# Derivation Self-inductance of two current-carrying wires

Level 4 (for very advanced students)
Level 4 requires the knowledge of vector calculus, (multidimensional) differential and integral calculus. Suitable for advanced students.
Updated by Alexander Fufaev on

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

Formula: Magnetic field outside a conductor displaced to the left
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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
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## 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 $$x$$:

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

Formula: Magnetic field within a conductor displaced to the left
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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

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

Integral for magnetic flux through area between wires
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The antiderivative of $$1/x$$ is $$\ln(x)$$, so 12 becomes by substitution:

First calculated integral for magnetic flux
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Logarithm rules can be used to combine the two logarithm expressions in 13:

First combined integral for magnetic flux
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Insert the integration limits and simplify:

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

Integral for the magnetic flux of the left wire
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The antiderivative of 16 is easy to find. Then, insert integration limits, multiply out and simplify:

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

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

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

Inductance of two parallel wires
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