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XYLENE POWER LTD.

SPHEROMAK WALL

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

INTRODUCTION:
In a theoretical spheromak the electric field inside the spheromakwall is zero.
However, spheromak theoryindicates that the charge Q is uniformly distributed along the length Lhof the charge filament. That condition demands that the spheromak cross sectionbe non round. On this web page we attempt to find an equationf for the spheromak cross section.

The spheromak wall defines the filament path. This pathis highly symmetric. An element dLh of this pathis: Let Theta = angle about thespheromak minor axis R = ro, Z = 0.

Let Phi = angle about the major ais of symmetry R = 0.

Consider a point on the spheromak wall. Let X = distance from R = Ro, Z = 0 to the point Rw, Zw on the spheromak wall. An eelement of length along the charge motion path is:
[d(Phi)[Ro + X cos(theta)]^2 + [X d(Theta)]^2 + dX^2 = [c dt]^2 = (dLh)^2

During time period Lh / C the spheromak goes through one cycle during which time the poloidal angle advanced by:
d(Phi) = Np 2 Pi
andthe toroidal angle advanced by:
d(Theta) = Nt 2 Pi

Hence if the angle of advancement is constant:
d(Phi) / d(Theta) = Np / Nt

The surface surface charge per unit of filament length is Q / Lh

Hence the charge per unit area is:
dQ / dA = (Q / Lh) dLh / {[d(Phi)[Ro + X(Theta) cos(Theta)][x(Theta) d(Theta)]} In order for the electric field to be zero along the spheromak's equatorial plane:
{[dQ / dA]|R = Rc)} Rc = {[dQ / dA]|R = Rs)} Rs
or
{[dQ / dA]|Theta = 0)} Rc = {[dQ / dA]|Theta = Pi)} Rs
or
At Theta = 0, X = (Rs - Ro), cos(Theta) = 1, dX = 0
At Theta = Pi, X = Ro - Rc), cos(Theta) = -1, dX = 0
Hence:
{(Q / Lh) dLh / {[d(Phi)[Ro + X(Theta) cos(Theta)][x(Theta) d(Theta)]} |Theta = 0} Rc
= {(Q / Lh) dLh / {[d(Phi)[Ro + X(Theta) cos(Theta)][x(Theta) d(Theta)]} |Theta = Pi} Rs
or
{(Q / Lh) dLh / {[d(Phi)[Ro + (Rs - Ro) 1][(Rs - Ro) d(Theta)]} } Rc
= {(Q / Lh) dLh / {[d(Phi)[Ro + (Ro - Rc) (- 1)][(Ro - Rc) d(Theta)]}} Rs
or
{(Q / Lh) dLh / {[d(Phi)[Rs][(Rs - Ro) d(Theta)]} } Rc
= {(Q / Lh) dLh / {[d(Phi)[Rc][(Ro - Rc) d(Theta)]}} Rs
or
{(Q / Lh) dLh / {[(Np / Nt) d(Theta)[Rs][(Rs - Ro) d(Theta)]}} Rc
= {(Q / Lh) dLh / {[(Np / Nt) d(Theta)[Rc][(Ro - Rc) d(Theta)]}} Rs
or
Rc / {[(Np / Nt)[Rs][(Rs - Ro)]}
= Rs / {[(Np / Nt)[Rc][(Ro - Rc)]}
or
Rc / {[Rs][(Rs - Ro)]}
= Rs / {[[Rc][(Ro - Rc)]}
or
(Rc^2 (Ro - Rc) = Rs^2 (Rs - Ro) or
Ro (Rc^2 + Rs^2) = Rc^3 + Rs^3 or
Ro = [(Rs^3 + Rc^3) / (Rs^2 + Rc)^2]

This equation gives us the radial position of the spheromak peak Ho.

In order for the electric field to be zero vertically in the spheromak:
{[dQ / dA]|Theta = Pi / 2} Ro = {[dQ / dA]|Theta = - Pi / 2} Ro

This web page last updated April 6, 2026.

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