Magnetic vector potential from an infinite current sheet with thickness t

[superiority]

Homework Statement

An infinite sheet of copper conductor, thickness t, lies in the xz-plane. The sides of the sheet intersect the y-axis at $y=\pm\frac{t}{2}$. The current density in the sheet is given by:
$${\bf{j}}({\bf{r}}) = \begin{cases} j_0\left(\frac{y}{t}\right)^2{\bf{\hat{x}}}, & -\frac{t}{2} < y < \frac{t}{2} \\ 0, & \textrm{elsewhere} \end{cases}$$
Determine the magnetic vector potential A inside and outside the sheet and hence find B.

Homework Equations

$$\nabla^2A_x = \mu_0 j_x = j_0\left(\frac{y}{t}\right)^2$$
$$\nabla^2A_y = \mu_0 j_y = 0$$
$$\nabla^2A_z = \mu_0 j_z = 0$$
$$\nabla \cdot {\bf{A}} = 0$$
$$\nabla\times {\bf{A}} = {\bf{B}}$$

The Attempt at a Solution

I know the standard equation for finding magnetic vector potential is ${\bf{A}}({\bf{r}}) = \frac{\mu_0}{4\pi}\int\frac{{\bf{j}}({\bf{r'}})}{|{\bf{r}}-{\bf{r'}}|} \,d\tau '$, but that only applies when J goes to 0 at infinity. When the current is constant over an infinite region, it diverges.

I know that in calculating magnetic vector potential for an infinite wire, you can substitute in some finite limit c, calculate the integral, and then (I believe) take the limit as $c\rightarrow\infty$ and possibly add some constant term (allowed because of the freedom in A), but I can't figure out what the appropriate integral would be in this situation, or even if that same strategy would work at all. I'm not even sure which direction A would point (though B is fairly straightforward, I think: straight up at the edge on the positive y-side, diminishing as you move to infinity or to the origin, then changing direction and doing the same thing backwards).

 

[Kirjava]

Well you've written down the differential equations for A, so have you tried solving them?

 

[superiority]

Okay, I need boundary conditions to find the solutions, right? I know that A is continuous everywhere. (Incidentally, I realized that I forgot to write $\nabla^2{\bf{A}}_x = 0$ outside the sheet). A is 0 at $y = \infty$. So that gives me

$$\frac{\partial^2 A^{in}_x}{\partial x^2} + \frac{\partial^2 A^{in}_x}{\partial y^2} + \frac{\partial^2 A^{in}_x}{\partial z^2} = j_0\left(\frac{y}{t}\right)^2$$
$$\frac{\partial^2 A^{in}_y}{\partial x^2} + \frac{\partial^2 A^{in}_y}{\partial y^2} + \frac{\partial^2 A^{in}_y}{\partial z^2} = 0$$
$$\frac{\partial^2 A^{in}_z}{\partial x^2} + \frac{\partial^2 A^{in}_z}{\partial y^2} + \frac{\partial^2 A^{in}_z}{\partial z^2} = 0$$
$$\frac{\partial^2 A^{out}_x}{\partial x^2} + \frac{\partial^2 A^{out}_x}{\partial y^2} + \frac{\partial^2 A^{out}_x}{\partial z^2} = 0$$
$$\frac{\partial^2 A^{out}_y}{\partial x^2} + \frac{\partial^2 A^{out}_y}{\partial y^2} + \frac{\partial^2 A^{out}_y}{\partial z^2} = 0$$
$$\frac{\partial^2 A^{out}_z}{\partial x^2} + \frac{\partial^2 A^{out}_z}{\partial y^2} + \frac{\partial^2 A^{out}_z}{\partial z^2} = 0$$
$$\frac{\partial A^{in}_x}{\partial x} + \frac{\partial A^{in}_y}{\partial y} + \frac{\partial A^{in}_z}{\partial z} = 0$$
$$\frac{\partial A^{out}_x}{\partial x} + \frac{\partial A^{out}_y}{\partial y} + \frac{\partial A^{out}_z}{\partial z} = 0$$
$${\bf{A}}^{in}\left(y = \tfrac{t}{2}\right) = {\bf{A}}^{out}\left(y = \tfrac{t}{2}\right)$$
$$\lim_{y \to \infty}{\bf{A}}^{out}(y) = 0$$

But the problem is that I don't actually know how to solve for the components of A from this point.

 

[Kirjava]

In problems like these it pays enormously to think about the symmetries of your setup. You have a sheet which is infinite in the x and z directions. Would it make sense for the answer to depend on these coordinates?

 

[paranormal]

I would first start by breaking down the problem into smaller parts. The first step would be to determine the magnetic vector potential inside the sheet.

Using the given current density, we can solve the Laplace equation for the x-component of the magnetic vector potential, A_x, using the expression \nabla^2A_x = \mu_0 j_x. This gives us A_x = \frac{\mu_0 j_0}{6t^2}(y^3 - \frac{t^3}{2}) + c, where c is a constant of integration. We can then use the given boundary conditions at y = \pm\frac{t}{2} to solve for the constant c.

Next, we can use the fact that the magnetic vector potential is continuous at the boundary of the sheet (y = \pm\frac{t}{2}) to solve for the y-component of A, A_y. Since the current density is zero outside the sheet, A_y must also be zero outside the sheet. Therefore, we only need to solve for A_y inside the sheet, which can be done using the Laplace equation \nabla^2A_y = \mu_0 j_y = 0. This gives us A_y = c_1x + c_2z, where c_1 and c_2 are constants of integration. Again, we can use the boundary conditions to solve for these constants.

Finally, we can use the fact that the magnetic vector potential must be divergence-free (\nabla \cdot {\bf{A}} = 0) to solve for the z-component of A, A_z. Since A_z must also be zero outside the sheet, we only need to solve for A_z inside the sheet using the Laplace equation \nabla^2A_z = \mu_0 j_z = 0. This gives us A_z = -\frac{\mu_0 j_0}{6t^2}(y^3 - \frac{t^3}{2}) + c_3, where c_3 is a constant of integration. Again, we can use the boundary conditions to solve for this constant.

Now that we have determined the magnetic vector potential inside the sheet, we can use the expression \nabla \times {\bf{A}} = {\bf{B}} to calculate the magnetic field, B. This gives us B =