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• When I expose material to an external magnetic field,

• then we learned last time that the field inside that material is modified.

• And we expressed that in terms of an equation, that the field inside the material is kappa of M,

• which is called the relative permeability, times the external field,

• and I will refer to that all the time as the vacuum field.

• And when we have diamagnetic material, kappa M is just a hair smaller than one;

• with paramagnetic material, it is a hair larger than one;

• but when we have ferromagnetic material it can be huge.

• It can be thousands, ten thousands and even higher.

• Now in the case of para- and ferromagnetic material,

• the kappa of M is the result of the fact that the intrinsic dipoles of the atoms and the molecules

• are going to be aligned by the external field.

• And today I want to raise the question,

• how large can the magnetic dipole moment of a single atom be?

• And then comes the logical question, how strong can we actually, then,

• have a field inside ferromagnetic material?

• That means if we were able to align all the dipole moments of all the atoms,

• what is the maximum that we can achieve?

• To calculate the magnetic dipole moment of an atom,

• you have to do some quantum mechanics and that's beyond the scope of this course.

• And so I will derive it in a classical way

• and then at the very end I will add a little pepper and salt,

• which is quantum mechanics, just to make the result right.

• But it can be done in a classical way and it can give you a very good-- good idea.

• If I have a hydrogen atom, which has a proton at the center,

• has a charge plus e, e is the charge of the electron but this is the plus charge.

• And let this have an orbit R, circular orbit.

• And the electron here e, I'll give it a minus sign to make sure that you know that it's negative.

• Say the electron goes around in this direction.

• This is the velocity of the electron.

• That means that of course the current around the proton would then be in this direction.

• If an electron goes like this, the current goes like that,

• that's just by convention.

• The mass of an electron, you should know that by now,

• is approximately nine point one times ten to the minus thirty-one kilograms.

• The charge of the electron, one point six times ten to the minus nineteen coulombs.

• And the radius of the orbit in a hydrogen atom, it's often called the Bohr radius,

• by the way, is approximately five times ten to the minus eleven meters.

• We're going to need these numbers.

• That's why I write them down for you.

• If you look at this current running around the proton,

• it's really a current which the current, say, goes in this direction.

• And here is that proton, trying to make you see three dimensionally.

• Then it creates a magnetic field in this direction, and so the magnetic dipole moment mu is up.

• And the magnitude of that magnetic dipole moment, as we learned last time,

• is simply this current I times the area A of this current loop.

• Now the area A is trivial to calculate.

• That's pi R squared, R being the radius of the orbit.

• And so A, that's the easiest, is pi R squared.

• And if I use my five times ten to the minus eleven,

• then I find that this area is eight times ten to the minus twenty-one square meters.

• So that's easy.

• But now comes the question, what is I?

• What is the current?

• So now we have to do a little bit more work.

• And we have to combine our knowledge of 802 with our knowledge of 801.

• If this electron goes around,

• the reason why it goes around is that the proton and the electron attract each other.

• And so there is a force in this direction.

• And we know that force, that's the Coulomb force.

• It's an electric force.

• That force is this charge times this charge, so that's E squared,

• divided by our famous four pi epsilon zero

• and then we have to divide it by the radius squared.

• So that's Coulomb's Law.

• But from 801, from Newtonian mechanics,

• we know that this is what we call the centripetal force that holds it in orbit, so to speak,

• and that is m v squared, m being the mass of the electron,

• v being the speed of the electron, m v squared divided by R.

• And so this allows me to calculate, as a first step,

• before we get into the current, what the velocity of this electron is.

• It's phenomenal.

• It's an incredible speed.

• So v then becomes, I lose one R, so I get the square root,

• I get an e squared upstairs here, my m goes downstairs,

• I have four pi epsilon zero and I have here an R.

• And I know all these numbers.

• I know what E is, I know what capital R is, I know what four pi epsilon zero is,

• one over four pi epsilon zero is the famous nine to the power-- nine times ten to the power nine.

• And so I can calculate what v is.

• And if I stick in the numbers and if I did not make a mistake,

• then I find about two point three times ten to the six meters per second.

• It's an immensely high speed, five million miles per hour.

• If this were a straight line, you would make it to the moon in three minutes.

• Five million miles per hour this electron goes around the proton.

• Now I have to go to the current.

• I have to find out what the current is.

• So the question that I'm going to ask now

• is how long does it take for this electron to go around.

• Well, that time, capital T, is of course the circumference of my circle

• divided by the speed of the electron.

• Trivial.

• Even the high school students in my audience will understand that one, I hope.

• And so I know what two pi R is, because I know R and I know v

• and so I can calculate that time, just by sticking in the numbers.

• And I find that it is about one point one four times ten to the minus sixteen seconds.

• Just imagine how small that time is.

• You cannot even, we cannot even imagine what it's like.

• It goes ten to the sixteen times per second around,

• because it has this huge speed.

• [tape slows down]

• The one point one four times ten to the minus sixteen

• really should have been one point four times ten to the minus sixteen.

• Of course, it doesn't make much difference, but in case you substitute in the numbers,

• it is one point four times ten to the minus sixteen.

• [tape resumes normal speed]

• Now, we still haven't found the current, but we're almost there.

• Because when you look here, there is this electron going by

• and every one point one four ten to the minus sixteen seconds, that electron goes by.

• So the current I, that's the definition of current, is the charge per unit time.

• And so every capital T seconds, the charge E goes by

• and so this is per definition the current.

• And so this current, then, that you have,

• which is simply due to the electron going around the proton,

• is about one point one times ten to the minus three amperes.

• And that is mind-boggling.

• A milliampere.

• One electron going around a proton represents a current of a milliampere.

• And now of course I have the magnetic moment mu, that is I times A.

• We already calculated A and now we also have the current I

• and so we now get that mu is approximately nine point three,

• if you put in all the decimals correctly, times ten to the minus twenty-four

• and the unit is of course amperes square meters.

• This is area and this is current.

• This A has nothing to do with that A, hey.

• This is amperes.

• Be careful.

• And this is square meters.

• But these are the units.

• And this has a name.

• This is called the Bohr magneton.

• Bohr magneton.

• What we cannot understand with our knowledge now,

• but you can if you ever take quantum mechanics,

• that the magnetic moment of all electrons in orbit

• can only be a multiple of this number, nothing in between.

• Quantum mechanics, the word says it, it's quantization.

• It's not in between.

• It's either, or.

• It includes even zero, which is even harder to understand,

• that it could even be zero.

• In addition to a dipole moment due to the electron going around the proton,

• the electron itself is a charge which spins about its own axis,

• and that also means that a charge is going around on the spinning scale of the electron.

• And that magnetic dipole moment is always this value.

• And so the net magnetic dipole moment of an atom or a molecule is now the vectorial sum--

• of all these dipole moments, of all these electrons going around,

• means orbital dipole moments and you have to add these spin dipoles.

• Some of these pair each other out.

• One electron would have its dipole moment in this direction

• and the other in this direction and then the vectorial sum is zero.

• The net result is that most atoms and molecules have dipole moments

• which are either one Bohr magneton or two Bohr magnetons.

• That is very common.

• And that's what I will need today to discuss with you how strong a field we can create

• if we align all those magnetic dipoles.

• The magnetic field that is produced inside a material when I expose it to an external field,

• that magnetic field B is the vacuum field that I can create with a solenoid,

• we will discuss that further today, plus the field which I will call B prime

• which is that magnetic field that is the result of the fact that we're going to align these dipoles.

• The external field wants to align these dipoles,

• and the degree of success depends on the strength of the external field

• and of course on the temperature.

• If the temperature is low, it's easier to align them,

• because there is less thermal agitation.

• If, and that's a big if, today you will see why it's a big if,

• if B prime is linearly proportional to B vacuum, if that is the case,

• today you will see that there are situations where that's not the case,

• then I can write down that B prime equals xi of M,

• we called that last lecture the magnetic susceptibility, times B vacuum.

• The linear proportionality constant.

• If I can do that, because now I can write down that B is one plus xi M times B vacuum.

• And that, for that we write kappa of M times B vacuum,

• which is the equation that I started out with today.

• And so that is only a meaningful equation if the sum of the alignment of all these dipoles

• can be written as being linearly proportional with the external field.

• And this is what I want to explore today in more detail.

• With paramagnetic material, there is never any worry that the linearity doesn't hold.

• But with ferromagnetic material, that is not the case,

• because with ferromagnetic material, it is relatively easy to align these dipoles,

• because they already group in domains, as we discussed last time,

• and the domains flip in unison.

• And so with ferromagnetic material, as you will see today,

• we can actually go into what we call saturation,

• that all the dipoles are aligned in the same direction.

• And now the question is, how strong would that field be?

• I'm going to make a rough calculation that gives you a pretty good feeling for the numbers.

• It depends on what material you have.

• I will choose a material whereby the magnetic dipole moment is two Bohr magnetons,

• so this is, I told you it's either one or two or three, I pick one for which it is two.

• And I have them all aligned.

• So I take the situation that they're all aligned.

• So here is the current going around the nucleus, here's another one,

• here's another one, here's another one.

• This is a solid material, so these atoms or these molecules are nicely packed.

• And here we see all these currents going around

• and all these magnetic dipole moments are nicely aligned.

• And so these magnetic fields are supporting each other.

• And the question now is, what is the magnetic field inside here?

• Well, that's an easy calculation, because this really looks like a solenoid,

• like you have windings and you have a current going around.

• And you remember, or should remember, that if we have a solenoid

• and we run a current through a solenoid

• that the magnetic field in the solenoid is mu zero,

• this mu zero, is not this mu, this mu zero is the same one that,

• oh no, it's no-- we don't have it on the blackboard.

• You know, that's the famous four pi times ten to the minus seven.

• And then we have the current I, and then we have N,

• if that's the number of windings that we have in the solenoid

• and then we have L, which is the length of the solenoid.

• So this is the number of windings of the solenoid per unit length,

• the number of windings per meter.

• So if we could figure out, for this arrangement, what this quantity is, then we're in business.

• I take a material, which is not unreasonable, whereby the number density of atoms,

• I call that capital N, written in the subscript way, is about ten to the twenty-nine.

• So this is atoms or molecules, whatever may be the case, per cubic meter.

• That's not unreasonable.

• And now I have to somehow manipulate, massage the mathematics,

• so that I get in here this magnetic moment, this Bohr magneton.

• The two Bohr magnetons.

• And there are several ways of doing that.

• I have chosen one way and that's the following.

• I take here a length of one meter.

• So this is a solenoid and I pick only one meter.

• Could have taken three or five meters, makes no difference.

• I take one meter.

• And each one of these loops here has an area A.

• So we would agree, I hope, that the area, that the volume, the volume of this solenoid,

• this has a length one meter, that that volume is A square meters times one.

• And so the volume is A cubic meters.

• A times one meter is A cubic meters.

• But the number of atoms per cubic meter is ten to the twenty-ninth,

• and so the number of atoms that I have in this solenoid per meter is this A times that N.

• So this is the number of windings, if I call this one winding,

• the number of windings per meter.

• Or you can think of it the number of atoms per meter, the way they're lined up.

• And so now I am in business, because this now is my N divided by l.

• And so I can write now mu zero times the current I times that area A times N,

• which is ten to the twenty-nine.

• But look now.

• Now you see why I did it this way,

• because I times A is the magnetic dipole moment of my atoms.

• And that was two Bohr magnetons.