop current.

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

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

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

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

Thus:
(Eb - Ea) / Bz = 2 M

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

However in terms of photon emission or absorption:
Eb - Ea = h Fp
where Fp = photon frequency. Note that Fp and Fh are not simply connected due to energy contained in the externally applied magnetic field Bz.

Thus: h Fp = 2 M Bz
or
Fp / Bz = 2 M / h

Hence:
M = (h / 2) (Fp / Bz)

This formula is used to measure the magnetic moment M of various atomic particles by measuring the frequency of the photons emitted or absorbed from a sample in a uniform strong magnetic field. Note that for each particle the parameter that is actually measured is (Fp / Bz).

The observed value of (Fp / Bz) for a proton is (42.5781 X 10^6 Hz / T).

The corresponding calculated value of M for a proton is:
M = 14.10626497 X 10^-27 J / T

APPROXIMATE SOLUTION:
Assume a spheromak has an effective radius Ro. Then:
A = Pi Ro^2
and
I = Q Fh Np

Hence:
M = I A
= Q Fh Np Pi Ro^2

(Eb - Ea) / Bz = h Fp / Bz
= 2 M
= 2 Q Fh Np Pi Ro^2

Hence:
Fp / Bz = 2 Q Fh Np Pi Ro^2 / h
= 2 Q Ro Np Pi (Fh Ro) / h

From the web page titled: PLANCK CONSTANT:
(Fh Ro) = [C / (Pi Nt)] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]
giving:
Fp / Bz = [2 Q Ro Np Pi / h][C / (Pi Nt)] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]
= [2 Q Ro Nr C / h] [So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5]
which allows solving for Ro.

PRECISE CALCULATION OF M FOR A SPHEROMAK:
For the more complex geometry of a spheromak the individual elemental ring 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 geometry indicates that:
- 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 / Ro {[Np Rw / Ro]^2 + [((Rs / Ro) - (Rc / Ro)) Nt / 2]^2}^0.5}
{2 dRw / [Ro ([(Rs / Ro) - (Rc / Ro)]^2 - [(Rs / Ro) + (Rc / Ro) - 2 (Rw / Ro)]^2)^0.5]}

= Integral from Rw = Rc to Rw = Rs of:
(Q Np Fh So^2) {Np Rw^3 / Ro {[Np So Rw / Ro]^2 + [(So^2 - 1) Nt / 2]^2}^0.5}
{2 dRw / [Ro ([So^2 - 1]^2 - [So^2 + 1 - 2 (So Rw / Ro)]^2)^0.5]}

= 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}
{2 (dRw / Ro) / [([So^2 - 1]^2 - [So^2 + 1 - 2 (So Rw / Ro)]^2)^0.5]}

= 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}
{2 (dRw / Ro) / [([So^4 - 2 So^2 + 1] - [(So^2 + 1)^2 + 4 (So Rw / Ro)^2 - 4 (So^2 + 1)(So Rw / Ro)])^0.5]}

= 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}
{2 (dRw / Ro) / [( - [ 4 So^2 + 4 (So Rw / Ro)^2 - 4 (So^2 + 1)(So Rw / Ro)])^0.5]}

= 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 + 1)(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) / So^2] {X^3 / {[X]^2 + [(So^2 - 1) / (2 Nr)]^2}^0.5}
{dX / [[(X - 1)(So^2 - X)]^0.5]}

PRECISE MAGNETIC MOMENT SOLUTION:
Hence the exact magnetic moment is given by:
M = Integral from X = 1 to X = So^2 of:
[(Q Np Fh Ro^2)/ 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 Ro^2 / So^2] Im
where:
Im = Integral from X = 1 to X = So^2 of:
{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 = 2.025950275
So^2 = 4.104676
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:
{X^3 / {[X]^2 + [(So^2 - 1) / (2 Nr)]^2}^0.5}
{dX / [[(X - 1)(So^2 - X)]^0.5]

= Integral from X = 1 to X = 4.104676 of:
{X^3 / {[X]^2 + 4.548480189}^0.5}
{dX / [[(X - 1)(4.104676 - X)]^0.5]}

= Integral from X = 1 to X = 4.104676 of:
[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 = 19.9349

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 Ro^2 / So^2] Im
= [(Q Np Ro / So^2)] Im [C / (Pi Nt)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
= [(Q C Ro / So^2)] Im [Nr / Pi] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
which gives:
Fp / Bz = (2 / h) M
= (2 / h)[Q C Ro / So^2] Im [Nr / Pi] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
= [(Q C Ro) / (So^2 h)] [Im] [2 Nr / Pi] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Rearranging this equation gives:
(1 / Ro) = [Q C / So^2] [Im] [Nr / (M Pi)] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

This equation allows precise determination of Ro for a charged particle spheromak.

The ratio of the precise solution for (Fp / Bz) to the approximate solution for (Fp / Bz) is:
[(Q C Ro / So^2) / h] [Im] [2 Nr / Pi] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]
/ [Q Ro C / h] [2 Nr] {So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}
= (Im / So^2 Pi)
= 1.545915

Note that a plot of the Im integrand shows that most of Im is due to circulating current near the outer rim of the spheromak.

SPHEROMAK ELECTROMAGNETIC FIELD 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]

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

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

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

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

The experimentally measured magnetic moments of some particles are:
PARTICLE    MAGNETIC MOMENT (X 10^-27 J / T)SPIN QUANTUM NUMBER
Electron-9284.7641 / 2
Proton14.1060671 / 2
Neutron-9.662361 / 2
H-24.33073461
H-315.0460941 / 2
He-3-10.7461741 / 2
He-40.00

A very important observation is that in atomic nuclei the magnetic moments of protons and neutrons tend to cancel each other.
Let:
Mp = 14.106067 = relative proton magnetic moment
Mn = -9.66236 = relative neutron magnetic moment
Mp + Mn = 4.443707 ~ 4.3307346 = H-2 relative magnetic moment
Mp + Mn - Mn = 14.106067 ~ 15.046094 = H-3 relative magnetic moment
Mp - Mp + Mn = - 9.66236 ~ -10.746174 = He-3 relative magnetic moment
Mp - Mp + Mn - Mn = 0.0 ~ He-4 relative magnetic moment

Clearly protons and neutrons behave as distinct particles within an atomic nucleus. These particles magnetically bind together but can spontaneously move to minimize the total nuclear energy. We need to develop an expression that relates the magnetic moment of a neutron to its quasi-spheromak geometry.

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

Im = 19.9349
Q = 1.60217656 X 10^-19 coul

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

NUMERICAL EVALUATION OF Ett:
[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

[(Q C^2 Im h Nr) / (So^2 Pi^2 Nt)]
= [1.60217656 X 10^-19 coul X (2.99792458 X 10^8 m / s)^2
X Im X 6.62606597 X 10^-34 m^2 kg / s X 0.7278688525] / [(4.104474517)(3.14159)^2 X 305]
= 0.0056208792 X 10-37 Im coul m^2 m^2 kg / s^3
= 0.0056208792 X 10-37 (19.9349) coul m^2 J / s
= 0.1120516657 X 10^-37 coul m^2 J / s

Thus the total spheromak field energy is given by:
Ett = [1 / M] [(Q C^2 Im h Nr) / (So^2 Pi^2 Nt)][So^2 / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}]
= [1 / M] [0.1120516657 X 10^-37][0.175091599] coul m^2 J / s]
= [1 / M] [0.0196193053 X 10^-37] coul m^2 J / s]
= [1 / M] [1.96193053 X 10^-39 coul m^2 J / s]

For a proton published experimental data indicates that:
M = 14.10626497 X 10^-27 J / T

Hence for a proton the spheromak field energy is:
Ett = [1.96193053 X 10^-39 coul m^2 J / s] / [14.10626497 X 10^-27 J / T]
= 0.1390822116 X 10^-12 coul (m^2 J T) / (s-J)
= 0.1390822116 X 10^-12 coul (m^2 kg) / (s^2 coul)
= 0.1390822116 X 10^-12 J
= 0.1390822116 X 10^-12 J X 1 eV / 1.60217656 X 10^-19 J
= 0.08681785 X 10^7 eV
= .8681785 MeV

The spheromak field energy calculated from the proton magnetic moment is only 0.09235% of the proton rest mass energy of 938.257 MeV. Clearly a proton's structure is more complex than a single simple spheromak. As a minimum a proton must consist of a confined photon that causes most of the proton's rest mass plus a spheromak that causes the proton's charge and magnetic moment.

Consider a deuteron (H-2 nucleus):
M = 4.3307346 X 10^-27 J / T

Deuteron spheromak static field energy content is:
Ett = [1.96193053 X 10^-39 coul m^2 J / s] / M
= [1.96193053 X 10^-39 coul m^2 J / s] / 4.3307346 X 10^-27 J / T
= 0.4530248817 X 10^-12 coul m^2 T / s
= 0.4530248817 X 10^-12 J
= 0.4530248817 X 10^-12 J X 1 eV / 1.60217656 X 10^-19 J
= 0.2827559041 X 10^7 eV
= 2.827559041 MeV

By comparison the rest mass of a deuteron is:
3.343 583 719 x 10-27 kg
or
3.343 583 719 x 10-27 kg X [2.99792458 X 10^8 m / s]^2
= 30.05063183 x 10^-11 J

Thus the deuteron's static spheromak energy as a fraction of the total rest mass energy is:
0.4530248817 X 10^-12 J / 30.05063183 x 10^-11 J
= 0.00150753862
= 0.150753862%

For an electron published experimental data indicates that:
M = (h / 2) (Fp / Bz)
= -9284.764 X 10^-27 J / T

The corresponding electron spheromak field energy is:
Ett = [1 / M] [(Q C^2 Im h Nr) / (So^2 Pi^2 Nt)][So^2 / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}]

= [1 / (9284 X 10^-27 J / T)] [1.96193053 X 10^-39 coul m^2 J / s]
= 2.113238399 X 10^-4 X 10^-12 T coul m^2 / s
= 2.113238399 X 10^-16 [kg / coul s][coul m^2 / s]
= 2.113238399 X 10^-16 J
= 2.113238399 X 10^-16 J X 1 eV / 1.602 X 10^-19 J
= 1.319125 x 10^3 eV
= 0.001319125 MeV which is only 0.25814% of the electron rest mass energy of 0.511 MeV. It appears that an electron likely consists of a spheromak which causes its charge and magnetic moment plus a confined photon that accounts for most of its experimentally measured rest mass.

PHYSICAL SIZE OF A PROTON SPHEROMAK AS INDICATED BY ITS MAGNETIC MOMENT:
Find the nominal radius Ro of a proton spheromak:
(1 / Ro) = [Q C / So^2] [Im] [Nr / M Pi] [So / {[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 Pi So^2 / (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 / (Pi So^2)) X (222 / 305))][1 / 0.418439]
= ( (Pi So^2) / 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)]
= ( Pi So^2 / 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)]
= (Pi So^2 / Im) 9.6437 X 10^-16 J s / T coul m
= (Pi So^2 / Im) 9.6437 X 10^-16 J s coul s / kg coul m
= (Pi So^2 / Im) 9.6437 X 10^-16 m
= (Pi So^2 / Im) 0.96437 X 10^-15 m

= (1 / 1.545) 0.96437 fermi
Ro = 0.62418 fermi
where:
Ro = (Rs Rc)^0.5

This figure compares to the published "proton charge radius" from scattering experiments in the range .84 fermi to 0.87 fermi. Hence the effective scattering radius lies between Ro and Rs.

QUANTIFICATION OF Bpo FOR A PROTON AS INDICATED BY ITS MAGNETIC MOMENT:
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 / {(1.22398 X 10^-30 m^2) Ln[1 + (1 /4.104474517)]}
= [3.290157699 X 10^-18 T m^2 / {(1.22398 X 10^-30 m^2) [0.2180397]}
= 12.328 X 10^12 T

Thus as indicated by its magnetic moment Bpo for a proton is about 12.328 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 July 16, 2018.