Recent content by sammydafish

Yes, that's it. I was missing the fact that the vector potential equation includes both the top and bottom automatically while in Ampere's Law I chose to enclose just the current in the top surface. Upon enclosing both the top and bottom surface I do of course get zero B-field.

Ok, I see. That's interesting. So this means there is something special about the slab/geometry? I applied my method above for the E field outside a uniformly charged sphere is given by gauss' law: $$\mathbf{E} = \frac{\rho R^{3}}{3 \epsilon_0 r^{2}} \hat{r}$$ Where again, $$ \mathbf{E} =...

The integral I mentioned is solved repeatedly in any introduction to electrodynamics course before students learn about gauss' law and potentials. See Griffiths 4th edition Chapter 2 for various configurations which require solving that integral. I understand it can be done the way you suggest...

The bound surface current is $$\mathbf{K}_b = \mathbf{M} \times \hat{n}$$ which gives $$\mathbf{K}_b = {M} \hat{x}$$ Applying ampere's law: ##\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}## to the top surface, which encloses the bound surface current yields: ##\mathbf{B} = \mu_0...