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So now we know what the magnetic field looks like for a straight wire, in the next step, we will consider



2

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loops of a wire or current loops.



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And in terms of this lecture, we will also establish the magnetic dipole.



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So he is warning these lectures quite long and also a bit more difficult from the mathematical side,



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00:00:22,740 --> 00:00:26,460

or maybe not difficult, but quite, quite long in the derivation.



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00:00:27,000 --> 00:00:32,460

So if you're really tired and don't want to deal with all of this mathematics, you could skip ahead



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00:00:32,820 --> 00:00:36,510

to the end of this lecture where I will discuss the results.



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And they will also see that the result is very, very similar to the electric dipole that, you know,



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00:00:43,140 --> 00:00:45,900

from the previous section of electrostatic.



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00:00:47,040 --> 00:00:50,810

However, the device is a bit different and quite interesting.



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So I would really encourage you to follow through the video.



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So this is here, the current loop that we look at.



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00:00:59,250 --> 00:01:05,880

So this is a top down view of our wire that is closed and the current is flowing in a circle.



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Now, if we look at this wire from this perspective, then it will look like this and we could now separate



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our circular wire into individual segments that are all basically straight.



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And we know already from the previous lecture what the magnetic field of such a straight wire looks



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00:01:26,910 --> 00:01:27,210

like.



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There will be field lines going along the Pollner direction, as you can see here in this cut.



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So this go along to pull our direction.



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Now, they are not circular anymore, but still you get the idea.



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And so this is the result that we can expect already from our knowledge of the previous lecture.



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And now we want to really calculate what the profile looks like.



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00:01:54,140 --> 00:02:01,910

So we start from our sketch from the top down view, and we have here again the equation for our vector



24

00:02:01,920 --> 00:02:02,570

potential.



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So this should be familiar from the previous lecture of this trade wir.



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Now, again, we apply the same approximation as in the previous lecture, we consider an infinitely



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thin wire, which means that this is here the square root of the knee area of the cross section.



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So you could say it's the radius.



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Also, the diameter and or distance here of our reference point is almost always much, much larger



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than this diameter.



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So basically, we consider it a quasi one dimensional layer.



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And what this does is it allows us to separate disintegration, which is three dimensional into a one



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00:02:46,440 --> 00:02:52,770

dimensional equation, integration along the loop that we have here, and a two dimensional integration



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00:02:53,580 --> 00:02:58,020

of the cross section that is the same for every individual cross section.



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So this is why I have already carried out this integration, just like in the previous lecture.



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And we get here not the current density, but the current flowing through the wire that we can even



37

00:03:09,480 --> 00:03:10,920

pull out of the integral.



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So this looks now much simpler than this expression here.



39

00:03:16,830 --> 00:03:22,800

Now, the next step is something very similar to what we have done for the electric dipole.



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00:03:24,390 --> 00:03:32,340

So we are not interested in the precise solution, but we want to make statements about the, you know,



41

00:03:32,340 --> 00:03:34,320

the loop far away from the loop.



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So this means that our distance here is much larger than the radius of the whole loop.



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And if this is the case, then we can use our Taylor expansion of this function here.



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00:03:48,900 --> 00:03:55,650

So Taylor expansion, you can do for every function and it's always correct, but we can even truncate



45

00:03:55,650 --> 00:03:56,850

this Taylor expansion.



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00:03:57,610 --> 00:04:02,280

So in the perfect Taylor expansion, you would have an infinite number of terms because it's zero.



47

00:04:02,850 --> 00:04:08,340

But since we are in this limit here, we could say, OK, we just take the first two terms and then



48

00:04:08,340 --> 00:04:08,910

that's enough.



49

00:04:08,910 --> 00:04:10,500

That's a good approximation already.



50

00:04:11,670 --> 00:04:15,360

So here you see, this is the first term of of this expression here.



51

00:04:15,390 --> 00:04:21,330

So this is just one of over and then the other term is related to this gradient of one over.



52

00:04:22,770 --> 00:04:28,440

So the gradient of one of our, you know, already from one of the previous lectures of the electric



53

00:04:28,710 --> 00:04:29,200

dipole.



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So this is so this looks a bit counterintuitive, but if you really do it component wise, you will



55

00:04:36,240 --> 00:04:37,200

see that it's correct.



56

00:04:37,500 --> 00:04:42,000

It's just minus the position vector divided by R to the power of three.



57

00:04:43,380 --> 00:04:48,270

So this means we can write down our vector potential here in these two terms.



58

00:04:48,660 --> 00:04:57,000

So this first term, one of our and that is here, the second term which has these are not times R and



59

00:04:57,000 --> 00:04:59,550

then divided by R to the power of three.



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00:05:02,380 --> 00:05:11,290

OK, so since this is here a loop and we integrate just of this back to this position vector and there



61

00:05:11,290 --> 00:05:16,410

is not really an integration here, this is a zero because, yeah, it's a loop.



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The starting point is the same as the end point.



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So this compensates and the zero.



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So we only have to calculate this expression here.



65

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So let's do this.



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We so the prefect is quite easy.



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The only difficult thing that we have to calculate is this one here, this integral of this loop.



68

00:05:40,630 --> 00:05:44,500

So this may look quite easy in the beginning, but is really not that easy.



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So we have to use some really clever tricks.



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And if you follow along with these slides or if you even try to do it yourself, please don't be sad



71

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if it doesn't work out, because it's really it's really unintuitive.



72

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You need to have a really good idea.



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And I think the first physicist who has done it took not just some minutes, it took him probably many



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days or even weeks to get these ideas.



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So what we have done here is actually nothing.



76

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We have just written it down in a different way.



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So he you can see this term is exactly the same term as this one here, but we have one half in front



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of it, so we can take it twice.



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You will soon see why I did this.



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This may look a bit counter-intuitive at this moment.



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And then here we just add a zero.



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So we add here minus some term and plus some term.



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00:06:37,960 --> 00:06:40,990

So, yeah, if we would if you would add them up, they would cancel out.



84

00:06:40,990 --> 00:06:44,680

And this means we just get twice this term here, which is this one.



85

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So that's all correct.



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00:06:47,430 --> 00:06:52,140

Now, you can see I've already written it down in these two lines, and the reason is that we will now



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use two very cool tricks.



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The first one is the rule for such a double vector product that can be expressed in such a way.



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And you can see we can relate these two expressions to these two expressions.



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00:07:07,170 --> 00:07:11,280

So this means we can write this one here in terms of such a double vector product.



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Now, the other thing is something like a product rule.



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I mean, this is not really a derivative, but you can see that you already have here this deal.



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So if I show you the solution, first of all, we can express the first line in terms of this double



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00:07:28,930 --> 00:07:33,980

effect, a product so you can post video and really see that this is true, that this makes sense.



95

00:07:34,740 --> 00:07:37,500

And then here we can write it down like this.



96

00:07:37,590 --> 00:07:43,770

So if you think of it in terms of a derivative, which is it is actually not, but it follows the same



97

00:07:43,770 --> 00:07:49,310

mathematical rules, you have to apply this D to all the Arnotts here.



98

00:07:49,320 --> 00:07:54,250

So you get two terms and the one term it acts on this one and the other one, it acts on this one.



99

00:07:54,510 --> 00:07:55,770

So these are these two terms.



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00:07:57,490 --> 00:08:03,040

Now, the next thing that we can do is we can find out that this one here has actually zero, and this



101

00:08:03,040 --> 00:08:08,700

is because we integrate over this expression here and we have a loop.



102

00:08:08,890 --> 00:08:12,830

So, again, our starting point is the same as our end point.



103

00:08:13,030 --> 00:08:16,180

So it gives zero so we can get rid of this.



104

00:08:16,600 --> 00:08:21,520

And this means this whole integral can be expressed as such an integral.



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00:08:23,780 --> 00:08:29,130

And you can even see that we integrate only over this are here and not over this one here.



106

00:08:29,390 --> 00:08:31,680

So this means we could pull it out of the integral.



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00:08:32,450 --> 00:08:35,360

So this is what we have ended up with.



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This may not look so much easier right now, but you will soon see why it is, in fact, easier.



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00:08:42,110 --> 00:08:47,900

OK, and the next step, we just take our integral here and put it back into the vector potential.



110

00:08:48,410 --> 00:08:50,840

So the vector potential now looks like this.



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00:08:51,710 --> 00:09:00,110

And with that, we are done because this whole bracket here, we identify with the magnetic dipole moment.



112

00:09:00,910 --> 00:09:07,550

So this expression here may look a bit difficult, but if we go to two dimensions or let's say if we



113

00:09:07,790 --> 00:09:14,420

look at a plane, our current loop, as I have drawn here, then this whole expression just becomes



114

00:09:14,720 --> 00:09:21,400

the current times, the surface area times, the normal vector of the surface.



115

00:09:21,590 --> 00:09:23,390

So it points here out of the plane.



116

00:09:25,040 --> 00:09:30,770

And so this means our whole vector potential is just as prefecture, and then we have here a vector



117

00:09:30,770 --> 00:09:37,340

product of a magnetic dipole moment with the position vector divided by our to the power of three.



118

00:09:39,820 --> 00:09:47,260

Now that we know the potential, we can just calculate its rotation to get the magnetic fields, but



119

00:09:47,650 --> 00:09:52,870

that is not so easy because the rotation of this term is not easily calculated.



120

00:09:53,580 --> 00:09:55,210

I will show you how it is done.



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00:09:56,860 --> 00:10:03,970

So for this derivation, I will use several identities for the novela operator that I will.



122

00:10:03,970 --> 00:10:04,570

All right.



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00:10:04,570 --> 00:10:10,320

Down here on the right hand side so you can find in math textbooks or even on Wikipedia.



124

00:10:10,750 --> 00:10:15,850

And of course, you can also go ahead and verify all of these identities yourself.



125

00:10:16,210 --> 00:10:21,430

And it might be even rewarding because you then better understand how another operator works.



126

00:10:22,570 --> 00:10:29,500

So first of all, we need an identity that calculates the rotation of a product of a scalar function



127

00:10:29,500 --> 00:10:30,530

and the vector function.



128

00:10:31,000 --> 00:10:36,340

So these are not here, the electrostatic potential and the vector potential, but just some scalar



129

00:10:36,340 --> 00:10:38,050

function and some vector function.



130

00:10:38,740 --> 00:10:45,460

In our case, the vector is M cross R and scalar is one of our to the power of three.



131

00:10:46,690 --> 00:10:49,810

So if you apply this law here, you get this expression.



132

00:10:51,370 --> 00:10:58,360

Now the next step I calculate the gradient of one divided by R to the power of three.



133

00:10:59,470 --> 00:11:02,790

And this is also something we have done already for the electric dipole.



134

00:11:03,070 --> 00:11:04,450

So this is the result.



135

00:11:04,450 --> 00:11:11,680

We get minus three from this exponent here we are from the derivative and we get to your R to the power



136

00:11:11,680 --> 00:11:18,730

of five because here we have to power our to the power of minus three divided by two.



137

00:11:18,880 --> 00:11:23,530

We have to decrease it by one so we get ah to the power or we get Yeah.



138

00:11:23,710 --> 00:11:27,780

This expression to the power of minus five divided by two.



139

00:11:28,390 --> 00:11:32,680

And so since R is the square root this is here are to the par five.



140

00:11:34,000 --> 00:11:40,390

OK, so I put it in here already and I have changed his mind assigned to the positive side because of



141

00:11:40,390 --> 00:11:41,230

this minus here.



142

00:11:42,960 --> 00:11:48,690

Now, the next step, we need even two identities, the first one is quite easy.



143

00:11:48,690 --> 00:11:50,680

That's something we have used already before.



144

00:11:51,150 --> 00:11:54,780

This is this ABC rule for this double vector product.



145

00:11:56,040 --> 00:12:00,350

Does we apply here to the second term to so that we get these two terms?



146

00:12:02,190 --> 00:12:08,010

Now, the other law is a bit more difficult because it includes or it involves, does not the operator.



147

00:12:08,340 --> 00:12:15,000

So here we do not only get to terms but four terms because remember is Novela is an operator and it's



148

00:12:15,000 --> 00:12:16,370

something like a derivative.



149

00:12:16,800 --> 00:12:21,200

So you have to for both of these terms, used the product rule.



150

00:12:21,360 --> 00:12:25,290

So you get two times two terms, which are these four times here.



151

00:12:26,370 --> 00:12:32,940

But luckily we have one of these two vectors, A and B, which are emond are one of these vectors to



152

00:12:32,940 --> 00:12:35,600

them here is not dependent on the position.



153

00:12:36,060 --> 00:12:41,990

So all of these terms where it is derivative acts on M or zero.



154

00:12:42,690 --> 00:12:46,830

So this means we do not get to term.



155

00:12:46,840 --> 00:12:49,540

We get this one here and we do not get this one here.



156

00:12:49,560 --> 00:12:52,540

These are both zero and we only get these two terms here.



157

00:12:53,340 --> 00:12:57,420

So these are these two terms and now we are almost done.



158

00:12:57,420 --> 00:13:02,460

We just need to calculate I calculate this expression here and this expression here.



159

00:13:03,210 --> 00:13:08,760

And the divergence of are is three, that something you can calculate yourself.



160

00:13:09,240 --> 00:13:12,680

And then this term here I have written down in terms of components.



161

00:13:12,690 --> 00:13:18,810

So it is a scalar product, Kadem X times the derivative with respect to X and the other two terms.



162

00:13:19,170 --> 00:13:23,270

And you let all of this act on the position vector, which is of course X, Y, Z.



163

00:13:24,090 --> 00:13:30,420

And so you get here, you get one, you get one, you get one, and then you multiply by M, X and Y



164

00:13:30,420 --> 00:13:30,790

and Z.



165

00:13:31,620 --> 00:13:39,600

So this is the vector and this means we end up with this expression here and you can see this term and



166

00:13:39,600 --> 00:13:40,890

this term cancel out.



167

00:13:41,040 --> 00:13:43,500

So we only have these two terms here.



168

00:13:45,310 --> 00:13:52,960

So our solution or total solution for the magnetic field is that in this prefecture here and multiplied



169

00:13:52,960 --> 00:13:56,710

by these two terms and I can write it down in this manner here.



170

00:13:59,320 --> 00:14:02,150

OK, so this is our solution, this is our magnetic field.



171

00:14:02,170 --> 00:14:04,120

This is our magnetic dipole moment.



172

00:14:04,120 --> 00:14:11,470

And if we have a plane, our current loop, which is circular, then this is just a current multiplied



173

00:14:11,470 --> 00:14:18,340

by the area, multiplied by the vector along the perpendicular direction.



174

00:14:19,360 --> 00:14:29,440

And so now we can understand what this does mathematically idolized, um, magnetic dipole means.



175

00:14:29,610 --> 00:14:34,260

So this year is the correct solution of such a closed current loop.



176

00:14:34,540 --> 00:14:41,200

And this is also what we had expected based on our previous considerations of the trade wire, where



177

00:14:41,200 --> 00:14:47,620

you have this magnetic field lines that follow the right hand rule and have these circular shapes around



178

00:14:47,620 --> 00:14:48,190

the wire.



179

00:14:49,450 --> 00:14:55,780

Now, what we have done here is we have fused a Taylor expansion that we have truncated, which means



180

00:14:55,810 --> 00:15:02,230

that the distance and use to position vectors is always much larger than the radius of our current loop.



181

00:15:02,800 --> 00:15:06,090

And essentially, this means we just zoom out.



182

00:15:06,670 --> 00:15:12,130

So overall, our wire just becomes a point, so to say.



183

00:15:12,490 --> 00:15:19,210

And it is just identified by this magnetic dipole moment that, as you can see, it always points perpendicular



184

00:15:19,210 --> 00:15:20,930

to the plane of this wire.



185

00:15:21,910 --> 00:15:25,520

So this is what we have derived in this whole lecture.



186

00:15:25,960 --> 00:15:33,260

This is the mathematically ideal magnetic dipole and this is the corresponding magnetic field.



187

00:15:34,570 --> 00:15:40,180

And if you remember the result from the electric dipole, you may say, OK, this looks very, very



188

00:15:40,180 --> 00:15:40,700

similar.



189

00:15:40,930 --> 00:15:42,340

And in fact, this is true.



190

00:15:43,000 --> 00:15:50,200

If you zoom out the electric dipole and bring it to a single point, then you can see the electric field



191

00:15:50,200 --> 00:15:54,250

lines look exactly like the magnetic field lines for this magnetic dipole.



192

00:15:55,250 --> 00:16:00,580

So this means if we zoom out far enough, both of these systems look exactly the same.



193

00:16:01,280 --> 00:16:04,880

Just, of course, here we have the magnetic field and here we have the electric field.



194

00:16:06,290 --> 00:16:12,950

Just when we zoom in, when we really separate these two charges or if we go away from this point and



195

00:16:12,950 --> 00:16:18,800

make it really a loop, then it looks different here in the middle, but far away from the loop and



196

00:16:18,800 --> 00:16:23,810

far away from these two point charges, the line profile looks exactly the same.



197

00:16:25,010 --> 00:16:29,990

And this is, of course, the case, because when we look at our results for the electric field, we



198

00:16:29,990 --> 00:16:34,910

just have a different factor here, which is based on the Maxwell equations that we have here, this



199

00:16:34,910 --> 00:16:35,500

music row.



200

00:16:35,520 --> 00:16:38,180

And here we have this one over Epsilon zero.



201

00:16:38,900 --> 00:16:41,540

And then here we have the same functional dependent's.



202

00:16:41,870 --> 00:16:48,320

It's just now here we have the electric dipole moment and here we have the magnetic dipole moment.