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So the good results of the last lecture was that we have now at the back to potential and we know that

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it's correct and we have the vector potential in terms of an array, we have an array capital eight,

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which includes points in space that you show that you see here.

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And then we have the corresponding value of the vector potential.

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And we've also seen that only to Z component of the vector potential is different from zero.

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So it only has this a z value that we plot here and what we will consider in the following.

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So according to the Maxwell equation, we can calculate the magnetic fields based on the kernel of the

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vector potential, and we have our vector potential readily available.

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And if you remember in the previous section, we have also learned about the kernel.

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So this was the section about the derivatives and we have introduced another operator and we have even

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programmed to call for a continuous function.

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So now here, it's a tiny bit different because we have an array and so we must reprogram this kernel,

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but it's not really that much different.

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And if you remember, I told you back in this lecture that we will have an exercise later on.

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And now is the time.

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Now you will calculate really a kernel which has a physical meaning.

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And so you know how to calculate the kernel.

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And the only thing that we can use now that we have to consider is that we have considered here basically

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an infinitely long wire, which therefore has the same vector potential in all the layers.

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This is a bit tricky here because in our example, when the wire is not infinitely long, it's not actually

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constant with Z, but it's only very tiny changes here.

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So we will consider that these are zero here because we have only considered a single layer, so we

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wouldn't be able to calculate these derivatives with our current version of the code and principle.

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It's possible, but then you would have to calculate a for different layers.

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And then in the end, you will see that it's just very, very close to zero.

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So this means using this approximation, these two terms here will be zero.

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And so this is what we have to calculate.

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And first, we have to create an empty list like we did before for the vector potential for the magnetic

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field.

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And this will be B is equal to and now I will, of course, save a bit of typing and scroll up to this

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point where we have to vector potential and it will be exactly the same thing here.

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Oops.

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Like this?

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OK, so now we have the empty magnetic field and now we could, of course, loop again over all the

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positions.

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But actually here it's pretty, pretty easy to just work with the arrays.

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So let us get started and let us first program the X component, which would be something like this

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X component, all the values in X, all the values and Y and of course, only two single value in Z.

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And now if you remember how we calculate a derivative, there was always a good compromise, I would

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say, when we use the central difference as methods, because if you remember, this had a better order

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of the error than the forward or backward differences.

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And also it was quite easy to program because it only contains a single term.

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In this case, we go one step in the right direction, forward and backward, and we subtract two terms

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and divide by the difference in space.

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So this means we run into a bit of a problem here at the edge of our list, which is why I will just

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disregard the edge and I will just start from the second value, which is the index one.

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And I go to the second last because this one is excluded.

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It will be the second last and the same for the wider region.

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So we could, of course, also program an equation for the last and the first values.

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But then we would have to use them forward and backwards differences methods.

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But I will just disregard these values and we don't care here because as you will see, the magnetic

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fields will be very close to zero at these values anyway.

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So now comes the central differences methods.

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So we need to use here A and the Z components, and this will be a two and then we will use here all

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the so basically the same indices along the X direction and for the Y direction we go one step forward.

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So basically, we take this one, but we go one step forward, which means like this, we start from

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the third value, which is in X two.

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And yes, and then we go to the very last one, which is programmed like this and also take the only

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value on the Z direction.

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Now we must subtract from here the same thing where we go into the negative direction, which will be

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programmed like this.

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We go start from the first and go to the third last year because minus two means here, it's excluded.

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It's not included in the list anymore.

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And then we just have to divide by the total difference, which will be on the difference and risk base,

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which will be two times.

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OK, so this is already this term and the other terms, of course, very similar.

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So I will just copy and then fixed the indices.

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So here it will be now the X sorry, not the X, but the Y coordinates so index one.

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And we have to consider here again the Z component, but the derivative along the X direction and a

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minus sign.

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So first of all, add here the minus sign and then I basically just have to swap these indices here

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and also here.

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OK, that's already it.

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And now we must use the Z component.

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This one is a bit more to write because they are now two terms.

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So the Z component considers the X derivative of a Y, so we can basically take this one, but we must

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use a Y.

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So we change to two, two to one here in the first index, and then we must subtract the Y derivative

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of X so we can take this one and change the index to zero for eight X.

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And that's already it.

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Now we have calculated all the B values except for the outermost values of our.

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But as I told you, we simply do not care about this.

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We could, as I told you, just use central differences, but we do not care here.

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Now I want to pull up the individual coordinates and so we can just copy this code because I want to

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use again a control plot.

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So I use here now instead be and this will be the magnetic fields and magnetic fields.

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And the first one will be the X, which has the index zero, of course, and we can just plot it.

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And though I want to plot also the others.

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So now you could make subplots right next to each other, which which would look a bit nicer.

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But I want to do it really quickly and save a bit of time.

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So I just try to appeal to shows so know I can use several plot commands in this cell and I just copy

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this twice.

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Use here do Y components or index one.

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B Y and index to be easy.

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And so now you see we have here magnetic field in the X direction, which is basically very small or

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close to zero everywhere, except close to the middle.

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And same for B y.

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But you see there is some sign change going on here and there is a rotation of this funny looking thing.

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And you do see component is just zero.

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So if you have already worked with such quantum plots, then maybe you can already imagine what the

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solution looks like.

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But if not, I want to show you here another way of plotting, and this will be, first of all, the

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absolute value.

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And for this, I will just copy once again such a code here and here.

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I will use now the absolute value, and for this we must use, of course, and p towards square root

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of

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B X squared plus.

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And then I copied this b y squared plus p z squared.

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And you see, this is now the absolute value of the magnetic fields.

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And even though there is a tiny artifact here from the plotting or for the from the from the letters

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from the coordinate grid that we are using, I can tell you that actually when you saw that analytically,

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it is radial, symmetric and despite these points here, also in our numeric solution, it is radial,

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symmetric and it has quite a large value in the middle.

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And then it declines when you go away from the wire and then you can look at the individual components

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here.

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You see, the X component is positive and here at the bottom and it's negative here and the Y component

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is positive here and negative here.

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So you can already imagine that we really have such a scenario here where the current rotates, because

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here also in the negative y direction, the magnetic field is along a positive X and so on is exactly

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the same situation as in the image, and we will now go ahead and create such an arrow plots that you

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have just seen in this figure.

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And we will also compare our result to the analytical solution.

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So for the analytical solution, I have basically taken this vector potential that I've shown you earlier

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that I have derived in the course and then I have just analytically calculated the curl.

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So Knobloch cross the vector potential.

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So this is what the analytical solution looks like.

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It is, in a sense, similar to the vector potential.

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But instead of the logarithm, you have now won over the square root of x squared plus y square and

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you have not the vector potential along the Z direction, but along this particular, you call it the

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toroidal orientation, which means that the magnetic field circulates around the z axis.

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So this is really what you can also see above.

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So the X component is proportional to minus Y and the Y component is proportional to positive X.

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So you see Y component proportional to positive x x component proportionate to negative y.

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And if I program this and compare it with our solution here, I have chosen to show B y and cut along

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the x axis.

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So this will be B y cut along this axis.

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Then we will get the following profile this one.

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And I think we have once again a nice agreement of our calculated the numerical solution and blue points

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and the analytical result and ret.

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Sorry.

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The only problem that we see here is that the outermost point looks a bit odd.

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This is for the reason that I have explained to you that I did not consider these for either of a tiff.

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So they are just chosen to be zero here and we have not actually calculated them.

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We could fix this by forward or backward differences.

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Then it would probably go here closer to the red line.

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And there's another problem pretty close to the center, which is basically due to the fact that here

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we would need more points to get a better result.

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So if you want, you can try to get a better results by improving the parameters and improving the convergence

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of the integral.

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But I think it's not really necessary because as a result looks really good.

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So now the last thing I want to show you is that we want to recreate such a figure in a sense, so we

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will generate such an arrow plot for the magnetic fields around our charged wire.