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Derivation Self-Inductance of Two Current-Carrying Wires

Opposite current carrying wires
Level 4 (for physics pros)
Level 4 requires the knowledge of vector calculus, (multidimensional) differential and integral calculus. Suitable for advanced students.
Updated by Alexander Fufaev on
Table of contents
  1. Magnetic field outside the wires
  2. Magnetic field within wires
  3. Magnetic flux between two wires
Opposite current carrying wires
Illustration : 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 \(I\):

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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