Derivation: Self-Inductance of Two Current-Carrying Wires

Table of contents

Two current-carrying wires arranged in parallel.

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 $I$:

$$ \begin{align} L ~=~ \frac{\Phi_{\text m}}{I} \end{align} $$

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:

$$ \begin{align} \Phi_{\text m} ~=~ \int_A B \, \text{d}a \end{align} $$

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:

$$ \begin{align} B(x) ~=~ \frac{\mu_0 \, I}{2 \pi} \, \frac{1}{x} \end{align} $$

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:

$$ \begin{align} B_1(x) & ~=~ -\frac{\mu_0 \, I}{2 \pi} \, \frac{1}{x - \frac{d}{2}} \\\\
& ~=~ \frac{\mu_0 \, I}{2 \pi} \, \frac{1}{\frac{d}{2} - x} \end{align} $$

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

$$ \begin{align} B_2(x) ~=~ \frac{\mu_0 \, I}{2 \pi} \, \frac{1}{\frac{d}{2} + x} \end{align} $$

The total magnetic field $ B_{\text e} $ outside both conductors is the sum of the partial fields 4 and 5:

Magnetic field outside the wires

$$ \begin{align} B_{\text e} ~=~ \frac{\mu_0 \, I}{2 \pi} \, \left( \frac{1}{\frac{d}{2} + x} ~+~ \frac{1}{\frac{d}{2} - x} \right) \end{align} $$

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 $x$:

$$ \begin{align} B(x) ~=~ \frac{\mu_0 \, I}{2 \pi \, R^2} \, x \end{align} $$

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

$$ \begin{align} B_3(x) &~=~ -\frac{\mu_0 \, I}{2 \pi \, R^2} \, \left( x - \frac{d}{2} \right) \\\\
&~=~ \frac{\mu_0 \, I}{2 \pi \, R^2} \, \left( \frac{d}{2} - x\right) \end{align} $$

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

$$ \begin{align} B_4(x) ~=~ \frac{\mu_0 \, I}{2 \pi \, R^2} \, \left( \frac{d}{2} + x\right) \end{align} $$

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 $A$.

$$ \begin{align} \Phi_{\text m} ~=~ \int_A B \, \text{d}a \end{align} $$

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:

$$ \begin{align} \Phi_{\text m} ~=~ l \int B \, \text{d}x \end{align} $$

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:

$$ \begin{align} \Phi_{\text m} & ~=~ l \int_{-d/2 + R}^{d/2 - R} \frac{\mu_0 \, I}{2 \pi} \, \left( \frac{1}{\frac{d}{2} + x} + \frac{1}{\frac{d}{2} - x} \right) \, \text{d}x \\\\
& ~+~ l \int_{-d/2}^{-d/2 + R} \frac{\mu_0 \, I}{2 \pi \, R^2} \, \left( \frac{d}{2} + x\right) \, \text{d}x \\\\
& ~+~ l \int_{d/2}^{d/2 - R} \frac{\mu_0 \, I}{2 \pi \, R^2} \, \left( \frac{d}{2} - x\right) \, \text{d}x \end{align} $$

Let us look at the first integral in 11. The formula specifies the total magnetic field outside the conductor:
1st Integral:

$$ \begin{align} \Phi_{\text m1} ~=~ \frac{\mu_0 \, I \, l}{2 \pi} \int_{-d/2 + R}^{d/2 - R} \left( \frac{1}{\frac{d}{2} + x} ~+~ \frac{1}{\frac{d}{2} - x} \right) \, \text{d}x \end{align} $$

The antiderivative of $1/x$ is $\ln(x)$, so 12 becomes by substitution:

$$ \begin{align} \Phi_{\text m1} ~=~ \frac{\mu_0 \, I \, l}{2 \pi} \left[ \ln\left(\frac{d}{2} + x\right) ~-~ \ln\left(\frac{d}{2} - x\right) \right]^{d/2 - R}_{-d/2 + R} \end{align} $$

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

$$ \begin{align} \Phi_{\text m1} ~=~ \frac{\mu_0 \, I \, l}{2 \pi} \left[ \ln\left(\frac{d/2 + x}{d/2 - x}\right) \right]^{d/2 - R}_{-d/2 + R} \end{align} $$

Insert the integration limits and simplify:

$$ \begin{align} \Phi_{\text m1} &~=~ \frac{\mu_0 \, I \, l}{2 \pi} \, \left[ \ln\left(\frac{d - R}{R}\right) ~-~ \ln\left(\frac{R}{d-R}\right) \right] \\\\
&~=~ \frac{\mu_0 \, I \, l}{\pi} \, \ln\left(\frac{d - R}{R}\right) \end{align} $$

Let's move on to the second integral in 11, in which we consider the internal magnetic field in the left conductor.
2nd integral:

$$ \begin{align} \Phi_{\text m2} ~=~ \frac{\mu_0 \, I \, l}{2 \pi \, R^2} \int_{-d/2}^{-d/2 + R} \left( \frac{d}{2} + x\right) \, \text{d}x \end{align} $$

The antiderivative of 16 is easy to find. Then, insert integration limits, multiply out and simplify:

$$ \begin{align} \Phi_{\text m2} &~=~ \frac{\mu_0 \, I \, l}{2 \pi \, R^2} \, \left[ \frac{d}{2} \, x + \frac{1}{2} \, x^2\right]_{-d/2+R}^{-d/2} \\\\
&~=~ \frac{\mu_0 \, I \, l}{4 \pi} \end{align} $$

The same procedure is performed for the third integral in 11. The result is the same as for the second integral.
3rd integral:

$$ \begin{align} \Phi_{\text m3} &~=~ \frac{\mu_0 \, I \, l}{2 \pi \, R^2} \int_{d/2 - R}^{d/2} \left( \frac{d}{2} - x\right) \, \text{d}x \\\\
&~=~ \frac{\mu_0 \, I \, l}{4 \pi} \end{align} $$

Add results 15, 17, 18 and substitute them into 10. Then we get the total magnetic flux:

Magnetic flux of two parallel wires

$$ \begin{align} \Phi_{\text m} ~=~ \frac{\mu_0 \, I \, l}{\pi} \, \left[ \frac{1}{2} + \ln\left(\frac{d-R}{R}\right) \right] \end{align} $$

Insert the magnetic flux 19 in 1 to get the inductance of the parallel current-carrying wires:

Inductance of two parallel wires

$$ \begin{align} L ~=~ \frac{\mu_0 \, l}{\pi} \, \left[ \frac{1}{2} + \ln\left(\frac{d-R}{R}\right) \right] \end{align} $$