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 five ET, 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. yes, there you see this,
Let this be a solenoid and this is a closed circuit,
and we want 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 *Empea's* Law,
that the magnetic field inside the
solenoids is new 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,
file B, is simply the area by 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,
new zero I divided by L. And this we call L times I.
That's our definition for self-inductors.
And so the self-inductions L is purely geometry.
It's pi level R squared, capital N squared divided by L times new zero. yes, there you see this,
Let me check this. Pi little R squared,
I have a capital N squared, new 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 point six 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 point 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 sum-, simple current,
just one wire that goes around -- whether it's this,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 nine nano Henrys, maybe only micro Henrys, but it's never zero. yes, there you see this,
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
EMFV. 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,
these 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 the switch the
current wants to increase, but the self-inductance says
uh-huh, uh-huh, take it easy,
land slow, 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 volume,
which you can find with Ohm's Law, because the self-inductance
itself
has no resistor. Think of the self-inductance as
made of super conducting 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 I 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 move 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
*Kipshof'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
*Kipshof'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
five ET. I advise you to go to the eight
0 two 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 EL, if you go around the
circuit, is not zero, is minus D five ET -- Faraday's
Law -- so it's minus L the IDT. So we have to go around to
circuits and we have to apply Faraday's
law and not *Kipshof'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 fields 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 VL. 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 if 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 LVIDT.
If you go in the direction opposed to the current,
then it is plus LVIDT 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 EL -- 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 IR E and
the L are in the same direction. Ohm's Law tells me it's IR.
In the battery, I go against the electric
fields, and so I get minus V. That now equals minus L,
DIDT, and this is the only 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 the L,
and L, to one side -- so I can write down that V minus LDIDT
equals IR. It's the same equation when you
look, and the nice thing about writing it this way is that
since the IDT 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 *Land's* Law is all
about. It's not until the IDT has
become zero that V equals IR, and that happens,
of course, if you wait long enough.
And so we have to solve the differential equation,
and what is often done that you bring all the terms to the left
side and that you get an, zero on the right side.
And so what you often see is that LVIDT plus IR minus V
equals zero. And because we have a zero
here, some physicist thinks that this is an application of
*Kipshof's* Rule. This is nonsense.
You can always make it zero here by bringing all the terms
to this side. The closed loop integral of E
DEL TL is not zero, the closed loop integral of E
DEL TL is minus LDIDT, but when I shift minus LDIDT 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
*Kipshof'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 eight 0 one
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 IMX -- 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.
Substituting here T equals zero, you get one minus one.
So you find, indeed, that I equals zero.
Substituting 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 IMX, because if T is L over R, then you get one minus one
divided by E, and that is about 0 point six
three. 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 point six three 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? yes, there you see this,
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 IMX 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 re-define 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 IMX -- 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 square RDT,
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 IMX outside and I can
bring the resistance outside and the I get the integral from zero
to infinity of E to the minus R over L times TDT.
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 square 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 IMX 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 LI squared, then I have here I,
and so I can replace that I there by B divided by new 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 here -- 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 new 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 magne-,
magnetic field energy density is then B squared divided by two
new 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 whatit
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 in-, inside of there,
whereby I have a twelve volt car battery and I have two light
bulbs. 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 our 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 percentof 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 seventy five percent of maximum,
and even if you wait nine seconds, then the light that
comes out here is still only ninety 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. What you're seeing here is a
remarkable example that the self-inductance is fighting
itself. 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 *Kipshof'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 DIDT'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 is 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. Because remember,
you always have the EMF proportion to the IDT.
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. yes, there you see this,
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 six point
three 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 frequently 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. There's no music either
[laughter] OK. [Playback of classical music
continues, without violin notes.] OK.
It's different music. 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 cannot 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
twenty-five 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 with you, and so the magnetic field is
coming down like this -- magnetic dipole field --
produced by this coil. And then we have here,
we had a conducting place 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 imp-, oppose that change,
and so the induced current that is going to run,
which we call eddy currents, 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: *Lendz* 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 Lendz 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 *raf* 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 each other, 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 effect that there
is a finite self-inductance [tapping noise] in here.
If R is zero, then of course we have a super
conductor, 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-, this 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. In 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 . yes, there you see this,
this ring. Maybe I should first show you
the whole set up. So this ring goes over here --
this is an aluminum ring -- and I'm going to make it levitate by
simply running sixty hertz under 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 and 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.