Inductance of the two-wire transmission line per unit length (Индуктивность двухпроводной линии передачи на единицу длины)
The
total inductance is infinitely big because line is infinitely big,
that’s why we should find the inductance per unit length. First of
all, we have to define inductance of the two-wire line itself without
influence of the cylinder. Then we add the additional flux which is
induced by ferromagnetic cylinder. As usual the total inductance in
such line (diameter of each wire is much less than the distance
between this two wires and much less than the distance between wire
and ferromagnetic cylinder) may be splitted in two parts:
,
where
(cause we have 2 wires)
External fluxes (Внешние потоки)
Our
system of two wire lines and ferromagnetic cylinder may be replaced
by four long wires. Two of them are exactly the same as initial wires
(A’ A), but there are two mirror images (B B’), they have
current, that differs from the current in primary wires and these
currents are defined by this expression:
.
We want to find a flux which is coupled with two wire lines (A’
A). That means the flux which crosses the line which connects centers
of wires (A’ A). The most important part of this flux is the flux
which is induced by the line itself.
, ФA
– flux which is induced by the wire A; xA
– the distance between A’ and A; ФA=
ФA’
(absolutely symmetrical system), R – radius of wire.
The
magnetic field which is induced by the mirror image B also has a form
of
,
where r – distance between the center of the wire and the point of
interest. This magnetic field should be integrated over segment A’
A (really we should integrate over the area, but we consider a
two-wire line with the length of one unit, so area is simply a
product of this one unit and the distance between the centers of
wires A’ A). We should integrate the flux density which is induced
by the wire B and will get exactly the same expression but the limits
will be different:
ФB – flux
which is induced by the wire B; xB
– the distance between wires A’ and B;
ФB’ –
flux which is induced by the wire B’; ФB’=
-ФB
(opposite currents)
For
simplicity we should use
(relative magnetic permeability and magnetic permeability of vacuum)
so we have such expressions:
Total inductance (Общая индуктивность)
We
should summarize all the fluxes, after that we divided it by the
current:
Finally we have:
Forces. The first line. (Силы. 1ая линия)
We
can consider the force which is applied to the left wire with the
coordinates of zero (A’), may be calculated as a sum of three
components:
,
where
- the force from the sight of a second wire;
– the force from the sight of first image;
- the force from the sight of second image. If we have two wires with
the same directions of the current the force will be attractive. And
if we have two wires with the opposite directions of the current the
force will be repulsive. According to these principles we have such
expressions:
,
it has minus because force is repulsive
, it has minus because force is repulsive
,
it has positive sign because force is attractive
Finally:
