I made some research on the self-inductance of a straight wire, and many sources cited “Inductance Calculations” of F.W. Grover. So I read a bit of that book but I couldn't find the answer to my question.

As far as I know, the self-inductance is given by the sum of internal and external inductance of the wire. Since the first one is constant with respect to the radius, I assume that the latter decreases when the radius increases.

For the external inductance, I found here (see Figure 1) the formula \$\frac{\mu l}{(2\pi)}ln(D1/D2) \$ with D2 the diamater of the wire, and D1 the diameter of a fictitious wire. It seems indeed that when D1 gets larger, the inductance decrease. However, in my opinion, with a single wire D2 should tend to infinity. So, how does this formula make sense?

Maybe I have this doubt because I don't really have clear what is the external inductance of a single wire. In fact, I don't see what the flux is linked to when the magnetic field is external to the wire and how it can generate an emf across the wire.

Also, other than that formula, the real question is what is the physical phenomenon that makes the self-inductance decrease when the radius increases?

  • \$\begingroup\$ The short, very incomplete answer is that HF tends to travel in the outer 'skin' of a conductor, sometimes only the first1 mil is carrying the majority of the signal. Hence, the more diameter the more surface area to travel in. Google "Litz-wire" for more info on conductor surface area. \$\endgroup\$ – user105652 Jan 5 '19 at 23:08
  • \$\begingroup\$ It should be logarithm of (length/diameter) * length * ... \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 6 '19 at 0:30
  • \$\begingroup\$ @Sparky256 I know about the skin effect, however, the phenomenon that I'm asking about seems to be true at any frequency, even when the skin effect is negligible. \$\endgroup\$ – Elia Jan 6 '19 at 18:11
  • \$\begingroup\$ @TonyEErocketscientist Hi Tony, I found that formula as well. I'm not sure which is more accurate. However, I didn't find an intuitive explanation for the one you mentioned, so I preferred to ask about the other one. \$\endgroup\$ – Elia Jan 6 '19 at 18:11
  • \$\begingroup\$ I dont see the link for the other one \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 6 '19 at 18:12

Imagine a very thin long wire of inductance L. If it were comprised of two very thin long wires that were perfectly magnetically coupled (i.e. shared 100% the same flux) then the net inductance would still be L. That means that the individual inductance of each strand is 2L and, when paralleled, the net becomes L.

Take this a step further and have a large bundle of strands making up the long (but now not so thin) wire. Then ask yourself if one of the strands on one edge of the bunch will couple its flux 100% to a strand on the opposite edge of the bunch.

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  • \$\begingroup\$ Hi Andy, I liked your answer, it's very intuitive and it makes sense! Now I'm convinced, but I was wondering if there is an analogy between what you said and the theory of internal and external inductance. I would like to understand the phenomenon the other way as well! \$\endgroup\$ – Elia Jan 6 '19 at 18:01
  • \$\begingroup\$ I’m not sure I could attack it properly from the other direction with either ease or clarity. I think maybe try your luck on physics SE. \$\endgroup\$ – Andy aka Jan 6 '19 at 19:26

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