So now we know what the magnetic field looks like for a straight wire, in the next step, we will consider
loops of a wire or current loops.
And in terms of this lecture, we will also establish the magnetic dipole.
So he is warning these lectures quite long and also a bit more difficult from the mathematical side,
or maybe not difficult, but quite, quite long in the derivation.
So if you're really tired and don't want to deal with all of this mathematics, you could skip ahead
to the end of this lecture where I will discuss the results.
And they will also see that the result is very, very similar to the electric dipole that, you know,
from the previous section of electrostatic.
However, the device is a bit different and quite interesting.
So I would really encourage you to follow through the video.
So this is here, the current loop that we look at.
So this is a top down view of our wire that is closed and the current is flowing in a circle.
Now, if we look at this wire from this perspective, then it will look like this and we could now separate
our circular wire into individual segments that are all basically straight.
And we know already from the previous lecture what the magnetic field of such a straight wire looks
like.
There will be field lines going along the Pollner direction, as you can see here in this cut.
So this go along to pull our direction.
Now, they are not circular anymore, but still you get the idea.
And so this is the result that we can expect already from our knowledge of the previous lecture.
And now we want to really calculate what the profile looks like.
So we start from our sketch from the top down view, and we have here again the equation for our vector
potential.
So this should be familiar from the previous lecture of this trade wir.
Now, again, we apply the same approximation as in the previous lecture, we consider an infinitely
thin wire, which means that this is here the square root of the knee area of the cross section.
So you could say it's the radius.
Also, the diameter and or distance here of our reference point is almost always much, much larger
than this diameter.
So basically, we consider it a quasi one dimensional layer.
And what this does is it allows us to separate disintegration, which is three dimensional into a one
dimensional equation, integration along the loop that we have here, and a two dimensional integration
of the cross section that is the same for every individual cross section.
So this is why I have already carried out this integration, just like in the previous lecture.
And we get here not the current density, but the current flowing through the wire that we can even
pull out of the integral.
So this looks now much simpler than this expression here.
Now, the next step is something very similar to what we have done for the electric dipole.
So we are not interested in the precise solution, but we want to make statements about the, you know,
the loop far away from the loop.
So this means that our distance here is much larger than the radius of the whole loop.
And if this is the case, then we can use our Taylor expansion of this function here.
So Taylor expansion, you can do for every function and it's always correct, but we can even truncate
this Taylor expansion.
So in the perfect Taylor expansion, you would have an infinite number of terms because it's zero.
But since we are in this limit here, we could say, OK, we just take the first two terms and then
that's enough.
That's a good approximation already.
So here you see, this is the first term of of this expression here.
So this is just one of over and then the other term is related to this gradient of one over.
So the gradient of one of our, you know, already from one of the previous lectures of the electric
dipole.
So this is so this looks a bit counterintuitive, but if you really do it component wise, you will
see that it's correct.
It's just minus the position vector divided by R to the power of three.
So this means we can write down our vector potential here in these two terms.
So this first term, one of our and that is here, the second term which has these are not times R and
then divided by R to the power of three.
OK, so since this is here a loop and we integrate just of this back to this position vector and there
is not really an integration here, this is a zero because, yeah, it's a loop.
The starting point is the same as the end point.
So this compensates and the zero.
So we only have to calculate this expression here.
So let's do this.
We so the prefect is quite easy.
The only difficult thing that we have to calculate is this one here, this integral of this loop.
So this may look quite easy in the beginning, but is really not that easy.
So we have to use some really clever tricks.
And if you follow along with these slides or if you even try to do it yourself, please don't be sad
if it doesn't work out, because it's really it's really unintuitive.
You need to have a really good idea.
And I think the first physicist who has done it took not just some minutes, it took him probably many
days or even weeks to get these ideas.
So what we have done here is actually nothing.
We have just written it down in a different way.
So he you can see this term is exactly the same term as this one here, but we have one half in front
of it, so we can take it twice.
You will soon see why I did this.
This may look a bit counter-intuitive at this moment.
And then here we just add a zero.
So we add here minus some term and plus some term.
So, yeah, if we would if you would add them up, they would cancel out.
And this means we just get twice this term here, which is this one.
So that's all correct.
Now, you can see I've already written it down in these two lines, and the reason is that we will now
use two very cool tricks.
The first one is the rule for such a double vector product that can be expressed in such a way.
And you can see we can relate these two expressions to these two expressions.
So this means we can write this one here in terms of such a double vector product.
Now, the other thing is something like a product rule.
I mean, this is not really a derivative, but you can see that you already have here this deal.
So if I show you the solution, first of all, we can express the first line in terms of this double
effect, a product so you can post video and really see that this is true, that this makes sense.
And then here we can write it down like this.
So if you think of it in terms of a derivative, which is it is actually not, but it follows the same
mathematical rules, you have to apply this D to all the Arnotts here.
So you get two terms and the one term it acts on this one and the other one, it acts on this one.
So these are these two terms.
Now, the next thing that we can do is we can find out that this one here has actually zero, and this
is because we integrate over this expression here and we have a loop.
So, again, our starting point is the same as our end point.
So it gives zero so we can get rid of this.
And this means this whole integral can be expressed as such an integral.
And you can even see that we integrate only over this are here and not over this one here.
So this means we could pull it out of the integral.
So this is what we have ended up with.
This may not look so much easier right now, but you will soon see why it is, in fact, easier.
OK, and the next step, we just take our integral here and put it back into the vector potential.
So the vector potential now looks like this.
And with that, we are done because this whole bracket here, we identify with the magnetic dipole moment.
So this expression here may look a bit difficult, but if we go to two dimensions or let's say if we
look at a plane, our current loop, as I have drawn here, then this whole expression just becomes
the current times, the surface area times, the normal vector of the surface.
So it points here out of the plane.
And so this means our whole vector potential is just as prefecture, and then we have here a vector
product of a magnetic dipole moment with the position vector divided by our to the power of three.
Now that we know the potential, we can just calculate its rotation to get the magnetic fields, but
that is not so easy because the rotation of this term is not easily calculated.
I will show you how it is done.
So for this derivation, I will use several identities for the novela operator that I will.
All right.
Down here on the right hand side so you can find in math textbooks or even on Wikipedia.
And of course, you can also go ahead and verify all of these identities yourself.
And it might be even rewarding because you then better understand how another operator works.
So first of all, we need an identity that calculates the rotation of a product of a scalar function
and the vector function.
So these are not here, the electrostatic potential and the vector potential, but just some scalar
function and some vector function.
In our case, the vector is M cross R and scalar is one of our to the power of three.
So if you apply this law here, you get this expression.
Now the next step I calculate the gradient of one divided by R to the power of three.
And this is also something we have done already for the electric dipole.
So this is the result.
We get minus three from this exponent here we are from the derivative and we get to your R to the power
of five because here we have to power our to the power of minus three divided by two.
We have to decrease it by one so we get ah to the power or we get Yeah.
This expression to the power of minus five divided by two.
And so since R is the square root this is here are to the par five.
OK, so I put it in here already and I have changed his mind assigned to the positive side because of
this minus here.
Now, the next step, we need even two identities, the first one is quite easy.
That's something we have used already before.
This is this ABC rule for this double vector product.
Does we apply here to the second term to so that we get these two terms?
Now, the other law is a bit more difficult because it includes or it involves, does not the operator.
So here we do not only get to terms but four terms because remember is Novela is an operator and it's
something like a derivative.
So you have to for both of these terms, used the product rule.
So you get two times two terms, which are these four times here.
But luckily we have one of these two vectors, A and B, which are emond are one of these vectors to
them here is not dependent on the position.
So all of these terms where it is derivative acts on M or zero.
So this means we do not get to term.
We get this one here and we do not get this one here.
These are both zero and we only get these two terms here.
So these are these two terms and now we are almost done.
We just need to calculate I calculate this expression here and this expression here.
And the divergence of are is three, that something you can calculate yourself.
And then this term here I have written down in terms of components.
So it is a scalar product, Kadem X times the derivative with respect to X and the other two terms.
And you let all of this act on the position vector, which is of course X, Y, Z.
And so you get here, you get one, you get one, you get one, and then you multiply by M, X and Y
and Z.
So this is the vector and this means we end up with this expression here and you can see this term and
this term cancel out.
So we only have these two terms here.
So our solution or total solution for the magnetic field is that in this prefecture here and multiplied
by these two terms and I can write it down in this manner here.
OK, so this is our solution, this is our magnetic field.
This is our magnetic dipole moment.
And if we have a plane, our current loop, which is circular, then this is just a current multiplied
by the area, multiplied by the vector along the perpendicular direction.
And so now we can understand what this does mathematically idolized, um, magnetic dipole means.
So this year is the correct solution of such a closed current loop.
And this is also what we had expected based on our previous considerations of the trade wire, where
you have this magnetic field lines that follow the right hand rule and have these circular shapes around
the wire.
Now, what we have done here is we have fused a Taylor expansion that we have truncated, which means
that the distance and use to position vectors is always much larger than the radius of our current loop.
And essentially, this means we just zoom out.
So overall, our wire just becomes a point, so to say.
And it is just identified by this magnetic dipole moment that, as you can see, it always points perpendicular
to the plane of this wire.
So this is what we have derived in this whole lecture.
This is the mathematically ideal magnetic dipole and this is the corresponding magnetic field.
And if you remember the result from the electric dipole, you may say, OK, this looks very, very
similar.
And in fact, this is true.
If you zoom out the electric dipole and bring it to a single point, then you can see the electric field
lines look exactly like the magnetic field lines for this magnetic dipole.
So this means if we zoom out far enough, both of these systems look exactly the same.
Just, of course, here we have the magnetic field and here we have the electric field.
Just when we zoom in, when we really separate these two charges or if we go away from this point and
make it really a loop, then it looks different here in the middle, but far away from the loop and
far away from these two point charges, the line profile looks exactly the same.
And this is, of course, the case, because when we look at our results for the electric field, we
just have a different factor here, which is based on the Maxwell equations that we have here, this
music row.
And here we have this one over Epsilon zero.
And then here we have the same functional dependent's.
It's just now here we have the electric dipole moment and here we have the magnetic dipole moment.