Here are the topics the way I see them.
They're on the web. Look on the lecture supplement
of today and you can download them.
I want to point out that, uh, the discount is very
reasonable because these magnets are broken, and when you break a
magnet, you end up with two monopoles, so you get fifty
percent off. You should know better by
looking at the key equation there,
that magnetic monopoles don't exist, but that's a detail.
I want you to appreciate that I cannot possibly cover today
these topics in any depth, nor can I cover all of these
during a fifty minute exam. So please understand that this
review is highly incomplete, and what is not covered today
can and will be on the exam. I'm interested in concepts.
I'm not interested in math. There will be seven problems.
Five of the seven problems have only one question.
Two problems have two questions.
I don't think that time with the length of the exam is going
to be an issue. This exam was taken by several
instructors. It took them fifteen to
eighteen minutes, and that's normally my
objective. Courtesy of Professor Belcher,
we have old tests on the website.
I don't have the solutions. There's only so much I can do.
I'm very grateful to Professor Belcher that he made these exams
available. If you can't do some of the
problems, I would suggest you see your tutors or you see your
instructor. I will also be available all
day afternoon, not tomorrow,
because tomorrow afternoon I'll be
in twenty-six one hundred all afternoon to work on
demonstrations for the next week.
Many of these problems are straightforward.
You may also want to consult your study guides.
All right. Let's first start with
Biot-Savart. There are not too many problems
that one can do with Biot-Savart.
DB equals mu zero divided by four pi
times the current DL cross R divided by R squared.
That's the formalism. A classic problem that you
probably have done. We have point P here at a
distance D from a wire, and the current through the
wire is I. If you want to know what the
magnetic field at P is, you can use Biot-Savart.
It would be a stupid thing to do, but you can do it.
You take then a small element DL here of the wire.
This distance is R. This is the unit vector R,
which you have in this equation.
And you can calculate, then, what the contribution to
the magnetic field right here is -- it comes out of the
blackboard -- due to this section DL.
This angle is theta, and the sine of theta is D
divided by R.
And then you have to do an integral over the whole wire,
theta zero to pi. And then you get the magnetic
field here. Not very smart thing to do.
A waste of time. Because clearly the way to do
this is to use Ampere's Law, which is the one at the bottom
there. In which case,
you would construct a closed loop with radius D.
This loop is perpendicular to the blackboard.
I'll try to make you see it three dimensionally.
You have to attach an open surface to that closed loop.
Any open surface will do. Let's make the open surface
flat. And then we apply Ampere's Law,
which is the one at the bottom there.
We don't have the second portion because there is no
changing electric flux. We don't deal with kappa M at
all. So we simply have that B times
two pi D, that is going around in this
circle, because I know that B is tangentially to that circle.
I also even know the direction according to the right-hand
corkscrew rule, coming out of the blackboard
here. So as I go around the circle,
B and the DL that you see there at the bottom are in the same
direction. So I get two pi B times D
equals mu zero times the current that penetrates that surface,
and that is I. And so the answer is very
simple, mu zero I divided by two pi D.
That's the way you would do this problem,
and you would stay away from Biot-Savart.
There is one particular problem whereby Ampere's Law will fail.
Of course, Ampere's Law in general works when we have
cylindrical symmetry. This is cylindrical symmetry.
There is one problem where Ampere's
Law bitterly fails and where Biot-Savart is highly superior.
I have here a conducting loop. It's a circle.
And you're being asked, it runs a certain current I,
and you're being asked, what is the magnetic field
right at the center? It only works for the center.
You could not find what the magnetic field is here.
We did that in class. You probably also did that for
your homework. Biot-Savart will immediately
give you the answer. And I will leave you with that.
And Ampere's Law won't work. So let's now turn to Ampere's
Law and do a few problems with Ampere's Law.
We need cylindrical symmetry, with very few exceptions.
I have a hollow cylinder here, radius R1.
Concentric another cylinder with radius R2.
These are very long cylinders. And assume that there is a
current flowing, I, in this direction on the
surface of the inner cylinder, and the current I is returning
on the surface of the outer cylinder, and the two currents
are the same in magnitude. And I want to know what is the
magnetic field everywhere in space.
I will make a, a drawing whereby we only see
the cross-section. So this is R1 and this is R2.
And let's first calculate the magnetic field for R being
larger than R2. It's immediately obvious that
your closed loop that you choose itself is going to be a circle.
That is a must. Radius R.
We use a symmetry argument. Whatever the magnetic field is
at that distance R, little R, it must be the same
everywhere. It cannot be any different here
in terms of magnitude than there because of the symmetry
of the problem. We have cylindrical symmetry.
We also know that if there is any magnetic field,
that it is going to be tangential, either in this
direction or in that direction. And so we go around.
We make the closed loop integral, use the equation that
we have there at the bottom, and so you get B times two pi
little R equals mu zero. And now I have to attach an
open surface to this loop.
I will use the surface in the blackboard.
I could use any surface, but I might as well use a flat
surface. And now I have to know what is
the current going through that surface.
Right here, on this surface, the current is going into the
blackboard, and right here, on this surface,
the current is coming out of the blackboard.
The two magnitudes are the same, so the net current is
zero. And so the magnetic field
outside the second cylinder is zero.
So let's now look at the area in between the two cylinders.
These are hollow cylinders, now, this is completely open
here and this is completely open.
They are thin, thin material,
thin shells, are both cylinders.
Now of course the closed loop is going to be one inside the
opening between the two cylinders.
And again, I pick radius R. And here I go.
I get B times two pi R equals mu zero, but now there is
current going through this surface, and the current that is
going through is I. Not this one,
but this one. And so we get I.
And so now we get that B is mu zero times I divided by two pi
R. One over R field.
The direction you will find from the right-hand corkscrew
rule. That's the way I normally do
it. You take a corkscrew and you
turn it clockwise, it goes into the blackboard,
so the magnetic field here is in this direction.
If you're not used to turning corkscrews in corks,
think about when you try to tighten a screw with a
screwdriver. If you go clockwise,
the screw goes in. At least, that is the case with
ninety-nine point nine nine percent of all screws that you
find in this country. They have a right-handed
thread. You could make one that has a
left-handed threa- thread, but that's not done in general.
So now we can go to the area R less than R1.
So that's inside this thin shell cylinder,
and again, you would take a surface, a closed loop,
that is a circle with radius little R,
and this is your surface attached, open surface attached
to that closed loop. And you will find for the same
reason that you found here that B is zero, because there is no
current going through that surface.
And so if you now make a plot of the magnetic field B as a
function of R, here is R1, here is R2,
then it is zero here, it's zero there,
it has some value here, which is this value if you
substitute for little R R1 and right here it is this value,
if you substitute for little R R2.
And this is a curve that is proportional to one over R.
And so this is the magnetic field, and inside,
seen from where you were sitting, it is clockwise.
There is one case, and we did that in lectures,
whereby we wanted to calculate the magnetic field inside a
solenoid, where Ampere's Law works very well,
even though we don't have then closed loops which are circles,
we chose a rectangle. I want you to revisit that.
It's undoubtedly in your book, and I covered it during my
lectures. We assumed that the magnetic
field was uniform inside the solenoid and
zero outside the solenoid, which was not a bad assumption,
and that allows you then, with Ampere's Law,
even though you go rectangular and not circular,
to get the magnetic field inside a solenoid.
Very classic. Check your lecture notes or
watch my lecture again, which you can do on the web.
So now I want to turn to Lorentz force.
The Lorentz force, F, is Q times E plus V cross B.
This is the charge which is moving with velocity V in a
magnetic field B and at the location of that charge there
happens to be also an electric field E.
Let's take a situation that I have an electron.
I put a minus sign there to remind you that the charge is
negative. And let this be the velocity of
that electron. And let the magnetic field be
uniform, perpendicular to the blackboard in this direction,
going into the blackboard. There is no electric field,
so we only deal with Q times V cross B.
V cross B -- you should be able to do that cross product -- is
in this direction, but since the charge is
negative, the force on this charge is in this direction.
And so the electron is going to turn over in this direction and
is going to get a circle if the magnetic field is uniform.
And the force is right at the center of that circle.
A little later in time, when the electron is here,
going with the same speed, the force is like so.
The speed cannot change. The force cannot do any work,
cannot change the kinetic energy, because F is always
perpendicular to the plane through V and B,
because it's a cross product. And since the force is always
perpendicular to the velocity, you cannot change the kinetic
energy, because you're never doing any work.
The only thing that the force is doing is make it go around.
It changes the direction of the velocity.
It doesn't change the magnitude of the velocity.
It doesn't change the speed. All right?
So let this radius be little R. Combining eight oh two with
eight oh one, we now have that M V
squared over R -- M being the mass of the electron -- is that
force. And since V is perpendicular to
B, the sine of the angle between them is one, and so I simply get
here Q V B. This is the magnitude of the
force. We already know the direction.
And so one V goes, and so I get that the radius of
that circle is M V divided by Q B.
This being the momentum of the electron.
So V is the velocity of the electron, speed.
Q is the charge. B is the magnetic field.
M is the mass of the electron. And as long as this velocity is
reasonably smaller than the speed of light,
we don't have to make a relativistic correction here.
If it's not much smaller than the speed of light,
we have to make a relativistic correction.
I discussed the relativistic correction in my lectures,
but I also mentioned that that would not be part of exams,
so I will not further elaborate on that.
So we will just assume that this is completely
non-relativistic. The time for this electron to
go around, capital T, is two pi R divided by V.
But I know R, so I get two pi times M V
divided by Q B. And then I have this V,
and I lose this V, and that is not so intuitive,
but you see that the time is independent of the velocity of
the electron, assuming that it's
non-relativistic. And we discussed that in the
framework of cyclotrons, time for these particles to go
around is independent of the velocity.
Suppose I have a magnetic field B which is seven point eight
times ten to the minus four tesla.
Now I know what you're thinking.
You're thinking, why the hell do you take this
crazy value? There is a reason for that,
because I have a demonstration here.
And the magnetic field is seven point eight times ten to the
minus four tesla. If that is the field,
I can calculate how long it takes to go around,
because I know the mass of the electron, I know the charge of
the electron, I know the magnetic field.
So in that case, T is about forty-six
nanoseconds. And so the number of times that
it goes around,
F, which is the frequency in hertz, is one over the period,
is about twenty-two megahertz. So it goes around,
this electron, twenty-two million times per
second. The demonstration that I will
show you. We have a, a glass container,
spherical. It's about yea big.
You can see it there. There is low pressure gas in
there, so that some of the electrons can go around in a
circle before they ionize the gas.
We will accelerate the electrons over a potential
difference, which I can vary from zero to three hundred
volts. And so I can give them a
certain velocity. Suppose I gave them a potential
difference of hundred volts, then Q times delta V,
which is the potential difference,
would then be one half M V squared, and so if I make this
one hundred volts, I can calculate what V is,
and I've found that the velocity of the electrons then
is about five point nine times ten to the six meters per
second. So that's the velocity if I
have a potential difference of one hundred volts.
And so I can use that velocity, substitute that in this
equation, and then I get the radius of these electrons in
that magnetic field, and that radius turns out then
to be about four point three centimeters.
And that is a little smaller than the sphere that I have.
How can I create a uniform magnetic field of that
magnitude, that's indeed perpendicular to the orbit of
the electrons? We do that using Helmholtz
coals- uh, coils. We never discussed that in
detail during our lectures.
You see here coils, and you see here coils.
And when you have two sets of coils like that,
and you have a proper distance between them,
you can create between them a magnetic field which is almost
constant. In this case,
this is the field, fifteen times larger than the
Earth's magnetic field. Perhaps some of you remember
that I linked you on the eight oh two
website a few weeks ago to a very nice program that I
received from Professor Belcher which allowed you to calculate
magnetic field configurations with current loops.
And one of the things I asked you on the website is,
try to create what we call a magnetic bottle.
This is a magnetic bottle. It's also referred to as
Helmholtz coils, whereby you can create between
two coils a near-uniform magnetic field.
And so that's what we have here, and I'm going to show you
now how beautiful these electrons go
around in circles. They ionize the gas inside,
and then through de-excitation of the ionization,
you will see light. Just the same idea that you see
from Aurora. So in that sense,
you're seeing an artificial Aurora inside a glass tube.
I have to make it very dark for this, because the light is not
very bright. So we have to turn this off.
And I have to turn all this off.
And I have to turn this off. And some of you who are close
can actually see the light immediately in the,
in the sphere right here in front of me.
But if not, then you will see it there on the screen.
So this is the -- let me make sure that I can have my -- so
this is the electron gun here. It has a potential difference
which I set roughly at a hundred volts.
And here you see these electrons going around.
And they go around roughly twenty-two million times per
second. And the gas pressure is low so
that some of them make it all the way before they ionize the
gas. And I can change the delta V.
I now make the speed lower by making the potential difference
over which I accelerate the electrons, by making that lower.
The radius is proportional to the square root of that
potential difference. So now you see here,
when I increase the potential difference, the radius goes up.
See how beautiful that circle is?
It's absolutely amazing. Very, very nice demonstration.
And then we reach the edge of the -- we are now very close to
the diameter of the, of the glass sphere,
so we can't go much further. So now I want to discuss with
you Faraday's Law. Faraday's Law comes in many
different ways. Faraday's Law -- I will get up
the, get the Maxwell's equations up again.
Faraday's Law -- Faraday's Law tells you that
the induced EMF is minus D phi DT.
D phi DT is the magnetic flux change through an open surface
attached to a loop. The loop could be somewhere in
your brains or it could be a real loop.
You can imagine any loop in space.
That is always correct, that statement.
So the EMF induced equals minus D phi DT, minus D DT of the
integral of B dot DA,
open surface. You can think of many ways of
changing the magnetic flux. One way is a stationary loop
and having a changing magnetic field.
Another way is of having B constant but changing the
geometry so that the magnetic flux is changing through the
surface, if you rotate it around or you move things around.
Let's look at both today. Let's first take the stationary
loop. I spent a major part of one
lecture on that. Say this is a conducting wire
and it has a resistance capital R.
This is net resistance of the entire wire.
And suppose here I have a changing magnetic field,
and this area right here is A. And the magnetic field,
let's say, is perpendicular to the blackboard.
And the strength of the field is B, and so the magnetic flux
phi of B is then simply A times B.
And so D phi DT is then A times DB DT.
Remember we were going to keep the geometry constant.
It's a s- it's a stationary system, but we're going to
change the magnetic field. So this magnetic field is
changing. Perhaps it's coming out of the
blackboard now, and maybe it's increasing.
And so you induce then in this conducting wire an EMF,
and the EMF has this magnitude. This is the magnitude of the
induced EMF. And so the current that is
going to flow, that is the induced current,
is the induced EMF divided by the resistance of that whole
conducting loop. The direction is never an
issue, because you all can handle Lenz's Law.
Notice I didn't even put a minus sign here,
because the EMF is of course minus D phi DT.
That's not important for me if you put a minus sign here,
but I put these bars there, so get rid of minus signs.
Because if the magnetic field is coming out of the board and
if it were increasing, then Lenz's Law will run a
current in this direction to oppose
that increase. And so minus signs in general
are for the birds. You can always reason in which
direction the current is going. So that's a case whereby we
have a stationary loop and whereby the magnetic field is
changing but not the geometry. Now we'll have a case whereby
the magnetic field, say, is constant.
So we have here a conducting wire and we have a magnetic
field, for instance, coming out of the
blackboard, uniform -- uniform is always nice when we do
integration to get this flux. It's always nice to have it be
uniform. And we have a bar here which
has length L. This is length L here.
And we move the bar with velocity V to the right.
So this is a very simple case. V is here perpendicular to the
bar. That makes it always easier.
And B is perpendicular to the plane through V and L,
so that makes all sines of theta that we may have,
or cosines of theta, all one.
Magnetic field is constant, and so the magnetic flux,
phi of B, is the area times the magnetic field.
The angles are just wonderful, so that is B times X times L if
this distance is X. So the magnetic flux change in
time, D phi B DT -- notice I don't
care about minus signs. I don't need n- minus signs.
That D phi DT is there for L times B times DX DT,
and DX DT is the velocity. So the magnitude of the EMF is
L B V. And so the current that is
going to flow, the induced current,
is that value, L B V, divided by the
resistance R, and that is the resistance,
then, that is in this entire loop.
Whatever there is and where it is I don't care,
but that is the magnitude of the current.
The direction is again easy, it's non-negotiable.
If I move this bar to the right, the magnetic flux is
increasing. It's pointing in this
direction, the magnetic field, so the current is going to flow
in such a direction that it opposes that change,
and so the current will flow in this direction.
That's the induced current, and this is the magnitude.
In nineteen ninety-six, NASA attached a twenty
kilometer conducting wire called a tether to the shuttle,
so L was twenty kilometers. The magnetic field of the Earth
is about half a gauss. Even at a distance of two
hundred miles, that is not much different from
here. So that is five times ten to
the minus five tesla. And the shuttle,
as you should know from eight oh one, like any near-Earth
satellite, they all fly with a speed of about eight kilometers
per second. If they go much faster,
then they will leave the gravitational field of the
Earth. So V is about,
in circular orbit, is about eight kilometers per
second. So here you have it in meters
per second. If I calculate for this tether,
moved around, if I calculate L B V,
I would get eight kilovolts. However, keep in mind that that
is only correct if B were perpendicular to the plane
through L V, and also if the velocity of the shuttle were
perpendicular to the direction of the wire.
None of that is the case. The magnetic field where they
were is not exactly perpendicular to the plane
through V and L. And so what they observed was
three and a half kilovolts, about half of the maximum that
you could achieve. You may say,
gee, this is strange, because if you just drag a
conducting wire through space, you don't have a closed loop
circuit, so you can't talk about the idea of a magnetic flux,
because you have no surface. Well, at the altitude where the
shuttle is flying, there is still a teeny-weenie
little bit of air, very little but some,
and that is highly ionized because
of the ultraviolet light from the sun.
We call that the ionosphere. It's a plasma.
And so in the surrounding of this wire, here and here,
you have a conducting medium. And so current can flow through
that medium, this way or this way, and that's exactly what
will happen. So that you don't know
precisely the path. Current will flow,
so you do have closed loops. And so it is meaningful to talk
about magnetic flux change and about the EMF,
the induced EMF that is generated as a result of that.
The thing that NASA could not predict very well was the
current, because you do not quite know what the resistance
is of that closed loop. You know what the resistance is
of the cable, but you don't know what the
resistance is of the currents as they flow through the
ionosphere. But the net result was that
they had a current I in their wire
of about one ampere. It was very tragic,
because this current was so high that the conducting wire
melted, and the tether broke off, and very early on in that
experiment, the tether separated from the shuttle.
But this is a marvelous example of where you have motional EMF
in space. I want you to think about it at
home. Where does the energy come
from? Because you're generating
current. You could have lit a light
bulb. The energy must come from
somewhere. Think about that.
Conceptually interesting question.
By the way, I linked you on the website to the tether.
All you have to do is click on it and you will get some more
information on this incredible experiment.
All right. Let's continue with the most
important contribution, Faraday to our economy.
And that is the situation whereby we rotate a loop,
a current loop, conducting loop -- could be
rectangular, could be circular -- and we rotate it round with
angular frequency omega. And for simplicity,
let's have the magnetic field just straight up.
This is supposed to be three dimensional.
That's my idea. And let this side be length A
and this B. So if you look at it from this
direction, you will only see the line.
B here is rotating like this, with angular velocity omega.
And a little later in time would be here.
And so this length here is B. And so this length is B cosine
theta. And let's assume -- we call
this angle theta -- let's assume that it is here at
T equals zero when theta is zero.
I can choose that arbitrarily any way I want to,
of course. And there could be a resistance
R in this entire loop. And so the magnetic flux,
phi of B, through a surface attached to this loop is B --
that's constant, that's not changing -- and we
have A times B, but remember since it is a dot
product between B dot DA then you get a cosine theta in there,
so you get A times B times cosine theta.
So that's the flux. But cosine theta is changing
with time. Theta is omega T,
constant angular frequency omega, so the magnetic flux is A
B B cosine omega T. So D phi DT.
Take the derivative, I get an omega,
I get A, I get B, I get B, I get sine of omega T.
If you're really interested in that minus sign,
you get that automatically, there's nothing I can do about
it. You get a minus sign there,
but I'm not too much interested in minus signs.
But since I get it, I can't shove it under the rug.
So I put a minus sign here and I change that into a plus one.
The advantage is then that I have immediately the induced
EMF. Notice that the induced EMF
itself is linearly proportional with omega.
And so my induced current that is going to run in that loop is
the induced EMF divided by the resistance of that loop.
And so if the induced EMF is proportional to omega,
so is the induced current. So if you go faster around,
you get a higher current for the same resistor.
You built motors, and during the prize ceremony I
told you that the faster your motor rotates,
the larger an induced EMF is which is generated in the loop.
You had a magnetic field which let's assume is more or less
constant. Of course, it wasn't in your
case, but let's assume for the sake of argument.
And as your motor was going to rotate, there was an induced
current generated by Mr. Faraday, which is proportional
to omega, so the current is larger when your motor goes
faster. And that induced current
opposes the current which was produced by your battery.
And perhaps you remember that I did a demonstration with the
winning motor. I blocked the rotor and I
showed you that the current through the rotor when it was
blocked, when omega was zero, was one point six amperes.
Simply Ohm's Law. Battery, voltage V,
resistance R, the current is V divided by R.
Then we ran the motor, and the current went down
enormously, by a factor of forty.
It was only forty milliamperes, the time-averaged current,
when the motor was running. The reason is what you see
here. You get an induced EMF,
an induced current, which opposes the current from
the battery. L R circuits.
I prefer to stay on this center board, although I could go
there, but I don't think we need these
anymore. Can I cover this?
Most of you are happy with that.
I could work there, but I prefer to stay at the
center. L R circuits.
With L R circuits, we have the embarrassment --
not for us but for some others -- that almost all textbooks,
college physics, do not understand Faraday's
Law, and therefore they treat the subject incorrectly.
Embarrassing, but of course since they know
the answer, their answers are correct.
But their physics is totally wrong.
I will address a problem which I found on, on the website when
I looked at Professor Belcher's exams.
He has one very nice problem of an L R circuit.
And I will go through that with you.
We have an AC power supply, V equals V zero cosine omega T.
The frequency is sixty hertz. It just comes out of the wall.
And so omega, which is two pi F,
is about three hundred seventy-seven radians per
second. And V zero in this case,
hundred volts. And we have here a
self-inductor, and here we have a light bulb.
And the light bulb has a resistance which is hundred
ohms. And this self-inductor is
variable. We don't know yet how you make
a variable self-inductor, but we will learn that very
shortly, either Friday or next week, I forgot.
This self-inductance can be increased over a certain range.
And the first question that Professor Belcher is asking is,
what is the energy dissipation in this light bulb?
The closed loop integral of E dot DL is not zero,
because Kirchoff's loop rule does not apply here.
There is a self-inductor. If you attach an open surface
to this closed loop, there is a magnetic flux going
through there, so you must deal with the third
equation there. You must deal with Faraday's
Law, no matter what your textbooks tell you.
Kirchoff's loop rule does not hold.
And if you do that correctly, you end up with the right
differential equation, which happens to be the same
one that those people find who do not understand the physics,
but they get the same equation. They massage things in such a
way that they get the same answer.
And the I- answer then is that the current that is going to
flow as a result of this voltage, variable voltage,
that current I is a maximum value times the cosine omega T
minus phi. That the maximum current itself
is V zero divided by the s- square root of R squared plus
omega L squared. And that the tangent of phi is
omega L divided by R. I spent quite some time on this
during one of my lectures. And so the maximum current that
will flow depends on omega and
depends on L, on the self-inductance -- of
course on the resistance as well.
In Professor Belcher's problem, he starts off with L equals
zero, and he asks you what is the power dissipated in this
light bulb. If L equals zero,
this term is not there, and so you simply have Ohm's
Law, I equals V zero divided by R.
So I max is hundred divided by hundred, is one ampere.
But then he asks you something that would give you the hiccups,
and that is what is now the time-averaged power dissipated
in that light bulb? And you will remember,
or should remember, that one half I squared R is
the power dissipated in the light bulb if I is the current
through the light bulb and R is the resistance of the light
bulb. And we will assume that the
resistance is independent of the current, independent of
temperature. But I is changing with time,
in a cosinusoidal fashion. And so now you have to be able
to evaluate the time average of a cosine squared omega T
function. And the time average of a
cosine squared or sine squared omega T function is always one
half. You'd like to remember that,
rather than spending five minutes on deriving that.
And so this, time averaged,
is then going to be one half times I max squared times R --
and by the way, sorry, the -- I was,
I was ahead of myself with my, my -- I squared R is the energy
that is dissipated, right, in the,
in the resistor, not one half I squared R.
It is I squared R. And because of the time-average
of the cosine squared function I get my one half there.
And so it's very easy now. I max is one ampere,
R is one hundred, and this factor of one half,
which comes from the time average of the square of this
function -- the phi has nothing to do with that -- the average
of co- of the cosine squared of this function is one half,
and you get fifty watts. And now he calls this device a
light dimmer. Now he is going to increase L
to three hundred millihenries. So L now becomes three hundred
millihenry. At three hundred millihenry,
omega L is hundred and thirteen ohms.
I multiply three hundred by the three seven seven.
And now I find that this time-averaged value,
which is now one half times this, I find that I max -- I put
in for omega L hundred thirteen squared, hundred squared,
I max is now oh point six seven amperes.
I put that in this equation and I find that the power is now
about twenty-two watts. So that's what a light dimmer
is doing for you. So you turn up the
self-inductance, and now your light is dimmer.
Now a very deep, very deep, maybe nasty,
conceptual question. Why would one want to build a
light dimmer with a self-inductor?
Why not put in here a variable resistor and then turning up the
resistor so that the current, which, when the resistor is
zero, the current is one ampere, so you get you fifty watts,
but then you increase that resistor so that the current
goes down to oh point six seven amperes, and then the light bulb
will dissipate twenty-two joules per second.
Why would you not put here a variable resistor,
but why a variable self-inductor?
If you can answer that question, that shows that you
have a deep insight already in eight oh
two. I promise you I will not ask
this question Wednesday, but I may ask it on the final.
Displacement current. Displacement current is always
a little bit problematic in the sense that there are not too
many problems that you can do with displacement current.
This is here this displacement current term,
that Amp- that, uh, Maxwell added to Ampere's
Law. The only application at this
stage in eight oh two that I can think of
is the one that I hit very hard during my lectures.
You have plate capacitor disks. They are circular plates.
And you charge the capacitor or discharge the capacitor,
and you can calculate now, using this law,
what the magnetic field is in between the capacitor plates.
I advise you to check your notes from that lecture or watch
the lecture again on the web. The lecture is on the web.
Then some fatherly advice when it comes to the exam itself.
I would advise you to read a problem at least twice to make
sure that you fully understand the text.
If you read fast -- at least that happens to me -- you
sometimes misread. At least, I often misread.
I would then advise you to do the easiest problems first.
What may be the easiest for you may not be, be the easiest for
you. But do the easiest for you
first. If you get stuck on a problem,
never spend more than ten minutes on one problem.
Immediately abandon in then, when you see that it's grinding
yourself into a hole, and go to another problem.
Between now and Wednesday, I would suggest you see tutors
if you need them. You can see your instructor.
I have office hours this afternoon.
I'll make myself as much available today as I can.
Tomorrow afternoon I will be in twenty-six one hundred working
on the demonstrations for you for
next week, so I'm not available tomorrow.
I wish you luck, and I'll see you on Wednesday.