XYLENE POWER LTD.

SPHEROMAK WALL

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

INTRODUCTION:
This web page shows that a spheromak wall has a cross section centered at R = Ro, Z = 0 which is the minor axis of the quasi-toroid.

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.

Assume Z(R) is a symetrical quasi-toroid where the detailed shape is unknown./P>

A cross section of the spheromak wall can be described by a function:
Z(R) where Z is in the range:
- Ho < Z < Ho

Due to spheromak symmetry, in the range Rc < R < Rs:
Z(R) = - Z(R)
Z(Ro) = Ho, - Ho
Z (Rc) = 0
Z(Rs) = 0

Alternatively:
R(z) has two positive real solutions.
R(z) = R (-Z).
R(Ho) = R(-Ho) = Ro

At Z = 0:
R(Z) = Rc, Rs

(dZ /dR) = 0 at R = Ro

|(dZ / dR)| goes to infinity at R = Rc and at R = Rs

Let So be the surface charge density where S = So at R = Ro.
Then:
S(R) = So (Ro / R)^2
giving:
[S|(R = Rs)]= So (Ro / Rs)^2,
and
[S|(R = Rc)] = So(Ro / Rc)^2

At R = Ro: S = So(Ro /Ro)^2 = So
 

An important issue is the total charge Q on a spheromak which should be independent of spheromak's linear size and hence independent of Ro.

Let S(R) be the spheromak's surface charge density.

So = Q / Lp Lt
where Lp and Lt each scale with Ro.

Np, Nt are integer constants independent of Ro.

At a particular value of Ro: S = So (Ro / R)^2
= [Q / Lp Lt][Ro / R]^2 dLt = [(dZ)^2 + (dR)^2]^0.5
= [(dZ / dR)^2 + 1]^0.5 dR

Hence:
Q = Integral from R = Rc to R = Rs of:
[4 Pi R So (Ro^2 / R^2)] [(dZ / dR)^2 + 1]^0.5 dR

Hence:
Lt = Integral from R = Rc to R = Rs of:
Ro {(2 / R) [(dZ / dR)^2 + 1]^0.5 dR}

In order for Lt to scale properly, the quantity:
Integral from R = Rc to R = Rs of:
{(2 / R) [(dZ / dR)^2 + 1]^0.5 dR
must be a constant independent of Ro.

The function Z(R) must also meet the boundary conditions at Rc, Ro and Rs.

Try: [(dz / dR)^2 + 1]^0.5 = K R Ro / (Rs- R)(R - Rc)

Then:
(1 / R)[(dZ / dR)^2 + 1]^0.5 dR = Kc Ro dR / (Rs- R)(R - Rc)

This integration results in a constant.
(Rs - R)(R - Rc) = - R^2 + R (Rs + Rc) - Rc Rs
a = -1
b = (Rs + Rc)
c = - Rs Rc b^2 = (Rs + Rc)^2
4 a c = 4 Rs Rc
For b^2 > 4 a c:
Lt = Integral from R = Rc to R = Rs of:
Ro {(2 / R) [(dZ / dR)^2 + 1]^0.5 dR}

Try:
[(dz / dR)^2 + 1]^0.5 = [Kc R Ro / (Rs- R)(R - Rc)]

Lt = Integral from R = Rc to R = Rs of:
(2 Ro / R)[(dZ / dR)^2 + 1]^0.5 dR

Lt = Integral from R = Rc to R = Rs of:
(2 Ro / R)[[Kc R Ro / (Rs- R)(R - Rc)] dR
= Integral from R = Rc to R = Rs of:
[2 Kc Ro^2 / (Rs - R)(R - Rc)] dR
= Lt.

Lt = Ro Integral from R = Rc to R = Rs of:
[2 Kc Ro / (Rs - R)(R - Rc)] dR

We need to determine K from spheromak parameters.

Recall that:
[(dz / dR)^2 + 1]^0.5 = [Kc R Ro / (Rs- R)(R - Rc)]

[(dZ / dR)]^2 = {[Kc R Ro]^2 / [(Rs- R)(R - Rc)]^2} - 1

At (dZ / dR) = 0:
Kc R Ro / [(Rs- R)(R - Rc)] = 1
or
Kc R Ro = R(Rs + Rc) - R^2 - Rs Rc
or
R^2 + R(Kc Ro - Rs - Rc) + Rs Rc = 0
or
R = -{(Kc Ro - Rs - Rc) +/- [(Kc Ro - Rs - Rc)^2 - 4(1) Rs Rc]^0.5} / 2

Assume that:
(Rs + Rc) = 2 Ro
then at (dZ / dR) = 0::
R = - {(Kc Ro - 2 Ro) +/- [(Kc Ro - 2 Ro)^2 - 4(1) Rs Rc]^0.5} / 2
= Ro {- (Kc / 2) +/- [(Kc - 2)^2 / 4 - Rs Rc]^0.5

(dZ / dR) = +/- {[Kc R Ro]^2 + [(Rs- R)(R - Rc)]^2}^0.5 / [(Rs- R)(R - Rc)]

If Kc^2 = -1 then at R = Ro -(Ro Ro)^2 + [(Rs - Ro)(Ro - Rc)]^2 = 0 at Ro = (Rs - Rc) / 2

In reality K may differ from -1 . This issue needs more analysis.
 

Quantification of Lt:

Lt = Ro Integral from R = Rc to R = Rs of:
[2 K Ro / (Rs - R)(R - Rc)] dR

Note that Lt is proportional to theproduct [K Ro].

For K = +/- 1:
Lt / Ro = Integral from R = Rc to R = Rs of:
[2 Ro / (Rs - R)(R - Rc)] dR
= _______________________
 

This web page last updated April 15, 2026.