Closed Helix Flux

When I initially read this problem, I found it absurd. Because parameterising the surface area element of a closed helix seems a daunting task, to say the least. After some pondering, I realised that the only way to tackle this problem is to use the concept of a magnetic vector potential. We know that:

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

$B_x = \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} = 0$ $B_y = \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} = 1$ $B_z = \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} = 1$

On inspection, the following vector satisfies the above equations:

$\vec{A} = z \ \hat{i} + x \ \hat{j} + 0 \ \hat{k}$

Now, Using Stoke's theorem, the flux through the closed helix is:

$\Phi=\oint_{C} \vec{A} \cdot d\vec{r}$

The position vector of a point on the helix is:

$\vec{r} = \cos{\theta} \ \hat{i} + \sin{\theta} \ \hat{j} + \frac{\theta}{2 \pi} \ \hat{k}$ $d\vec{r} = \left(-\sin{\theta} \ \hat{i} + \cos{\theta} \ \hat{j} + \frac{1}{2 \pi} \ \hat{k}\right) \ d\theta$

The line segment joining the start and endpoints of the helix can be ignored as that line is along the Z direction and its contribution in the line integral computation will be zero due to the nature of the magnetic vector potential.

So after simplifications, the required line integral is:

$\boxed{\Phi = \int_{0}^{2 \pi} \left(\cos^2{\theta} - \frac{\theta\sin{\theta}}{2 \pi}\right) \ d\theta = \pi +1}$

using the concept of area vector: