XYLENE POWER LTD.

SPHEROMAK WALL

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

INTRODUCTION:
In a theoretical spheromak the electric field inside the spheromak wall is zero.
Spheromak theory indicates that the spheromak charge Q is uniformly distributed along the length Lh of the charge filament. That condition demands that the cross section of the spheromak looking along the spheromak's minor axis is not round.

Tthe local surface charge density S is an element of charge dQ spread over an element of spheromak wall surface area dA.

The surface charge density of a spheromak is S(Theta), where Theta is measured around the minor axis ring R = Ro, Z = 0.

The total sphermak charge Q is given by:
Q = 2 Integral from Theta = 0 to Theta = Pi of:
2 Pi R X(Theta) d((Theta)S(Theta) where: R = Ro + X(Theta)cos(Theta) or
Q = Integral from Theta = 0 to Theta= Pi of:
4 Pi [Ro + X(Theta) cos(Theta)]X(Theta) d((Theta) S(Theta)

However, to reduce the horizontal electric field to zero:
S(Thetas)[Ro+ X(Thetas) cos (Thetas)] = S(Pi - Theta)[Ro + X(Thetac) cos(Thetac)]

At R = Rc, Z = 0:
s(Theta) = S(Thetac)
X(Pi -Thetac) = (Ro - Rc)
cos(Thetac) = -1

At R = Rs, Z = 0:
S(Theta) = S(Thetas)
X(Thetas) = (Rs - Ro)
cos(Thetas) = 1

Hence on the Z = 0 plane:
s(Thetas)[Ro + (Rs - Ro) = S(Thetac) [Ro - (Ro - Rc)]
or
S(Thetas) Rs = S(Thetac) Rc

The fraction of the spheromak surface charge that causes a horizontal radial electric field and hence needs radial weighting is:
cos(Alpha)
where Alpha is the angle between the horizontal vector R and a vector normal to the spheromak surface.

A weighting factor of R / Ro is required to reduce the internal horizontal electic fields to zero.

Hence, at every point on the spheromak wall surface:
(surface charge density) cos(Alpha) (R / Ro) = constant

The surface charge is a function of Theta

(R / Ro) = [Ro + X(Theta) cos(Theta)] / Ro

The total surface charge is:
2 Integral from Theta = 0 to Theta = Pi of
[Ro + X(Theta) cos(Theta)] 2 Pi X(Theta) d(Theta) (surface charge (Theta))

The filament path defines the spheromak wall. This wall is highly symmetric.

The cross section of a spheromak wall can be expressed as a function of the form: Zw(Rw).

Due to spheromak symmetry, on the spheromak wall:
Zw(Rw)= Zw(-Rw) = -Zw(Rw) = -Zw(-Rw)

The charge on the filament per unit of filament length is:
Q / Lh
Hence, dQ = (Q / Lh) dLh

Let Theta = angle about the spheromak minor axis defined by R = Ro, Z = 0 and the Z = 0 plane.

Consider a point (Rw, Zw) on the spheromak wall. Let X = shortest distance from circle R = Ro, Z = 0 to the ring (Rw, Zw) on the spheromak wall.
An element of length dLh along the charge motion path tangent to the spheromak wall 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 rate of angle advancement is constant:
d(Phi) / d(Theta) = Np / Nt

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

Recall that:
The total surface charge is:
Q = 2 Integral from Theta = 0 to Theta = Pi of
[Ro + X(Theta) cos(Theta)] 2 Pi X(Theta) d(Theta) (S(Theta))

An element of surface charge is:

dQ = 2 [Ro + X(Theta) cos(Theta)] 2 Pi X(Theta) d(Theta) (S(Theta))

Hence the surface charge per unit of Theta is:
dQ / d(Theta)= 4 Pi [Ro + X(Theta) cos(Theta)] X(Theta) (S (Theta))

Now consider the special case of Theta = (Pi / 2).
Then:
X(Theta) = Ho
Cos(Theta = 0
Hence:
dQ / d(Theta) = 4 Pi Ro Ho S(Pi / 2)

At this point:
dQ = 2 Pi Ro Ho d(Theta) S(Pi / 2)
or
dQ / d(Theta) = 2 Pi Ro Ho S(Pi / 2)

Recall that:
(surface charge density) cos(Alpha) (R / Ro) = constant

Hence:
(Q / Lh) dLh (R / Ro) cos(Alpha) / {[d(Phi)[Ro + X(Theta) cos(Theta)][x(Theta) d(Theta)]} = constant

This expression can be evaluated at both R = Rs and R = Rc to give:
(Q / Lh) dLh (Rs / Ro) cos(Alphas) / {[d(Phi)[Ro + X(Thetas) cos(Thetas)][x(Thetas) d(Theta)]}
= (Q / Lh) dLh (Rc / Ro) cos(Alphac) / {[d(Phi)[Ro + X(Thetac) cos(Thetac)][x(Thetac) d(Theta)]}

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

Example: Rs = 4, Rc = 1 Ro = [65 / 17] = 3.82

Note that the peak Ho is close to the outside wall.
&nbs;

NOW FOCUS ON THE CIRCLE R = Ro, Z = Ho
This point occurs at Theta = (Pi / 2) where:
cos(Theta) = 0
X(Theta) = Ho.
dX = 0

At the circle: R = Ro, Z = Ho the surface charge does not affect the horizontal electric field.

At R = Ro, Z = Ho
the surface charge is unmodified.
At R= Rs, Z = 0 the surface electric field is reduced from its R = Ro, Z = 0 value by the ratio (Ro / Rs). Hence: Ho = (Rs / Ro)Rs CHECK

This web page last updated April 6, 2026.