2
votes
4answers
132 views

Why does $E=\nabla\phi$ follow from $\nabla\times E=0$?

I understand that using one of Maxwell's equations, $$\vec{\nabla} \times \vec{E}(\vec{x})=0,$$ it can be said that $$\vec{E}(\vec{x})=-\vec \nabla \phi(\vec{x}).$$ However, I can't find or ...
2
votes
2answers
188 views

Applying $\nabla\times\mathbf{B} = \mu_0\mathbf{J}$ in the presence of magnetic shielding

2012-06-13 - Revised question in experimental format (This is a thought experiment for which RF experts may have an immediate answer.) I'll assume (I could be wrong) the possibility of creating a ...