Subtitles section Play video Print subtitles 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.