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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 VIV) $A\in\mathbb{R}^--(-1,1),\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+1\right)\rightarrow$ RH Circularly Polarized.
11. Case VV) $A\in\mathbb{R}^+-(-1,1),\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+1\right)\rightarrow$ 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?

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?

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) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\mathbb{R}-\left\{\frac{\pi}{2}\mathbb{Z}_O\right\},A\phi<0\rightarrow$ RH Elliptically Polarized.
7. Case VI) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\mathbb{R}-\left\{\frac{\pi}{2}\mathbb{Z}_O\right\},A\phi>0\rightarrow$ LH Elliptically Polarized.
8. Case VII) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+3\right)\rightarrow$ RH Circularly Polarized.
9. Case VIII) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+1\right)\rightarrow$ 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?

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 VIV) $A\in\mathbb{R}^--(-1,1),\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+1\right)\rightarrow$ RH Circularly Polarized.
11. Case VV) $A\in\mathbb{R}^+-(-1,1),\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+1\right)\rightarrow$ 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?

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) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\mathbb{R}-\left\{\frac{\pi}{2}\mathbb{Z}_O\right\},A\phi<0\rightarrow$ RH Elliptically Polarized.
7. Case VI) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\mathbb{R}-\left\{\frac{\pi}{2}\mathbb{Z}_O\right\},A\phi>0\rightarrow$ LH Elliptically Polarized.
8. Case VII) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\frac{\pi}{2}\mathbb{Z}_O\rightarrow$ LH Circularly Polarized.

I know there is room for improvement, however, because I couldn't identify a case where the wave could be RH circularly polarized. If you know when that happens, please tell me...

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?

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) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\mathbb{R}-\left\{\frac{\pi}{2}\mathbb{Z}_O\right\},A\phi<0\rightarrow$ RH Elliptically Polarized.
7. Case VI) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\mathbb{R}-\left\{\frac{\pi}{2}\mathbb{Z}_O\right\},A\phi>0\rightarrow$ LH Elliptically Polarized.
8. Case VII) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+3\right)\rightarrow$ RH Circularly Polarized.
9. Case VIII) $A\in\mathbb{R}-\left\{(-1,0)\cup(0,1) \right\},\phi\in\frac{\pi}{2}\left(4\mathbb{Z}+1\right)\rightarrow$ 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?