So let us now come to the solutions to the exercises in this section about electromagnetic waves, so
I hope you took the time and solved these tasks.
So if you haven't done so, please take a sheet of paper and read the tasks in the previous lecture.
Now I'm going to present you the solutions.
So first of all, we had a task where we should plot the real part of these two electromagnetic waves,
especially of the electric fields, for four different times.
And from that, you should then determine the polarisation and the corresponding magnetic field.
So let us begin.
We have these two functions, A and B, and I want to start with A..
So what I'm going to do here is, first of all, I want to explain to you what this coordinate system
on the left.
So we have here our base vector K and so in general, this is a dot product of the key vector and the
position vector are so since here it is K times X, we know that our character is along the X direction,
so this means our wave propagates along X..
This is why I've chosen this axis here.
Furthermore, here we have to position X equal to zero.
And then here I want to show you where X is the space vector divided by four, ten by two or three over
four.
And then here the full, uh, the full lambda full wavelength.
OK, so now let's calculate, um, our real part of the electric field.
So in a the real part of E.
Is just one of square root of two Epsilon zero, and then we have here a zero, of course, and then
for the Y component, we have one times cosine of this argument and then one times I times sign of this
argument.
So the real part is just as cosine function.
So we have a cosine of K, X minus Hormiga T and for the for Z component we get three times cosine.
The argument plus I times I times sine which is minus sine.
And this is the report minus sign Kate X, minus Omega T.
Now let's go ahead and calculate this wave for the time T equal to zero.
So here we have four four X equals zero and T equals zero.
We get cosine of zero, which is one.
So are our Y component is one and our Z component is zero because the sign of zero zero.
Then if we go one quarter of a period, the cosine turns into a zero and the sign turns into one.
But we have a minus sign so we get minus one.
So this means we have here a minus one.
And then for half a period, the cosine turns into minus one designed to zero.
And then for another quarter period we go here.
And then finally we are back at the beginning.
So this means let me try to draw the wave on this tablet here.
It's a bit difficult, I think on your sheet of paper, it's much more easy.
So let's go ahead.
Try to connect these these dots.
So we get something like like this one.
And that's OK.
Yeah, it doesn't look look too good, to be honest, so here we have, um.
Yeah, we we can try it like it.
Like this.
And then we got here.
Um, yeah, like this and.
OK, so I think you you can see how this works and you can really see how the electric field wraps around
in the circle.
So we already know of a solution for for the second part.
So this is a circular polarization.
For the other times, let's think for a second.
So if our time propagates or is continued by one period over four, then we increase this argument here.
So this means our callsign and our sign are shifted into the negative direction.
So this means that we start not at this point here for zero, but we start at this point here for zero.
So this means our whole wave is then shifted, so to say.
So you can you can really see how the wave propagates along this direction, because this is our character.
And this means you have to shift this wave along this direction when you propagate over time.
So I will not draw the other images here.
I think you can really imagine what it looks like.
And also we have these animations in the previous sections.
So let's continue with no, it would be with the second wave.
So here once again, we have to use the real part, of course, because this is the physically relevant
quantity.
So, by the way, yeah, this is here.
This is a vector.
So what we get in this case is one of five is zero and then a vector.
So this time our Y factor is obviously along the Z direction.
So this is why I've drawn this coordinate system like this.
This time we have here X and here Y and Kate is along.
This is the direction, which is the direction of the propagation.
All right.
So now let's consider the real and the imaginary parts.
So here this time, it's a bit more difficult because we have in this vector one minus two AI, which
is a real and imaginary part.
So this means we will also get two parts here.
The first one will be one times the cosine.
So this is just one times the cosine of casette minus Omega T, then we get another term, which is
minus two times I times I times sign of the argument.
So this means we get um.
Yeah, we get two times.
The sign.
Yeah, exactly, because minus two times I squared plus two times, that was a sign of Casey minus Omega
T. And then for the Y component, we get two times cosine.
And of course, here we got four times sine Casey and we get T zero and that one.
OK, so now we can write this down a bit of a different way, we write down that this is equal to one
over five e0.
Zero two, and then we can use a relation which allows us to tell what is cosine of this argument plus
two times sign of this argument, and this is something that you do not have to know, but you can look
up on the Internet, for example.
So I use Wolfram Alpha and look at what this is.
In this particular case, this is just the square root of five times the sign of this argument, Casey,
minus Omega T, and then we get another constant, which in this case is if you look it up, it is the
arcus tangles of one half.
But for us, it doesn't really matter what it is.
Let's just say it is some constant because our cost of one half is just some number.
It doesn't really matter what it is.
And here, of course, we get the same thing.
So Casey minus Omega T plus this constant.
OK, now we can start drawing and for for a moment to make it a bit easier.
I just want to show you here that you get a general idea what this wave looks like, look looks like.
So I will not consider this arcus tangency.
I will get to zero in the end.
It is just a phase shift.
So what what we get here in this case is we get for the time equals zero and the Z component zero,
we get an argument of zero.
So the sign in both cases is zero.
So we get really zero here.
And then one quarter of the wavelength further we get a one because the sign has done his maximum.
So it turns into one.
So this means we get for the X and for the Y component, the maximum.
So let's say we get this one here.
Then for half the wavelength we get again zero because design is zero.
And then four.
Yeah, three of four of the wavelengths we get minus one and again, zero.
So let me try to connect this.
I hope it doesn't look too ugly.
OK.
Looks a bit ugly, but OK, and yeah, and so this time you can really see how the wave is linearly
polarized and how it the electric field just oscillates in this plane that is here and yeah.
Spent by this latest line between the X and Y axis.
And of course, there's the axis.
And now what we only have to consider here is this Arcus Tongans are four and half, which means we
get an additional phase shift.
So this means actually our whole wave is a bit shifted.
And then when we consider the different times, you can once again see how this whole sine function
is shifted.
Yeah.
So this means here for this position, we do not get zero, but we get a negative value.
So this means our whole wave has propagated along the Z direction.
So we have already discussed this.
The wave and A is circularly polarized and the wave and B is linearly polarized.
So we know our answer for you for the second task.
So we know that determine the polarization we get in a circular and and B, we get linear.
All right.
Now test three in test three, we have to calculate the corresponding magnetic field.
So how is that done?
You may ask because we only have the electric field.
How can we do it?
And what I want you to do is I just want to, uh, want you to think about the orientation of the magnetic
field.
And what we have learned is that the magnetic field is perpendicular to the character and it's perpendicular
to the electric field and also B and perpendicular to each other.
So this means we must be able to express to be field.
So this is proportional to the vector product of K and E, given the fact that both of them are perpendicular,
which must always be the case.
And it's it is the case here.
So all we have to do here really is to calculate the um.
Yeah.
The vector product.
So in a what we get is OK vector, we have discussed this already, it's along the X direction so we
can just write one zero zero times and then we must just take this vector here in the beginning of the
end of, uh, characterizing the direction of this EESA, we have one over the square root of two and
then we have zero one and I and now if we calculate this, we get for the X component zero for the Y
component, we had zero minus I and for the Z component we get one.
So this means our B of our A. is I'm sorry I forgot one of a square root of two.
It is one of a square root of two and then the vector zero minus I one times B zero to account for the
unit and then the exponential function.
So that's the same argument, of course, as previously for the electric field.
And what we get in this case is, again, a circularly polarized wave, which is always perpendicular
to the electric field that we have drawn on the previous slide.
So if you want, you can go ahead and add this magnetic field with a different color in your sketch.
Now, for B, we must do the same thing.
So again, the cross product of the vector and of this vector characterizing the polarization or the
the orientation of the electric field.
So this time the vector is oriented along the Z direction.
So we get here zero zero one and what we calculated, we get one of five and then for the X component
we get zero minus two minus minus four.
I for the Y component we get one times one minus two ie minus zero.
And for the Z component we get zero point zero.
All right.
So this time the B field is given by one over five, minus two plus four.
I one minus two I zero times be zero times exponential function I K, Z, minus God.
OK, so these are the two solutions.
And also you can see here that like for the electric field we have here, this vector, when we take
the real part and the imaginary part, we see that both of them are parallel to each other.
So this means we get a linear polarization also for the magnetic field.
So once again, you can go ahead and at the magnetic field into your sketch using a different color.
And also something that I have forgot to to mention here in task number B or task number two, we are
just looked at our sketches to determine the polarization.
But of course, you can also look at this vector here where you see that when you take the real part
of this vector and the imaginary part of this vector, that they are perpendicular.
So this is why it's circular.
And here you take the rear part of the vector and the imaginary part of the vector and you see that
they are parallel, which is why they are linearly polarized.
And here it's the same thing.
All right, so this brings us to our last task, which is very easy once you have solved the third task.
So here again, we need the relation that the electric field is always perpendicular to K. and is always
perpendicular to be.
And we must explain why the following electric field cannot correspond to an electromagnetic wave.
And I can already tell you the answer.
It is because in this case, the electric field is not perpendicular to K, and we can check this by
just calculating the orientation where we can just take one eye minus one and some factors which we
do not have to write down here.
And then let's calculate the DOT product with the K vector, which in this case is along.
Why?
And this is of course one time zero zero plus eight times one is EI plus minus one time zero zero.
So we get EI, which is obviously different from zero.
So this means the electric field is in this case not perpendicular to K and therefore it does not correspond
to an electromagnetic wave.