# Electromagnetic wave phasor conversion 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 ?

I'm a couple months late ...

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 $z$.