Derivation of self-inductance of a long wire: Is my interpretation correct?

Question

Derivation of self-inductance of a long wire

Flux linkage inside of a conductor

Flux linkage inside conductor

So many people asked the same question, yet no satisfactory answers. I tried to develop my own solution to the problem. I seek to know whether my answer is correct, and also to help other people who are also confused by this problem. First, we should remember what is self-inductance and flux linkage:

Here 5 turns (in total) induce a flux of $\phi$ which then again cuts the 5 turns. Each turn induces an EMF of $\phi'$ but since there are 5 turns, each one of them acts like a battery connected in series and sums to EMF = $5\phi'$. Here the concept of flux linkage appears, we use $\lambda=N\phi$ instead of $\phi$ and restate Faraday's law of induction as follows: $\lambda' = (N\phi)' = N(\phi') = \Delta V.$ Hence, the EMF can be directly calculated as the derivative of the flux linkage. I prepared the following (better) illustration:

So to find the internal self-inductance of a conductor, we should calculate the flux linkage. Not flux itself. In all three questions above, OPs calculated flux spinning horizontally inside the conductor. And they did it without a mistake:

$$\phi=\frac{\mu}{8\pi}i$$

But they never did calculate the $\lambda=LI$ to extract the self-inductance. So how shall we do it, as the number of turns is obscure here? First, let's come to the step they are stuck.

$$B_x = \mu I\frac{x}{2 \pi r^2}$$ $$d\phi = \mu I\frac{x}{2 \pi r^2} \times 1 \times dr$$

Here they are stuck. They have to find $d\lambda = Nd\phi$ after this. But how? Simple as that: We will drive a vertical circle for the horizontal flux to flow into, an artificial turn inside the conductor, containing our flux. In practice, these artificial paths of current, correspond to the eddy currents causing the skin effect:

See that $I_w$? It will be our "turn". This turn will start from the outer edge of $B_x$ and then end in the origin. Indeed this is why current condenses on the shell of the conductor. The turn will be a circle taking $\frac{x}{2}$ as the origin.

Here $dr$ is drawn a little bit big for an infinitesimal, and I couldn't either draw the half-bottoms of the circles for a better depiction. As you can see for $x=r$, we have the biggest turn. We will accept that turn as $N=1$ and weigh other turns by their relative areas. The area for $x=r$ is $\pi(\frac{r}{2})^2 = \pi(\frac{r^2}{4})$. This is $N=1$. The area for an arbitrary $x$ will be $\pi(\frac{x^2}{4})$, and we will accept this as $N=\frac{\pi(\frac{x^2}{4})}{\pi(\frac{r^2}{4})} = \frac{x^2}{r^2}$ turns. For example, if we take $x=\frac{r}{2}$, number of turns will be $N=\frac{1}{4}=0.25$.

After all, we can calculate the flux linkage by taking the integral of the following equation:

$$d\lambda = Nd\phi = N(x)[B(x)dr] = \frac{x^2}{r^2}B(x)dr$$

So this is where the $\frac{x^2}{r^2}$ coefficient confusing people come from, and the rest is straightforward. Is my interpretation correct?

Answer 1

I don't think that it makes sense to try to understand the concept of self-induction in a long wire using concepts such as imaginary turns, or flux linkage.

The bizarre weighing of magnetic flux contribution by square of distance from the center line can be motivated in a better way.

First, the very idea of calculating self-inductance of a very long wire, assuming it is determined by the field inside the wire, is wrong. The wire has magnetic field outside the wire as well. This outside field produces flux which makes important contribution to self-inductance and it makes self-inductance superlinear function of the wire length. For infinite wire, self-inductance per unit length comes out infinite. So infinite wire is not a realistic enough simplification for real wires. Real circuits have finite length wires and finite self-inductance per unit length, which can't be calculated based on the flux just inside the wire.

The standard calculation that considers only field inside the wire can make sense if we assume the wire is enclosed in a cylinder of almost the same diameter (the cylinder being electrically insulated from the wire), and in the cylinder there is a current of the same magnitude but opposite direction, so magnetic field outside the cylinder vanishes. Only then it makes sense to relate self-inductance to the magnetic flux in the wire.

In general, self-inductance of a circuit is defined as the constant $L$ in the formula for induced EMF in the circuit, when current changes:

$$ \mathscr{E}_i = - L\frac{dI}{dt}. $$ $\mathscr{E}_i$ is usually defined for a closed path in space, thus an infinitely thin closed curve. So, we need to imagine such closed curve, then we can determine magnetic flux, and then we can determine $L$.

With standard solenoid, this is easy - the curve is inside the wire, no matter where exactly, as long as it follows the turns. Magnetic flux for such curve does not depend substantially on whether the curve sticks to the wire center curve, or is more close to the wire surface.

But when the magnetic flux is limited to the body of the wire, now it does matter whether the closed curve goes through the center or near the surface of the wire. Different choices of the curve will result in different magnetic flux, and thus different $L$. We have a problem - there is no single curve, and thus no single $L$ which could be the obvious answer.

However, it is still a single wire and (we imagine) a single current circuit, which should behave as if some inductor was in the path. So it makes sense to guess that in practice, as long as current stays uniform throughout the cross-section of the wire (thus we limit ourselves to low enough frequencies, where there is no strong skin effect), the circuit will behave as if it had some inductor with effective self-inductance $L_{eff}$ in the path. This should be close to some "average" value of all those $L$'s for different curves in the wire.

$$ L_{eff} \approx L_{av}. $$

One good criterion for such average value $L_{av}$ is that it should give the same magnetic energy in the circuit via the standard formula $E_m = \frac{1}{2}LI^2$, as the magnetic energy defined by the Poynting formula:

$$ E_{m,Poynting} = \int \frac{1}{2\mu_0} B^2 dV, $$ where the integration goes over the whole space where magnetic field is, thus in this case, the body of the wire.

It turns out that the procedure where the flux contributions are weighed by squared distance from the center produces just such $L_{av}$. So the weighed flux $\Phi_{weighed}$ divided by current $I$ yields $L_{av}$ that gives the correct magnetic energy in the wire using the general formula

$$ E_m = \frac{1}{2}LI^2 = E_{m,Poynting}. $$