So I was given the electromagnetic E field equation in phasor form and I converted it to sinusoidal form. Is it correct ?
Also will it be a reflected wave since we have (wt+Bz) and not (wt-Bz) ?
Also will it be circularly polarized since it is constant at all angles ?
Best Answer
I'm a couple months late ...
Your math checks out, but your words are imprecise.
When we describe a quantity in phasor notation, we drop the \$e^{+j\omega t}\$ time dependence. That's because the frequency and time dependence should be implied. (Note that typically electrical engineers use the \$e^{+j\omega t}\$ time dependence convention, while physicists often use \$e^{-j\omega t}\$.)
The relationship between the electric field in the time domain \$\mathcal{E}(x,y,z,t)\$, and the electric field phasor \$\mathbf{E}(x,y,z)\$ is
$$\mathcal{E}(x,y,z,t) \equiv \text{Re}\left\{ \mathbf{E}(x,y,z) e^{+j\omega t}\right\} $$
in your case, you should write
$$\mathbf{E}(x,y,z) = 10^{-4}\left( \mathbf{\hat{x}} e^{j(20z)} + \mathbf{\hat{y}} e^{j(20z+\pi/2)} \right)$$ $$\mathcal{E}(x,y,z,t) = 10^{-4}\left( \mathbf{\hat{x}} \cos(\omega t+20z) - \mathbf{\hat{y}} \sin(\omega t+20z) \right) $$
Reflected vs transmitted, depends on the direction of an incident field, which you haven't told us. What I can tell you is that your wave is propagating in the \$-z\$ direction.
It is circularly polarized, but I don't know what you mean by "constant at all angles". It's circularly polarized because the polarization is rotating in a circle. At \$t=-\frac{1}{\omega}20z\$ the field is entirely polarized in the \$+\mathbf{\hat{x}}\$ direction. Increase the time to \$t=-\frac{1}{\omega}20z + \frac{1}{\omega}\frac{\pi}{2}\$, and it's polarized in the \$-\mathbf{\hat{y}}\$ direction. Increase in time another \$\frac{1}{\omega}\frac{\pi}{2}\$ and it's \$-\mathbf{\hat{x}}\$ polarized, another one and it's \$+\mathbf{\hat{y}}\$. Circular.
Also, if we want to be super picky, we should note that \$20\$ isn't unitless. It's \$20 \text{meters}^{-1}\$, or whatever length unit you're using for \$z\$.