0
$\begingroup$

The authors of Power Systems Analysis calculate the inductance per unit length (henrys/meter) of a transmission line attributed only to the flux inside the conductor as "flux linkages per ampere." The flux is drawn in the x-y plane (i.e., in the page). Is the flux linkage also in the x-y plane?

enter image description here

"In the tubular element of thickness $dx$" the flux per meter of length is $d\phi$. "The flux linkages $d\lambda$ per meter of length, which are caused by the flux in the tubular element, are the product of the flux per meter of length and the fraction of the current linked" (p149): \begin{align} \mathrm{d\lambda}=\frac{πx^2}{πr^2}{d\phi} \end{align} What is the direction of $d\lambda$? Given that the flux is directed concentrically (as shown in the figure), is the corresponding flux linkage also directed concentrically in the cross section of the conductor?


Note: The figure above assumes uniform distribution of current throughout the cross section (p145).

$\endgroup$
9
  • $\begingroup$ Why are you looking for "flux linkage" inside a conductor? That is a dubious idea. Flux linkage is just a term referring to magnetic flux through a closed path that is not a simple circle but has many turns. $\endgroup$ Commented Feb 24, 2023 at 22:54
  • $\begingroup$ Flux linkage seems important to understanding inductance in a transmission line (electrical4u.com/inductance-in-power-transmission-line) $\endgroup$ Commented Feb 24, 2023 at 23:07
  • $\begingroup$ Calculation in section "Calculation of Inductance of Single Conductor" on that website is bizarre - in addition to using the term "flux linkage" in a perplexing way, it lacks explicit description of the closed path for which the magnetic flux is being calculated, and the resulting formula for self-inductance is achieved only after introducing an arbitrary limit of integration at $x=D$. Thus the resulting formula is useless - inductance depends on arbitrary choice of $D$ and diverges to infinity as $D$ goes to infinity. $\endgroup$ Commented Feb 25, 2023 at 1:21
  • $\begingroup$ Author of that text assumes magnetic field is that of straight current-carrying wire of infinite length. However, infinite wire has infinite self-inductance per unit length, which means realistic calculation has to take into account real length of the wire $\ell$, and the resulting $\lambda$ depends on $\ell$. $\endgroup$ Commented Feb 25, 2023 at 1:23
  • $\begingroup$ They did get finite result for the first contribution to self-inductance (due to region of the wire), but there is no way to check that a part of self-inductance is right. Experiments with self-induction only manifest total self-inductance, nature does not care how we split calculation on paper. $\endgroup$ Commented Feb 25, 2023 at 1:39

1 Answer 1

0
$\begingroup$

My initial thought on this seems to be consistent with Wikipedia.

In circuit theory, flux linkage is a property of a two-terminal element.

Basically, the term "flux linkage" is an idea in circuits and more of an engineering aid. The actual field lines and flux for a particular surface (which must be defined before one can integrate) are the more physical ideas.

Full disclosure, the second sentence says [citation needed] in that article, so it may be a bit of a point of controversy. The article definitely needs improving, but my understanding is consistent with it. I checked a few more places and everyone refers to linkage in terms of wires, so this seems to be the norm.

The interior magnetic field in a conducting wire does lead to observed effects of course, and those will be predicted at the very least by Maxwell's equations and the Lorentz force law. There may be a way to define "equivalent flux linkage" for a wire and simplify the analysis, I'm not sure. I have never seen this and am skeptical of it.

There is a bit of technical point here. Since the $\vec{E}$ field is no longer irrotational when there is a time varying $\vec{B}$ field, the "total voltage" is no longer path independent. This is why you get a concept like flux linkage. It is the path enforced by the wires, with a highly conductive path and highly resistive boundaries, that lead to a particular "total voltage" or "EMF" as it can be called:

$$ EMF = \oint \vec{E} \cdot d\vec{r} $$

I personally don't like the whole EMF thing and just think of it as voltage. Anyway, in the interior of a conductor, the path of the contour integral is not defined, so I don't know how a "flux linkage" would be defined. The behavior of the fields will be predicted by solving Maxwell's equations in a material.

$\endgroup$
9
  • $\begingroup$ > "I personally don't like the whole EMF thing and just think of it as voltage." This will cause you trouble when analyzing perfect inductor in AC circuits, where induced EMF acts against potential gradient, so that net electric field in the inductor coils can be zero. It is better to reserve the use of "voltage" to differences of potential (which are often of interest and thus the thing we want to measure), and not use it for induced EMF (which is a general concept for any closed loop and it is not common to measure its value). $\endgroup$ Commented Feb 24, 2023 at 22:40
  • $\begingroup$ Setting the V / EMF philosophical discussion aside, do you agree with my actual answer? I have never heard the term "flux linkage" used any other way. $\endgroup$ Commented Feb 24, 2023 at 22:42
  • $\begingroup$ I don't think it's a good answer. The question is about a dubious word group "flux linkage inside a conductor", I don't think we know what is meant. $\endgroup$ Commented Feb 24, 2023 at 22:51
  • $\begingroup$ Well, my answer is that it isn't a defined idea as far as I can see. Flux linkage is defined for a defined path (space curve) which is enforced usually by a wire coiled into a shape. A conductor is a 3D continuum, there is no space curve. Maybe I should say it that way? $\endgroup$ Commented Feb 24, 2023 at 22:54
  • $\begingroup$ That's better, and close to how I understand it as well. Flux linkage is just a term for magnetic flux through a closed path that has more than one turn. $\endgroup$ Commented Feb 24, 2023 at 22:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.