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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.

  • And so this now also equals mu zero times twice mu Bohr times N.

  • And I'm finished, because I know what mu zero is and I know what moo- mu Bohr is,

  • we calculated that, it's still here, this number and I know what my capital subscript N is,

  • ten to the twenty-nine atoms per cubic meter.

  • And so I just shove those numbers in my equation,

  • and I find that this is approximately two point three tesla

  • for the numbers that I have chosen.

  • It's not for all materials this way,

  • because I have adopted two Bohr magnetons for the magnetic dipole moment of each atom

  • and I have adopted a density of ten to the twenty-nine atoms per cubic meter.

  • And for that situation, you get two point three tesla.

  • Now I want to use this number and understand what's going to happen in ferromagnetic material.

  • I take ferromagnetic material and I expose it to an external field,

  • I call it the vacuum field.

  • So I stick it in a solenoid and I can choose the current through my solenoid.

  • And I'm going to plot for you the vacuum field.

  • This vacuum field is linearly proportional to the current through my solenoid.

  • This is an actual physical solenoid.

  • I have a wire, it goes around like this.

  • I don't have one here now.

  • And I run a current through there.

  • And that vacuum B field equals mu zero times I times N divided by l,

  • except that this N divided by l is now the number of windings of my current wire

  • and this is the length of my solenoid.

  • So don't confuse it with the one we had there, because that was on an atomic scale

  • and this is on a macroscopic scale.

  • How many windings you have could be, where you had a big one here in class,

  • twenty-eight hundred windings and we had sixty centimeters long

  • and so that's what this number is all about.

  • So the moment that I know what the current is through my solenoid,

  • I immediately know what my vacuum field is.

  • There's a one-to-one correspondence.

  • And now I'm going to measure inside the ferromagnetic material,

  • which I stick in the solenoid.

  • I'm going to measure there the magnetic field.

  • And what's going to happen now?

  • Well, first of all I cannot at all plot this curve on a one-to-one scale,

  • the reason being that kappa of M for ferromagnetic material is so large,

  • let us adopt for now a number of one thousand, say.

  • Or it could be larger, even.

  • Let's say kappa of M is one thousand.

  • That means that if the magnetic field in terms of a vector is this big,

  • one centimeter in the length of the vector,

  • that the field inside the ferromagnetic material is a thousand times higher.

  • If this is one centimeter in length, a thousand times higher is ten meters.

  • So this is the length of the vector of the magnetic field inside the ferromagnetic material.

  • That's why I cannot do it to scale.

  • So when I draw this line here, keep in mind that if I did it to scale,

  • if I did, but I can't, then the tangent of alpha would be kappa of M,

  • which in this case would be ten to the third.

  • And so this angle alpha is something like eighty-nine point nine something degrees.

  • So I cannot do it to scale.

  • Keep that in mind.

  • So in the beginning I get a nice linear curve,

  • but now slowly I'm beginning to reach saturation,

  • that all these dipoles are going to be aligned

  • and what you're going to see is that this curve bends over and bends over and bends over

  • and the magnetic field that you finally achieve here

  • is the famous two point three tesla,

  • which I calculated for that imaginary material, plus B vacuum.

  • The two point three is now the field which I will call B prime.

  • This is the field that is the result of the alignment of all those dipoles.

  • And so when I increase the vacuum field, this goes into saturation

  • and settles for two point three and can no longer increase

  • because I have aligned all these magnetic dipoles.

  • And so if your vacuum field is this strong,

  • then this field is no longer thousand times stronger than the vacuum field.

  • You're no longer in the linear part.

  • So you could also think of it as kappa of M being smaller than a thousand.

  • Whichever way you prefer is fine.

  • But it's no longer proportional to the value of one thousand.

  • If the temperature is lower of the material, it's easier to align them,

  • and so you will achieve saturation earlier and so your curve would go like this.

  • So the curve is also a function of temperature.

  • So this is if the temperature is low relative to this one.

  • So these curves depend on, on temperature as well.

  • The lower the temperature, the easier it is to align them.

  • If I reach this point here, when my B prime goes into saturation,

  • I can only increase the field, the B field in the material,

  • by increasing the vacuum field, because B prime is not going to go up again.

  • And so I can only get a higher field by increasing this current so that this v-- this v-- vacuum goes up.

  • And that goes up very slowly, because this huge magnification factor of one thousand

  • is gone now.

  • So the slow, it's very slow, the growth, and that's why you see that I drew it like this,

  • that it increases very slowly.

  • But my plot is not to scale anyhow.

  • Now I want to discuss with you what happens with the material

  • once I have driven it into saturation.

  • What happens if now I change the current and I make my vacuum field zero again?

  • And now you get a very unusual behavior.

  • Let me do that here on the blackboard.

  • So I'll make a new drawing.

  • I could have continued with that one, but let me make a new one.

  • So I'm going to do the following experiment in my head.

  • I have a solenoid and I run a current through that solenoid.

  • And if the current is in clockwise direction, my vacuum field will be in this direction.

  • And when the current is in counterclockwise direction,

  • I will assume that my vacuum field is in this direction.

  • So there is ferromagnetic material in here.

  • If the current flows in clockwise direction, the vacuum field is in this direction.

  • If I run it in counterclockwise direction, the vacuum field is in that direction.

  • And here is going to be my vacuum field.

  • Easy for me to know what that is, because if I know the current through my solenoid,

  • this equation will immediately tell me what the vacuum field is.

  • So I have never any problems with the vacuum field.

  • I stick a probe in here and I measure the magnetic field inside that material.

  • How we do that is not so easy, but we can do it.

  • There are a few things that I can't tell you.

  • This is one of them.

  • So here is the magnetic field inside the material.

  • All right, so there we start.

  • We do the same thing that we did here.

  • So we approach the saturation.

  • But now when I'm here, I am reducing the current and go back to zero.

  • Remember that when we are here, all these domains that we discussed last time

  • have all flipped in the direction of the vacuum field.

  • So this field is enormous.

  • But now I make the current go back to zero.

  • And what happens now is I end up here, at this point P here.

  • The current now is zero.

  • There is no current going through the solenoid.

  • Notice the vacuum field is zero.

  • I can take the material out, the ferromagnetic material.

  • The material itself is now magnetic.

  • And you see there is a magnetic field inside it.

  • Why is that? Because some of those domains remain aligned, they don't go back.

  • And so we have created permanent magnetism.

  • And so in the location, at the location P, we have B vacuum, is zero, but B prime,

  • which is the result of those aligned magnetic moments, is still in this direction.

  • Nothing is to scale here, of course.

  • And so you still have a magnetic field.

  • Now I reverse the current.

  • I go counterclockwise.

  • So I'm creating a magnetic field vacuum now in this direction.

  • And so now what will happen with this curve?

  • I come up here.

  • And look now here, at this location Q.

  • What do I have now? I have something very bizarre.

  • I have now a situation whereby the vacuum field is in this direction

  • but there is no magnetic field inside the material.

  • The magnetic field inside is zero.

  • So when we have point Q, so we have B vacuum is in this direction, but B inside, B prime--

  • oh no, it's not B prime, it's B.

  • It's the total field inside, is zero.

  • The reason being that B prime is still in this direction.

  • The reason being that the domains are still aligned in this direction

  • and so the vacuum field plus the B prime field, which has to be vectorially added,

  • adds up with a net field zero.

  • Quite bizarre, isn't it.

  • Now I increase the current, but I keep going counterclockwise

  • and so the magnetic field of the vacuum remains in this direction.

  • I go into saturation again, in a similar way that I went into saturation here.

  • And now I stop here-- oh, I don't want to lose my brooch.

  • And now I stop here and I say to the current, go back to zero again.

  • So my current now goes back to zero.

  • There we go.

  • And now I arrive here, point S.

  • And again, I have a situation that my vacuum field is zero.

  • I could take the material out of the solenoid, just walk around with it on the street.

  • It will be a permanent magnet.

  • But now the magnetic field inside this material in point P it was in this direction.

  • If I take it out here, then it is in this direction.

  • Now some domains stay aligned in this direction.

  • The reason was that I had counterclockwise current

  • and so those domains flipped over.

  • And they're not all willing to flip back again.

  • So I've also made a permanent magnet here.

  • Vacuum field zero, but B prime is now in the opposite direction.

  • And then if I continue now to go clockwise with current again and increase the current,

  • I end up there.

  • And this is a bizarre curve.

  • We call this the hysteresis curve.

  • If you look at this curve, it's really amazing.

  • It's actually hard to-- to digest this.

  • You-- you have to give it a little bit of thought.

  • Because for one particular value of the current, for instance here,

  • once I have a particular value of the current, I know that B vacuum is a given,

  • I have two possibilities here for the magnetic field.

  • And for the current here, I have two possibilities for the magnetic field.

  • And so I cannot even know when I take this material and I expose it to an external field,

  • I can't even calculate what the magnetic field inside will be.

  • It depends on the history of this material.

  • Look at this point here and at this point there.

  • If I asked you what is kappa of M, pff it's almost a ridiculous question.

  • Because what is kappa of M?

  • I have, um, I have a vacuum field, but I have no field inside.

  • So the field inside, which is B, is zero and the vacuum field is not zero.

  • So you would have to answer, kappa M is zero.

  • That's the only thing you could say.

  • Quite bizarre, right.

  • But right here, there is a vacuum field but no field inside.

  • So kappa M is zero here and kappa M is zero here.

  • And remember, here it was one thousand.

  • Look at the situation, take this point of the curve and this point of the curve.

  • Kappa M is less than zero, is negative, because here the vacuum field is in this direction,

  • but B prime is in that direction, so they are in opposite directions,

  • so the net field is in that direction, but B vacuum is in this direction.

  • And here it's also reversed.

  • So there's a bizarre situation that you are effectively having situations whereby

  • kappa of M is zero and kappa of M can also be negative for those s-- points that I have there.

  • I can show you this hysteresis curve, and I do it exactly the way that I explained to you

  • except that I will not be able to run this current very slowly up and down.

  • I do it with sixty hertz alternating current, just get it out of the-- the wall.

  • And so I run through this solenoid n-- sixty hertz alternating current.

  • That means we go through this curve very quickly back and forth.

  • Between this point maximum current and this point maximum current.

  • The current clockwise counterclockwise clockwise counterclockwise

  • and we change that sixty times per second.

  • And then I will show you this curve, again, highly distorted.

  • I cannot plot it one-to-one, for the reasons that I explained to you.

  • And you will see then the hysteresis curve.

  • That's called the hysteresis curve.

  • And for that, I have to do several things.

  • And I always forget what I have to do, but just don't worry about it, I will find out.

  • My TV goes on.

  • This light goes off and this light goes off.

  • Look there on the screen and you will see within seconds there is the hysteresis curve.

  • And as I said to you earlier, I cannot start here, unfortunately,

  • because we switch it so fast back and forth that this part of the curve, by the way,

  • which is called the virginal curve, virginal, because it's once and never again.

  • Once you have reached that point, from that moment on, you always stick to this--

  • I know, you should know about that.

  • [laughter]

  • And so here you see a striking example of a hysteresis curve.

  • And so you can ask yourself now the question, can we make this material virginal again?

  • And then the answer is yes.

  • There are various ways you can do that.

  • One way is you could take the material out, so that means you take it out, for instance,

  • when there is remnant magnetism here, so it's a magnet

  • and now you heat it up above the Curie point, as we did last time in my lecture

  • and then the domains completely fall apart and then you cool it again below the Curie point

  • and then it is virginal material again.

  • And then you could start here again.

  • That's one way you could do it.

  • There is another way you could try to do it.

  • You take a hammer and you just bang on it.

  • So you take it out and you have permanent magnetism, either here or there,

  • and you bang on it and you hope for the best.

  • And maybe you can get it back to here.

  • There is another way, which we call demagnetization,

  • and I think that's what happens when you steal a book in the library

  • and you try to get away with it and the alarm goes off.

  • Someone hasn't demagnetized the magnetic strip in the book.

  • You may have noticed that when you take it out, that someone under the table goes like this.

  • And what they are doing then is they are demagnetizing that strip.

  • And the way you do that is as follows.

  • Here the current goes back and forth between this value and this value.

  • That means B vacuum and I are directly coupled to each other.

  • So I think of this as being current.

  • And now I go up to here and then back and then here and here and here and here...

  • ...and here and here and end up there again.

  • And I can show you that.

  • So again you have alternating current, but the amplitude of the current you decrease

  • and decrease and decrease and you're going to see that there.

  • And you see that I can turn this material-- see the hysteresis curve changes.

  • So the amplitude of the current is not as large,

  • so the amplitude of the B vacuum is not as large and I slowly go back.

  • And I can change a non-virgin back into a virgin.

  • And that's the way we do it as physicists.

  • [laughter]

  • Demagnetization.

  • Did I turn that off? Yes, I did.

  • I have here a-- a coil.

  • You can't see that there is a coil inside here, but there is.

  • I can, uh, power this coil, putting a current through it.

  • Ferromagnetic material, I don't know what kappa is, but, uh, at least a thousand.

  • I get an enormously strong field inside here.

  • And this field is so strong that this piece of ferromagnetic material will be attracted.

  • The field is non-uniform outside.

  • We discussed last time that it will go clunk, it will stick there.

  • So let's do that.

  • So I power this electromagnet, you can't see that I did,

  • you have to take my word for it, but you believe it now.

  • [ctick]

  • There it goes.

  • Oh boy.

  • The force is so large that I would ask two people to come

  • and see whether they can pull it apart.

  • The force that the two are together is so large that you may not even be able to separate it.

  • Do we have two strong people?

  • One strong woman, one strong man.

  • You look very strong to me.

  • [laughter]

  • Come on.

  • It's not going to be a tug of war between the two of you.

  • That's not my plan.

  • Be very careful, because if you touch this, you get electrocuted.

  • So don't do that.

  • [laughter]

  • But make sure that everyone can see you.

  • Because there is a current running through this solenoid now, yeah? OK.

  • Now just in case that you succeed, I don't want you to get hurt.

  • Student: OK.

  • So make sure that you secure yourself.

  • [laughter]

  • Because suppose the current stopped running all of a sudden.

  • It's possible, it's MIT right?

  • Anything could happen.

  • [laughter]

  • Then, of course, it's no longer a strong magnet.

  • It's only a strong magnet as long as the current is running through the solenoid.

  • So you secure yourself too.

  • OK, three, two, one, zero, go!

  • [laughter]

  • Don't worry.

  • I knew that in advance.

  • But thank you very much.

  • [laughter]

  • Very kind of you.

  • [applause]

  • Now comes something that you will understand.

  • If I make the current go to zero, then the vacuum field goes to zero.

  • That means the field that is generated by the solenoid goes to zero.

  • I will do that now in front of your own eyes.

  • No more current, right?

  • Why is this still hanging there?

  • Yeah?

  • Student: ...the solenoid... [inaudible]

  • ...magnetize the ferrous... [inaudible].

  • You hear?

  • You've taken the vacuum field out,

  • but the domains to a certain degree are still aligned

  • and so the whole thing is still a magnet.

  • Not as strong as magnet-- if I were to invite you to come now and take it apart, you could.

  • But it still takes sub-- substantial force.

  • And I will show you how large that force is.

  • I'm going to load it down now.

  • There's now one kilogram hanging on it.

  • Ooh, let me make sure I secure that, otherwise I can get killed for a change.

  • OK, three kilograms is hanging on it now.

  • Five kilograms is hanging on it now.

  • Seven kilograms is hanging on it now.

  • Nine kilograms is hanging on it now.

  • Oh boy, we may never make it.

  • Ten kilograms is...

  • [loud bang]

  • ...there it goes.

  • [laughter]

  • [loud racket, weights falling, more laughter]

  • Ten kilograms.

  • Now the show is not over yet.

  • What is very interesting and I want you to think about it,

  • that if now I take these two pieces of--

  • [again loud noises, laughter].

  • If I take these two pieces of ferromagnetic material, that--

  • nothing.

  • Do you know why?

  • I dropped it on the floor.

  • [laughter]

  • That's really why.

  • I dropped it on the floor and that is like banging it with a hammer

  • and then the domains go away.

  • Had I not dropped it on the floor...

  • [laughter]

  • ...there would have been something left,

  • but very little, which is interesting by itself.

  • The shock of the separation, when that happened,

  • already makes many of the domains flip back and there would be very little left.

  • Not enough to carry this weight anymore,

  • but that's largely because I dropped it on the floor.

  • I did that purposely so you can see when you drop things on the floor...

  • [laughter]

  • If I bring ferromagnetic material in the vicinity of a magnet,

  • I change the magnetic field configuration and that's very easy to understand now.

  • Suppose I have here a magnet, north pole, south pole

  • and the magnetic field, magnetic dipole field is sort of like this.

  • And now I bring in the vicinity here a piece of ferromagnetic material.

  • Could be a wrench.

  • What happens now is that this ferromagnetic material will see this vacuum field--

  • this is called the vacuum field, is an external field.

  • And so these domains in there trying to align a little bit.

  • Degree of success depends on how strong the field is,

  • depends on the temperature, depends on the kappa M of that material.

  • But certainly this will become sort of a south pole, this will become a north pole.

  • That's the way that these dipoles are going to align themselves.

  • They are going to create themselves a field in this direction.

  • They're going to support that field.

  • And so the net result is that the field inside here becomes very strong.

  • And so what happens with these field lines, they go like this.

  • They're being sucked into this ferromagnetic material.

  • It's very hard to know exactly how they go.

  • And the field here will weaken.

  • And I'm going to demonstrate that to you.

  • This is actually very easy to demonstrate.

  • And the way I'm going to demonstrate that is as follows.

  • I have there a setup whereby we have a-- a magnet

  • and we have a nail and we have a string.

  • And the nail wants to go to the magnet.

  • The mail its-- the nail itself is, uh, ferromagnetic.

  • So the nail would love to go in this direction, but it can't.

  • So it just sits there, hangs there in space.

  • First of all, what I want to show you is that if I bring paramagnetic material in the vicinity,

  • that that magnetic field configuration here is not going to change at all.

  • Paramagnetic material has a kappa of M so close to one that nothing is going to happen.

  • But the moment that I bring ferromagnetic material, for instance, here,

  • then you get a field configuration change and if I do that just the right way,

  • then the nail will fall.

  • In other words, there is not enough magnetic field here

  • in order to hold the nail in that direction.

  • And I'm going to show that for you-- to you there.

  • Need some power here, I believe and we have to make it dark here.

  • I'm going to shadow project it for you.

  • And the shadow projection you will see coming up very shortly.

  • This is a carbon arc.

  • You have to give it a little bit of time to start.

  • There is the carbon arc.

  • So there you see the nail.

  • And here you see the magnet.

  • You see that?

  • That's exactly the way I drew the picture.

  • And here, you hav-- I have here a piece of aluminum, which is, paramagnetic.

  • I can bring that through the field here.

  • Nothing happens.

  • My hands, believe it or not, are definitely not ferromagnetic,

  • so I can also bring my hands here.

  • Nothing.

  • Nothing.

  • So magnetic field is not disturbed in any way, in any serious way,

  • either by paramagnetic material, aluminum, or my hands,

  • which I think are also paramagnetic, but I'm not sure.

  • I'm not sure whether I'm diamagnetic or paramagnetic,

  • but it doesn't make any difference,

  • because in both cases there is no significant change of the magnetic field.

  • But now I have a wrench here.

  • Here, there's a wrench.

  • You see it?

  • [laughter]

  • OK.

  • And now I'll bring the wrench close to that magnet.

  • My major worry is that magnetic field is so strong that once the wrench go--

  • there's no way I can get it off again.

  • So I get only one shot at it.

  • And there goes the nail.

  • So what I-- what you saw now in front of your own eyes is that I--

  • changed the magnetic field configuration in such a way that the field was not strong enough

  • to pull in the-- the nail.

  • Now comes an important question, a big moment in our life.

  • And that is, what now is the effect of magnetic material on Maxwell's equations?

  • And let's take a look at Maxwell's equations.

  • Here we have Maxwell's equations the way we know them.

  • [laughter]

  • And let's first look at number one.

  • That's Gauss's Law.

  • Gauss's Law says that the closed surface integral of E dot dA--

  • that's the electric flux through a closed surface--

  • E is equal to all the charge inside divided by epsilon zero,

  • but you have to allow for the kappa, for the electric, dielectric constant.

  • The kappa, by the way, always lowers the field inside the material

  • when we deal with electric fields.

  • It never increases it like magnetic fields.

  • It always lowers it.

  • Kappa is normally a few-- except for water, it is eighty, it's quite large

  • and there are some ridiculous substances whereby kappa can be as large as three hundred.

  • I think strontium titanate, I just looked it up this morning,

  • has a ridiculous value of kappa of three hundred.

  • So that's Gauss's Law.

  • Nothing's going to change there as far as I can see.

  • And then we have the second one.

  • The closed surface integral of B dot dl equals zero.

  • Oh, it says an, it says an l.

  • That shouldn't even be an l.

  • Ho, I hope you caught that.

  • This is an A, of course.

  • How could I?

  • This is the closed surface integral of B dot DA is zero.

  • What this is telling me is that magnetic ma-- magnetic monopoles don't exist.

  • At least we think they don't exist.

  • Don't think that people are not trying to find them.

  • And if you find a magnetic monopole and if you put that inside a box,

  • then the closed surface integral of the magnetic flux through that box would not be zero.

  • And so then this is not true.

  • But as far as we know, it's always true,

  • because we don't think that magnetic poles-- monopoles exist.

  • And so then we come to the Faraday's Law.

  • Faraday's Law runs our economy.

  • Faraday's Law tells you when you move conducting loops in magnetic fields that you create electricity.

  • This equation runs our economy.

  • And now we come-- none of these, by the way, require any adjustment in terms of kappa of M.

  • But now we come to Ampere's Law.

  • Ampere's Law, which was amended by Maxwell himself,

  • tells me what the magnetic field is and all these results are for vacuum.

  • But now we know that that's not true anymore.

  • So this has to be adjusted now by a factor of kappa of M,

  • which is the relative permeability.

  • And kappa of M is perfectly kosher for paramagnetic and diamagnetic materials.

  • There's never any problem there.

  • So here it comes.

  • For diamagnetic materials, it's a little less than one.

  • For paramagnetic materials, a little larger than one.

  • But when we deal with ferromagnetic materials, you have to be very careful,

  • because we have seen today this hysteresis phenomenon,

  • that there are even situations whereby kappa of M is negative, whereby kappa M is zero,

  • and whereby kappa M can be huge can be ten to the third.

  • So there you have to be very, very careful when you apply this equation without thinking.

  • Maxwell's equations are so important that I'm sure you want to see more of them.

  • So you see them there again.

  • And maybe that's not enough.

  • [laughter]

  • Maybe you want to see even more of them.

  • So look at them.

  • Inhale them.

  • Let them penetrate your brains.

  • [laughter]

  • I don't care in which direction you look now.

  • It's hard not to see them.

  • Today is therefore very special, because today we have all four Maxwell's equations in place.

  • And this was one of the main objectives of 802.

  • So we have completed the long journey and on April five we have reached the summit.

  • Now I realize that the view is not spectacular for all of you yet,

  • because often at the summit there is some fog.

  • But the fog will clear.

  • And I can assure you that from here on on, it's climbing downhill.

  • I think this moment is worth celebrating

  • and therefore I bought six hundred daffodils for this occasion.

  • [laughter]

  • And I would like you to come at the-- at the end of the lecture

  • and pick up one of these daffodils and take it back to your dormitory.

  • I don't know whether I have enough for all these high school students and all their parents,

  • but why not, give it a shot and take one.

  • And when you look at it tonight at home and tomorrow,

  • remember that you only once in your life go through this experience,

  • that for the first time you see all four Maxwell's equations complete, hold it,

  • and that you're capable of appreciating them, at least in principle.

  • This will never happen again.

  • You will never be the same.

  • [laughter]

  • To put it in simple terms, as far as eight 802 is concerned,

  • you are no longer unspoiled virginal material.

  • [laughter]

  • You've lost your virginity, congratulations!

  • [laughter]

  • [applause]

  • [crowd noise]

When I expose material to an external magnetic field,

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B1 中級 美國腔

8.02x - 第22講--麥克斯韋方程--600水仙花儀式 (8.02x - Lect 22 - Maxwell's Equations - 600 Daffodil Ceremony)

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    黃安慧 發佈於 2021 年 01 月 14 日
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