Magnetic Flux (5-28-2021)

First, the magnetic vector potential is computed using the relation:

$\nabla \times \vec{A} = \vec{B}$ $\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}=1 \ ; \ \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}=1 \ ; \ \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}=1$

With a little bit of trial and error, it is seen that the following vector satisfies the above relations.

$\vec{A} = z(\theta) \ \hat{i} + x(\theta) \ \hat{j} + y(\theta) \ \hat{k}$

Now, by applying Stokes' theorem

$\Phi= \int \int_{S}\vec{B} \cdot d\vec{S} =\int \int_{S} (\nabla \times \vec{A}) \cdot d\vec{S} = \oint_{C} \vec{A} \cdot d\vec{R}$

$\vec{R} = r(\theta) \cos{\theta} \ \hat{i} + r(\theta) \sin{\theta} \ \hat{j} + z(\theta) \ \hat{k}$ $\frac{d\vec{R}}{d\theta} = \frac{d}{d\theta} \left(r(\theta) \cos{\theta}\right) \ \hat{i} + \frac{d}{d\theta} \left( r(\theta) \sin{\theta}\right) \ \hat{j} + \frac{dz(\theta)}{d\theta} \ \hat{k}$ $d\vec{R} = \left(\frac{d}{d\theta} \left(r(\theta) \cos{\theta}\right) \ \hat{i} + \frac{d}{d\theta} \left( r(\theta) \sin{\theta}\right) \ \hat{j} + \frac{dz(\theta)}{d\theta} \ \hat{k}\right) \ d\theta$

With the parameterisation complete, the next step is to evaluate the integrand. Here, simplifications have been left out and may be attempted by anyone interested.

$\Phi = \int_{0}^{2 \pi} \vec{A} \cdot d\vec{R} = \int_{0}^{2 \pi} f(\theta) \ d\theta \implies \boxed{\lvert \Phi \rvert \approx 7.3566}$

The integral can be solved by outsourcing to Wolfram-Alpha or by using a script of code.