# (Direct current) Inductance formula for twin lead

We consider a return circuit of a twin lead, or a two cylindrical conductor.

```   2*a       2*a
->| |<-   ->| |<-
( )       ( )___
+<-- D -->+

where
D = distance between conductors (m)
```

The inductance of this circuit per unit length is usually described In the following way:

```  L = Le + Li                                       (1)

with
L = inductance of cable (H/m)
Le = external inductance (H/m)
Li = internal inductance (H/m)
```
The inductance is separated into an external inductance part from magnetic energy outside the conductors and an internal inductance part from magnetic energy inside the conductors. However, in the simplest case (direct current), most electrical engineering and physics texts assume
```  1 << D/a                                          (2)
```
and show these formulae:
```  Le = u/PI*log(D/a)                                (3)
Li = u/4/PI                                       (4)

with
u = permeability (H/m)
= 4e7*PI (for non-magnetic materials such as air, copper, aluminum, etc)
PI = 3.14159265..
```
Of course, log() is the natural logarithm. The derivation is usually conducted as shown in footnote 1, and is based on the assumption that the currents in the two conductors do not affect the current distribution in the other conductor.

However, in reality, magnetic flux from the current in one conductor exists in the other conductor. This effect is stronger inside from the axis of the conductor and weaker outside. As a whole, the flux from a self-induced current will be somewhat cancelled out, and the internal conductance will be smaller than the value in equation (4).

Furthermore, since the current distributes over the whole cross-section of the conductor, flux linkage exists not only between a and D - a, but also between D - a and D + a, therefore the external conductance will be larger than the value in equation (3).

The conditions in equation (2) disregard this interaction by placing the other conductor very far away. However, in real cables, D/d is usually around 2, and apparently equation (2) is not valid. This is just like when Japanese courts think the value of one vote in Japan is the same even though some votes are weighed a few times more heavily than others.

We stumble upon the following question: If we apply equations (3) and (4) to a real cable where equation (2) does not hold, how much error is there, and is it possible to conduct exact calculations? This is the question for this session.

Footnote 1 - Standard derivation of equations (3) and (4)

For external inductance, taking I as the total current in the conductor, the magnitude of the magnetic field, a distance r from the center of the conductor outside a single cylindrical conductor, is easily determined as

```  He = I/(2*PI*r)                                     (A)

Therefore, the flux linkage between two conductors per unit cell
due to the current from one conductor is

D-a
P = u/2/PI*I*Integral((1/r)*dr
a

= u/2/PI/*log(D/a)
```
The flux linkage from the other conductor has the same direction and magnitude, and therefore the flux linkage of this circuit is double this amount. From the definition of inductance,
```  Le = P/I
= u/PI*log(D/a)
```
For internal inductance, the magnitude of the magnetic field, a distance r from the center of the conductor inside a single cylindrical conductor, is
```  Hi = I*r/2/PI/a^2
```
The magnetic energy inside the conductor Ui is calculated as
```  Ui = (1/2)*Li*I^2
```
and therefore, it is easy to calculate the internal inductance as
```          a
Ui = Integral((1/2)*u*Hi^2*2*PI*r)*dr
0
a
= I^2*u/4/PI/a^4*Integral(r^3)*dt
0
= u/16/PI*I^2
```
As such, usually
```  Li = u/8/PI
```

Kouichi Hirabayashi, (C) 1999