Magnetic Vector Potential

Based on the given information:

$B_x = \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} = 0$ $B_y = \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} = 0$ $B_z = \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} = y^2 -1$

Parameterising in polar coordinates:

$\vec{B} = \left(r^2 \sin^2{\theta} -1 \right) \hat{k}$ $d\vec{S} =\left(r \ dr \ d\theta\right) \ \hat{k}$ $d \Phi = \vec{B} \cdot d\vec{S}$

Integration of the above evaluates to:

$\boxed{\lvert \Phi \rvert = \frac{3\pi}{4}}$

As for the bonus question, yes there are multiple vector potentials because:

$A_x = x^2+y+C_1 \ ; \ A_y=xy^2 + C_2 \ ; \ A_z = C_3$

Where the additional constants are arbitrary. The use of the above vector potential will lead to the same magnetic field and the constants can take be any real number. So there are infinite possibilities of feasible vector potentials.

One way to solve this problem is to reconstruct $\vec{B}$ , and then compute the surface integral of $\vec{B}$ over the disk. The solution provided by @Karan Chatrath shows this method. The other way is to use Stoke's theorem:

$\phi = \int \int_S \vec{B} \cdot \vec{d S} = \int \int_S (\nabla \times \vec{A}) \cdot \vec{d S} = \oint_C \vec{A} \cdot \vec{d \ell}$

Here, the curve $C$ is the circle which bounds the disk. Both methods yield the same answer. As for the bonus, we can add any curl-free vector field to $\vec{A}$ and get the same physics. This gives us an entry-level glimpse at a physics property known as "gauge invariance".

$\vec B=\vec \nabla \times \vec A=(\hat i\frac{\partial }{\partial x}+\hat j\frac{\partial}{\partial y})\times \left (\hat i (x^2+y)+\hat j xy^2\right )=\hat k (y^2-1)$ .

The elementary magnetic flux over an element of area $dxdy$ is $|d\Phi_{mag}|=(1-y^2)dxdy$ , so that the overall magnetic flux linked with the disc is

$|\Phi_{mag}|=\displaystyle \int_{-1}^1 \displaystyle \int_{-\sqrt {1-y^2}}^{\sqrt {1-y^2}} (1-y^2)dxdy=4\displaystyle \int_0^1 (1-y^2)^{\frac{3}{2}}dy=\dfrac{3π}{4}\approx \boxed{2.35619}$ .

Bonus Since $\vec \nabla\times (\vec \nabla \chi) =0$ , therefore the flux density remains unchanged if $\vec A$ be replaced by $\vec A+\vec \nabla \chi$ .