# Simplified forms of Maxwell's curl equations for special case of $\vec{J}$

From Maxwell's curl equations, obtain the particular differential equations for the case of $$\vec{J} = J_z(y,t)\hat{z}$$.

The solution provided for this question shows something like this:

$$\begin{vmatrix} \vec{a_x}&\vec{a_y}&\vec{a_z}\\ 0&\frac{\partial}{\partial y}&0\\ {E_x}&{E_y}&{E_z}\\ \end{vmatrix} = -\frac{\partial \vec{B}}{\partial t}$$ and $$\begin{vmatrix} \vec{a_x}&\vec{a_y}&\vec{a_z}\\ 0&\frac{\partial}{\partial y}&0\\ {H_x}&{H_y}&{H_z}\\ \end{vmatrix} = \vec{J}+\frac{\partial \vec{D}}{\partial t}$$

Why does the $$\nabla$$ have components with 0 value?

$$\nabla$$ does not have zero components. I'm guessing that this is just a short-hand notation for the fact that derivatives of the fields w.r.t $$x,z$$ should vanish, as the sources have a translational symmetry along the $$x,z$$ coordinates. So of course $$\partial_x \neq 0$$, but $$\partial_x E_x = \partial_x E_y = \partial_x E_z = 0$$ etc.
• Why does it have a symmetry along the $x$, $z$ coordinate? How do we define $\vec{E}$'s direction from the given $\vec{J}$? Oct 25 '20 at 10:12
• The currents are independent of $x,z$, thus the problem has the translational syymetry, and thus the fields also have this symmetry, which means that they are independent of $x,z$. I did not say that we know something about the direction of $E$. Just the fact that the $x,z$ derivatives of any component vanish. Oct 25 '20 at 12:05