- #1

goodphy

- 216

- 8

Maybe this is the one of the most typical example in Electromagnetism textbook.

There is a single round wire carrying current I (in D.C) with radius of R and length is infinity.

From Ampere's law, the B field inside the wire is

B = μIρ/2πR

^{2}a

_{θ}

where ρ is radial distance from symmetric axis to point of field and a

_{θ}is unit vector in azimuthal coordinate in cylindrical coordinate.

The magnetic flux through the region of infinitesimal area of width dρ and unit length is simply

dψ

_{1}= Bdρdz = μIρ/2πR

^{2}dρdz.

It is very straightforward until this step.

My question arise from here; According to textbook, the

**flux linkage**is not equal to flux but flux multiplied I

_{enc}/I. It seems flux linkage counts only contribution of current which actually induces B in flux above. It seems reasonable since the flux linkage should not count the current which doesn't contribute.

Physically I feel it make sense, but I need a more rigorous

**definition of flux linkage**to redrive this mathematical expression.

Could you please help me how to understand the flux linkage in clear view?

And in addition inductance L for single round wire is only calculated with the flux inside the conductor. But still there is a field outside the wire! Why is external field not counted?? Maybe this field is associated to external inductance L

_{ext}and total inductance L = L

_{int}+

_{ext}where L

_{int}is linked to flux linkage inside the wire as described above?