# Cylindrical conductor with an off center cavity

1. Apr 11, 2007

### arunma

I have a quick question. I recently did an E&M problem in which I was given a cylindrical conducting wire of radius a, in which someone bores an off-center cylindrical cavity of radius b (the center of the smaller circle is offset from the center of the wire by distance d). I'm asked to find the magnetic field inside the cavity. I've attached a crude diagram to illustrate the problem. Anyway, I understand how to do the solution: find the magnetic field of a normal cylindrical wire, superimpose another wire with current flowing in the opposite direction, where the cavity is supposed to be, and then add the magnetic fields. And of course there's a simple coordinate transformation involved.

There's just one part of this solution that I don't understand. When calculating the magnetic field from the larger wire by using Ampere's Law, it's necessary to find the current density. I assumed that it would be $$J = \dfrac{I}{\pi a^2}$$. But it turns out that I need to subtract off the area of the smaller circle, so that $$J = \dfrac{I}{\pi (a^2-b^2)}$$. Can anyone explain qualitatively why this is necessary? I thought the cavity was already accounted for by superimposing the current in the other direction.

#### Attached Files:

• ###### untitled.GIF
File size:
1.4 KB
Views:
253
2. Apr 11, 2007

### Staff: Mentor

That assumes that the current through the wire is uniformly distributed over the full area--but it's not, since there's a hole!
The current that would have gone through the part of the wire where the cavity is must instead go through the rest of the area--increasing the current density. (Current only flows where the wire actually is. )
Right, but only if you use the correct current density.

3. Apr 12, 2007

### arunma

Well...that sort of makes sense. Thanks Al!