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**CONCENTRIC SPHEROMAKS:**

The mathematical model for a single spheromak with a quantum charge can be applied to electrons and protons. In each case if the nominal radius Ro is chosen so that the theoretical and experimental magnetic moments agree then the calculated spheromak mass is much smaller than the experimentally measured particle mass. Hence a single quantum charge spheromak does not adequately model a real proton or a real electron.

To make theoretical calculations match experimental data it is necessary to assume that quantum charged particles such as electrons and protons consist of at least two concentric spheromaks. The larger outer spheromak provides most of the particle magnetic moment and a smaller inner spheromak provides most of the particle mass. These spheromaks share a common main axis of symmetry. The value of Ro for the outer spheromak, which is the main source of the particle magnetic moment, is about 200X larger than the value of Ro for the inner spheromak, which is the main source of the particle rest mass.

**STATIC CHARGE:**

The inner spheromak(s) and outer spheromak must have static charges such that the net static charge for an electron is - Q and the net static charge for a proton is + Q.

If the spheromak static charges conform to quark theory (Standard Model) then for a proton there are two positively charged concentric spheromaks each with static charge + (2 / 3) Q and one negatively charged spheromak with static charge - (1 / 3) Q.

If the spheromak static charges conform to quark theory (Standard Model) for a neutron then there must be one spheromak with static charge + (2 / 3) Q and two concentric spheromaks each with charge - (1 / 3) Q.

**IGNORING QUARK THEORY:**

In trying to match spheromak static charge to Standard Model quark theory there is a problem with the Planck constant which contains one whole quantum charge. Any spheromak that emits or absorbs photons requires an integral quantum static charge to conform to the Plank constant. That condition seems to require that the other spheromaks also have integral quantum charges. For example:

Electron = (-ve outer spheromak) + (neutral inner spheromak)

Proton = (+ve outer spheromak) + (neutral central spheromak)

Neutron = (+ve outer spheromak) + (-ve charge inner spheromak)

**COMMENTS:**

1) The neutral inner spheromaks are all highly unstable outside the stabilizing fields of their outer spheromaks. Hence these neutral inner spheromaks are never seen as stable free particles.

On decay the inner spheromak of a neutron emits a -ve portion that forms a free electron and the remainder becomes the neutral central spheromak of the proton.

**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

**NUMERICAL 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

= 2.67193 X 10^-4 X 10^12 coul / kg

= 2.67193 X 10^8 coul / kg

Thus:

**dFh / dBx** = [2 Q Nr / Pi Nt Me] [So^2 / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}]

= [2.67193 X 10^8 coul / kg] [0.175091599]

= 0.467833 X 10^8 coul / kg

= 46.7833 X 10^6 coul / kg

= 46.733 X 10^6 (1 / T-s)

= **46.733 MHz / T**

This value compares to the experimentally measured value for protons of:

**(dF / dBx) = (42.5781 X 10^6 Hz / T)**

even though the mass Me of an electron was used as the electromagnetic energy content of the proton spheromak in the above calculation.

**PHYSICAL SIZE OF A CHARGED PARTICLE SPHEROMAK:**

**Find the nominal radius Ro for proton spheromaks:**

Assume that the total proton spheromak field energy is approximately equal to the rest mass energy of an electron. Then from the web page titled: SPHEROMAK ENERGY

Efs = (Mu C^2 / 2)[Qa / 4]^2 [1 / Ro]

and

Ett = (Ett / Efs) Efs

= Me C^2

Thus:

Efs = Me C^2 / (Ett / Efs)

= (Mu C^2 / 2)[Qa / 4]^2 [1 / Ro]

Hence:

(Mu C^2 / 2)[Qa / 4]^2 [1 / Ro] = Me C^2 (Efs / Ett)

or

Ro = (Mu / 2)[Qa / 4]^2 [1 / Me] [Ett / Efs]

From the web page titled: PLANCK CONSTANT

(Ett / Efs) = {1 - [(So -1)^2 / (So^2 + 1)]^2}

So = 2.025950275

So^2 = 4.104474517

Numerical substitution gives:

(Ett / Efs) = {1 - [(So -1)^2 / (So^2 + 1)]^2}

= {1 - [(1.025950275)^2 / (5.104474517)]^2}

= **0.959602867**

Then:
Ro = [4 Pi X 10^-7 T^2 m^3 / J][1 / 2][1.602^2 X 10^-38 coul^2 / 16][0.959602867] / 9.109 X 10^-31 kg

= 0.10613 X 10^-14 T^2 m^3 coul^2 / J kg

= 1.0613 X 10^-15 (kg / coul s)^2 m^3 coul^2 / J kg

= 1.0613 X 10^-15 (kg m^3 / s^2 J)

= 1.0613 X 10^-15 m

= **1.0613 fermi**

This figure compares to a published "proton charge radius" in the range .84 fermi to 0.87 fermi.

**QUANTIFICATION OF Bpo:**

Recall from web page titled: NUCLEAR 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 / {(1.0613 X 10^-15 m)^2 Pi Ln[1 + (1 /4.104474517)]}

= [3.290157699 X 10^-18 T m^2 / {(1.12635769 X 10^-30 m^2) Pi [0.2180397]}

= **4.264 X 10^12 T**

Thus Bpo for a charged particle spheromak with electromagnetic field energy equal to an electron rest mass is about **4.264 X 10^12 T**

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

An important conclusion from the above work is that assuming that protons are sensed by conventional NMR much of a proton's rest mass is not directly due to its electric and magnetic fields whereas with an electron its rest mass is almost entirely due to its electric and magnetic fields.

This web page last updated March 2, 2018.

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