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.

  • $\begingroup$ Why does it have a symmetry along the $x$, $z$ coordinate? How do we define $\vec{E}$'s direction from the given $\vec{J}$? $\endgroup$
    – SharonZh
    Oct 25 '20 at 10:12
  • $\begingroup$ 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. $\endgroup$
    – Rd Basha
    Oct 25 '20 at 12:05

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