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 \(I\):

Inductance is magnetic flux divided by current

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

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:

Magnetic flux is integral of magnetic flux density over area

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

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:

Formula: Magnetic field outside two adjacent wires

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:

Formula: Magnetic field outside a conductor displaced to the right

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 \(x\):

Formula: Magnetic field inside a conductor

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

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

Formula: Magnetic field within a conductor displaced to the right

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 \(A\).

Integral of magnetic flux density over area

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

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:

Integral of the magnetic flux density over the length

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

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:

Magnetic flux with inserted magnetic field into the integral

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