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Jul 26, 2014 at 11:54 comment added Robert Seifert That would mean $\Delta R = \frac{R}{N} ~\rightarrow ~ \Delta R = \frac{R-\epsilon}{N}$ and $r ~\rightarrow ~ r + \epsilon$ - if you insert that into my formular you won't get much of a change. It is basically right, that the magnetic flux goes to infinity at the origin. That's why you can't calculate that geometry with field theory or even basic relations involving $\Phi$, $B$ or $H$-Fields. For static fields the reluctance should be constant along every flux path.
Jul 26, 2014 at 11:12 comment added Andy aka Maybe it won't converge because at very short lengths of "r" the reluctance will likely be close to zero - maybe you have to accept a minimum distance for r so that r extends from the wire radius plus (say) 1mm all the way to the outer radius of the torus?
Jul 25, 2014 at 11:53 comment added Robert Seifert As I understand you're suggesting what I have tried. But the resulting series does not converge.
Jul 24, 2014 at 17:05 comment added Robert Seifert $N$ is the number of divisions: $Ae = A/N$
Jul 24, 2014 at 16:55 comment added Andy aka I'm not following what N is - I've added a picture to my answer
Jul 24, 2014 at 16:54 history edited Andy aka CC BY-SA 3.0
Added a picture to help
Jul 24, 2014 at 16:37 comment added Robert Seifert I tried it the way you suggested (see my edited question) - but it is a little weird that I just get results as expected for small $N$, so guite big partitions, but not for large $N$. Do you see a reason for that? Is this the way you thought about?
S Jul 23, 2014 at 20:17 history edited Andy aka CC BY-SA 3.0
deleted 9 characters in body
Jul 23, 2014 at 20:16 review Suggested edits
S Jul 23, 2014 at 20:17
Jul 23, 2014 at 20:13 history migrated from electronics.stackexchange.com (revisions)
Jul 23, 2014 at 20:09 history answered Andy aka CC BY-SA 3.0