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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.