# What is the internal linkage flux of a wire with radius a(shaped like a cylinder)

How can we clarify that the relation between linkage flux an flux in a cylinder shaped wire is $d\lambda=\frac{x^2}{a^2}\ d\phi$?can we illustrate this by electromagnetic formula?I think I would be able to explain this if I knew the exact definition of linkage flux,I think linkage flux is not a real flux that exists(like flux itself),any help would be appreciated.

• The flux linkage is the product of the flux $\Phi$ and the number of coils, N, exist in the wire. Now are you able to explain that? Oct 11 '17 at 14:43
• I know this ,but here we don't have a coil!it is a cylinder! Oct 11 '17 at 15:06
• Here you should think of N as the fractional volumes. Oct 11 '17 at 16:08
• Oct 11 '17 at 16:10