Mutual Inductance - Wavy Loop

Consider a point on loop 1:

$\vec{r}_1 = \cos{\theta_1} \ \hat{i} + \sin{\theta_1} \ \hat{j} + 0 \ \hat{k}$

And that on loop 2:

$\vec{r}_2 = \cos{\theta_2} \ \hat{i} + \sin{\theta_2} \ \hat{j} + (3+2\sin{\theta_2}) \ \hat{k}$

The arc length elements on each loop can be computed by evaluating the total differentials:

$d\vec{r}_1 = \left(-\sin{\theta_1} \ \hat{i} + \cos{\theta_1} \ \hat{j} + 0 \ \hat{k} \right) \ d\theta_1$ $d\vec{r}_2 = \left(-\sin{\theta_2} \ \hat{i} + \cos{\theta_2} \ \hat{j} + 2\cos{\theta_2} \ \hat{k} \right) \ d\theta_2$

Finally, we can compute the elementary magnetic vector potential at any general point due to loop 1:

$d\vec{A} = \frac{\mu_o I}{4 \pi}\left(\frac{d\vec{r}_1}{\lvert \vec{r} - \vec{r}_1\rvert}\right)$

The total magnetic potential vector due to loop at any point can be evaluated as such:

$\vec{A} = \frac{\mu_o I}{4 \pi} \oint_{L1} \frac{d\vec{r}_1}{\lvert \vec{r} - \vec{r}_1\rvert}$

Now, we know that the magnetic field at that point can be computed as such:

$\vec{B} = \nabla \times \vec{A}$

The elementary surface area through the loop is say: $d\vec{S}$ . This means that elementary flux is:

$d\Phi = \vec{B} \cdot d\vec{S}$

The total flux can be computed by integrating over the surface area of loop 2 as such:

$\Phi = \oint_{S2}\vec{B} \cdot d\vec{S}$ $\implies \Phi = \oint_{S2} \left(\nabla \times \vec{A}\right) \cdot d\vec{S}$

Applying Stoke's theorem leads to:

$\Phi = \oint_{S2} \left(\nabla \times \vec{A}\right) \cdot d\vec{S} = \oint_{L2} \vec{A} \cdot d\vec{r}_2$ $\Phi = \oint_{L2} \oint_{L1} \frac{\mu_o I}{4 \pi} \frac{d\vec{r}_1}{\lvert \vec{r}_2 - \vec{r}_1\rvert}\cdot d\vec{r}_2$

Substituting expressions and simplifying the integrand gives a function $f(\theta_1,\theta_2)$ and the integral looks as such:

$\Phi = \int_{0}^{2 \pi} \int_{0}^{2 \pi} f(\theta_1,\theta_2) d\theta_1 \ d\theta_2$

I performed this entire calculation process using a script of code and have only provided an outline. The answer evaluates to:

$\boxed{\Phi \approx 0.1106}$

$\textcolor{#BA33D6}{FunProblem}$ As always, it is a pleasure to solve your problems. Thank you for another good one.

The wavy circle is changing its position according to the curve $sin \theta_{2}$ . Its range is from $z=1$ to $z=5$
Take a $z=\gamma$ a coordinate . From the centre of loop $2$ move $\beta$ rightwards and take a elementary strip of width $d\beta$ Find field at this point $(\beta, 0,\gamma)$ . Multiply it with $dA=2π \beta d\beta$ Now by doing this , average it over $\gamma$ , $\gamma=1$ to $\gamma=5$
After that you will reach this expression $\large \phi = 8^{-1}\int_{1}^{5} \int_{0}^{1} \int_{0}^{2\pi} \frac{\beta(1+\beta cos \alpha)}{(\gamma^{2}+sin^{2} \alpha+(cos \alpha-\beta)^{2})^{1.5}} d \alpha d\beta d\gamma \approx 0.11$
I will elaborate my solution, if asked by anyone.