Determining plane wave polarization given the magnetic-field vector phasor

Given the following magnetic-field vector phasor:

$$\vec H(\vec r)=\left[\hat x - j\hat y\right]H_o e^{jkz}$$

I need to find the associated E-field vector phasor so that I can determine the polarization, i.e., linear, circular, or elliptical, and whether it is left-handed (LH) or right-handed (RH).

I've devised and utilized the following classification system on past problems with much success. Here are the steps I use:

1. Put the E-field vector phasor in the form $$\vec E(\vec r) = \left[\hat x + Ae^{j\phi}\hat y\right]E_oe^{-jkx}$$.
2. Case I) $$A=0,\phi\in\mathbb{R}\rightarrow$$ Linearly Polarized.
3. Case II) $$A\in\mathbb{R},\phi\in\pi\mathbb{Z}\rightarrow$$ Linearly Polarized.
4. Case III) $$A\in\left\{(-1,0)\cup(0,1) \right\},\phi\in\mathbb{R}-\pi\mathbb{Z},A\phi<0\rightarrow$$ RH Elliptically Polarized.
5. Case IV) $$A\in\left\{(-1,0)\cup(0,1) \right\},\phi\in\mathbb{R}-\pi\mathbb{Z},A\phi>0\rightarrow$$ LH Elliptically Polarized.
6. Case V) $$A\in\mathbb{R}-(-1,1),\phi\in\mathbb{R}-\left\{\frac{\pi}{2}\mathbb{Z}_O\right\},A\phi<0\rightarrow$$ RH Elliptically Polarized.
7. Case VI) $$A\in\mathbb{R}-(-1,1),\phi\in\mathbb{R}-\left\{\frac{\pi}{2}\mathbb{Z}_O\right\},A\phi>0\rightarrow$$ LH Elliptically Polarized.
8. Case VII) $$A\in\mathbb{R}^+-(-1,1),\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+3\right)\rightarrow$$ RH Circularly Polarized.
9. Case VIII) $$A\in\mathbb{R}^--(-1,1),\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+3\right)\rightarrow$$ LH Circularly Polarized.
10. Case IX) $$A\in\mathbb{R}^--(-1,1),\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+1\right)\rightarrow$$ RH Circularly Polarized.
11. Case X) $$A\in\mathbb{R}^+-(-1,1),\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+1\right)\rightarrow$$ LH Circularly Polarized.

So I know the solution goes like this...

$$\vec H(\vec r)=\left[\hat x - j\hat y\right]H_o e^{jkz} \rightarrow\vec E(\vec r) = \frac{\nabla\times\vec H(\vec r)}{j\omega\epsilon_o}\implies\vec E(\vec r)=\left[\hat y + e^{j\frac{\pi}{2}}\hat x\right]\eta_oH_oe^{jkz}\rightarrow \style{font-family:inherit;}{\text{LH Circularly Polarized.}}$$

But the first step implies that $$\nabla\times\vec H(\vec r)=j\omega\epsilon_o\vec E(\vec r)$$ when it should be $$\nabla\times\vec H(\vec r)=\vec J(\vec r) + j\omega\epsilon_o\vec E(\vec r)$$ for the vector phasors of time-harmonic fields. Why is $$\vec J(\vec r)$$ assumed to be zero?

• Unless you were told that the wave was in a vacuum or other non-conducting medium, then you can't assume $\vec{J}=0$. Incidentally, 9 is IX and 10 is X in Roman numerals. Aug 31, 2020 at 17:22
• @RobJeffries Thanks! Aug 31, 2020 at 17:55

I suspect you were either told that the wave was in a vacuum or another non-conducting medium, in which case it can be assumed that $$\vec{J}=0$$.
If not, then I think you can still assume $$\vec{J}=0$$ if you are told $$k$$ is a real number, because solutions to Maxwell's equations in conductors are of the form $$\vec{H}(\vec{r}) = H_0 e^{jkz} = H_0 e^{-k'z}e^{jk''z}\ ,$$ where $$k = k'' + jk'$$. i.e. They have a dissipative factor which $$\rightarrow 1$$ when $$k$$ is real. So, if $$k$$ is complex then you cannot assume $$\vec{J}=0$$. Or to put it another way, if $$\vec{J} \neq 0$$ then $$k$$ has an imaginary component.