That was all very theoretical.
So in this lecture, we will consider a real example, we will calculate the magnetic field for a current
in a wire.
So this is the set up that we are considering here.
So we have such a straight wire and it has the length L and here this is the equation that we have established
previously for the vector potential.
And you can see that it includes an integral of the current density.
So our current flows in this wire and at the position are prime.
And then we have here two different position vectors.
And so these are the different position vectors.
So this one here includes all the positions that are basically inside of the wire and the other one
includes the whole space.
So because we want to know the vector potential at every possible point in space.
So let us now go ahead and calculate this integral.
So the first thing I do is I apply an approximation.
I want to consider here an infinitely thin wire.
So this means that the square root of this area, so basically you could say the radius or the diameter
is very, very small.
It's infinitely thin.
So what this means is that at every position outside of the wire.
So when we are when this R is outside of this wire, we can say that this distance to the wire is much
larger to the area.
Of course, this is not really true if we are very, very close to the wire, but we are here only interested
in the vector potential and in the magnetic field far away from the wire.
Or we could also say we consider an infinitely thin wire.
So what this does is it allows us to separate disintegration of the volume into an integration over
the cross section of an area and an integration over the Z component.
So along the wire.
So this means we have here is integration of all of the whole length and here this integration of this
cross section, surface area.
And now we can really apply or we can carry out this integration of a D over this area because we know
that this term here is much larger than the surface.
So this means we can calculate integral over F then this G of R and D.
S, and this just gives us the current that is flowing through the wire.
And this current is not dependent on the position because our wires straight and has the same cross
section area along all the length so we can pull it out of the integral.
So this is what we are left with.
Please keep in mind this is an approximation where we say that the current does not change.
Yeah, 30 the area.
So we can just separate these two integration's.
Now, what this means is we have simplified our expression for eight, so that means we can pull out
here.
It is current as well, and we are left only with this integration over the length of the wire.
So what we have done essentially is we have simplified our problem to a quasi one dimensional problem,
which is because our wire is infinitely thin.
So what we have to do is we have to integrate here from minus L to L.
So let's say here zero and here is minus Alah divided by two and here is plus Al divided by two.
And these are the boundaries for our integration.
Now, this distance in these two position vectors is actually given by this distance here, and you
can apply now this Pythagoras theorem where you have here the square root of L Square and square, and
then you integrate over the L and we know that this wire is always along the Z direction.
So this D-R vector here, I have expressed in terms of the L, it was just one dimensional scalar,
you could say, and then some vector easy.
Now we carry out this integration here and the integral of this one of a square root thing here can
be expressed in terms of this logarithm.
So we have as a solution this logarithmic function with the boundaries that we still have to put in
for our.
Now, I apply another approximation, which is that we consider an infinitely long wire, so this means
that it doesn't really matter at which position we are looking at.
It doesn't matter if we are right in the middle or bit above because our wire is infinitely long.
So we are basically always sitting in the middle, roughly speaking.
So this means we always have an infinitely long wire above us and below us, which allows us to simplify
this expression.
Because now we can see that this vector potential is given by the same prefecture, but the multiplied
by two, then we have our current again and is natural logarithm simplifies just to al the length of
our wire divided by S, which is this distance here.
And then we have the orientation, which is a long this direction.
And just so you know what it is, this s is of course the square root of X squared plus Y square, because
remember, even though this looks two dimensional, here are real problems, of course, three dimensional.
So now we know what our vector potential looks like is the solution.
And we have applied here to the approximation of this infinitely long wire and also the approximation
of an infinitely thin wire.
Now, what we do is we calculate the magnetic fields, and for this we only have to calculate the rotation
of this expression here.
So, of course, you could now apply polar coordinates and do some fancy tricks, which I think would
be very elegant.
But sometimes I think you understand more if you just do it straightforwardly and don't use any fancy
tricks and just write down all of the individual components of your vector and just carry out these
differentiations.
So here I have written down.
Yeah.
The derivatives of the sector potential in terms of the three components.
So here can see our vector potential only has a Z component.
So this means we only get two entries here for the calculation of this vector product.
So here I have calculated the derivative of the Z component with respect to the Y coordinate and here
the derivative of the Z component with respect to the X component with a negative sign.
And yeah, I can I can go through it for one of these components.
So d that revetted of the natural logarithm is one divided by this argument here.
So we get, you know, basically the reciprocal value of this one here, which is written down here.
Next, we have to calculate it in a derivative, which is the derivative of this term here.
So we have basically this X squared plus Y square in brackets to the power of minus one half.
So we get the factor of minus one half and then we have to decrease the exponent by one.
So we have now an exponent of minus three 1/2.
So this is here, this one.
So the square root, which is exponent of one half to the power of three.
And then we also have to consider it in a derivative of this thing here, which is if we derive for
the Y component, which is just to Y.
So we have these three different parts that we have to multiply.
And for the X component, you can see the first two parts are exactly the same.
And then for the third part, we just get to X instead of two Y.
So now we can see that most of this expression of yours is the same for these two components, so we
can pull this whole thing out of the of the vector and write it in front of it.
And also, you can see here we have the square root of X squared was twice grand.
Here we have the same thing, but to the power of three.
So we are left with the power of two, which is just one, which just compensates the square root.
So this is why we get this expression here.
And for the vector we just have here, minus Y, X and zero.
Now, the next thing or we could say, OK, this is our solution, but we could also make it a bit more
useful.
So you can see here this is pointing along the Y directions depending on X, and here we have basically
one over the square root of X squared plus Y square squared and the square root of X squared plus Y
square is basically the, you know, the length in the X Y plate.
So this means we can re express this in terms of X Square and we can take one of these S's to this vector
here and this will give us the unit vector in terms of our indirect and pull our direction.
So just to pull our unit vector.
So this is our solution.
We have here a magnetic field that is determined by the current and by the distance, and it is oriented
always along the polar direction.
And this is, of course, something that you already know from high school, where you just use this
right hand rule, where you have your wire and the current flowing along this direction.
You take your thumb and then you take the fingers of your right hand to see when which directions the
magnetic field lines go.
And this is something that we have derived from scratch.
We have not relied on any phenomenological explanations.
We have really considered the Maxwell equations we have derived abusable are also the law for calculating
the vector potential based on currents.
And then we have gone through all of these mathematical derivations and have ended up with this simple
looking expression here.
Now, as a final comment, I want to discuss these two approximations that we have used.
So these approximations led to the fact that this expression looks so simple.
So, for example, if we would not have considered an infinitely long wire, then we would, of course,
get some XY dependence because then it would make a difference where we cut the wire, where we look
at the magnetic field, for example, if the wire would really end here, then of course our magnetic
field would look very different if we would look here at the top or the bottom compared to the middle.
But since our wire is infinitely long, we are, so to speak, always in the middle.
So every cut through the wire looks the same in terms of the magnetic field profile.
And then for the other approximation here, we have considered that our wire is infinitely thin, which
means that the wire is not three dimensional anymore, but basically just one dimensional.
And this means that when we are very close to the wire, we would see different effects.
We will get we would get additional terms that look more complicated than this one here, because then
there would be really a different distance from this position to the right end of the wire and to the
left end of the wire.
But due to our approximation, all the points in the cross section of the wire have the same distance
to our reference point.
So this is why this looks so simple.
But it turns out it's really a good approximation, because also this is the equation that you learn
in school.
Maybe not this if I hear about this magnitude at least, and also this right hand rule.
So this really turns out to be correct.
And these approximations may seem a bit strange at first, but they are really useful and really sensible
also.