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So in the previous lecture, we have set up diaries for coordinates and for defector potential, which

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is still empty, so only zeros.

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And now we can actually carry out the integral here.

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So we will do this now, and I told you already that we will not integrate over the whole space with

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just over the wire.

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So just over the sea access from zero negative to positive other n0.

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So first of all, we need a number for the integration steps.

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So basically, how many points do we put into the wire?

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And they check that it gives us a nicely converged result for five thousand or five thousand and one

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when we have one to have a nicely spaced list here.

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So you see the length is two times one thousand and the numbers of steps that we have is five thousand

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and one, and now we can do our loop.

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So for index for the X coordinate in and P Dot arrange num points and the same for Y, of course.

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So this will be then in the loop over the X and Y coordinate of the normal position vectors of this

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one here or also of this one.

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So we will go now through the whole space and there we will ask, OK, what's the value of a what's

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the value of B later on?

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So for example, here this will be one of the points in this theory that we try to fill now.

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So here we go.

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This is now the position of port position are for which we are currently calculating the vector potential

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eight and so are is given by MP Dot Perry.

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We could now call it just from from courts, but we can also reconstruct that it doesn't really matter

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in terms of calculation times or much.

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So I just tried minus court max plus i x times D.

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And then we have minus courts.

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Basically the same thing here, just with the high wire city index for why.

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And we said that U.S. values should always be zero.

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So I don't remember, did we define already the value of these?

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No, we forgot it.

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So the value of that I'm using here is the step size and the step size, of course, is directly calculated

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from these two numbers.

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It's calculated by DH is equal to two times called MAX, divided by number of points minus one.

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So in our case, it is two times four point nine, divided by 49.

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So it is 0.2 and you can expect these leaders.

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This is true here.

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Step sizes 0.2.

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OK.

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So, of course, I have to run these cells now again because we changed them.

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And now we can continue writing the code here.

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So we have no defined the position vector and now we must start the integration for this particular

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position vector.

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So for C.J., so this will be the z coordinate of the position vector that corresponds to the current.

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So this will be the z component of this, our dash and also this one here.

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So I labeled it Z J.

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This particular index z j in and P totally in space going from minus L0 to positive at zero in steps

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of no end.

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So basically, this will go the number of steps that we have in this range here and now we finally do

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the integration.

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So we integrate over all

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our

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dash in the wire.

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So we called his Ardeshir RJ because I don't like the dash here for programming, so let's call it RJ

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for our current density.

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And this will be ENPI Dot three

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zero zero and then C j and now we can do it, we can write a.

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And here we write all the coordinates and then the value of the index for the X Coordinate Index for

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the Y Coordinate Index, for the Z coordinate, which is zero.

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And now we can just write is equal to the old one because we integrate.

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This will be basically a sum here, plus the change, which will be what is written down.

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Where is it here?

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This one?

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So j at this position divided by the absolute value of the difference you.

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So we write J of our J divided by and then we could write the appeal and I can all or we can also just

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do it ourself and fiddle with the square root,

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our zero squared.

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So now it's pretty easy to calculate the absolute value of the difference because this one here only

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has an X and Y coordinates.

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And this one here only has a Z coordinates, so there's actually not really a difference.

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We can just write down like this

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and then RJ to square.

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But of course, you could also write here and in our top norm of army notes, RJ would give you the

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same result.

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OK, so now this should work.

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I think because we integrate over this, oh, I forgot to write the return.

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So no, it's no sorry.

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I was confused here for a second.

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I thought I would have forgotten to write the return.

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But this is not a function.

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This is just a loop, so we don't need any return.

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So this is just good to go.

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And you see, because I already used quite a large number.

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The calculation takes some time.

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So if you have a bit of a weak computer or laptop, then please reduce the number, maybe go to five

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hundred and one because otherwise you will sit forever.

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And if you take a lower value here, then your result will not be as accurate as mine.

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But that's not really a big problem.

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It's just here.

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So we get the general results correct.

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So it's not a big deal if you don't have accurate numbers.

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So you can use here a bit of a smaller value as well.

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And especially when you first test these code.

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And also this is what I did when I prepared the notebook.

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I, of course, started here with a much smaller value.

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So now one thing that we are still missing is while this is running, I can explain this to you.

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We are still missing the prefecture and we did not account for the size of this integration element

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DV.

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So by the way, now the calculation is finished, but we need to still calculate something.

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So a must actually be multiplied with the prefecture, which is mile zero, which was one and then divided

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by four times PI.

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And then we must calculate the DV element.

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This would be basically d x times d y times.

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The Z and the Z is pretty easy.

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This is just two times the length of a wire divided by the number of our integration steps.

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So now and minus one and the X and Y are both the same.

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This would be pretty similar, actually two times code MAX divided by NUM points minus one and then

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the same for d y.

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However, I don't want to do it like this.

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This would be like the you could see the brute force way of calculating this, because this would really

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correspond to the numerical integration.

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But actually, what this would do when you think about it, it would put our wire here in cubes so you

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would have no cubes or Q boards that are stacked on top of each other.

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And so the cross-section would then just be a square where the edges are given basically by the value

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of of the of the size here D.

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And this is, of course not what we want.

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We want that.

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We have really our a cross section, which is a circle and where the radius is given by our value.

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So we are maybe cheating here a bit in terms of numerics.

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But you could also say we are just clever because we are physicists and we know what the radius will

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be.

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So.

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Let me comment off these lines here because because this is not what we will use.

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Instead, we will introduce a new element, which will be the element of the of the area.

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So I call this d f and this will be, of course, the radius or the area of the circle with the radius

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r zero.

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So and p told pi times r zero squared.

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And then we can calculate a by writing eight is eight times the prefect.

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Our times the easy times t f.

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OK, so then we will get a much better result.

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So this calculation has finished and now I run this cell for the normalization of the result.

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And now we can finally go ahead and plot this whole thing.

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So as typically, we will use some of these options here for plotting, I will make the aspect ratio

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equal to one.

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And I will add here a label for X and Y, and then I will go ahead and plot the vector potential.

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So for this, I will use a control plot.

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So basically, we are the value of the vector potential is given by the color.

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And so the X, the list of the X well use will be co-ords zero and all X or Y, and only the plane where

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Z is equal to zero, then the wireless will basically be the same, but of course, with the Y component

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here.

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So we change this to one.

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And now we must include the value of eight.

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And for a I can first use the X component of A.

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So then all X or Y and only if the plane was equal to zero.

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And when I run this, it gives me just a plain color, which indicates that probably all the values

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o zero, we cannot cross chapters by adding here a color bar.

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So kielty that color bar.

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And you see, yes.

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So it's very small, so all the values here are zero.

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Now we can check for the Y coordinate with Y component better to say also zero.

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But the Z component is on zero and that's the one that we are interested in.

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So this is now the z component of the vector potential.

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Let me write this down here.

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So this is the C bar and we will change to C Bar set label.

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And this is the vector potential A to Z components.

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All right.

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So this is the result of our first integration or this is the integration that we want to do.

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So we integrated this equation here over divides and divides included all the points in the wire and

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we did the approximation that the wire is only on the z axis and we considered a radius, which was

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the 0.001.

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And you see that the vector potential is largest here in the middle and then it declines when you go

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away from the z axis.

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So not just some comments about the analytical solution.

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So I told you already earlier that in my other course on the Maxwell equations, we solved this problem

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analytically.

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So you don't have to understand what's shown here, but I just want to show it to you that it's possible

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to solve this analytically.

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And basically, we do hear a similar thing.

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We integrate, of course, over the current density, which we assume to be constant in the wire so

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we can pull it out of the integral.

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And so we basically only integrate of army knows our dash, and this will be something like a square

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root.

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And if you integrate over such a thing, it will give you the logarithm.

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And so I have programmed this equation here.

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So the analytical solution is written down here again.

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This I programmed and I cut basically through the through our plot here along this line.

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So for Y equals zero or close to zero because we didn't have the value of zero in our list.

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But I've chosen zero point one and I've cut through this line and plot now the vector potential to Z

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component.

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And you see as you see the blue values, all the ones from our numerical calculation, which is the

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same that are shown here.

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So just cut through this and then look at the color.

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This will give you the same values.

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And then the red line that I've programmed using this equation, which is the same as I've just shown

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you this one here.

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This gives you these red lines and you see we have a pretty nice agreement.

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There is maybe a small problem here directly in the middle, but that's pretty clear because, yeah,

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there are we have maybe not enough points for our integration and for considering all the points in

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this space, but overall, it's a pretty remarkable result.

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And we got the vector potential correct.