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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 dBx 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
 

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 uniform strong magnetic field. Note that for each particle the parameter that is actually measured is (dFh / dBh).

The observed value of (dFh / dBx) 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
 

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]}
 

APPROXIMATE MAGNETIC MOMENT 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
leading to:
dF / dBx = 2 M / h
= 2 Q Np Fh Pi Ro^2 / h
= [Q Pi Ro / h] [2 Np (Fh Ro)]
= [Q Pi Ro / h] [2 Np][C / (Pi Nt)] [So / [(Nr (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}
 

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: dF / dBx = (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 (dF / dBx) to the approximate solution for (dF / dBx) 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.
 

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 is not just one spheromak. As a minimum a proton must consist of a small radius spheromak that contains the proton rest mass and a larger radius spheromak that causes the proton magnetic moment.

For an electron published experimental data indicates that:
M = (h / 2) (dFh / dBx)
= -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 must consist of at least two concentric spheromaks to account for both its rest mass and its magnetic moment.
 

ALTERNATE ANALYSIS:
A much simpler but less precise a methodology of analyzing a spheromak in an externally applied magnetic field is to assume that the dominant effect of the external magnetic field is to change the spheromak peak core magnetic field Bpo. This assumption is poor because most of the spheromak magnetic moment is due to current circulating at the spheromak outer rim which is far from the center of the spheromak where Bpo is known. Recall from the web page titled: ELECTROMAGNETIC SPHEROMAK that:
Bpo = (Mu C Qs) / (4 Pi Ro^2)
or
Ro^2 = (Mu C Qs) / (4 Pi Bpo)
which implies that Bpo determines Ro.

Further recall from the same web page that:
Uo = Bpo^2 / (2 Mu)
and that:
Efs = Uo Ro^3 Pi^2

Hence combining these three equations gives:
Efs = Uo Ro^3 Pi^2
= [Bpo^2 / (2 Mu)] Ro^3 Pi^2
= [[(Mu C Qs) / (4 Pi Ro^2)]^2 / (2 Mu)] Ro^3 Pi^2
= (Mu C^2 Qs^2) / (32 Ro)
or
dEfs = [(Mu C^2 Qs^2) / (32)] d(1 / Ro)

Recall that:
Bpo = (Mu C Qs) / (4 Pi Ro^2)
or
(1 / Ro)^2 = (4 Pi Bpo) / (Mu C Qs)
or
(1 / Ro) = [(4 Pi Bpo) / (Mu C Qs)]^0.5
or
d(1 / Ro) = [(4 Pi) / (Mu C Qs)]^0.5(1 / 2)Bpo^-0.5 dBpo
= [Pi / (Mu C Qs Bpo)]^0.5 dBpo

Hence:
dEfs = [(Mu C^2 Qs^2) / (32)] d(1 / Ro)
= [(Mu C^2 Qs^2) / (32)] [Pi / (Mu C Qs Bpo)]^0.5 dBpo
= [(Mu C^2 Qs^2) / (32)] [Pi / (Mu C Qs {(Mu C Qs) / (4 Pi Ro^2)})]^0.5 dBpo
= [(Mu C^2 Qs^2) / (32)] [(Pi^2 4 Ro^2) / ((Mu C Qs)^2)]^0.5 dBpo
= [(Mu C^2 Qs^2) / (32)] [(Pi 2 Ro) / ((Mu C Qs))] dBpo
= [(C Qs) / (32)] [Pi 2 Ro] dBpo
= [(C Qs Pi Ro) / (16)] dBpo

However, when a spheromak changes from aligned to non-aligned the change in magnetic field that the spheromak experiences is:
dBpo = dBx - (- dBx) = 2 dBx.
Thus:
dEfs = [(C Qs Pi Ro) / (16)][2 dBx]
or
(dEfs / dBx) = [(C Qs Pi Ro) / 8]

Thus:
Ett = (Ett / Efs) Efs
or
dEtt = (Ett / Efs) dEfs
= (Ett / Efs)[(C Qs Pi Ro) / (8)] dBx
giving:
dEtt / dBx = (Ett / Efs)[(C Qs Pi Ro) / 8]

However:
dEtt = h dFh
giving:
dFh / dBx = (Ett / Efs)[(C Qs Pi Ro) / (8 h)]

The web page: PLANCK Constant gives:
(Ett / Efs) = {1 - [(So -1)^2 / (So^2 + 1)]^2}

Hence:
dFh / dBx = (Ett / Efs)[(C Qs Pi Ro) / (8 h)]
= [(C Qs Pi Ro) / (8 h)]{1 - [(So -1)^2 / (So^2 + 1)]^2}
= [(C Qs Ro) / h] [Pi / 8] {1 - [(So -1)^2 / (So^2 + 1)]^2}

The coefficient of [(C Pi Ro) / h] is:
[Pi / 8] {1 - [(So - 1)^2 / (So^2 + 1)]^2}
= [Pi / 8] {1 - [(2.025950275 - 1)^2 / (4.104474517 + 1)]^2}
= [Pi / 8] {1 - [(1.025950275)^2 / (5.104474517)]^2}
= [Pi / 8] {0.9574790294}
= 0.3760
 

METHODOLOGY COMPARISON:
Recall that the precise magnetic moment analysis gave:
dFh / dBx = 2 M / h
= [(Q C Ro) / (So^2 h)] [Im] [2 Nr / Pi] [So / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}^0.5]

Numerical evaluation of the coefficient of [(C Qs Ro) / h] in the precise calculation gives:
[2 Nr / Pi][Im / So^2] {So / [(Nr (So^2 + 1))^2 + (So^2 - 1)^2]^0.5}
= [2 ( 0.7278688525) / 3.14159][19.9349 / 4.104676]{2.025950275 / [(0.7278688525 (5.104474517))^2 + (3.104474517)^2]^0.5}
= [0.4633760946][4.856631802] [2.025950275 / [(13.80410806) + (9.637762027)]^0.5}
= [0.4633760946] [4.856631802] {2.025950275 / [4.841680503]}
= 0.94167

The ratio of the two coefficients of [(C Qs Ro) / h] is:
0.94167 / 0.3760 = 2.504
 

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

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) (0.62418 fermi)
= 0.795 fermi

We calculate Rs to be:
Rs = So Ro = 2.026 (0.62418 fermi)
= 1.265 fermi

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

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 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 February 26, 2018.

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