Inductance per unit length of coaxial transmission line

[Abdulwahab Hajar]

1. Homework Statement
An air coaxial transmission line has a solid inner conductor of radius a and a very thin outer conductor of inner radius b. Determine the inductance per unit length of the line.

Homework Equations

the book states the methodology to find the inductance as follows:
1) Choose an appropriate coordinate system for the given geometry.
2) Assume a current I in the conducting wire.
3) Find B from I by Ampere's circuital law, if symmetry exists; if not Biot-Savart law must be used.
4)Find the flux linking with each turn, Φ, from B by integration:
Φ = ∫B.ds
Where S is the area over which B exists and links with the assumed current.
I couldn't really grasp quite understand what S is in this question.
5)Find the flux linkage Λ by multiplying Φ by the number of turns.
6) Find L by taking the ratio L = Λ / I

The Attempt at a Solution

I have the solution uploaded as it is in the book, I understand what happened first cylindrical coordinates were used, Current I was assumed and B was found using Ampere's law (steps 1,2, and 3).
In the 4th step which is finding the flux he integrated over the surface (1(unit length) * dr)... But then he said that represents only dΦ in other words the entire flux and commenced to integrate over dr again...
We already integrated over thus measuring the flux over the whole wire why did he integrate again??/ Thank you very much

 

[haruspex]

Confession: this topic is entirely new to me, just going by the text you reproduced.

The linkage is between current at any two radii, r and s say. So it is a double integral $\int_{s=0}^b\int_{r=0}^s$.

 

[rude man]

View attachment 111715 View attachment 111716 View attachment 111718 But then he said that represents only dΦ in other words the entire flux and commenced to integrate over dr again...
We already integrated over thus measuring the flux over the whole wire why did he integrate again??/
Thank you very much

dΦ is the flux over a thin rectangular strip of unit length times dr. So to get the total flux you need to cover the entire area which is a rectangle of unit length times b.

Carefrul when you compute I in the region 0<r<a; there the current is a function of r (whereas for a<r<b it's constant).

Here's a very nice diagram of the area to be integrated over:
https://www.physics.byu.edu/faculty/christensen/Physics%20220/FTI/32%20Inductance/32.11%20The%20inductance%20of%20a%20coaxial%20cable.htm

But they assume the current in the inside conductor is all on its surface, not uniformly distributed within 0<r<a as in your problem.

 

[rude man]

An alternative & perhaps clearer view is go with energy:

Energy dU in a differential volume dV of a B field in vacuo is 1/2 BH dV = 1/2B²dV/μ₀. Integrate B over the volume of the field, then equating U to the energy of an inductor = 1/2 LI², getting the same answer. The total field volume is πb² times unit length. Again, care must be taken in computing the B field in regions 0<r<a and a<r< b.