We have seen last time
that using Biot and Savart's formula that if you have a current
going straight into the blackboard perpendicular to the blackboard
that we get a magnetic field at a distance R. The magnetic field tangentially to the circle,
B here, B here and that the strength of that magnetic field
equals mu zero times I divided by two pi r. If you walk around this circle,
just walk around and you carve up this circle
in little elements dl-- and you calculate the closed circle integral,
so the closed circle of B dot dl, so everywhere locally
you dot B with dl, the B and dl are in exactly the same direction everywhere, then you would find that this
obviously is B times two pi r. But B times two pi r equals also
mu zero times I. This dl here has nothing to do
with this dl here. Don't confuse the two. This dl is a small amount of length
in the wire that goes into the blackboard
which carries a current. This dl is simply your dl
when you walk around this current wire. And it doesn't matter at what distance
you walk around. You always get mu zero times I. You see it right in front of your eyes because
B is inversely proportional to R. And it was Ampere who first recognized that
you don't have to walk around in a circle to get the answer mu zero I, but that you
could walk around in any crooked path as long as it is a closed path,
something like this. And now you have here the local B, which of
course is perpendicular to this radius and here you have your local dl and if now you
go around, closed circle of any path, it doesn't have to be a circle, dot dl. That now becomes mu zero times I,
which is known as Ampere's Law and I then is often given
an index and closed. It is the current which is enclosed
by that path. It is actually easy to prove this using
Biot and Savart's formalism. This is almost the third Maxwell's equation. We already had two out of four. This is almost number three,
not quite. We're going to amend it in the future. What is ill-defined a little bit in this equation
is what we mean by enclosed and I'm going to define that now so uniquely
that there is never any misunderstanding. If I have a very strange looking closed path
that I have chosen, that's the path I walk, I have to attach to that closed loop a surface,
an open surface. That's mandatory. You can make it flat. That's fine. You're free to choose it. You can also make it sort of a plastic bag
so it's open here. You can put your hands in here and here,
like a hat. Any surface is fine, but you must attach
to that loop a surface, so here I have some paths
that you could be walking and this would be perfectly fine open--
open surface. Could be flat, but it could also be open,
so it's like a hat. And now I can define uniquely what it means by--
what it means by this I enclosed, because if now I have a current that goes through
this surface and pokes out here, then I have a current penetrating the surface
and that is uniquely defined and if I have another one coming in
through the surface, call it I two, this is penetrating that surface. By convention, if you go clockwise around,
we follow the same notation that we had before, in the right-hand corkscrew notation, the
connection between magnetic field and current. If you go around clockwise seen from
this side, so you go clockwise, then I one as I have it here, the new equation
would have to be larger than zero. I two is then smaller than zero. But if you decided to go counterclockwise,
which is perfectly fine, Ampere's law doesn't at all dictate
in which direction you have to march around, then I one would be negative
and then I two would be positive. So we follow the right-hand
corkscrew notation. And so if you want to amend now Ampere's Law
to do me a favor, but you don't do books a favor
because all the books use the word enclosed. I would like to see this replaced
by penetration. It is the penetration of the surface
of the current that is uniquely defined. But a current enclosed by a loop
is ill-defined. Because where possible,
when we apply Ampere's Law, we will try to find easy passes around,
circles sometimes, sometimes rectangles and since you are free to choose the surface
that you attach to the loop if you can get away with it
you use a flat surface, but you can not always get away
with a flat surface. So the recipe is as follows. You choose your closed loop. Any loop is allowed. It may not help you very much if you choose
the wrong loop. Any loop is allowed. You then attach an open surface
to that loop. And I penetrate is now the current that penetrates
through that surface, according to this convention. And the direction of rotation is free to you. How you go around to pass,
it's your choice, but that defines then the sign
of the penetrating of the curve-- of the-- of the current,
according to the right-hand corkscrew. So now we can, for the first time,
calculate the magnetic field inside a wire that draws a current using Ampere's Law. I have here a wire that has a radius capital R
and a current is coming to me, I, and let's assume that the current
is uniformly throughout the wire, so it has a uniform current density. And I would like to know what the magnetic
field is everywhere. The cylindrical symmetry, I want to know outside
the wire and I want to know inside the wire. Let's first look at radius
which is larger than R and so here we have the cross-section
of that wire, radius R. A current I is going through this surface. I now have to choose a closed path. Since we have cylindrical symmetry
it is clear that we would choose a circle, with radius little r, so we can be sure
that the magnetic field strength is the same everywhere
because of reasons of symmetry. Since the current is coming towards me and I am free to choose
in which direction I'm going to march, I know that the magnetic field
is in this direction, so I might as well also march in this direction
so that my dls are all in this direction. I don't have to do that. I could march the other way around,
but if I march counterclockwise then both terms left and right
of Ampere's Law will be positive. I now have to attach an open surface
to my path. Well, this will be, the blackboard will be,
that open surface. And so now I apply Ampere's Law,
so I get B times two pi little r, because dl and B are in the same direction
so it's a trivial integral. That now equals mu zero times I,
which now penetrates my surface. Uniquely determined, all this current
from this wire that comes to me penetrates my surface,
so times I. And so B equals mu zero
times I divided by two pi r and that's the same result
that we found last time, when we applied
Biot and Savart. So that's no surprise that you see this. But now we have a way of finding
the magnetic field also inside the wire, so here we have now the wire again,
the cross-section, current coming out of the blackboard and now I want a radius
which is smaller than capital R and of course my closed pass,
again for reasons of symmetry, is going to be a circle with radius R. And my surface that I attach is a flat surface
and so here I go, B times two pi little r equals mu zero times--
ah, now I have to be careful, because now not the full current I
is now penetrating my surface, but it is only a fraction
that penetrates the surface, and the fraction that penetrates the surface
is now little r squared divided by capital R squared
times I. You see, because the total current
comes through the radius capital R, but I only have now a circle
with radius little r. And so I lose one r here
and so we get a very different result. You get now that the magnetic field equals
mu zero times I is now linear in little r divided by two pi capital R squared. And this grows linearly with r,
whereas this falls off as one over r. And if you substitute in this equation
r equals capital R, which then would be the magnetic field
right at the surface of the wire, you find exactly the same result here. Little r becomes a capital R. If little r becomes a capital R,
you lose one capital R, you get the same result. And so if you make a plot of the magnetic
field as a function of little r, then it looks like--
like so, so this is little r, this is capital R
and this is the magnetic field strength because we know that it is tangentially
to the circles. It would be straight line and then here
it falls off as one over R, and the maximum value here
is the value that you find there if you substitute little r
equals capital R. I will now show you that we can,
using Ampere's Law, also come very close
to calculating the magnetic field inside what we call solenoids. Solenoids is like a slinky current that goes
around in a spiral, one loop after another. I want to remind you that if we had a loop,
a nice current loop coming out of the blackboard here
and the current going into the blackboard so there's a circular wire
but I only show you the cross-section. I want to remind you that the magnetic field
as we discussed last time, would be clockwise here,
would be counterclockwise here. In the middle, remember,
it was like this. And then in between
it was like so. That was sort of the magnetic field configuration
in the vicinity of a-- a loop through which
we have a current going. But now imagine that you put
another loop here, current again coming out of the blackboard
going into the blackboard, and another one,
and so on, several. What do you think is going to happen with
these magnetic field lines which now diverge? They're going to be sucked in here. This loop also wants the field lines to come
through its circle, so to speak, and this one too and so you're beginning to get
a near-constant magnetic field and the more tightly
these loops are wound, the more accurately will your magnetic field be approximately constant, And I have some transparencies which
will show that in more detail. Here we have a figure. You see five windings, a spiral. If you look from the left, the current is
going in clockwise direction, and so the magnetic field is going
from this side to that direction. And when you look here you see that the magnetic
field is approximately constant inside, and outside these current loops,
outside the solenoids, we call them solenoids,
magnetic field is extremely low. And if you start winding these loops
very tightly, then you get a configuration
looks like this. You get an almost perfect constant
magnetic field inside the solenoids and the magnetic field outside the solenoid
is extremely weak. And now I would like to calculate with you
using Ampere's Law what that magnetic field
inside such a solenoid would be. And we have to make a few assumptions. Let this be my solenoid-- and the length of
the solenoid is capital L. A current I is going through like so. Eh, I. And I assume that if I look from the left side
that the windings are wound clockwise, so I know that the magnetic field
is then in this direction. I make the assumption that the magnetic field
outside the solenoid is approximately zero. I will show you later with a demonstration
that that's a pretty good approximation. And so the question now is,
what is the magnetic field there. And I assume I have N loops,
N windings, capital N. So now I have to choose a path. I have to apply Ampere's Law. I choose a path and you may be surprised
the path I'm going to take. This is the path I choose. It's a rectangle. And the length of this side
inside the solenoid is l. And I think of this as four different passes. Number one, number two, number three
and number four. Let's first look at number two. We have assumed that the magnetic field
is practically zero, so clearly if you'd go the integral,
if you go around then the contribution here
must be zero. If the magnetic field is zero then the integral
B dot dl is zero, so that's easy. But if you look at one and three,
there is no magnetic field, very small magnetic field outside. The magnetic field inside is in this direction. But dl is like this if I march like this,
and B is like this so there's ninety degree angles
and so the dot product is zero. And so the only paths that contribute to my
closed-loop integral of Ampere's Law is only path four and that tells me then
that B times little l, because B is constant, I have assumed that it is constant,
and I integrate it over a length little l. Now I have to agree on my surface. What surface am I going to choose
to attach to that closed loop? Well, why not using a flat surface
just like the blackboard? So now, I have to calculate the current
that penetrates that surface. The current that penetrates that surface... I have to know how many times this winding
pokes through that surface. If the length of my rectangle is little l
and if the length of the solenoid is capital L and if there are N windings on capital L,
this is the number of times that the current pokes through that surface,
uniquely defined. I have a surface now. There is no such thing as I enclosed. There is I penetrating,
through that surface. It's a soap film
and I poke straight through it and I do it so many times
that I poke through it and then I have to multiply this
by mu zero and then I have my I, but each time that it pokes through
I have current I. And so what you see now is that B equals
mu zero times I times N divided by L, and so that is our prediction for approximately
constant magnetic field inside a solenoid, and this actually is a very good approximation
as long as L, the length of the solenoid, is substantially larger than the radius of
the solenoid. The radius would be
the radius of these loops. I'll work out a numerical example which is
aimed at a demonstration that comes shortly. We have here a solenoid
whereby N is about 2800 and L is about 60 centimeters,
that is 0.6 meters and I'm going to run through there a current
which I really don't know yet but it's going to be close to
four and a half amperes, we will see when we do
the demonstration. And so I can calculate now what the magnetic
field is going to be. So the magnetic field strength is going to
be four pi times ten to the minus seven, because that's what mu zero is
and then I have to multiply it by 2800. I have to multiply it, divide it by 0.6
and then I multiply it by the current, 4.5 and when I do that I find
about 0.026 Tesla. 0.026 Tesla, that is about 260 Gauss. And when we do the experiment,
the current will be a little different but you will see that indeed the field
will be very close to 260 Gauss. Why is it that the magnetic field is not
proportional to the number of loops, but proportional to the number of loops
per unit length? You may say, "Well, if I have one loop
I have a certain magnetic field, two loops I have
twice that magnetic field, ten loops I have ten times
that magnetic field." Well, imagine that we start a solenoid
in lobby seven. And so here is the sole--
here is lobby seven and here is that solenoid. There it goes, all the way,
all the way, thousands and thousands
and housands and thousands of loops and we end up here in 26-100. Look at this loop. Think of it at the first loop. Creates a magnetic field. What is the shape
of that magnetic field? Well, it is a current loop
and as we discussed last time, the magnetic field that one loop
produces is like a dipole field. So do you really think that here in 26-100
we can sense the magnetic field that is produced
by this one dinky toy loop? Practically nothing! It falls off so rapidly,
the magnetic field, that we don't notice it--
notice it here. So it's immediately obvious
that the magnetic field is not proportional
to how many loops you have. If, however, you put all those loops
on top of each other, then of course you can add
the magnetic fields. And so it is natural that you get how many
windings you have per unit length. So now I want to first show you
the magnetic field configuration of a very loosely wound loop,
with seven windings and I will do that by sprinkling magnetites,
this iron file, in the vicinity. We've done this before
for other current configurations. Now we'll do it for this solenoid
with seven windings and I'm going to run a few hundred
amperes through there, have to first get this
car battery . All right. And so we put some iron files on here and what I want you to see now
is that the magnetic field inside, even though it's very loosely wound,
begins to look nicely uniform and that there's almost no electric-- er,
magnetic field outside. Look at this, isn't that wonderful? Isn't that incredible? You see how these iron files
line themselves up very nicely horizontally inside the loops and when you look outside the loop
here or there, where we assume the magnetic field
was about zero, you don't see the iron file being where we ended
in any preferred direction, which indicates that the magnetic field
is very low. Now, I want to show you what magnetic field
we can get with this baby, which is exactly what I had here
on the blackboard. It has 2800 windings-- and we're going to run a current
which is something like 4.5 amperes, but I'm going to tell you
what that current is, because I have a current meter there for you
and I also have a-- a meter which indicates
the magnetic field. The lower one is the current meter
and the maximum current that you see there at the three would be six amperes. And the upper one is calibrated
in such a way that if it is full scale, you would have three hundred Gauss,
so a three is three hundred Gauss. And I can-- I have a probe, a magnetic probe,
we never discussed how that works. We call it a Hall probe-- and this Hall probe allows me to measure
the magnetic field in the vicinity of this solenoid. It's even sign sensitive. If the magnetic field is like this,
it would go to the right. If the magnetic field is like this,
it would go to the left. And so this allows us, then,
to be actually quite quantitative and evaluate the magnetic field
near the opening of the solenoid. Then we can go in there
and we can also probe the outside. So I'm running now a current. Let's look at it,
that's the bottom meter. So that is about 4.8 amperes. I assumed it was 4.5. It's a little higher. And here comes this probe and I'm now about a foot
away from the entrance and you see nothing. And I come closer to the entrance
and the magnetic field begins to show. Nowhere nearly constant yet. I'm now entering one hundred Gauss. I'm going in deeper. Two hundred Gauss. Even deeper. And deeper, and now I have about 240
Gauss and notice, as I go in farther, it doesn't increase. It's more or less constant. Amazing, that it's more or less constant. And when I come out here, I move it back and
forth about twenty centimeters. If I came in from the other side,
you would simply see a reversal in the sign, which is not so interesting, so you see
240 Gauss with the other direction because this probe is sign-sensitive. I can now also show you that if I come on
the outside of the solenoid, you see nothing. So indeed, our assumption that
the magnetic field is very low outside a tightly wound solenoid
was a very good assumption. Very well. You've been asked and the deadline is
Friday four pm. to explain the behavior of
the Kelvin water dropper. And I decided to give you
a little bit of help on that. Most of you may already have figured it out,
but those who haven't probably won't figure it out
between now and Friday anyhow so I might as well tell you. That water dropper, called the Kelvin water
dropper, is an amazing battery. We've seen it before. We know what it's doing,
but I will go over that again. We have here buckets A and B. Water comes down from above,
water runs through. You see the water there, it runs out. And we collect these water drops here
in bucket D, it's a conductor, and this water is collected in bucket C,
it's also a conductor, and the paint can A is connected to C. That's crucial. And the paint can B is connected with D. And here there are two bowls,
which I can bring close together. I run water and after a while
I see a spark here. I can even see a spark when the distance is
something like six millimeters, which would be about potential difference
of about twenty kilovolts. How does it work? Well, water has a pH of
seven. And that means one in ten to the seven molecules
is ionized. So I have OH minus and I have H plus. And the- those are going to be
the current carriers. The ions are doing the work here,
are doing the job. Let us make an enlargement here of can A. And can A, let us assume that purely by chance,
it has a little bit of positive charge on it. It could be negative, but I'll just assume
it's positive for now. In either case it will work,
you will see. Just by chance, like you have a little bit
of net charge. You're not completely neutral. And so this can has a little bit of positive
extra charge. Now here is the drop from the spout. What's going to happen? Through induction, through polarization, you get a little bit of extra-- excess
negative charge here and a little bit of excess positive there, because the positive repels each other
and the negative is being attracted. So the H plus goes a little bit up
and the OH minus comes a little bit down. But now the drop breaks
and there goes the drop. So it's a little bit negative. So now a little bit of negative drops come
down and so this becomes negative. But this is connected with B,
so B becomes negative. But what do you think is going to happen now
with the drops that fall through B? They are going to become positive, because if B is negative
then of course this will be reversed, the bottom will be positive,
the top will be negative and so those drops now
that are going to fall through are going to be positive. So C becomes positive. But C is connected with A,
so A becomes more positive and so A can do even a better job now
on these water drops and polarize them even more,
and so you get a runaway process. And the whole system feeds on itself until
the potential difference here becomes so high that you exceed the three million volts
per meter and there you get a breakdown
and you see a spark. Now, you can think of a continuous stream
of water as a stream of individual drops, so it also works if you just have a regular
stream of water going down. Who is doing the work here? Someone has to do the work. You have a battery. The battery's being charged and then it is
discharged to the spark. Who is doing the work? Any idea? Have you thought about that? Yeah? Gravity. Very good. It's gravity that is doing the work. We can see that very easily by identifying
the current that is flowing and the electric field. How is the current flowing? If negative charge is going down,
would we all agree that the current is going up? If positive charge is going down,
do we agree that the current is going down? This side of the spout, here,
will be slightly positive and this is slightly negative, because the H plus is more abundant
here than there and so we're going to get a current
through the water, in this direction. The water has a low but
sufficient conductivity, because it's ionized one out of
ten to the seven molecules and so once in a while if you see a spark here,
then you get a current there, but that's intermittent of course. How about the electric fields? Well, electric fields we know go from plus to minus,
so that's easy. We can put the electric fields just in like
that. Electric field must here be in this direction. Electric fields here goes from plus to minus. C is plus charge, B is minus charge
and so the electric field is in this direction. Electric field is from plus to minus. This is plus, this is minus, so the electric
field is in this direction. Electric field is from plus to minus. The can A was positive, remember? So the electric field
is from plus to minus. So that's the electric field configuration. But now look what's happening. Here, the E field and the current
are in the same direction. That's fine. Here the E field and the current
are also in the same direction. That's great, but now look at these
poor negative ions. These negative ions don't
want to go in the direction of E. Negative charge wants to go
against the direction of E. But gravity says, "Sorry, you can't do it,
I force you down." And so these negative drops
are forced by gravity to go down. Look at this positive charge. These poor water drops which are positively
charged go against the electric field. They don't want that. Positive charges want to go
with the electric field. Gravity says, "Sorry, it's too bad,
I force you down." And so gravity is doing the work, so to speak,
against the will of the charges. And then the battery charges up
and charges up until the potential difference becomes
so high there that you see a spark, and you deal with potential differences
of something like twenty kilovolts. Remember last time, and you will see that
again today, that as the water goes through
and as the system charges up, that you begin to see that the water
which starts running like so, begins to spread out. It fans out. You can hear it by a change in the sound,
but you can also see it, and I'll make you see it again today. Why is that? Well, that's immediately obvious. If this can is positively charged
but if the water is negatively charged, the negative charge wants to go
through the positive can, and so it spreads out. So it's clear that by the time
that you reach almost a spark, that that water will spread out quite
substantially and you will see that. And then when there is a discharge,
it will start running again narrow stream, and then slowly in time
the water will spread. And so now I will do
a series of demonstrations which will support what we know
and what we perhaps don't know. Let's first give a little bit of light here,
because we will need that and then I'm going to make it
completely dark, so that you get maximum
pleasure for your money. And I will first show you here the-- the gap those two bowls
and this time we have a real treat, which we owe to Marcos, who is standing
behind the instrument very modestly. This time we have a gauge,
a meter, which measures the electric fields
very close to can A, and the reason why that is nice is
that as this system charges up, we just don't have to wait now
just until we see the spark, but we can look at the
galvanometer there and slowly see it being charged up and then
we get a spark and then it discharges, so you get a lot more for your money. So I propose that the first thing we do
is what we did last time, that we simply let it run
and see whether we get a spark. Uh, there's always air bubbles in the system
which I have to get out. OK, I think I did that. OK, let's just be patient. Ah, it starts already. Look at the, look at the e-field, Bang! First spark. Can you see the spark on the screen? Look at it again. There's a spark. And at the same time you see how the electric
field goes away, charges up, bang. Charges up, bang. So it cha-- it starts the whole system charged
because of a random positive or negative charge that would be present
on one of the cans. What I want to do now is I want to increase
the gap between the two bowls and I make it so large that you will never achieve
a magnetic-- an electric field there of three million volts per meter. But somewhere else on the unit,
it has lots of sharp edges and sharp points, somewhere else the electric field
will reach the breakdown voltage and so the system will go into-- dis-- into discharge,
into corona discharge. You will never see sparks
but it goes into discharge. And you can see that because now,
I will open the gap now. If you now look at the electric field
it will reach a maximum value and now it goes into discharge,
you see. It begins to sort of sputter a little bit
back and forth, so now there is no longer
a spark on the bowl, so you don't see anything
between the bowls, but it is somewhere else,
I don't know where, where there is a continuous stream now,
of charging, leaving the system and so it's discharging
to corona discharge. And so in this mode, you expect the water
to be spread all the time. And I will show that to you
by switching now to the water. Maybe Marcos, you can improve on the--
on the-- on the light. I could turn this off,
maybe then you'll see it better. But you see that the water is spread and I will now bring the two bowls
closer together-- ah, you also want to see of course
the electric fields, to see the discharge. I bring them closer together,
stopping the corona discharge. There's a spark. And now watch the water. See, the water is now just not interesting,
and there it spreads. I'll turn the light off. You look at the water. So slowly the system is charging up. You see the water? Bingo! You can even tell by the water
when it sparks. There it goes. Now I want to do something real mean. What I want to do now is to raise the spout
so high that A can not reach out
all the way to the spout. It's too far away
and can not polarize the water. So the poor battery can not start. That's a pretty mean thing to do
to a battery. But on the other hand,
I'm not all bad. What I can do is I can start the system,
I can help the system. I use my extrophorus disk and I'm going to hold my extrophorus disk
very close to one spout, temporarily allowing it to polarize
the water and once it starts, chances are that it will feed on itself
and go into the runaway. It's not very predictable
but I will make an attempt. So the first thing that I will have to do
is maybe you can zero the E field. Yeah, thank you. So the first thing I want to do is to make
sure that if we start running water now, that the system doesn't start by itself,
because then I can't make-- then I can't be nice to the system anymore. So let's just run some water. I hope that Marcos brought it high enough,
and let's see-- and you can really tell
by looking at the-- Oh, there's the air bubbles
in the system. I have to get rid of the air bubbles. There we go. OK. So look at the electric field
doing nothing. The system can't start. A is desperately reaching out
to this water, but it's too far away. It can't polarize it
and the battery can't get going. Ah, ay, ay, it does get going. Oh, Marcos, we didn't bring it far enough! You have to bring it higher! Oh, by the way, this is interesting. Let's see where it actually--
oh, boy, this is incredible. This is a distance of about fifty centimeters,
and look at it! That is amazing! The little sucker,
I had never expected that. Let's give it a fair chance. Let's give it a chance. It's only reasonable. Goes very slowly now. You can see why it goes so slowly,
because A very far away from the spout. But boy, and-- and you can see the water. Still-- the water is still pretty normal. Oh, man, it's almost cheating on me! Uh-oh, uh-oh, we'll have to raise it
a little further, er, aft-- but I--
I want to give it a chance to, uh, to spark. Yeah, the two balls are close enough,
so I will probably get there. Boy, this is almost torture for me. Aw, look at that, look at that. I begin to see the water already. Look at the water, it's already spreading it--
ah! there it is. Marcos, can you run it even higher and then
we'll see whether we can actually-- stop it altogether,
bring it a lot higher-- and then we will make an attempt
to start it with some help from a friend. You ready? Oh, man, I can hardly-- OK. Don't tell me that it's going to--
it can go to the other direction, of course, because it's a random choice
that it has at the start. Aah, aah, it's letting me down! Let me see whether
I can help it a little. My electr-- electrophorus disk. I'm going to hold it here. You know why I held it there? Because I was hoping
it would reverse the direction. It doesn't even want to do that. It's really recalcitrant today, isn't it? Hold it here, negative charge. The reason why you see the sudden change in
the electric field is because of the e-field probe which senses the electrophorus disk
when I held it near the spout above B. It has decided, no matter what--
well, has it really? Ah, ah, ah, ah. Has it really made any decision? Ah! It's thinking now what it's going to do. Oh, man. Yeah, it's going in the direction
that I wanted it to go because I held negative charge
close to B and I knew that would force it into the
direction that it's going now. Oh, man. Yeah. Well, look, the distance is now,
what is it, now, something like eighty-- eighty-five centimeters? The system is desperate, but it's,
it's, it's doing what it's supposed to do. Physics works again. And we will get a spark because--
is this close enough? Ah, boy you got a lot
for your money this time. You really did, didn't you? You saw for one thing
that the system first started up going in one direction and later in the other because there was a random charge
which changed polarity, which is something that
we can not always control, but by holding this negatively charged
disk close to the spout above B, I forced the polarization in the way
that I wanted it and that's what you're seeing now. We've got to give it a fair chance,
and then I'll give you light and then you have four more minutes left
to fill out the evaluation. But let's at least give this battery a charge now,
to lay its egg. The water, it still doesn't show much
of a sign of spreading, but I expect that
that will happen very shortly because we've seen
this now before. It's slowly increasing the charge on the cans,
A and B. It's creeping so slowly that if we wait
you may never able to fill out your-- i'll give you a little light so that you can
fill out the evaluation and then we can still let it run
and see what happens. So please leave the evaluations here
as you leave and I will keep that going
for some of you who have the patience
to see what will happen.
