# Self-Inductance of Two Current-Carrying Wires

## Table of contents

Here we want to derive the **inductance** \(L\) of two conductors arranged in parallel. Consider two current-carrying conductors of **length** \(l\) and **radius** \(R\) (see illustration 1). The two conductors are at **distance** \(d\) from each other. A **positive current** \( +I \) flows through the left conductor and the **negative current** \(-I\) flows through the right conductor. The two currents are therefore equal in magnitude, but *opposite*.

The **inductance** \(L\) is the constant of proportionality between the **magnetic flux** \(\Phi_{\text m}\) penetrating the area \(A\) which is parallel to the two currents. The inductance is the ratio of \(\Phi_{\text m}\) to \(I\):

To find out the inductance, the magnetic flux through the surface \(A\) must be calculated. The surface parallel to the two conductors has the length \(l\). The current \(I\) is assumed to be known.

The magnetic flux is defined as follows:

Here \(B\) is the **magnetic flux density** (or magnetic field for short) that penetrates the area \(A\). We need to determine \(B\). The magnetic field \(B\) is composed of the magnetic field *outside* the two conductors and the magnetic field *inside* the conductors.

## Magnetic field outside the wires

We know that the magnetic field *outside* of a current-carrying conductor passing through the origin depends reciprocally on the distance \(x\) to the conductor:

Our conductors, on the other hand, do not pass through the origin, but are offset. The external magnetic field of a conductor which is at \(x= d/2\) and through which the current \(-I\) flows (this is the right conductor) is given by:

& ~=~ \frac{\mu_0 \, I}{2 \pi} \, \frac{1}{\frac{d}{2} - x} \end{align} $$

The external magnetic field of the conductor located at \(x = -d/2\) and traversed by the current \(I\) (this is the left conductor) is:

The *total* magnetic field \( B_{\text e} \) outside both conductors is the sum of the partial fields 4

and 5

:

## Magnetic field within wires

The magnetic field *inside* a current-carrying conductor, which is located at the coordinate origin and has the radius \(R\), increases linearly with the distance \(x\):

The magnetic field inside a current-carrying conductor located at \(x=d/2\) and through which the current \(-I\) flows is:

&~=~ \frac{\mu_0 \, I}{2 \pi \, R^2} \, \left( \frac{d}{2} - x\right) \end{align} $$

The magnetic field inside a conductor located at \(x=-d/2\) and through which the current \(I\) flows is:

The total magnetic field \(B\) in 2

is the sum of the partial fields 6

, 8

and 9

.

## Magnetic flux between two wires

So we have found out the magnetic field for the integral 2

. Now we have to integrate it over the area \(A\).

The area \(A\) has length \(l\) in the \(z\) direction and the magnetic field \(B\) is independent of the \(z\) coordinate, so we only integrate the \(B(x)\) along the \(x\) coordinate:

- The integration region between the conductors goes from \(-d/2 + R\) to \( d/2 - R\).
- The integration region within the left conductor goes from \(-d/2\) to \(-d/2 + R\).
- The integration region within the right conductor goes from \(d/2-R \) to \(d/2\).

Substituting the total magnetic field and the integration limits into the integral 10

, gives:

& ~+~ l \int_{-d/2}^{-d/2 + R} \frac{\mu_0 \, I}{2 \pi \, R^2} \, \left( \frac{d}{2} + x\right) \, \text{d}x \\\\

& ~+~ l \int_{d/2}^{d/2 - R} \frac{\mu_0 \, I}{2 \pi \, R^2} \, \left( \frac{d}{2} - x\right) \, \text{d}x \end{align} $$

Let us look at the first integral in 11

. The formula specifies the total magnetic field *outside* the conductor:

**1st Integral**:

The antiderivative of \(1/x\) is \(\ln(x)\), so 12

becomes by substitution:

Logarithm rules can be used to combine the two logarithm expressions in 13

:

Insert the integration limits and simplify:

&~=~ \frac{\mu_0 \, I \, l}{\pi} \, \ln\left(\frac{d - R}{R}\right) \end{align} $$

Let's move on to the second integral in 11

, in which we consider the internal magnetic field in the left conductor.

**2nd integral**:

The antiderivative of 16

is easy to find. Then, insert integration limits, multiply out and simplify:

&~=~ \frac{\mu_0 \, I \, l}{4 \pi} \end{align} $$

The same procedure is performed for the third integral in 11

. The result is the same as for the second integral.

**3rd integral**:

&~=~ \frac{\mu_0 \, I \, l}{4 \pi} \end{align} $$

Add results 15

, 17

, 18

and substitute them into 10

. Then we get the total magnetic flux:

Insert the magnetic flux 19

in 1

to get the inductance of the parallel current-carrying wires: