So let us now come to our exercises, and if you had problems calculating these examples, please don't
feel discouraged, because the truth is in theoretical physics, the exercises are most often quite
difficult and you have to spend a lot of time solving them, and especially you have to spend the time
to get the right idea.
So now I will present the solutions for the three examples.
So the first task was about the electric field and point charges, and we had three different tasks.
First of all, we had to write down the electric field of the positive charge and we had to draw the
electric field lines that after we had to do the same thing with a negative charge, and then we had
to consider a combination of two charge that half a distance of the length A.
So first of all, let us write down the electric field for A and B, and in both cases, the result
is directly given by the Cullum's law.
So we have a one over R-squared dependence and here we have a positive charge plus.
Q And here we get the same results but with a negative charge minus Q.
And in both cases, we also want to know about the orientation, so here the orientation is always along
the radial direction.
So from the center where our charge is located, the orientation E.R. goes along the radial direction
and it also determines the direction of the electric field.
Now, for see, the result is actually quite simple to write down, because we want to consider our
first charge plus queue at the position of zero.
So it is just the field of eight and then the other charge is negative.
So it's similar results to be, but it's displaced by a.
So we have to account for this by shifting our vector here so we get our minus a square.
And also we need to be very careful because the orientation is now different.
It is still a radial orientation, but not with respect to zero, but with respect to the position of
the second charge, which is displaced by a.
So while this is written down, quite simply, it's actually very hard to really write down the exact
form because of this R minus a vector here, which is different from E r, so we have to be really careful.
And you will see later in this course we will determine the electric field of such a dipole.
This is what it is called.
But we can already draw the field lines, which is quite easy because we know that for a the field lines
look like this.
So it's along the radial direction.
So they pointier away from the center of this CU charge.
And for B it's the same thing.
But just with it, with the different sine.
OK, so now let's really consider these two charges as pears.
So this is the solution for see where these are displaced by a distance of a.
So what we have to do is we have to add up all of these vectors here.
So, of course, they will also be vectors in this wide space here.
So if we do this, we get something like this, which is the electric field of the dipole.
Now, the second task was about the induction law, so we had to consider a current loop with an area
of a square and it is dragged from outside of the magnetic field to the inside of a magnetic field,
which is constant and is pointing along the perpendicular direction compared to just plain here.
So we could say it's pointing along the Z direction.
And what we want to do is we want to bring this current look inside of the field.
So as you have just seen, we've brought it inside and then out again.
But here in this example, we just want to bring it inside and we want to calculate the voltage that
emerges.
And as we have learned, the induction voltage is given by the temporal change, this derivative of
the magnetic flux.
So basically, all we have to figure out is the magnetic flux of this example.
So the magnetic flux is given by this integral here where we have to integrate the magnetic fields in
the area of this current loop in this area.
And this this here is a vector that points along the Z direction.
So it's parallel to be.
So this means we can just calculate B times day.
And since our yeah, our area does not change, we can also write it down as B zero times the area of
aid that is inside of the field region.
So now we just have to think about how does the field rate region change over time when we move this
this loop here with the velocity of V?
So of course the area that is included is given by the length A, which was this one here.
And then we have another length which I called A in, which is the length of this side here that is
inside of the field.
So right now it is zero.
But if you bring this loop completely into the field, then it is a and this of course, given by the
velocity times the time because we considered a motion with a constant velocity.
V So we have our flux which is given by that one, and we can easily calculate the partial derivative
versus back to T.
So it's just minus B zero times.
Eight times V.
Now, of course, you also have to consider that once this whole loop is inside of the field and you
keep on moving it inside of the field, then the voltage will drop to zero.
This is because then the flux will be constant.
It will just be be zero times a day and then the time derivative will be zero.
Now, the third task is a bit more difficult.
So here we have to consider a straight wire that goes along the direction and we have the magnetic field
that has a profile that is given by this by this polar direction.
There's a certain cylindrical coordinates.
So these we have discussed in the mathematical tutorial section.
But even if you're not familiar with the with cylindrical coordinates, you can still solve this example
if you just use this representation of this vector here in Cartesian coordinates.
So it's it's helpful to know about cylindrical coordinates in this example, but it is not necessary.
So what we have to do here is we have to take amperes law.
And I just want you to check that it is correct so that this can be a correct solution.
So we have to verify that the Kerl or the rotation of B is given by zero times to the current density.
And here I have already given you a hint.
It's very important to realize that our current density is zero everywhere, except for the position
of the wire, which is for real equal to zero.
So this means we have to verify that this one here is zero for every point except for row zero.
So let's do it.
And here you can see the whole solution.
I mean, I already show you that result.
But let me briefly go through this calculation.
So what I have done here is just I have just written down the Knobler vector and then here we have the
magnetic fields.
And here I have used this F.I. vector, which is this one here, and I have used that row is given by
the square root of X squared plus Y square.
So this is sincere.
We have row and here we have one overall, we get one overall square, which is why we have here X squared
plus Y square.
Now the next step I have calculated this vector product here and if you look at it carefully, you will
see that this expression here only has a Z component.
This is where you calculate the X derivative of the Y component of this one here, minus the Y derivative
of the X component, which is this one here.
So we get these two terms and in both cases we have to use the product rule because we have a product
of X times one over X squared plus Y square.
And both of these functions are X dependent here and here.
So this means for both of these terms we get two additional terms.
So, of course, the first one is very easy.
We just have to differentiate X, which gives us one times the other function, which is one of X squared
plus Y square.
And then for the other derivative, we have to take X times the derivative of one over X squared plus
Y square and one of X equals Y square is the same as saying in brackets X squared plus Y square to depart
from minus one.
So we have to multiply by minus one, increase the Powerball, but I just say decrease the power, which
gives us power of minus two, which is this one here.
And then we have to account for the inner derivative, which is 2x X, which is this one.
And then we do the same thing for the other component.
We have to I have to apologize.
Here's a mistake.
This is not X, this is, of course, Y and this is X.
There's a I've made a typo when I wrote down the equation.
So this one must be y times to Y divided by X squared plus Y square squared.
OK, so if we have done this, we have this term here twice, so we have to over X squared plus Y square
and then we have here two Times Square over X squared plus by square.
And then the other term where we have this typo here we have two Times Square over X squared plus Y
square squared.
And then you can see that you can here get rid of this power too and get rid of this one.
So we have to over X squared plus Y square, minus two of X squared plus Y square, which is zero.
So this means we have verified that this is indeed zero.
But please be careful.
We have divided here by row, which means our whole calculation is only true for row different from
zero, which in this case is all we wanted to know.
We will learn later, underscores how we can really determine this relation here.
So we will really derive it.
We will go the other way around.
We will start from the umpire's law and then we will derive that the field looks like this.
And at the moment we cannot really do this because this calculation here is not valid for the position
are equal to zero.
So please stay tuned and continue watching the course and then you will see that we can also solve the
whole problem later on.