Prof: All right,
I'm going to begin by summarizing stuff done near the
end. Usually stuff done near the end
is maybe hurried, or also you may have forgotten
where we were, so it's helpful.
The first of what I did last
time is called Lenz and Faraday law,
and it says the following - in any circuit made with a real
conductor, if you integrate the electric
and magnetic force on a charge, unit charge around a closed
loop--this is called the electromotive force--
that = - the rate of change of flux through that loop.
And the flux is defined as
follows. The flux is just the integral
of the magnetic field over an area whose boundary is this
loop. So this loop is a boundary of
that, so if you have a loop,
and there is some flux coming out of it,
you integrate that, and you take the rate of
change. That covers a variety of
different phenomena. It's what I was trying to
explain last time. First of all,
if you ask, "Why does this thing change?
Why does the flux change?"
it can change for two reasons.
It can change because
B itself is changing with time.
Or even if B is not
changing with time, but changing with space,
if a loop happens to be moving, that can also change the flux
through it. For whatever reason,
it will change. So this total derivative means
the total rate of change of flux for the two reasons I mentioned,
or both. You can have a time dependent,
space dependent field in which a loop is dragged along.
Then at any instant,
electromotive force, namely the total force on the
charge, on the unit charge, is given by rate of change of
flux. Then I said we can write this
integral, this rate of change is made up of two parts.
One was the rate of change due
to explicit time dependence of the field on the old surface
another one that looks like v x
B⋅dl. This was what I did the last
time. And we got that result--I'll
remind you briefly, but I certainly don't want to
do it again, is that if you had a loop that
was doing this initially, it was doing that a little
later, the loop is moving, then if you want to find the
change in flux between that one and this one,
the natural thing to do is to integrate the flux on that loop
later, subtract from it the original
loop earlier. But calculationally,
it's easier to use the fact that if you want to find the
flux through a surfacing bounded by this loop,
you don't have to simply take the easiest loop you see.
You can take any loop with that
as a boundary. And we cleverly chose the
surface to be the one containing the original surface and the two
walls, and the circular walls,
if you like, that add onto it and produce a
new surface. So if you take one flat sheet
and glue the sides of the cylinder, you get a rim here,
and that's the final surface. The advantage of doing it is
that when you find the flux at a later time using this joint
surface, one contribution will come from
exactly the old surface, and the difference there is
only because of explicit time dependence.
The other will come from the
sides, and the sides, you can see,
are spanned by little rectangles.
This side of the rectangle is
v times dt. This side is dl.
And the dot product with the
flux will be that term. And you rearrange the product
you get this. Therefore now we can balance
that with that, or if you take a problem where
nothing is changing with time, then you can just forget this
term, and this term will match that
term. Then we are left with the
following result - the line integral of the electric field
on any loop = the surface integral of the rate of change
of B over that fixed loop.
So here is the fixed surface,
and here is the boundary of the fixed surface.
You have to understand the
difference. We started out with a statement
about a real physical loop. The physical loop was moving
and there is an emf driving charges around the loop.
It's got two parts.
One had to do with the motion
of the loop. The other had to do with the
changing of the field. This relation,
you see, has no need of a real loop to be present.
This one definitely needs a
real loop to be present, a real conducting loop,
because whose velocity is this? It's a velocity of each portion
of the loop that's moving. But this one is about a fixed
contour in space and has to do with the line integral of
E around the fixed contour with this one.
So it is true even if you
remove that loop. It's a basic statement about
electric and magnetic fields that tell you that if you have a
changing magnetic field, a time dependent magnetic
field, it will generate an electric field which,
unlike the electrostatic field, will have a line integral which
is not 0. That's the main point.
It's a non conservative
electric field, whose origin is not electric
charges, but really changing magnetic fields.
So this is called Faraday's law.
Faraday's law is a very
profound statement about electric and magnetic fields and
that's what's going to be more important.
But the original formula is
able to cover all situations where loops move in fields.
This - sign is due to Lenz and
the - sign is going to save your life.
The - sign tells you,
if you want to know when you drag a loop or when you change
the flux through it, what is it going to do,
the - sign tells you it's going to fight the change in flux.
Namely, if an emf generated,
which way will it point? Will it move charges this way
or that way? Answer is, it will move charges
in such a way that they produce their own current,
which will produce its own field.
That field will fight the
change that you're trying to produce.
In other words,
if you take a loop here and you've got a bar magnet sending
some lines of flux through it, you move the magnet closer to
the loop. That's an example of rate of
change of flux through the loop. Which way will the current flow?
The answer is very simple.
You're trying to increase the
flux through the loop. It will fight it by decreasing
the flux going upwards, so it will try to produce a
flux going downwards. Therefore it will have a
current that looks like this. If you have a current that
looks like this, let me see, you do that and the
thumb points down, and therefore the magnetic
field it produces will oppose the field.
So this rule is very important.
That's what distinguishes human
being from primates. The primates cannot do this
rule. In fact, there are lots of cave
drawings of apes trying to invent solenoids,
and they always kept saying this, and they thought the
magnetic field was parallel to the coil.
So one day, we figured this out.
That's the beginning of--it's
even better than fire, finding out that right hand
rule. Okay, so this is the rule you
should use. So whenever I do any
calculation of emf, I'm not going to worry about
the sign. In the end, we'll fix the sign
so it makes sense. That's going to be our formula.
All right, so now let's
continue with this one, and I remind you how we'd
explained everything regarding the other experiment.
Remember we had a magnetic
field going into the board, only up to some point.
To the right of this point,
it's everywhere uniform. Then you had a loop which we
were dragging to the right. And I put a light bulb here,
and I said if you drag it, the light bulb will glow.
Now we can understand because
there's an emf, and the emf is coming because
in this example, the term that really matters
for the emf is the v x B term.
There is no dB/dt.
It's fixed.
Somebody's holding a magnet,
shoving flux into the blackboard.
You're carrying the loop.
So it's the v x B
that gives you the emf. That's fine.
That's the part we're not very
impressed with in this law. And let's look at the - sign.
The - sign will be telling you
which way the current will flow. You are increasing the flux
going into the blackboard, therefore the current that is
generated will try to put flux coming out of the blackboard.
And that means the current will
flow that way. And that agrees with what you
expect from fundamental principles,
because if the wire is moving to the right,
B is into the board, v x B will
produce an emf this way. So we understand everything
here. I also explained to you that
the energy balance is that the work done on the wire,
on the light bulb, is paid for by the person
pulling this loop, because the minute you have a
current, that current doesn't like to be
dragged across a field. There is some BlI force
and when you overcome the force with your mechanical force,
you convert mechanical energy to electrical energy.
There's one subtlety that is
usually overlooked, which is the following - if you
go to this wire, here is the piece of wire.
There are some charges in here
the wire has moving at a velocity v.
And there is a B
somewhere, maybe like that. And we said v x B
is a certain force. But if there's a current
flowing in the wire, it also has a speed u along the
wire. This is something we saw even
in the loop problem. So the real force of a charge
on the wire is not simply v x B,
but v u x B⋅dl.
Because it's not just the
motion of the wire that we take for the velocity,
because charges are moving in the wire.
The net velocity is v
u, to some number like that.
But when you do the emf,
you don't have to worry about this part of the velocity along
the wire. I'll give you a second to
figure out why. But just take that part.
Why am I allowed to ignore it?
Because u x B is
perpendicular to u, and dl is parallel to
u. u is the direction of
motion in the wire and dl is along the wire.
So if you take a vector,
cross it with something else and take the dot product with
another vector parallel to itself, you will get 0.
That's why you don't worry
about the extra term in computing the emf.
This is a subtle point for
those of you who wake up with a sweat in the middle of the
night. I'm just trying to tell you,
calm down, it's okay. It's really present as a force,
but not as an emf. In fact, it's going to push the
charge perpendicular to the wire and not do any work going around
the loop. So this is one topic.
I'm going to illustrate the
reality of this result by describing to you the operation
of what's called a betatron. You remember the cyclotron,
what it does, right?
You had these Ds and the charge
was jumping around from one D to the other when you put a certain
voltage, so this is positive and that's negative.
It bends in a magnetic field
which is going in the plane into the blackboard.
Then it gets another kick and
when it comes here, you reverse the polarity,
so that this becomes and this becomes -.
It picks up another kick in
velocity and it keeps doing that.
And the remarkable thing that
was noticed is if you write the equation mv^(2)/R,
which is the force you need to bend into a circle,
that's going to be qvB. So if you cancel the velocity
here, you find v/R to be qB/m.
v/R is the angular
frequency with which it rotates and that's independent of the
radius. Therefore you have to reverse
the polarity with the same regularity, with the same
frequency, even as the particle picks up more and more speed.
It's not that you have to keep
changing the rate at which you flip the voltage.
Then you don't have any means
of doing it. But if it's flipping at a
definite rate, then of course you can get a
generator that generates voltage at that rate and you can make it
work. The key to all of that was this
remarkable fact that omega does not depend on the radius.
But that has a weakness that as
the particle picks up speed, eventually you will have to use
the fact that the real momentum of a particle is not mv,
but mv divided by this factor,
which comes from relativistic effects.
Then you can show the
centrifugal force you need. The centripetal force you need
is not given by mv/R,
but rather it's given by this momentum times omega,
where momentum is given by this formula and not by this.
Now that's a homework problem
where you'll get enough time to think about it.
That will show you that once
the momentum formula deviates appreciably from p = mv
to p equal to this, then condition for the
cyclotron orbit, the frequency being independent
of radius will fail. Essentially,
as the particle gets more velocity, it's harder to push
it. Your ability to accelerate it
changes and therefore this condition fails.
So the cyclotron can accelerate
particles only up to velocities where the relativistic
corrections are unimportant. But the betatron,
which I'm going to describe to you,
is a device that actually manages to accelerate particles
even as their motion becomes relativistic,
even as the momentum is given by that formula.
So I will tell you how that
works. You've got these poles of some
magnet, the side view, producing a magnetic field.
Let's look at the top view.
In the top view,
you've got some magnetic field. Let's say it's coming out of
the blackboard. So if you put a particle here,
some velocity v, v x B is to the
right and it will bend. But here's what we want to do.
We're going to have it go in a
circle of definite radius and yet pick up speed.
I'll show you how that happens.
So what you do is,
this is not a fixed bar magnet. It's an electromagnet,
the current in which you're changing, and therefore the
field is in fact time dependent. Furthermore,
the field will be strong near the middle and weak near the
edges. Now if that field is changing,
then if you draw any loop of radius r,
the flux through that loop is changing.
Therefore the electric field
will obey this condition, 2Πr times E.
The electric field will then
also be going around in circles. It will encircle the magnetic
field. If the magnetic field is coming
from ceiling to the floor, electric field will be
horizontal, horizontal circles. And by symmetry,
they'll be circles centered at the center of the magnet.
That's the direction of the
electric field. This field coming from top to
bottom, if it changes in strength, will generate a field
in a horizontal plane that's going round in circles.
So if you put a charge there,
you'll push it and it will speed up.
How big is the field?
Well, the field is constant on
the circle of radius r. It's pointing azimuthally.
So its line integral is just
2Πr times E at r.
That's going to be = (forget
the - sign) d/dt of B⋅dA
inside that circle. Now that
B⋅dA I'm going to write as
Πr^(2) ties some average B.
That's how we define the
average. The total flux through the
radius when the field is varying with distance can always be
written as the area times the average B and the time
derivative of that is, if you like,
d/dt of B_bar.
So I'm going to write now the
equation I get. E of r,
if you solve for it, is going to be r/2 times
d/dt of the average B.
Average is inside the circle of
radius r. Now if that's the electric
field you multiply both sides by q, which is the charge of
the object, that's the force. That force is the rate of
change of momentum. So rate of change of momentum
is qr/2 times rate of change of the average field.
Now if this is the rate of
change and that's equal to the rate of change of B,
then p of t = qr/2 times the average
B at time t. Actually, I'm making one
assumption here. Do you know what that is?
I'm assuming the initial
momentum was 0, because if you know only the
rate of change, you can add a constant to it.
So I'm assuming that t =
0, p was 0. So that's the momentum of this
guy after time t. So as the average B is
growing, the momentum is growing, it has this value at
this instant at that radius. Now what else do we need to
make sure that solution makes sense.
What do you need?
Is there anything else to this
story? Yes?
Would you like to guess?
If a particle picks up speed,
why would it continue to be in that orbit that I've shown here.
What does it take to keep it in
that orbit. Pardon me?
Student: The force of
that entity. Prof: Which force,
I'm sorry? Student: If it's in an
orbit of constant radius, then the velocity's increasing
the force, and that's the increase.
Prof: Right,
but at a given velocity. Even at a given velocity,
you need a force to bend something into a circle.
You guys know that.
Remember that force from last
term? That force is mv^(2)
over r. That comes from saying that the
particle has a velocity momentum like that now,
momentum like that a little later.
There's a change in momentum
pointing towards the center and you can calculate that as
mv^(2) over r. So who's going to provide that
force? In other words,
things don't move in a circle unless you pull them into the
center. My question is,
who's going to do that for this guy?
Who's going to provide that
force? Any idea?
Don't worry about the sign or
anything. Is there anything that can push
this guy towards the center? Want to make a guess?
Student: The magnetic
field? Prof: It's the magnetic
force, because that's a charged
particle, moving in a magnetic field,
and the force of that is q times v times
B. But this B is the
B at the orbital radius. It's not the average B.
This is the B where the
particle really is. So we're going to keep the
particle at a fixed radius. It's the B at that
radius that matters. Therefore you cancel the
velocity, you find that mv = qrB_0 and that's
the momentum. But we also saw the momentum at
time t = this. They are both expressions for
the momentum at time t. This just says the magnetic
field hopefully is strong enough to bend it.
That's the field that you need,
but I'm just going to equate the momentum computed two ways.
That tells me this interesting
result, that B_0 is ½
B_average. So let's summarize what I'm
saying. I want you to visualize this.
You've got a magnetic field
from ceiling to floor. It's pointing down, let's say.
You put a little charge there,
it won't do anything. But if you change that field,
because of this law, Faraday's law,
the change in field will produce a circulating electric
field, that circles the changing
magnetic flux and going around in circles,
and it will accelerate the particle along a circle,
because it is bending around a circle.
But at the same time,
you need the right magnetic field to keep it in that circle.
But the particle is speeding up.
Therefore the force needed is
also increasing. So you're in addition to
increasing the average field over the loop over the circle,
to produce the emf, you need to have the right
field at the boundary to provide exactly the right force needed
to keep it on an orbit of that radius at that time.
Therefore the story is,
as time passes, the field increases in
strength, the particle's tangential momentum increases.
And the radial force to keep it
in orbit at the tangential momentum also increases,
and the demand when you look at it says that the field at the
periphery, where it's rotating,
must be ½ the average field.
So you'll have to build a
magnet very carefully. You can always build a magnet
whose field varies as you go off center that gets weak when you
go off the center. It should go off the center in
such a way that by the time you come to this radius,
the field strength there is half the average.
Once you design a magnet that
way you just crank up the current and let a particle go at
the radius, it will pick up more and more and more speed.
Even though it's going faster
and faster, the field at the orbit will be just right to push
it towards the center with the right amount.
One of the homework problems
was to show something I made a big deal about,
namely, here I've used relativistic kinematics.
Momentum was mv,
the force to the center is mv^(2)/r.
But the homework problem shows
you that even in the relativistic case,
the momentum is mv divided by all of that,
this is still true. So the betatron,
in spite of the simple example I've given here,
actually works, even if the particle is
relativistic. In fact, the only reason the
betatron eventually fails is that when particles start going
in a circle at very, very high speed and they are
charged, they start radiating energy.
That's another aspect that we
have not discussed in our course yet,
but accelerating charges radiate energy,
and eventually, you cannot put in enough
energy. It radiates more than you can
give it. So then you need other things
called synchrotrons. So that's a constant struggle
for people building accelerators.
It's easy to get more and more
velocity, but if you bend them into a circle,
they start radiating, emit gamma rays,
emit light. And that energy eventually is
so big that you cannot compensate it with your pushing
force. So the next topic I want to
discuss is in a totally different vein,
and that has to do with more practical issues like this one.
This is going to be a power
generator. Remember, I told you one way to
make electricity in your house is to take this coil and tell
somebody to carry it and run. Then the light bulb will glow
here. Or you can have somebody carry
the magnet the other way and the light bulb will glow in your
house. But there's a more clever way
to make the light bulb glow, which I think you guys have
seen. By the way, I think these are
all things you have seen in one form or another,
so I'm not going to put too much energy into these things.
I just want to tell you things
you may have missed. So here's some magnet.
Here's some field lines going
from here to there, north to south.
In that magnet you put a coil,
which looks like this. The plane of the coil that's
pointing in that direction of the magnetic moment,
or if you want, the area vector points that
way, the B field is horizontal.
Now if you spin that coil,
I think you know what will happen.
The flux through the coil is
going to change and the rate of change of flux will give you an
emf. So what is the emf?
First you've got to find the
flux through the coil. The flux through the coil is
the area of the coil, the magnetic field,
times cosine of the angle between them.
This would be angle theta in
the figure. But you have attached this to
an axle and you're going to just keep spinning it mechanically.
And you agree to spin it at a
uniform rate, omega.
So theta is omega t.
Then you can see that the emf,
which is -dΦ/dt
= ABsinωt times another ω.
That's the emf.
Now what about the - sign.
Forget the - sign.
We can figure out what the
emf--which way the current will try to flow.
Look at this coil here.
Let's decide how we want to
turn this guy. If you turn it in the manner
I've shown here, it's going from some angle like
that, eventually to an angle perpendicular to the magnetic
field. What's happening to the flux
through it? Is it increasing or decreasing?
Increasing.
Increasing in this direction.
So it will fight it by trying
to produce a flux going in the opposite direction.
Therefore it will try to
produce a current that looks like this.
That's the direction of your
emf. Emf, given a chance,
will drive a current as shown here, because that one by right
hand rule will have the flux going the opposite way.
So that's why I don't care
about the - sign. I know which way the current
will flow if I let it flow. But now if this is an open
circuit like this, the two wires sit here,
what do you think will happen? What do you think will happen
in that case? If you're an electric charge in
that wire, what will you do under the emf?
You will follow the emf and you
will run from this terminal to that terminal.
So let me blow up that picture
for you. If the current is trying to go
that way, then charges will leave like this and pile up
here. They cannot go very far because
it's an open circuit. At some point,
these guys piled up here--so this is the induced electric
field obeying that equation. In fact, it's not induced
electric field. This is just the v x
B force on the magnet, on the charges.
But then these charges will
produce and electrostatic force called the Coulomb field,
which is the electric field due to charges,
which will oppose it, till the two cancel inside the
wire. In other words,
this wire is assumed to be a perfect conductor.
You all know that you cannot
have an electric field in a perfect conductor.
You can say,
"Hey, why do you have an electric field now?"
The real statement is,
in a perfect conductor, you cannot have any net charge
on the charged particles, because then they will pick up
infinite speed. So if there's a v x
B force, that is actually canceled by an
electrostatic force. The two of them cancel inside
the wire. And the line integral of the
v x B, which is the emf,
will numerically equal the integral of the electric field
from this terminal to that terminal.
But now if you put the whole
thing in a box and you don't know anything and you come
outside, what you will find is this will
be positively charged, this will be negatively charged.
There'll be voltage between
those two which will numerically equal your emf.
This will be the polarity.
So I'm giving you an option.
If you don't want to look under
the hood, you simply say whenever you rotate a coil,
you get an emf. That emf is like a voltage.
Case closed.
But if you really want to know
what's happening inside the wire that made up that coil,
inside the wire the net force on the charges is actually 0,
or an infinitesimal amount left to make them move this way.
That's because the v x
B force is countered by an internal electrostatic force
due to pile up of charges that opposes it so that inside it's
free. It's just like the ski
experiment I told you, where you have a ski lift.
You come down the ski lift,
you come here, and here is the lift that
carries you up. But during the time gravity's
acting down and the force of the lift is exactly equal to mg and
gravity. And once you're outside the
lift, it's gravity that brings you down back here.
Similarly, inside the thing the
two guys are opposed to each other.
Outside this region there is no
v x B force, but the electrostatic force has
a line integral that's independent of the path.
So integral that way the same
as the integral that way. So if you put a circuit here,
it will produce a voltage difference with this being
positive, that being negative. So the bottom line is that if
you have open circuit, and if you put a volt meter
that measures voltage, you will get exactly that time
dependent voltage. So this voltage is not constant.
It will look like this,
t versus V. So in the linear loop,
when you drag it and run, you get a constant voltage.
Here you get a time dependent
one, and most of the supplies in the world are either 50 cycles
or 60 cycles, and they come from rotating a
coil in this magnetic field. Now how much is the work done
by the person rotating this coil right now?
I'm telling you the coil is
made of massless, perfectly conducting wire.
What work do you have to do to
spin it? Any ideas?
Student:
> Prof: Which one,
dE? Student: Edq.
Prof: Yeah,
but I'm saying the total force--you don't have to do any
work because there is no current flowing in the coil yet.
There's no current.
If there's no current,
it's BlI force. So what we want to do then is
do something more interesting where you bring it here,
connect it to a resistor. Then the current will flow.
Now we're getting something.
Till now, we were getting
nothing from the generator. In other words,
look, where's a socket here? That's a socket.
There's voltage there waiting
for you to use. But you don't pay,
because you're not drawing any current.
You don't pay just because
someone gave you the voltage. You take a battery in your
hand, you don't run a current, you don't pay anything.
That's what the situation is.
But if I stick my fingers into
that socket, then there is current.
Then we'll all pay.
The loss is I^(2)R,
right? That's when you've got to
explain to yourself, who is paying for this?
Because we can see that the
power in the resistor will be I^(2)R.
I, if you want,
is E^(2)/R, emf or voltage squared over
R. That gives me
ω^(2)A^(2)B^(2) sin^(2)ωt/R.
That's the rate at which this
power is consumed by the resistor.
Someone's got to pay for that.
The someone,
I think you can imagine now, now that I've closed the
circuit and a current is actually flowing,
there is a torque on that loop in a magnetic field.
The torque on any loop,
you'll remember, is μ x B.
μ is the magnetic
moment. That happens to be area of the
loop, current in the loop, B times the sine of the
angle between area and B. That is torque.
So that's the torque that
you've got to fight. So you will have to apply
mechanical force to turn that. If you're turning the crank by
hand, the minute you put a load, you'll find it's hard to turn
it. And the power,
just like power is force times velocity, for rotational motion
it is torque times angle of velocity.
That gives me ABωI
sinθ. That's the power,
the mechanical power, P_m.
But then that =
ABω sinθ times
I. What is I?
The current in the loop is the
emf divided by R. If you put the emf I got for
you somewhere, then you'll find it's equal to
A^(2)B^(2)ω^(2)/R sin^(2)θ.
So this is how you have to
balance. We did a similar calculation
here. This is also a generator and if
you find out the minute there's a current flowing here,
you have I^(2)R or E^(2)/R energy loss here.
But the minute the current
flows through the loop, you will not be able to pull it
to the right without paying the force.
That force times the velocity
will be the power and the two will balance.
Similarly, here is a rotational
problem, and the torque times angular velocity will exactly
equal the power here. So the generators,
you've got hydroelectric generators,
or you've got steam, turbines that are spinning,
initially, because the turbines in the real world have a real
mass, it's not easy to spin them.
You spend some power,
but if you ignore that, the minute you put a device in
your house into the socket, your toaster,
you start drawing current, that current's got to flow
right through the generator. It's going to make it that much
harder to rotate the generator. That's when the steam turbines
do their work. That's when they pay for it.
Okay, so this is the end of the
story about how to get power out of this and how it balances.
I'm going to a fairly different
notion called inductors and inductance.
So here is the following
phenomenon. I take some cardboard tube and
around the cardboard tube I wrap some wire and I connect this to
some alternating voltage. That means there is going to be
some magnetic flux going through that coil.
Then I bring a second wire,
maybe I wrap it round a couple of times or a few times,
and I leave it there. That's it.
The question is,
what will happen if I look at the two ends of this wire?
You can see what will happen.
If this current changes,
the flux through this solenoid changes.
That means the flux through
this little guy alsochanges. Every loop has flux going
through it and that also changes.
So there's going to be an emf.
For every turn of wire here,
they'll be an emf equal to the rate of change of flux.
And therefore that emf will try
to draw a current, just like before,
and the charges will pile up maybe like this at some instant.
Now if you are outside all of
this, you will just think there is a voltage available to you.
But I want you to understand
the origin of that voltage. This is like the ski lift here.
Charges, once they come into
this region, are pushed up the wire and later on,
if you're connected to a load, they can drive a current.
This is just like the generator
I had there, except the flux is changing not
due to any mechanical motion of a coil in a fixed field,
but it's fixed coils in a changing magnetic field.
So you understand,
when I change the current here, you will get a voltage here.
This is the first thing Faraday
noticed, is that he thought first that magnetic field may
produce a current. It didn't.
Then he found a changing
magnetic field is able to produce a current and this is
why this happens. Once again, I want you to think
about the following question - we are going to take such things
and put them in electrical circuits and so on.
We'll draw all kinds of
circuits later on. And what I will do in all those
calculations, I will say I start here,
I go around a loop and I come back and the change in voltage
should be 0. I'm going to use that principle.
But there is one flaw if you
don't think about it that says maybe I shouldn't do that.
You remember that you can
define a voltage only for a conservative force,
whereas you definitely have non conservative forces that work in
this problem. That's why they have a line
integral not equal to 0. And yet we apply the laws of
conservative forces, or a notion of a voltage in a
circuit. I want to explain why that's
allowed. If it never bothered you,
you can ignore this part. But it's important to
understand how we can have a notion of a voltage in the
presence of time dependent magnetic fields.
The line integral of E
is not 0. Once again, what happens is,
if you look at this coil, there may be an electric field
that is induced now due to the changing flux of that.
That will try to build up
charges here and take them out of this end.
After a while,
these guys will say "Enough"
and they'll start fighting you, till they set up an
electrostatic field, a Coulomb field,
inside the wire that cancels this.
You understand,
the integral of this electric field from top to bottom will
numerically equal the integral of the induced electric field
from bottom to top, because at every point they're
equal in magnitude. Because you cannot have a
non-zero field inside a wire. So the two fields have to be
canceled. But this electrostatic field,
built by these charges, is a conservative field,
so if it does any work going this way,
it will also do work going that way.
Therefore if you don't open the
black box, you don't know what's inside,
if you just took the two wires coming out of it,
the static field coming from this will be able to do work
going from to - terminal. And if you put a resistor on
its path, it will deliver some energy to you.
The trick is,
yes, there are changing magnetic fields,
but they're hidden inside the coil.
In the region outside the coil,
we don't have to worry about the changing magnetic field,
where an electric potential can be defined as the integral of
the electric field. The electric field outside is
entirely electrostatic. Electric field inside is a
combination of electrostatic and induced one, which cancel.
Once current begins to flow,
you may worry that these charges will go away.
There'll always be some
electric charges making sure that the field inside the coil
continues to be 0. Again, there's one abuse of
terminology here, because I'm saying
electrostatic, and yet this is a problem where
the electric field you need is actually changing with time,
because the rate of change of flux is changing with time,
induced electric field is changing with time.
The compensating electrostatic
field is also changing with time.
So you really should not use
the laws of Coulomb for this problem,
but it turns out, even though Coulomb's law is
valid only strictly for fixed electric charges,
as long as the velocities required is not too big,
you can continue to use Coulomb's notion of an
electrostatic force. Basic question is,
you're telling the charges to constantly rearrange back and
forth. First you want this terminal to
be . A short time later,
you may want that to be . How quickly can they rearrange?
So that's connected to
something called plasma frequency of these materials.
And unless the frequencies are
like a trillion, you don't have to worry about
it. But any normal problem,
like 50 cycles per second, the charges will be able to
keep up with this changing field.
In other words,
can the charges go to the edge of a conductor and screen any
internal field you're trying to produce?
If it's a static field,
they will kill it, because you're giving them all
the time in the world. But if you keep changing your
mind, one minute the external field goes this way,
so the charges in the metal go that way to kill it.
Suddenly you reverse it,they've
got to go this way. So how quickly can they do this
dance? There's a limit to how quickly
they can do it. But that's a very,
very high frequency, so you don't have to worry
about that. Now we saw that the emf in the
second coil = the flux in the second coil divided by
d/dt. Now you've got to be a little
careful about flux. Normally, the flux due to the
magnetic field is defined as the integral of
B⋅dA. But if you've got a coil made
up of two loops, for example,
and there's a magnetic flux going through them,
then the emf is not simply the rate of change of the flux
through one of them, but you've got to double it,
because it's like two batteries in series.
An emf is running round and
round and round, so really, this is
N_2 times the rate of change of the magnetic
flux to that. So this Φ
is not simply the magnetic flux.
It's the magnetic flux times
the number of turns the second coil.
That's what the real emf is.
A real emf is really rate of
change of the flux linking with your whole circuit,
dt. That equals number of turns
times the actual magnetic flux dt.
Do you follow that?
Emf around a single loop is the
rate of change of flux, but if your coil loops around
twice, the end-to-end voltage will be double that.
If I loops around three
times, it will be triple that,
assuming the same magnetic flux is going through all of them,
which is the assumption I made here.
So now you can see here that
the flux in the second coil is due to the current in the first
coil and the coefficient of proportionality is called the
mutual inductance. So mutual inductance is how
much flux you can get in the second coil per unit current in
the first coil. Then you can write the emf in
the second coil as −M_12
dI_1/dt. M_12 is called
the mutual inductance of the first coil with respect to the
second one. It's true, but very hard to
prove that M_12 is the same as
M_21. Not at all obvious that if you
drove a current in the first coil,
it's going to have a magnetic field that's threading the
second coil and you can show that the flux per current in the
first coil due to the current in the second,
it's given by the same number. M_12--yes?
Student: How did the
number of turns go away? Prof: Where did it go
away? In this one you mean?
That's all included in the
definition of M. I'm going to now calculate
M so you will see the number of turns coming in.
So let's do the calculation of
M for a simple problem. Here's one solenoid.
It's got N_1
turns. And there's another guy.
I'm just going to show you one
turn of it, but it can have N_2 turns.
And the flux is going through
this. Now you remember that the
magnetic field due to any solenoid is
μ_0 times n,
where n is the number of turns per unit length times
I. Therefore the magnetic field in
this solenoid = μ_0
n_2 I_2--I'm sorry
about that one. This is the first wire of the
first coil. It's n_1I
_1. And the flux of the magnetic
field is μ_0n _1I_1
times the cross sectional area of the coil.
Now the flux linking with the
second one = μ_0n
_1I_1-- sorry, n_1AN
_2 times I_1.
You see that?
So the first coil has some
wrapping density of wires, n_1 turns per
unit length. I've shown you long back from
Ampere's law, the magnetic field that
produced this has that flux, that B value.
The integral of the field,
which is the magnetic flux, is just area times that.
But the linking with the second
coil is that flux times the number of turns in the second
coil. That by definition = the mutual
inductance M_21I_1
(I'm not going to call it 21 or 12.
It's independent of the order)
= μ_0n _1N_2A.
That's the mutual inductance.
So if I give you two closed
loops and I say, "Find the mutual
inductance," here's what you're supposed to
do. Drive a current in the first
one and produce some flux lines, magnetic lines.
Some of them will penetrate the
second one. You count how many penetrate
the second one, multiplied by the number of
turns, if they exist in the second one, divide by the
current producing it. That's the mutual inductance.
So inductance is measured in
henries, another thing for you. And usually you may find
millihenries or microhenries for common use.
Notice a very interesting
result. The flux is going through both
coils, you understand that? The same flux is going through
both coils. Maybe you're happier if I drew
the coil like this. Here's a doughnut coil, right?
You bring the wire here.
I told you how you can wrap it
around many times. Then the other coil,
the secondary coil, is also wrapped around the same
doughnut. It must be clear to you that
the magnetic field is going through all these coils.
The emf on the first one is
proportional to the number of turns on the first one times the
rate of change of the magnetic field.
The emf on the second one is
N_2 times dΦ
_B/dt. Therefore
E_1/E_2 = N_1/N
_2. The same flux is going around
the doughnut. One guy has N_1
turns around it, other has N_2
turns around it. Emf, you remember,
is not simply rate of change of magnetic flux..
It's that multiplied by the
number of turns. They both have the same flux
going through them, but this has
N_1 turns, this has N_2
turns. You can see this ratio.
That's a very powerful result.
You know what this gadget is
called? It's a transformer,
because you put in one voltage and you get another voltage.
You can have a step up
transformer or a step down transformer.
If you drive a current from
here, and you pick it up here, that's a step down transformer,
because you're going down in the number of turns.
If you connect your power
supply to this one and you pull it out of that one,
it's a step up transformer. So you can step up or step
down, but you can only do it for AC.
You cannot do it for DC,
at least, not in any simple way.
You need the changing thing to
do a transformer. But I don't have the time or
techniques to convince you, in spite of the ratio of emfs,
you don't gain or lose energy by transformers.
In other words,
this is not a device for manufacturing energy.
You will find out that if you
connect a load here and it's drawing some power,
the same power has to be provided by the source.
So it's not a way to
manufacture energy. It's like a lever.
You have a long thing and
you're trying to lift some object here, and you're trying
to life with--I got it backwards.
So this is a huge object,
a tiny object, you can balance them.
By varying the distance,
you can have a tiny guy lifting a big one in the inverse ratio
of the distances. But you don't get any free
mileage out of this, because if you look at the work
done by similar triangles, this will have to move a lot
that will have to move a little. So the work done is the same,
but doesn't mean it's useless, because this is the only way
you can lift something very heavy.
Likewise in a transformer,
you may not have the ability to give 5,000 volts,
starting with 200 volts, but you can if you use this
transformer. Quite often you want to step it
down. In all the gadgets you use in
your house, you start with 110 volts, you want to step it down
to some number, so you use a step down
transformer. All right, so that's the stuff
on inductance. Now we are going to come--this
part is really a curiosity. I'm not going to use it very
much, the notion of mutual inductance.
Mutual inductance is one coil
trying to generate a voltage in a second coil,
because they share a flux. And when it's changing one of
them, it's also changing the other one.
They don't have to be really
coaxial. You can put a second loop way
over here, and maybe some other extra flux coming out is
penetrating this. That's the mutual inductance
between this guy and that guy also.
Any time the flux of one coil
can go through another one, you have a mutual inductance,
because if you change the current in the first coil,
you're going to generate a voltage in the second coil.
That's why you need to know
that proportionality. All right, so now we are going
to do the most important circuit element, which is an inductor,
and it looks like this. That's a coil of wire.
Say some current is coming in
like this. It's wrapped around some
solenoid, in the form of a solenoid.
This wire is a perfect
conductor, therefore it takes no voltage at all to drive a
current through it. You put a battery there and it
just burns immediately. But if you put a time dependent
current you will need a voltage, because a time dependent
current will produce a time dependent magnetic flux through
this. So let us say the current was
originally 0. You're trying to increase it
and produce a magnetic flux here.
Then an emf would be generated
that opposes it. And we can ask,
how much is the emf? The emf is the rate of change
of flux to that coil, and I'm going to assume that
the flux through that coil = the current in the very same coil
time a number called self inductance.
The self inductance is how much
flux do you produce by a current going through yourself?
Not on another coil,
on yourself. Every coil, when it carries
current, will have some flux threading through itself.
So that ratio is called the
self inductance, also measured in henries.
So this becomes
−LdI/dt. You will calculate L in
a moment, but I'm just telling you that
as long as dI/dt is not 0,
you will have to oppose that back emf with a voltage from
some other supply. So in this example,
the current is going like this. Let us say it's trying to
increase. If it is trying to increase,
then the back emf will be set up, an electro emf will be set
up to fight it. It will try to push charges
that way, from this terminal to that terminal.
It will pile up charges there.
But if you don't look under the
hood and just went outside, you'll have charges and -
charges and an electrostatic field here will be able to push
them like this, maybe through a circuit.
So it's the same story again
and again. There is no net field inside
the coil. The electromotive force is
canceled by a Coulomb force. But the Coulomb force,
if it has a line integral here, will have the same line
integral there, because it's independent of the
path. So if you don't look inside,
you will just find there is some field which has lines from
here to here. That's the voltage.
But it will be a voltage drop,
so that if you want another convention for the voltage,
it's like this. If the current is going this
way and it's increasing in this direction, this will be and that
will be - at the instant. Just like a resistor,
it flows downhill, this will also flow downhill,
meaning this is higher than this,
provided the current is trying to increase in the direction of
the current. So we can have the following
very simple circuit. We have a battery here.
You've got a resistor there and
we have an inductor there. This is some voltage V,
this is R, this is some L henries.
So let us write a circuit
equation and here is where I spend an inordinate amount of
time justifying what I'm about to do.
I say start here.
There's going to be a voltage
defined everywhere except inside the black box.
You understand?
You cannot define a voltage
inside the black box where there's an inductor,
because there is a non-conservative electric field
inside. But we promise not to go there.
Then from here to here,
you go up by V_0.
Then here, current flows
downhill, so you drop RI and here,
if this is the sense of the current and it's increasing,
your loop will go from there to there and it will jump this and
come to this end. The drop from here to here is
LdI/dt, and the whole thing should add
up to 0. So here's a one word summary
for those of you who have heard enough - we learned the notion
of voltage can be defined, or a potential,
only in a conservative problem. But a changing magnetic field
inductor definitely is producing a field which is not
conservative. So if you go deep into the
coil, you will have problems defining voltage.
But if you come outside the
coil, I've tried to show you over and over again,
all it looks like is there is a voltage difference between the
two plates, the two terminals,
and that value is LdI/dt.
So when you do your circuit
equation, you go from there,
all the way back here, you skip these funny elements,
and around them you still have the notion of a voltage.
So this is the equation to
solve at any time for an LR circuit.
Now prior to this,
let me give you a very interesting result,
which we will use a lot. If you take an inductor which
had current I = 0 and you managed to drive a current
through it and slowly built up the current,
it's got some other value I at the end,
what was the work done to do that?
I will now show you that when
you drive a current through an inductor,
you are doing some work, because when you start driving
the current, it's opposing you with a
voltage LdI/dt, and you're ramming it down
that, up in spite of that opposition,
so that the work you do, the power,
is I times LdI/dt,
which is, if you like the rate of change
of energy. But I times LdI/dt is
d/dt of ½ IL squared.
I'm sorry, LI squared.
You see that from the rules of
calculus, the derivative of this guy is LI times
dI/dt. Therefore the integral of the
power is simply ½LI^(2),
assuming you started at 0 current, started at I =
0. So it takes some energy to
build up a current in the inductor.
That's the point.
Just like it takes some energy
to charge up a capacitor. I showed you when you charge up
a capacitor, here's a capacitor. It's got some charges,
, -, you want to charge it even more, you're going to ram more
positive charges and more negative charges here.
You fight it harder and the
total work done is q^(2)/2C.
Similarly, when you have an
inductor and you're trying to increase the current through it
from 0 to some final value, this is the amount of work done
by you. That energy is stored in the
inductor. I want to look a little bit
about inductors, but first let's calculate
L. So let's calculate L for
a simple solenoid. Here is my solenoid.
Remember, L is defined
as the flux linkage divided by the current.
It's going to be a one line
calculation so it's going to be very easy.
The magnetic field =
μ_0 little nI.
Little n you remember,
always is the number of turns per unit length.
The flux of the magnetic field
is μ_0 nI times the area of cross section.
That's the total magnetic flux.
But the flux linkage is
μ_0 little n IA times big N,
because every loop of the coil links with its own flux.
But that is by definition
LI. So you can see that L = μ_0
little n big NA. That's the self inductance of
the coil. What does this mean?
If this = 5 henries,
it means that if you shove 1 ampere through this guy,
a flux equal to 5 tesla squared meters will be linked to that
circuit. So the thing I want to do now
is to equate this energy, ½ LI^(2),
to the magnetic field inside the coil.
So you've got 1 over 2.
L is μ_0
little n, big NAI^(2).
So let me rewrite that as
½ μ_0n^(2)
times lA times I^(2).
In other words,
I've written, using this formula to write
little n as big N over l.
times A,
is the volume in which there's a magnetic field.
So this looks like 1 over
2μ_0 times μ_0nI whole
squared, times l times A.
But who is
μ_0nI. μ_0nI is
the magnetic field. So this looks like
B^(2)/2μ_0 times the volume of the
solenoid. From that we learn that when
you have a magnetic field, there's an energy of
B^(2)/2μ_0 per volume.
So the energy density of the
magnetic field = B^(2)/2μ_0.
Let me remind you the electric
field energy, energy density for electric
field, is ε_0/2
E^(2). You might remember that formula.
So they are very similar
formulas, except μ_0,
which is normally upstairs in every formula,
comes downstairs here, and ε_0 which
is always downstairs in every formula,
comes upstairs here. So let me summarize what you
should remember from all of this.
When you have a circuit element
called an inductor, it's just a coil of wire that's
wrapped around some solenoid. And when you change the current
through the inductor, it's going to fight it.
It's not like a resistor.
A resistor fights any current.
An inductor fights only a
change in current, so that's all summarized in
this equation, the voltage = LdI/dt
RI. This is the circuit we're going
to look at. Maybe a switch is open like
that. There's R,
that's L, that's a switch,
that's the voltage. When you close the switch,
you've got to ask yourself, what's the current going to be?
What will be the current
infinitesimally after the switch is closed?
Yes?
Suppose it was not 0,
but .2 amps, what's the problem?
After all, you closed the
circuit. Yes?
Student: You have
energy stored in the inductor without ___________.
Prof: Yeah,
first of all, if you had .2 amps,
you'll have ½LI^(2) and you can ask,
"Who had the time to do that?"
Nobody.
More importantly,
if LdI/dt is the voltage across this, it will become
infinite if dI/dt is infinite.
A current that jumps from 0 to
something in no time, that's got infinite derivative.
So any quantity whose
derivative is bounded cannot jump in its value just from
calculus. So the current in the inductor
will never jump. Likewise if you have a
capacitor, with some charge on it,
and you close the circuit, the charge on the capacitor
initially was Q_0,
will remain Q_0 one
femto-second after you close it, because charge on it,
the rate of change of the charge,
is a current. The current is finite in any
real problem. So capacitors cannot abruptly
change the charge they have, and inductors cannot abruptly
change the current they carry. If you want,
they are connected to energy. The energy in the capacitor is
Q^(2)/2C, therefore if q changes
abruptly, the energy changes by a finite
amount in infinitesimal time. Nobody can deliver that energy
or take out energy at that rate. Similarly for the inductor.
So I'll tell you what's in
store on Wednesday. We're going to come back and
look at this LR circuits and look at LC circuits and look at
LCR circuits. That's the kind of stuff I
think you've all done before in high school,
but I think still have to do that, because that's the kind of
stuff that may be more useful than many of the other things
I'm talking about. But it won't be in the greatest
depth. I just want to hit the high
points.
