# Find the direction of rotation of the given elliptically polarized electromagnetic wave

$$E_x = \cos(wt-\beta z)$$ $$E_y = \cos(wt -\beta z + \frac{\pi}{4})$$

$$E_x$$ is of course in the $$x$$ direction and $$E_y$$ is in $$y$$ direction. Direction of propagation of EM wave is $$z$$.

I already know that the above EM wave is elliptically polarized, what matters is the direction of polarization: Is it Left Handed Elliptically Polarized; LHEP or RHEP?

To simplify my life, I let $$z = 0$$

This gives: $$E_x = \cos(wt)$$ $$E_y = \cos(wt + \frac{\pi}{4})$$

Let's start at $$t = 0$$. $$E_x = 1$$ $$E_y = \frac{1}{\sqrt{2}}$$

Next, let's allow $$t = \frac{T}{4}$$ which means $$wt = \frac{\pi}{2}$$

$$E_x = 0$$ $$E_y = -\frac{1}{\sqrt{2}}$$

The movement of $$E$$ vector starts from the first quadrant down to the negative $$y$$-axis. This clearly looks like clockwise or right-handed polarization.

However, the answer key suggests its left-handed. What am I doing wrong? I have a lot of faith in the answer key as this question belongs to a prestigious examination.

• Is the difference due to the handedness convention used? Commented Dec 31, 2023 at 12:44
• @Farcher The examination from which this question is from is meant for graduate engineers and as mentioned in your link; "From the point of view of the source" convention is used by IEEE, which is what the question probably thinks about. My convention is the same. By right hand rule on an x-y plane, the direction of propagation of the EM wave is towards me and the movement of the E vector is clockwise in this case. There were two more EM waves mentioned in the question that were labelled as correct, I'll check them out soon. Commented Dec 31, 2023 at 15:00
• It is clockwise when you are looking in the negative z-direction, ie towards the source. Your convention requires you to be looking in the positive direction, ie away from the source which means that the motion is counter-clockwise, ie left-handed. Commented Dec 31, 2023 at 15:38
• Right.. so "from the point of view of the receiver" is what I have described in my answer; clock wise / right handed. "from the point of view of the source" would be anti-clock wise / left handed. Okay I think I understand. But again, I have to assume that the exam setters followed the "from the point of view of the source" standard / IEEE standard.. which seems to be the right assumption. Thanks @Farcher Commented Dec 31, 2023 at 17:15

In my question, I mentioned that the movement of the $$E$$ vector is clockwise and right-handed is incorrect. It is certainly clockwise (from the receiver's perspective), but not right-handed. This is because the wave is traveling in the $$z$$ direction; thus the wave is emerging out of the standard $$x$$-$$y$$ plane - which means the wave is approaching the receiver which means when I point my left thumb away from the plane (and also, away from the source), I end up pointing to myself. Therefore the wave is indeed left-handed. It is also anticlockwise from the source's perspective and clockwise from the receiver's perspective.