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Mysterious Magnetic LENGTH lo



 
 
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Old November 24th 05, 04:13 PM posted to sci.space.policy,sci.philosophy.tech,sci.astro,rec.org.mensa
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Default Mysterious Magnetic LENGTH lo

$ The magnetic radius of the electron
(..NOT to be CONFUSED with "Mysterious Magnetic LENGTH"):
$ GUESS, Rydberg LENGTH loo & Mysterious Magnetic LENGTH lo:
[ Mysterious Magnetic LENGTH lo = loo / 4 = 1 / 4*Roo.!! ]

A. Using the (MYSTERiOUS MAGNETiC LENGTH)^2, lo^2 ..note;
Ra = 16*pi*lo^2*(Nsample*thickness) ..in 3-d
= 16*pi*lo^2*(NsampleMAX Number) ..in 2-d.

B. Using the (RYDBERG LENGTH)^2, loo^2 ..note, then;
Ra = pi*loo^2*(Nsample*thickness) ..in 3-d
= pi*loo^2*(Nsamplemax number) ..in 2-d.

1. MOLAR Ra = 2*(pi + 1) = no*Vm*k = Na*k = F*k / {e}= Cg*[S].
2. Rydberg LENGTH loo = 4*( Mysterious Magnetic LENGTH lo ).
= 1 / Roo = 4*lo.
3. Rydberg MASS Moo = (HARTREE energy eH) / c^2 = a^2*me.
4. Moo*loo / 4*pi = Mp*lp = Planck MASS*Planck LENGTH.
5. ("e") = Naperian ..or, Natural LOGaRiTHM base.
6. Cg = The GUESS iSS general COMPLEXiTY icon.
7. {e} = Ampere*second is elementary CHARGE.
8. [S] = The GUESS iSS ENTROPY.
9. Topic..
You're starting to catch on with the web-math. GOOD SHOW.!!

Note your... "Mu = Pi R^2 I = Q V R/2, so R = 2 Mu/(Q V).":
..whereas: Mu = Pi*R^2*I = Q*V*R/2, so R = 2*Mu/Q*V...!!

GUESS icon * is a multiplication SiGN that SEPARATEs icons.

OBSERViNG (LEFT side)/(No BOTTOM bracket) saved 2 SPACEs.!!

brian a m stuckless


Phil Gardner wrote:
If we have a charge, Q, uniformly distributed around a ring of
radius, R, and this rotates at a constant peripheral speed, V,
the current circulating is I = Q V/(2 Pi R). The magnetic moment
of this "spinning body" is then Mu = Pi R^2 I = Q V R/2,
so R = 2 Mu/(Q V).
For it we can define a "magnetic radius", Rm, as the lowest
radius attainable in our universe by the equation Rm = 2 Mu/(Q c).
We can very reasonably assume that this equation holds good for any
axially symmetric body for which we know only the values of Q and Mu
and that this includes all charged subatomic particles. For the
electron, Mu = 1.8566 x 10^-23 A m^2. It follows from this that
Rm = (1.8566 x 10^-23)/(1.60206 x 10^-19)(2.998 x 10^8) = 3.866 x
10^-13 m, ie approximately 137 x (classical electron radius). Since
this is a lower bound I can only assume from it that the electron is
not a point particle but an extended electron cloud of finite charge
density spinning at high speed.

Phil Gardner

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