Today, I will quantify the ability of a circuit
to fight a magnetic flux that is produced by the circuits themselves. If you have a circuit and you run a current
through the circuit, then you create some magnetic fields and if the currents are changing
then the magnetic fields are changing. And so there will be an induced EMF
in that circuit that fights the change and we express that in terms of a self-inductance:
L self-inductance, and the word self speaks for itself. It's doing it to itself. Magnetic flux that is produced by the circuit
is always proportional to the current. You double the current,
the magnetic flux doubles. And so it is the proportionality constant
that we call L that is the self-inductance and so therefore the induced EMF equals
minus d phi dt, that is Faraday's Law. And so that becomes minus L dI dT. L is only a matter of geometry. L is not a function of the current itself. I will calculate for you a very simple case
of the self-inductance of a solenoid. Let this be a solenoid
and this is a closed circuit and we run a current I through the solenoid
and the radius of these windings is little r. Let's say they're N windings
and the length of the solenoid is little l. Perhaps you'll remember that we earlier derived,
using Ampere's Law, that the magnetic field inside the solenoid is
mu zero times I times capital N divided by L. This is the number of windings per meter. If we attach an open surface
to this closed loop, very difficult to imagine what
that open surface looks like, we discussed that many times, inside this solenoid you have
sort of a staircase-like of surface. That magnetic field penetrates that surface
N times because you have N loops. And so the magnetic flux, phi of B,
is simply the area, pi little r squared, which is the surface area of one loop, because I assume that the magnetic field
is constant inside the solenoid and I assume that it is zero outside,
which is a very good approximation. So we get pi little r squared,
surface area of one loop then we have N loops and then we have to multiply it
by that constant magnetic field. So we get an N squared
because we have an N here, mu zero I divided by l. And this we call L times I. That's our definition for self-inductors. And so the self inductance L
is purely geometry. It's pi little r squared, capital N squared
divided by l times mu zero. Let me check this. Pi little r squared, capital N squared,
mu zero, that's correct, divided by little l. And so we can calculate, for instance,
what this self inductance is for a solenoid that we have used in class
several times. We had one whereby we had
twenty eight hundred windings, r, I think was something like five centimeters,
you have to work SI of course, be careful and we had a length,
was 0.6 meters. We had it several times out here
and if you substitute those numbers in there, you will find that the self-inductance
of that solenoid is 0.1 in SI units and we call the SI units Henry,
capital H. It would be the same as volt seconds or ampere,
but no one would ever use that. We call that Henry. Every circuit has a finite value
for the self-inductance, however small that may be. Sometimes it's so small that we ignore it,
but if you take a simple loop, a simple current, just one wire that goes around,
whether it's a rectangle or whether it is a circle it doesn't make any difference,
it always produces a magnetic field. It always produces a magnetic flux to the surface
and so it always has a finite self-inductance. Maybe only ni-- nano Henrys,
maybe only micro Henrys, but it's never zero. And so now what I want to do is to show you
the remarkable consequences of the presence of a self-inductor in a circuit
and I start very simple. I have here a battery which has EMF V. I have here a switch
and here are the self-inductor. We always draw a self-inductor in a circuit
with these coils and we also have in series a resistor,
which we always indicate with this... teeth. And I close this switch when there is
no current running. In other words, a time t equals zero
when I close the switch, there is no current. When I close this switch
the current wants to increase, but the self-inductance says, "uh-huh, uh-huh,
take it easy, Land's Law--" "I don't like the change of such occurrence." So the self-inductance is fighting the current
that wants to go through it. There comes a time that the self-inductance
loses the fight, if you wait long enough and then of course the current has reached
a maximum value, which you can find with Ohm's Law,
because the self-inductance itself has no resistor. Think of the self-inductance as made of
superconducting material. There's no resistance. And so without knowing much about physics,
you can make a plot about the current-- that is going to flow as a function of time. You start out with zero and then ultimately,
if you wait long enough, you reach a maximum current
which is given by Ohm's Law, which is simply V divided by R. And you slowly approach that value. And how slowly depends on the value
of the self-inductance. If the self-inductance is very high,
it might climb up like this, so this is a high value for L. If the self-inductance is very low,
then there's a low value for L. If the self-inductance were zero,
it would come up instantaneously, but I just convinced you that there is
no such thing as zero self-inductance. There's always something finite,
no matter how small. And so this is qualitatively what you would expect
if you use your stomach and if you don't use your brains yet. There's nothing wrong with using your stomach
occasionally, but now I want to do this in a more civilized way,
and I want to use my brains and when I use my brains I have to set up
an equation for the circuit. And if you read your book, you will find
that mister John Coley tells you to use Kirchhoff's Loop Rule. But mister John Coley doesn't understand
Faraday's Law and he's not the only one. Almost every college book that you read
on physics do this wrong. They advise you to use Kirchhoff's Loop Rule, which says that the closed loop integral
around the circuit is zero. That, of course, is utter nonsense. How can it be zero? Because there is a change in magnetic flux
and so it can only be minus d phi dt. I advise you to go to the 802 website
and download a lecture supplement that you will find in which I address
this issue and hit it very hard. So the closed loop integral of E dot dl,
if you go around the circuit, is not zero, is minus d phi dt, Faraday's Law,
so it's minus L dI dt. So we have to go around the circuit
and we have to apply Faraday's Law and not Kirchoff's Loop Rule,
which doesn't apply here. This is the plus side of the battery
and this is the minus side, so the electric field in the battery
is in this direction. The electric field in the self-inductance
is zero because the self-inductance
has no resistance, it's super conducting material
and so the electric field in the resistor, if the current is in this direction,
which it will be then the electric field in the resistor
will be in this direction. So now I am equipped to write down
the closed loop integral of E dot dL. I start here and I always go in the direction
of the current and I advise you to do the same. I don't care that you guess the wrong direction
for the current. That's fine. Later, minus signs will correct that, they will tell you that you really guessed
the wrong direction, but always go around the loop
in the direction of the current because then the EMF is always minus L dI dt. If you go in the direction opposed to the current,
then it is plus L dI dt and that could become confusing. So I always go in the direction of the current
and so I first go through the self-inductance. There is no electric field in the self-inductance,
so the integral E dot dL, in going from here to here, is zero. This is where the books are wrong. It is zero. Now, I go through the resistor
and so now I get plus I R, E and dL are in the same direction. Ohm's Law tells me it's I R. In the battery, I go against the electric fields
and so I get minus V. That now equals minus L dI dt
and this is the only sane [thing?] and the only correct way to apply
Faraday's Law in this circuit. You can write it a little differently,
which may give you some insight. For instance, you could write that I can bring
V and dL and L, to one side, so I can write down
that V minus L dI dt equals IR. It's the same equation when you look
and the nice thing about writing it this way is that since dI dt is positive here,
it's growing in time, the induced EMF,
which is this value, notice it's always opposing
the voltage of my battery and that's what Lenz's Law is all about. It's not until dI dt has become zero
that V equals IR and that happens, of course,
if you wait long enough. And so we have to solve
that differential equation and what is often done,
that you bring all the terms to the left side and that you get a zero on the right side. So what you often see is that
L dI dt plus IR minus V equals zero. And because we have a zero here, some physicists think that this is an application
of Kirchhoff's Rule. Which is nonsense. You can always make it zero here
by bringing all the terms to this side. The closed loop integral of E dot dL is not zero, the closed loop integral of E dot dL
is minus L dI dt, but when I shift minus L dI dt to this side
I get zero here. And of course the people who write these books
know that this is the right answer and so they manipulate it so they get this equation
and they call that Kirchhoff's Rule. Sad and also embarrassing. So, this is the equation that you have to solve. Some of you may have solved this equation
in 801 already. Surely you didn't have an I here. You may have had an X here for the position,
but you probably solved it or you had friction. Maybe you didn't. I will give you the solution
to that differential equation. It's a very easy solution. The current as a function of time is a maximum
value times one minus e to the minus R divided by L times t and I max, that is the maximum current,
is V divided by R. And let's look at this
in a little bit more detail. First, notice that when t equals zero
that indeed you find I equals zero. Substitute in here t equals zero,
you get one minus one. So you find indeed,
that I equals zero. Substitute in there t goes to infinity, then you find
that I indeed becomes V divided by R, which is exactly what you expect. If t becomes infinity, then clearly the self-
inductance has lost all it's power, so to speak and the current is simply V divided by R,
the maximum current that you can have. And so that's a must,
that's a requirement. If you wait L over R seconds and believe it or not,
if you have some time, convince yourself that L over R indeed,
as a unit, is seconds, then the current I is about sixty three percent
of I max, because if T is L over R, then you get one minus one divided by E
and that is about 0.63. And if you wait double this time, then you have about eighty six percent
of the maximum current. In other words, right here,
if I wait L over R seconds, this value here is about 0.63 times
the maximum value possible and it's very, it's climbing up and asymptotically
approaches then, ultimately, the maximum current which is V divided by R. Make sure you download that lecture supplement
that you'll find on the Web. Now, what I'm going to do is all of a sudden
I'm going to make this voltage zero. The way I could do that is
by simply shorting it out. Of course on the blackboard,
I can simply remove it. So it's not there. The current is still running and all of a sudden
at a new time, t zero, I define the time t zero again,
the voltage is zero. And now comes the question
what is now going to happen? Well, the self-inductance doesn't like the fact
that the current is going down, so it's going to fight that change and so you expect that the current
is not going to die off right away, but you expect that the current is going
to go down sort of like so. And you want, when you wait long enough,
you want that, at t goes to infinity, where t equals zero the current is still maximum,
so it's still V over R, but when you go to infinity, if you wait long enough,
then, of course, the current has to become zero. And as the current dies out, heat is being produced in that resistor
at a rate of I square R joules per second and then there comes a time that the current
becomes almost zero and then the whole show is over. And so we can also calculate
the exact time behavior by going back to our differential equation
and make V zero. Where is that differential equation? Is it hiding? Oh, there it is. So I solved this differential equation,
but this is now zero. And the solution to that differential equation
is that I as a function of time-- is I max times e to the minus R over L times t, and that exactly has all the quantities
that you want it to have, because notice that at t equals zero,
when you put in t equals zero, the current is indeed maximum
and that's what you require. That's the moment that you make that
capital V zero, the current was still running. But notice that when t goes to infinity,
if you wait long enough, that indeed the current goes to zero. And if you wait L over R seconds, then you are down to about thirty seven percent
of your maximum current. So if you now go to-- I--
I redefine t equals zero here, so if now I wait L over R seconds then this value here is about
thirty seven percent of that value. So you've lost sixty three percent. And so you see, this is the consequence
of the fact that the circuit is capable of fighting-- its own magnetic flux that it is creating. When the current was running happily here,
with the battery in place, the current was, let's say, all the time I max,
at least very close to that value, V over R. And so all the time,
there was heat produced in the resistor. I square R joules per second. Who was providing that, uh, energy? Well,of course, the battery. But now, when I take the battery out,
there is still current running and that means while the current is dying
there is still heat produced in that resistor and that heat slowly comes out until
the current ultimately becomes zero. Now where does that energy come from? Well that energy must come
from the magnetic field that is present in the solenoid,
and this idea, that we have energy that comes out
in the form of heat, which really was there ear--
earlier in the form of a magnetic field, allows us to evaluate what we call
the magnetic energy field density. Let me first calculate
how much heat is produced as the current goes from a maximum value
down to zero. Hmmmm, I'll have to erase something. I'll erase this part here. So at any moment in time,
the current is producing heat in the resistor-- and so if I-- if my voltage becomes zero
at time t equals zero, then this is the amount of heat,
uh-uh, no square here. I squared R dt, integrated from zero to infinity, is the total heat that is produced
as the current dies out, but I know what this current was, I just erased it,
if I still remember it, so I can bring I max outside
and I can bring the resistance outside and then I get the integral from zero to infinity
of e to the minus R over L times t dt. And this is a trivial integral. This integral is, uhm,
L divided by two R. Oh, by the way, it is I squared,
so I have a two here. It's very important. Don't forget the two. And so that integral is L divided by two R and so if now I look at the product of I squared
maximum R L divided by two R, I get one half L times I maximum squared. So this comes out in the form of heat and I max is then the maximum current that we had
when the current was flowing after a long time. I can now by, by manipulating numbers,
I can now calculate how much-- energy there was in that field per cubic meter, because the magnetic field was exclusively
inside that solenoid. And if I know that the energy that is ultimately
coming out is one half l I squared, then I have here I, so I can replace that I there
by B divided by mu zero times l divided by N and here I have L. And so if I substitute in here the value for L
that we have on the blackboard there and we substitute for I the value that we have there,
you can drop the maximum now, this simply tells you, then, that any moment in time
that I have a current I running through a solenoid, that the energy that is available in the solenoid
in the forms of magnetic energy is one half LR squared. And so when you do that, you substitute capital L
and capital I, you will find that one half LI squared then becomes B squared over two mu zero
times pi little r squared times L. You check that at home. It's simply a substitution. But this is the volume of the solenoid
where the magnetic field exists, and we have assumed that the magnetic field
is zero everywhere outside. And if you accept that, then you see
that we now have a result for the magnetic field energy density,
that is how much energy there is per cubic meter, that is, of course, this value. Because this is the total energy
of the magnetic field, if we know the current and this is the volume of the magnetic field. So the magni-- magnetic field energy density
is then B squared divided by two mu zero and this is in joules per cubic meter. So in principle, if you knew the magnetic field
everywhere in space, then you can integrate over all space
and you can then calculate how much energy is, uhm,
present in the magnetic field. And earlier in this course,
we did something similar for electric fields. We calculated the electric field energy density. Perhaps you remember what it was. It was one half epsilon zero kappa
times E squared. It was also in joules per cubic meter. Now in the case of an electric field, this represents the work that I had to do to
arrange the charges in a certain configuration. In the case of a magnetic field, it represents the
work that I have to do to get a current going
inside a pure self-inductor. That means the resistance of the self-inductor
is zero and it takes work because the solenoid will oppose the building up
of the current and so I have to do work. So there's a parallel between the two. I can make you see, in a quite dramatic way,
how strong self-inductances can fight their own current and the way I'm going to do that
is with a setup there, whereby I have a twelve volt car battery
and I have two lightbulbs. I have here an enormous self-inductance L,
thirty Henry. We will learn later in the course how you make
such a high self-inductance and then here is a light bulb. The light bulb has a resistance of six ohm. This self-inductance, there is nothing
we can do about it. It happens to have four ohm resistance. We don't have a self, we don't have s--,
superconducting wires here, so it also has a resistance
of four ohms. Forgive me for that but there is nothing
I can do about it. I have here another resistance of four ohms
and a light bulb, which is the same one as that one,
also six ohms and then here is my car battery
plus a switch. And the car battery is twelve volts
and I'm going to throw the switch, turn it on. You will see that this light bulb
comes on almost instantaneously. There is no self-inductance in this loop,
well, maybe a few micro Henry or even less but in this loop here there is this huge
self-inductance and so the self-inductance says, "take it easy"
to the current, "take it easy, just wait." And you will very slowly see
that light bulb come on. And we can calculate how long it takes, because we have here ten ohms,
six ohms and four ohms, so L divided by R, we have thirty divided by ten,
so that is three seconds. So what that means is that even after six seconds,
which is twice this time, even then the current through this light bulb
is only eighty six percent of it's maximum. But that means since the light, of course,
is proportional to I square R in the light bulb, that the light is only 75 percent
of maximum and even if you wait nine seconds, then the light that comes out here
is still only 90 percent of it's maximum, whereas this one comes on immediately. The reason why we put the four ohm here,
we want this part to be, from an ohmic resistance point of view,
to be identical as this part. So that's why you artificially added here
the four ohm, because the four ohm is always there
in the self-inductance, there is nothing we can do about it. And so you are going to see
a remarkable example. This is one light bulb. That is the one that is here
and this is the light bulb that is there and the self-inductance is in
this incredible monster, here. I'll make the lights,
change the light setting a little bit so that we can see
the lights of the bulb. The bulbs are only eight-watt bulbs,
they're not very, very strong, not very bright bulbs,
but the effect will be very clear. So if you're ready for this, so this is the one
that I think comes on immediately and this is the one that takes it's time-- three, two, one, zero... One, two, three, four, five, six... You see how slow?
It's still not very bright. it's still getting brighter. It's still not as bright as this one. It's getting there, you can actually do
some timing by counting. By the way, the values of the resistance that
I gave you are when the light bulbs are hot. Three, two, one, zero... ...one, two, three, four, five, six, seven,
eight, nine, ten, eleven, twelve... I'm still seeing it getting brighter. So what you're seeing here
is a remarkable example that the self-inductance
is fighting the change in current. That's why the name self-inductance
is so nice. Now I want to go one step further and I want to
power the LR circuit with a AC power supply. If you have an AC power supply,
so it's changing all the time, the voltage, now of course, the self-inductance
is fighting back all the time, not just only in the beginning
as you saw in this circuit, but now, of course,
it is active almost all the time. So we can do away with this and so now
we replace the battery by a AC power supply, which we normally put just a wiggle there
and here is the self-inductance L and here is the resistor R. And let the voltage provided by this power supply
be V zero times the cosine of omega t, omega being the angular frequency. And now, I have to apply Faraday's Law,
not Kirchhoff's Rule, Faraday's Law, when I go around this circuit and I set up
the differential equation. And of course the differential equation
is going to be exactly like this, except that V, now, is V zero
times the cosine of omega t. And now I have to solve
for this differential equation. That's the only difference, so I don't have to start
from ground zero. And the solution to this differential equation
is quite remarkable and not so intuitive. The current, as a function of time, now,
is V zero divided by the square root-- of R squared plus omega L squared
times the cosine of omega T minus phi and the tangent of phi, that angle phi,
is omega L divided by R. And of course we need some time
to digest this. The first thing that you notice
is that there is a phase lag between the current and the driving voltage. If phi, as you're going to see, is, uhm,
ninety degrees, then the current is delayed by one quarter
of a cycle, though the fact that the current comes later
than the driving voltage perhaps is intuitive, because the self-inductance
is fighting the change in the current. So it's perhaps not so surprising that there's
going to be a delay, that the current comes a little later. If you look at this equation here,
then what you have in front of the cosine term is obviously the maximum possible current,
because the cosine term is just oscillating between plus one
and minus one and so this here is the maximum current
that you can ever get. In one full cycle you get positive
and you get negative net value. And notice here the role of omega L
really plays the role of a resistance, and in fact the dimension of omega L is ome--,
is--, is--, is ohms. It really plays the role of a resistance
and if omega is very high then the resistance here becomes very high
and so your current becomes very low. Well, that's intuitively pleasing
because if omega is high, then the changes, the dI dt's,
are very, very high and therefore if there are very fast changes
the induced EMF is going to be high and so the current will be low. Also, if L is very high, then the system also is capable of fighting back very hard and so it puts up a large resistance. So it's also pleasing that you see the L there,
downstairs. If omega is very low, in your mind you can make
omega zero. You don't even have alternating current
when omega is zero, then you have DC which is direct current. So when you make omega zero,
you simply get I is V zero divided by R. That's Ohm's Law,
that's obvious that you get that. Let's now look at the phase angle. The tangent of phi is omega L divided by R. If the self-inductor is very large, then the system
has a strong ability to fight back, so it can delay that current by a large amount,
and the same is true if omega is high. If omega is high then the time changes are very--
changes occur on a very small time scale and so the system can fight back. But remember, you always have the EMF
proportional to dI dt. And so it's also pleasing to see that omega
and L are upstairs here. Either one of being large it can fight back
and it can hold back the, the current. I have worked out a situation whereby
we have an LR circuit-- this is on the web, you can download that
so you don't have to copy it and the reason why I have these values is
because directly coupled to a demonstration that I will do shortly. You see, an L in series with an R,
the L is ten milli-Henry and the R is ten ohms and let V zero be ten volts. And here you see three frequencies
hundred hertz, thousand and ten thousand hertz, and here you see the values for omega. You have to multiply hertz with two pi. And look now at omega L. At low frequency, a hundred hertz,
omega L is 6.3 ohms. Compare that with the ten ohms. They'd be comparable, but now look for instance
at ten thousand hertz. The omega L is huge. It's six hundred and thirty ohms. So it entirely determines, so to speak,
the resistance of that circuit. And so the current that is going to run,
at least this is the maximum current is this value, which we also saw here
on the blackboard, that current at high frequency
is enormously reduced. It's fifty times lower than this current
at low frequency, even though they have the same value
for P zero. And then you see here the phase angles. And the reason why you have these values
is that I can make you listen to this, I can make you hear this, because your hearing
is very good at hundred hertz and since all of you are young you can probably
hear even ten thousand hertz, maybe some of you can even hear
twenty kilohertz. When you get older
you lose your high frequencies. In fact, my frequency cut off is somewhere near
four thousand hertz. I'm going to make you listen to music and there will be violins which produce probably
three, four, five thousand hertz, and then I will turn on, all of a sudden,
the ten milli-Henry. So first I will make you listen to music whereby
there is no ten milli-Henry in there and then I will turn on the ten milli-Henry
and what you will hear, that the violins disappear because the current
reduction is now huge on the high frequency but is very little on the low frequency and that's the idea of what a self-inductor
can do for you. So if you listen to this... [playback of classical music with violin] There's no self-inductor in now. [music continues but ends] There's no music either [chuckles] OK. [Playback of classical music continues,
with violin notes] [click] [music without violin notes] It's a different music. [click] No self-inductor. [Click] Self-inductor. The high frequencies are gone. [Playback of classical music continues,
again with violin notes] No self-inductor. [Click] You can turn a violin concerto
into a cello concerto. [laughter] You just cut the violins out. OK. I can not make you listen to the phase shift, not even in the case of the ninety degree
phase shift and that is quite obvious because what does it
mean that there is a ninety degree phase shift? It means that during one cycle
of ten thousand hertz, which takes only one ten-thousandth of a second, that the high frequencies are shifted
by only twenty-five microseconds and there's no way that your ears,
your ears can hear that. The fact that the composer wanted those violins
to come in 25 microseconds earlier than, than they do of course
is something you cannot hear. So I cannot make you listen to the phase shift,
but for the phase shift I have something else. And for that something else, I'm going to return
to the, to my last lecture, in which we levitated a woman,
magnetic levitation. And so I'm going to return to that idea
and grind a little deeper than we did when I gave that lecture,
just before spring break. I had a coil and I was driving that coal--
coil with sixty hertz AC. And let's assume that looking from above that
the current was running in clockwise direction, which is exactly what I assumed when I discussed
this with you and so the magnetic field is coming down like this,
magnetic dipole field, produced by this coil. And then we had here,
we had a conducting plate under there and I said to you when this magnetic field
is increasing in strength then there's going to be
an induced EMF here, which tries to oppose that change
and so the induced current that is going to run, which we call eddy current,
is going to run in this direction. If this is clockwise, this current
is going to be counter clockwise, so it's going to produce a magnetic field
in this direction. It opposes the change of the increase
in magnetic field: Lenz's Law. And since the two currents are in opposite
direction, the two repel each other. And you bought off on that
and we levitated a woman. However, no one asked me the question,
what happens a little later in time when the magnetic field is still in the downward direction, but it is decreasing, since it is an AC current, there comes a time that
the magnetic field will be decreasing in time. Now the EMF here must flip over because Lenz
says. "Sorry, we don't like the decrease." The moment that the EMF flips over,
this current will flip over. The two currents are now in the same direction
and they will attract each other and so there goes your magnetic levitation. Half the time attraction,
half the time they repel each other. But yet we did elev-- levitate a woman
and the secret lies in the self-inductance. This current that runs here runs over a pass--
which is very difficult for me to anticipate which has a certain resistance, R
and it has a certain self-inductance, L. We know what omega is;
that's about three hundred sixty. And so we do get in this conductor,
we get the current, the induced current here, is delayed by a phase angle
driven by this equation, is delayed over the induced EMF. The EMF immediately coupled to what this coil is
doing, but the induced current is delayed. And I have something that will allow that,
allow you to see that, perhaps in even more detail. This red curve is the current for the coil,
the coil that you see there above. And when the current is above the black line
it's clockwise and when it is below the black line
it's counter clockwise. The vertical scales are arbitrary. The green curve is the EMF,
which is induced in the conductor. Notice when the magnetic field increases,
when the current goes up in the coil, that the EMF in the conductor
is in such a direction that it opposes the change
of that magnetic field. But now when the magnetic fields go down,
when the current in the coil goes down, immediately the EMF flips over,
which is what I just mentioned to you and therefore if the induced current and the induced EMF were in phase with eachother, half the time you would have attraction and half the time you would have
that the two repel each other and that won't give you magnetic levitation. Here, what you see is a blue curve
which represents the induced current. I call it the eddy current. If there is no phase shift between the induced EMF and the induced current, notice that half the time the blue curve
and the red curve are in opposite direction. When they are in opposite direction
they repel each other. When they are in the same direction
they attract each other. But now, if I have a phase delay so that the induced current comes later than the EMF-- and I'm going to do something dramatic,
I'm going to shift it by ninety degrees so the current is now ninety degrees delayed relative to the induced EMF-- look now that the red curve and the blue curve
are always in opposite direction. And so now there is hundred percent
of the time a repelling force. The coil repels the conductor and the conductor
repels the coil. Now in the case when we levitated
the woman, I am sure that the phase delay
was not ninety degrees but maybe it was only thirty or forty degrees,
but the net result is-- here the shift is not ninety degrees, the net result is
that you get, on average, a repelling force. And so, the secret of the repelling force,
in the case of the levitation of this coil and therefore of the levitation of the woman,
lies in the fact that there is a finite self-inductance
[tapping noise] in here. If R is zero, then of course
we have a superconductor, then phi is always ninety degrees. When R is zero, this is infinitely high,
so now we get a ninety-degree phase shift, and I did a demonstration whereby I had
a little magnet floating above a super conductor. That was an ideal case. Phi was then ninety degrees,
so they always repel each other. Today I want to do
a more controlled demonstration whereby I can actually calculate
the self-inductance and I also can calculate the resistance. And what I will do today is I will have a coil
which is stationary and I will have a conductor
which is not stationary. Here is my coil. AC, sixty hertz. There's the coil. And I have a ring and the ring
is made of aluminum, and I know exactly
the dimensions of this ring. I know the radius, it's about five centimeters,
I know the thickness, I know everything. It's an aluminum ring and has a radius
of about five centimeters. Since I know all the dimensions,
I can calculate the resistance of that ring. You should be able to do that, too,
if I gave you the dimensions. And so the resistance of that ring very roughly is about seven times ten to the minus five ohms. It's very small. I can all--, that is at room temperature,
by the way, the lower temperature
the resistance is lower. I can also calculate very roughly
what the self-inductance is of that ring. Now that's not so easy because here when I calculated
the self-inductance, the magnetic field was constant. I assumed it was constant,
uniform inside. That's not the case when you have a ring. You have a dipole field. However, I just assumed that the magnetic field
was the same everywhere at the surface of the ring and with that assumption, admittedly I could be off maybe by twenty or thirty percent, with that assumption I find that the self-inductance
is ten to the minus seven Henry. I know what the omega is. It's three hundred sixty, roughly and so I find
that omega L over R for this ring is about one half, run at that frequency omega. And that gives me a phase angle phi
of twenty five degrees and therefore the ring is going to be repelled
by the coil and of course the coil is going to be repelled
by the ring. I'll put the ring here and the ring
is supported by this, so this ring cannot fall over. The only difference between this experiment
and that one is, first of all I can be very quantitative there,
I can actually calculate the phase angle, where here that's almost impossible. Here, it is the conductor that I'm going to make levitate and the coil is stationary. And here it was the conductor that was stationary and the coil is floating, but of course the idea is exactly the same. And so what I want to do now
is make you see there, actually, hmmmm... and we'll have to...
it should come up there... Yeah, there you see this--
this ring. Maybe I should first show you
the whole setup. So this ring goes over here,
this is an aluminum ring and I'm going to make it levitate by simply running
sixty hertz, hunderd ten volts, through this coil and I hope I do nothing wrong. Oh no, I have to turn on my AC. Oh my God, that was not, gee,
that was not my intention. A good thing we don't have a woman
sitting on the ring now. My idea was to have it levitate. By the way, you did see that it was repelled,
that was quite clear. I had the current too high. [laughter] I had the current too high
so we'll have it a little lower and I will make the current come up very slowly
and then I want you to see that it levitates. There it is, levitating. Oh, oh, off the screen. There it is. And I can turn it over and it's still levitating,
of course. And the secret is this phase delay
introduced by the self-inductance. I have another ring here which has a--
a slot, also aluminum, same ring, but it has a slot. Well, the EMF in this ring
is going to be identical. There's no difference. In fact, the self-inductance of this ring is identical, but the resistance of this ring is huge because there's a slot in there,
the resistance is almost infinitely high. And so if the resistance is infinitely high,
no matter what L is and what omega is, phi is going to be zero. It won't repel. Half the time it attracts,
half the time it repels. [metallic clanking sound] That means nothing happens,
no magnetic levitation. Same EMF, same self-inductance,
but an infinite resistance and here you see magnetic levitation. Since the induced current in the ring is extremely
small because of it's very high resistance, the force on the ring,
whether it's repelling or attracting, in any case is practically zero. So that alone is enough reason for the ring
not to move at all. All right, I hope to see all of you tomorrow
during our exciting testing of the motors.
