Magnetic flux through triangular ring

[Ngineer]

Homework Statement

Very long wire carrying current I is surrounded by a brass ring of a triangular cross section. (figure attached)

Show that ψ =

μ° I h 
------   (b - a ln ((a+b)/b) 
  2∏b

Homework Equations

A = (μ_°I/2∏ * ln x) a_z (according to one of the solutions, where x is the distance between wire and surface)

ψ = ∫A.dl

The Attempt at a Solution

First attempt:
My first approach was to integrate the magnetic vector potential over the volume of the the ring, I defined the distance from the wire to the surface as:
d(x,z) = a + b(z/h) + x {0<= z<= h and 0<=x<=b(1 - z/h)

where
z is the rise in the z axis
x is the distance through the surface of the ring (i.e. distance penetrated through the surface)then I took ψ as a triple integral of
z from 0 to h
x from 0 to b(1-z/h)
θ from 0 to 2∏,

of (μ_°I/2∏ * ln d(x,z)) a_{z * dxdydθ}

but I got a wrong answer

Second attempt
I took only the surface in consideration, i.e. removing all mention of θ. Still couldn't reach the original formula.

I spent almost 4 hours trying to figure this out. I looked up an online solution (out of genuine interest to know how the problem is solved, not academic dishonesty) and after a lengthy process they reach a different formula and claim that this is the solution!

This is really puzzling me because it looks rather simple, like I'm missing a basic formula or something.

Any help is greatly appreciated! Thanks!

 

[TSny]

Hello, Ngineer.

Note that finding "the total number of magnetic flux lines in the ring" is equivalent to finding the total magnetic flux through the triangular cross-section of the ring.

So, use the expression for the magnetic field B of a wire and integrate B over the triangular cross-section.

The answer given doesn't appear to me to be correct. You would expect the answer to go to zero as b goes to zero while keeping a and h constant.

[EDIT: I believe the b in the denominator of the logarithm should be a.]

 

[andymars]

Yeah. I also just did it using B=μ°I/2∏r and doing a double integral with z on the inside going from 0 to (h/b)(r-a) and r going from a to a+b. The result I got was nearly the same as expected except that I also found the denominator inside the logarithm is a and not b.