Self Inductance of "Thick" Wire Loop

Very thought provoking problem. I have not looked into the derivation of the formula yet, but I may share my thoughts on that in a following comment after I do read up a little.

To solve this problem, I used the concept of the magnetic vector potential, the definition of which can be found in this link .

A point on current carrying wire is:

$\vec{r}_1 = (\cos{t_1},\sin{t_1},0)$

Elementary arc vector in the vicinity of that point is:

$d\vec{r}_1 = (-\sin{t_1},\cos{t_1},0) \ dt_1$

A point on the boundary of the interior disk is:

$\vec{r}_2 = (0.99\cos{t_2},0.99\sin{t_2},0)$

Elementary arc vector in the vicinity of that point is:

$d\vec{r}_2 = (-0.99\sin{t_2},0.99\cos{t_2},0) \ dt_2$

The magnetic vector potential due to the elementary arc on the current carrying wire is:

$d\vec{A} = \frac{\mu_o}{4\pi} \frac{d\vec{r}_1 }{\lvert \vec{r}_2 - \vec{r}_1 \rvert}$ $\vec{A} = \frac{\mu_o}{4\pi} \oint_{C_1} \frac{d\vec{r}_1 }{\lvert \vec{r}_2 - \vec{r}_1 \rvert}$

Now, the flux through the interior disk is (as per Stokes' theorem):

$\Phi = \int \int_{S} \vec{B} \cdot d\vec{S} = \int \int_{S}(\nabla \times \vec{A}) \cdot d\vec{S} = \oint_{C_2} \vec{A} \cdot d\vec{r}_2$ $\implies \Phi = \oint_{C_2} \vec{A} \cdot d\vec{r}_2 = \oint_{C_2} \left(\frac{\mu_o}{4\pi} \oint_{C_1} \frac{d\vec{r}_1 }{\lvert \vec{r}_2 - \vec{r}_1 \rvert} \right) \cdot d\vec{r}_2$ $\implies \Phi =L_a= \frac{\mu_o}{4\pi} \oint_{C_2} \oint_{C_1} \left(\frac{d\vec{r}_1 }{\lvert \vec{r}_2 - \vec{r}_1 \rvert} \right) \cdot d\vec{r}_2$

Plugging in all expressions, simplifying and solving the integral by outsourcing to Wolfram Alpha is one way to go. I wrote a simple script of code to accomplish the task. Finally, the answer is:

$\frac{L_a}{L_f} \approx 0.9943$

As for the geometric constraints, I think that the thickness of the wire has to be much smaller relative to its diameter as is the case here, for the formula to hold good. But I think I will know better when I read up about its derivation.

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clear all
clc

% Numerical resoloution:
dt1   = pi/3000;
dt2   = pi/3000;

% Flux initialisation:
phi   = 0;

for t1 = 0:dt1:2*pi

    for t2 = 0:dt2:2*pi

        % Position vector and arc element vector of current carrying wire:
        r1          = [cos(t1);sin(t1);0];
        dr1         = dt1*[-sin(t1);cos(t1);0];

        % Position vector and arc element vector of interior disk:
        r2          = [0.99*cos(t2);0.99*sin(t2);0];
        dr2         = dt2*[-0.99*sin(t2);0.99*cos(t2);0];

        % Magnetic vector potential due to current carrying wire element:
        dA          = (1/(4*pi))*(dr1/(norm(r2 - r1)));

        % Flux due to the current carrying wire element (Stokes' theorem):
        dphi        = dA'*dr2;

        % Numerical integration:
        phi         = phi + dphi;


    end
end

D      = 2;
d      = 0.02;
Lf     = 0.5*D*(log((8*D)/d) - 2);

ANSWER = phi/Lf