Self Inductance of Square Wire Loop

This was a very good one.

So to solve this, I have exploited the symmetry in the geometry in order to avoid crunching too many integrals. Unit current is assumed.

$d_1 = 1$ $d_2 = 0.995$

Consider the right edge of the current carrying loop. A point on that can be parameterised as such:

$\vec{r}_1 = (d_1,y_1,0) \ ; \ -d_1 \le y_1 \le d_1$ $d\vec{r}_1 = (0,dy_1,0)$

Now, consider a point on the right side of the inner square

$\vec{r}_2 = (d_2,y_2,0) \ ; \ -d_2 \le y_1 \le d_2$ $d\vec{r}_2 = (0,dy_2,0)$

The magnetic vector potential due to the current carrying wire at a point on the inner square is: $d\vec{A}_{12} = \frac{\mu_o}{4\pi} \frac{d\vec{r}_1}{\lvert \vec{r}_2-\vec{r}_1\rvert}$

The line integral of the the magnetic vector potential along the right edge of the inner circle is:

$L_1 = \oint_{C_1} d\vec{A}_{12} \cdot d\vec{r}_2$

If this process were to be repeated for the top and bottom edge, the line integrals would evaluate to zero as the dot product between the current carrying element and the line elements are zero, as the two are orthogonal vectors.

This same process can be repeated for computing the line integral along the left edge of the inner square:

$\vec{r}_3 = (-d_2,y_3,0) \ ; \ -d_2 \le y_3 \le d_2$ $d\vec{r}_3 = (0,-dy_3,0)$

$L_2 = \oint_{C_1} d\vec{A}_{13} \cdot d\vec{r}_3$

Finally, the magnetic flux due to the right edge of the current carrying wire through the square is (By Stokes' theorem):

$\Phi = L_1 + L_2$

By the symmetry of the problem, the magnitude and sign of the flux due to the other three sides will be the same. Therefore:

$L_a = 4\Phi$

Finally,

$\frac{L_a}{L_f} \approx 0.997$

Numerical details left out. I may modify the solution to add the code used to solve the integrals, later.

The loop with side-length $S$ lies at the center of an $xy$ -coordinate system parallel to the axes, and a current $I$ flows around the loop counter-clockwise. We calculate the flux $\Phi$ through the (slightly smaller) area $A$ . By symmetry, each side of the loop contributes the same amount of flux, so we only need to consider the flux $\Phi'$ generated by the wire at the bottom $y=-\frac{S}{2}$ : $\begin{aligned} \Phi&=4\Phi', &&&\Phi'&=\frac{\mu_0 I}{4\pi}\int_{C_2}\int_{C_1}\frac{ <\vec{dl}_1,\:\vec{dl}_2> }{ \left|\vec{r}_1-\vec{r_2}\right| }&&&\left|\begin{aligned} &C_1\text{ is the wire-piece at the bottom along }I\\[.5em] &C_2\text{ surrounds }A\text{ counter-clockwise} \end{aligned}\right. \end{aligned}$ The vectors $\vec{r}_{1,2}$ lead to all points on the curves $C_{1,2}$ , respectively. The left and right side of $C_2$ are perpendicular to $C_1$ , so the scalar product and the integral over those sides vanish. We describe $C_1$ and the upper and lower pieces of $C_2$ : $\begin{aligned}\left.\begin{aligned} \text{bottom wire piece }C_1:&&&& \vec{r}_1&=\frac{S}{2}(u;\:-1;\:0)^T, & && d\vec{C}_1&=\frac{S}{2}\vec{e}_xdu\\[.5em] \text{lower piece of }C_2:&&&& \vec{r}_2&=\frac{S}{2}(v;\:-(1-c);\:0)^T, & && d\vec{C}_2&=\frac{S}{2}\vec{e}_xdv\\[.5em] \text{upper piece of }C_2:&&&& \vec{r}_2&=\frac{S}{2}(-v;\:1-c;\:0)^T, & && d\vec{C}_2&=-\frac{S}{2}\vec{e}_xdv \end{aligned}\quad\right|c:= \frac{2R}{S} >0 \end{aligned}$ The description of $C_{1,2}$ allows us to solve the flux integral. We begin with the lower piece of $C_2$ and substitute the parameter for the outer integral $v':=\frac{1\mp v}{c},\:v'\rightarrow v\quad (*)$ to combine both $\operatorname{arsinh}(..)$ -terms: $\begin{aligned} \Phi'_l &= \frac{\mu_0 I}{4\pi} \cdot\frac{S}{2}\int_{-(1-c)}^{1-c}\int_{-1}^1\frac{du\:dv}{ \sqrt{(u-v)^2 + c^2} } = \frac{\mu_0 IS}{8\pi} \int_{-(1-c)}^{1-c}\left[ \operatorname{arsinh}\left(\frac{u-v}{c}\right) \right]_{-1}^1\:dv\\[.5em] &\underset{(*)}{=}\frac{\mu_0 IS}{8\pi}\cdot 2\cdot c \int_1^V\operatorname{arsinh}(v)\:dv=:\frac{\mu_0 IS}{4\pi}c(G(V)-G(1))\qquad\left|V:=\frac{2-c}{c}\right. \end{aligned}$ The missing anti-derivative $G(v)$ is $\begin{aligned} \int\operatorname{arsinh}(v)\:dv&=v\operatorname{arsinh}(v) - \sqrt{1+v^2}+C=:G(v)+C \end{aligned}$ With the exact same steps we used for the lower piece of $C_2$ we consider the upper piece to obtain $\Phi'_u= \frac{\mu_0 I}{4\pi} \cdot\left(-\frac{S}{2}\right)\int_{-(1-c)}^{1-c} \operatorname{arsinh}\left(\frac{1-v}{2-c}\right) + \operatorname{arsinh}\left(\frac{1+v}{2-c}\right) \:du =-\frac{\mu_0 IS}{4\pi}(2-c)(G(1)-G(V^{-1}))$
We can now compute the total flux $\Phi = 4\Phi'$ and the self-inductance $L_a$ : $L_a I=\Phi=4\Phi'=4(\Phi'_l+\Phi'_u)=\frac{\mu_0 IS}{\pi}\left[ cG(V) + (2-c)G(V^{-1})-2G(1) \right]$ If we compare our result with the approximation $L_f$ , we get $\begin{aligned} \frac{L_a}{L_f} &= \frac{ cG(V)+ (2-c)G(V^{-1})-2G(1) }{ 2\left[ \log(\frac{S}{R}) - 0.774 \right] } \approx 0.9973&&&\Rightarrow &&&& \left\lfloor1000\frac{L_a}{L_f}\right\rfloor &= \boxed{997} \end{aligned}$

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Rem.: This was an interesting journey into approximation. I'm pretty sure the approximation given in this problem was derived this way - if we consider small radii $0<R\ll S$ and therefore $c\rightarrow 0,\:V\rightarrow\infty$ , all terms converge except for $cG(V)$ : $\begin{aligned} cG(V)&=(2-c)V^{-1}G(V)\rightarrow 2\left( \operatorname{arsinh}(V) -\sqrt{1+V^{-2}} \right)\rightarrow 2\left( \log(V)+\log(2)-1 \right)\rightarrow 2\left( \log\left(\frac{S}{R}\right)+\log(2)-1 \right) \end{aligned}$ Adding all converging constants, we get the approximation for small radii $\begin{aligned} L_a&\rightarrow \frac{2\mu_0 S}{\pi}\left( \log\left(\frac{S}{R}\right)+\log(2)-2-\log\left(1+\sqrt{2}\right)+\sqrt{2} \right)\approx\frac{2\mu_0 S}{\pi}\left( \log\left(\frac{S}{R}\right)-0.7740 \right) \end{aligned}$

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Rem.: For those unfamiliar with the formula, we derive the flux $\Phi$ from both "Biot-Savard's Law" and "Stoke's Theorem" (ST) via $\begin{aligned} &&\vec{B}(r)&=\nabla_{\vec{r}}\times\vec{A}(\vec{r}) \qquad\left| \vec{A}(\vec{r})=\frac{\mu_0 I}{4\pi}\int_{C_1}\frac{\vec{dl}_1}{|\vec{r}-\vec{r}_1|} \right.\\[.5em] \Rightarrow&&\Phi&=\int_A\vec{B}(\vec{r})\:d\vec{A}=\int_A\nabla_{\vec{r}}\times\vec{A}(\vec{r})\:d\vec{A}\underset{\text{ST}}{=}\int_{C_2}\vec{A}(\vec{r}_2)\:\vec{dl}_2=\frac{\mu_0I}{4\pi}\int_{C_2}\int_{C_1}\frac{<\vec{dl}_1,\:\vec{dl}_2>}{|\vec{r}_1-\vec{r}_2|} \end{aligned}$