Mutual Inductance (1-12-2021)

Essentially, the idea here is to pass one unit of current through one of the coils, and measure the resulting flux in the other coil. There are two ways to solve this, as indicated below:

$\phi = \iint_S \vec{B} \cdot \vec{dS} = \iint_S (\nabla \times \vec{A}) \cdot \vec{dS} = \oint_C \vec{A} \cdot \vec{d \ell}$

An integral is required to determine the field $\vec{B}$ or $\vec{A}$ from the source. And then integrating over a surface requires two additional integrals, whereas integrating over a boundary curve requires one additional integral. So the left side ultimately requires a triple integral, and the right side requires a double integral. The surfaces and boundary curves are all very well defined in this problem, so that isn't much of a concern. When integrating numerically, for a given amount of computation, you get a better result using the magnetic vector potential method. @Karan Chatrath has shown the triple integral method, and has remarked on the numerical difficulties. For reference, to get an answer close to that of the vector potential method, I divided each spatial parameter into $1000$ pieces, resulting in a billion loop iterations. Of course, it took forever to run. By contrast, with the vector potential method, I got good results with "just" $25$ million loop iterations. By "good results", I mean that the answers were all roughly equal with $25$ million, $100$ million, and $400$ million loop iterations.

I got lucky while solving this one, on the last try. This is because of my choice of numerical resolution and problem approach. My first approach is as follows:

$\vec{r}_1 = (1,r\cos{t_1},2+r\sin{t_1})$ $\vec{r}_2 = (\cos{t_2},\sin{t_2},0)$ $d\vec{r}_2 = (-\sin{t_2},\cos{t_2},0) \ dt_2$

Magnetic field due to an elementary arc of loop 2 at a point in the plane of loop 1:

$d\vec{B} = \frac{1}{4\pi} \frac{d\vec{r}_2 \times (\vec{r}_1-\vec{r}_2)}{\lvert \vec{r}_1-\vec{r}_2\rvert^3}$

The elementary flux through loop 1:

$d\Phi = d\vec{B} \cdot \left (r \ dr \ dt_1 \ \hat{i}\right) = f(r,t_1,t_2) dt_1 \ dt_2 \ dr$

$\Phi = \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{2\pi} f(r,t_1,t_2) dt_1 \ dt_2 \ dr \approx 0.0763$

Simplifications left out. Numerical resolution: $dr = 0.001 \ ; \ dt_1 = dt_2 = \pi/100$ .

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The approach that actually worked was:

$\vec{r}_1 = (1,\cos{t_1},2+\sin{t_1})$ $\vec{r}_2 = (r\cos{t_2},r\sin{t_2},0)$ $d\vec{r}_1 = (0,-\sin{t_1},\cos{t_1}) \ dt_1$

Magnetic field due to an elementary arc of loop 1 at a point in the plane of loop 2:

$d\vec{B} = \frac{1}{4\pi} \frac{d\vec{r}_1 \times (\vec{r}_2-\vec{r}_1)}{\lvert \vec{r}_2-\vec{r}_1\rvert^3}$

The elementary flux through loop 2:

$d\Phi = d\vec{B} \cdot \left (r \ dr \ dt_2 \ \hat{i}\right) = f(r,t_1,t_2) dt_1 \ dt_2 \ dr$

$\Phi = \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{2\pi} f(r,t_1,t_2) dt_1 \ dt_2 \ dr \approx 0.0741$

Simplifications left out. Numerical resolution: $dr = 0.001 \ ; \ dt_1 = dt_2 = \pi/100$ .