# Self inductance of a very thin conductor, is infinity?

1. Feb 20, 2012

### constfang

Hi, I was reading some text and came across this problem, the problem is also mentioned in this link from Wiki: http://en.wikipedia.org/wiki/Inductance#Self-inductance
They said that it is because the 1/R now becomes infinite, this is what I'm confused about. From my understand, there would be some point where R=0 while taking the integration, but in the thick conductor case, isn't there also overlapping point that make R=0? how come these points doesn't make the integration infinite? what made a thick wire so different from a very thin wire? and what's the physical meaning of this phenomena? Thank you.

2. Feb 20, 2012

### Philip Wood

I think the problem's even worse than you think it is! The inductance, L, of a length $\ell$ of straight wire is given by
L = $\frac{\mu_{0}\ell}{2\pi}$ ln$\frac{b}{a}$
in which a is the radius of the conductor and b is... Well, that's the problem: if you put it to infinity, your inductance goes off to infinity (just as badly as if you put a to zero. In fact the 'b' problem is easily understood: by putting b to infinity you're denying the possibility of a return path for the current. Surround the conductor by a return path in the form of a cylindrical shell (making a co-axial cable) and the problem is solved. There's no resultant field outside the cylinder, so you have to integrate only between the inner conductor and the cylinder.

As for your original problem, the field doesn't drop to zero as you penetrate the inner conductor, but is proportional to r, that is the distance to the central axis. You can easily derive this from Ampere's law. I think you'll find that the interior of the conductor contributes a finite amount, L0 to the inductance, namely
L0 = $\frac{\mu_{0}\ell}{4\pi}$.

Last edited: Feb 20, 2012
3. Feb 21, 2012

### Philip Wood

I should have pointed out that (as you can see from my formula for L0) the thickness of the wire makes no difference to L0. Physically, for a given current, if the wire is thinner, the current density is greater, and also the flux density inside the wire, and there is the same amount of flux linked with it – even within the thinner wire.

But making the wire thinner will increase the inductance due to the flux linkage outside the wire, the bit that depends on ln$\frac{b}{a}$.

Last edited: Feb 21, 2012
4. Feb 22, 2012

### constfang

Thank you very much, that cleared out the cloud a little bit for me.