Prof: Let's start.
 
I told you about Ampere's law,
which is key to what we are
going to do next,
so I'll remind you what the
deal is.
 
It says, if you have a closed
loop and around that loop you
integrate the magnetic field
that is equal to
μ_0 times all the
currents that penetrate this
loop.
 
So some current would be going
in, some current would be coming
out.
 
The guys who are outside the
loop don't contribute.
Guys who are trapped inside
this, they make a contribution.
That's Ampere's law.
 
Now you have to know the
convention.
The convention is that you do
the line integral
counterclockwise,
and if your fingers are
following the line integral,
your thumb points towards
outside the board,
and that current is positive.
Anything going into the board
therefore will be counted as
negative in this one.
 
So this has to be counted with
sign, do you understand?
If the contour goes one way,
all currents coming out of the
board are positive,
all those going in are
negative.
 
In particular,
if these 4 currents add up
algebraically to 0,
one is 2 amp,
another is -2 amp and so on,
then this integral will vanish,
even though there are currents.
 
So it counts not the absolute
value but the algebraic sum of
all the currents.
 
Now there's another subtlety I
didn't get into last time,
which is the following -
remember, this is the current
penetrating a surface,
bounded by this loop.
You guys with me on what that
means?
Now as long as I force you to
put this loop and to live in the
plane of the blackboard,
there's only one surface with
that loop as the boundary.
 
But in real life,
we are doing stuff in three
dimensions.
 
Imagine pulling this loop out
of the blackboard.
A lot of currents are coming
and going out.
What is the surface of which
this loop is the boundary?
The answer is not unique.
 
There are many,
many surfaces that have the
same loop as the boundary,
right?
Take a piece of wire and wrap
it in the form of this boundary
and dip it in soap,
and you pull it out,
you'll get a soap filament.
 
That soap filament has as its
rim, or the boundary,
this loop.
 
That may not even be planar.
 
And if you blow on it,
it's going to bulge outside,
right?
 
Still this rim will be the
boundary of that bulging
surface.
 
So actually,
even if the surface is coming
out of the blackboard,
the theorem is still valid.
So let me draw another picture
that may be helpful.
Suppose this is the surface and
you've got all these currents,
some going in and some coming
out.
You may have imagined all this
time I meant a flat planar
surface.
 
I don't.
 
You can imagine now a curvy
surface, like maybe a
hemisphere, of which this rim is
still the boundary.
You can ask,
"How can that be?
Where do you want me to count
the currents,
here or on that surface?"
 
The answer is,
it doesn't matter.
It doesn't matter because any
current coming in here has to
get out also through the other
one.
The only--or it may come in and
decide to come back here,
in which case it doesn't leave
the other one,
but it will make 0 contribution
to this contour because it's
coming in once and going out
once.
Because currents don't
terminate, just like magnetic
field lines, whatever crosses
this surface will also cross any
surface with the same boundary.
 
That's a fact that will become
important later on.
So you follow that?
 
The law, if you want to write
it formally,
I'm going to write it as some
surface,
and this is the surface
integral of the current density,
which is the current per unit
area.
And this loop is the boundary
of that surface,
and I told you that the
boundary of any surface is
written as partial S.
 
This is the correct way to
write Ampere's law.
And the point I'm making is,
this could be any surface with
that as the boundary.
 
Now once you've got Ampere's
law, I was trying to show you
how you can use it to solve
certain problems that might have
otherwise been difficult.
 
Just like Gauss's law.
 
Remember, Gauss's law was very
helpful when you took a solid
ball of charge.
 
The electric field of that is
usually hard to find,
but from spherical symmetry,
you know the field is radially
outwards.
 
You draw a Gaussian surface and
you know the field's magnitude
is constant on a sphere.
 
How big is it?
 
You get that from Gauss's law.
 
So certain problems which are
highly symmetric,
you can get using Ampere's law
also.
So one example I did,
I want to repeat it,
because one thing I said wasn't
right.
There's this current going from
- to infinity,
and we want to find the field
around it.
If you do an honest
calculation, you will say,
let me take a segment,
let me take a point,
let me join the segment to that
point and take the cross product
of the
dl with the R,
etc.
 
and so on, right?
 
And it will give you a
contribution coming out of the
blackboard, but then you have to
integrate it.
Now you don't have to do all
that if you use symmetry.
The symmetry argument would
tell you first of all that any
field configuration you have
will be the same if you rotate
it around a circle.
 
Because if you look at the wire
end on and you turn the wire,
it looks the same so the field
configuration should look the
same.
 
And I said there are only a
couple of things that have the
property.
 
Here's a good one.
 
Maybe it does that.
 
And the other one is this.
 
Here is the wire.
 
Notice if you turn the wire,
the wire looks the same,
configuration looks the same.
 
If you turn this wire around
its own axis,
wire looks the same and the
configuration looks the same.
Then I said we rule this guy
out, because magnetic field
lines cannot come and end at a
point.
Then I gave another
explanation, which is that if
you reverse the current,
the field lines are supposed to
go outwards,
but instead of reversing the
current,
if I grab it and turn it by 180
degrees around this axis,
then the lines will still go in
but the current would be
reversed.
So for whatever reason,
this is out and this is in.
Then as to the direction of the
field,
we also know that it has to be
perpendicular to the blackboard,
because the two vectors whose
cross product determines it lie
in the plane of the blackboard,
so the field has to come out.
It has to come out and the
amount by which it comes out
cannot change as you go around
the loop.
And the answer can only depend
on how far you are from the
wire.
 
Then I write Ampere's law as
2ΠR times B at
the radius r =
μ_0 times the
current.
 
In other words,
I look at the wire from the
end, and I take any contour I
like of radius r.
Then since B is
tangential to the circle,
the line integral is simply
2Πr times B,
and that's the current
enclosed, so you find B.
If you like,
B_
Φ,
meaning in the azimuthal
direction.
 
This angle is called phi.
 
B_Φ
= μ_0
I/2Πr.
 
That's the result we got from
integration, but you can also
get it from Ampere's law.
 
Then I did another variation.
 
I definitely don't want to
repeat that, which is,
what if the wire is thick.
 
Not a point wire but got finite
thickness with current coming
out of the blackboard.
 
What is going to be the
magnetic field?
Then the answer depends on
whether you pick a contour like
this or a contour inside the
wire.
And if you do the two things,
you'll find the field actually
grows linearly inside,
then falls like 1 over r
outside.
 
Anyway, this is stuff I did
near the end.
But now I want to consider two
new problems which I didn't
study last time.
 
The first one is going to be
very important,
is the solenoid.
 
If you imagine on a tube you
wrap some wire,
and the wire is wrapped like
this many, many times.
And you want to find the
magnetic field.
Now one loop we know produces a
field that goes like this.
You stack them up,
you expect the field to go like
that inside the coil,
then sort of loop around and
come back outside the coil.
 
What we want to find out is the
strength of the field inside the
coil, and also outside.
 
So the way you handle this
problem, you've got to take an
Amperian loop that looks like
this.
If this is the cardboard on
which you wound that thing,
see what this wire looks like
when you slice it vertically.
That's the tube on which you
wrapped it.
I want you to slice it
vertically.
On the left hand side,
I've current coming in.
On the right hand side--coming
out of the board and right,
it's going on,
you guys see that?
Wrap it around the solenoid and
slice it down the middle,
then on this end,
the current's coming out,
on this end,
the current's going in.
So I'm showing you the cross
section of the tube.
It looks like this.
 
So first thing I want to argue
is that the field outside does
not vary with distance,
because if I took a loop like
this--it's not a real loop;
it's a mathematical loop on
which I'm going to use Ampere's
law.
I know the field outside is
going to be coming down,
but maybe it's weaker here and
stronger here and that's what
I'm going to rule out.
 
You know how we can rule that
out?
What would go wrong if the
field was weaker here than here?
Something to do with Ampere's
law.
Well, let me explain.
 
Suppose you go around this loop.
 
Here B and dl are
perpendicular,
so there's no contribution to
B⋅dl.
Likewise here.
 
Here you have a B going
one way and you have the same
B but the path is going
the opposite way.
Now these two have to cancel.
 
These two have to cancel
because there is no current
coming out of this.
 
Because there's no current
coming out,
this line integral has
vanished, therefore B
times that length and B
times that length going the
opposite way must be equal in
magnitude.
That means the B here
must be as strong as the
B there.
 
So B doesn't vary when
you leave the wire.
Well, I can keep on repeating
this argument until I go all the
way to infinity,
but I know B is 0.
Therefore coming from infinity
inwards, B is going to be
0 everywhere here.
 
That's only true for an
infinite solenoid.
If you take a finite solenoid,
there will be lines coming in
here.
 
For an infinite solenoid,
is when the lines always go
perpendicularly.
 
See, in a finite solenoid,
you sort of know the lines will
do this, right?
 
They'll be perpendicular here,
but they could be at an angle
there.
 
But for an infinite solenoid,
every portion looks like you
are here.
 
The lines are only vertical,
they have no choice but to
vanish.
 
So the flux actually goes out
and returns on a sphere of
infinity.
 
So we only have to find the
magnetic field inside.
So then you've got to do the
right hand rule and the right
hand rule tells you the magnetic
field looks like that.
And the question is,
how big is it?
Once again, I want to argue
that the magnetic field is
constant inside the solenoid.
 
Now atleast now you should be
able to guess what argument I
will use.
 
The argument will be,
take an Amperian loop like this
one.
 
The two sides,
this and that,
do not contribute because they
are perpendicular to the field.
These two better make
cancelling contributions.
And that's going up and that's
going down.
For them to cancel,
the strength of the field here
must be the same as the strength
here.
So we have a result of the
magnetic field as some number
inside and 0 outside and we're
trying to find the one number.
This is going to be the case
whenever you use Ampere's law to
get anything.
 
In the end, Ampere's law is a
single statement about the
magnetic field.
 
You can get only one piece of
information from it.
If you reduced your problem to
the point where there's only one
thing you do not know,
you can find that one thing.
Here I've talked my way out of
the field in this region and
that region, and here it's
constant, and it's got a known
direction.
 
Only thing I don't know is,
how big is it?
To find how big it is,
you can take a loop like this.
Let me take a loop like this
one.
Let me repeat that for you.
 
So here is the current coming
out and I want to take a loop
like this.
 
I want the loop to have length
l.
So the magnetic field is some
number B there which I
don't know.
 
It is 0 on the outside and I
never care about these sides
because they're perpendicular.
 
So the line integral will be
B times l.going
up, nothing there,
0 times l.going down and
nothing here.
 
That's the whole line integral,
B⋅
dl..
 
That = μ_0 times
the current crossing the loop.
Now here's where you should not
make the mistake of saying
there's a current I
flowing in the wire and
therefore it's equal to
I.
Can you tell me what's wrong
with that logic?
Yes?
 
Student:  It's coiling
around so you have to count the
number of coils.
 
Prof: You understand
that?
First of all,
if you just walked into the
class now, first of all,
you're in trouble.
Secondly, if you just walked
into the class and you saw this
picture and you knew all about
Ampere's law,
did not know anything about
solenoid,
nothing, what will you do?
 
You'll say, "Here's a
loop, current's coming out.
I just count the current coming
in."
The fact that these guys all
belong to a big solenoid and
they wrap around is irrelevant.
 
All that matters for the
Ampere's loop is how much amps
are coming at me or going away
from me.
So the fact that they're all
part of one gigantic loop is
irrelevant.
 
So you've got to count the
number of turns that got trapped
in this length l.
 
So we use the symbol small
n, which is the number of
turns per unit length.
 
Then it's going to be number of
turns per unit length times
l.
 
So that depends on how tightly
you wound the coil.
If you wound 100 turns per
centimeter, well,
it's whatever,
10,000 turns per meter.
Now one nice thing is that this
l cancels.
Again, I want you to think
about why l should
cancel.
 
Why shouldn't the answer depend
on l?
This happens even when you do
Gauss's law, when you take some
surface, the details of the
surface cancel out always.
Yes?
 
Student: 
>
Prof: More turns of the
wire.
But the more important thing
is, the question that was given
to us was a solenoid.
 
It had some number of turns.
 
It had a certain current.
 
And the answer should only
depend on the current and how
densely it is wound.
 
This l.
 
was my own artifact.
 
The Amperian loop is not a real
loop.
I made it up.
 
So the answer I get for
B cannot depend on
whether I took that loop or that
loop or that loop.
They should all give the same
answer.
That's why the properties of
the loop should cancel out and
you should get in the end a
result that depends only on the
intrinsic variables in the
problem.
So this is a very useful result.
 
That's why we draw a box around
it.
That means important.
 
The field inside a solenoid is
μ_0 times the
number of turns per unit length,
times the current flowing to
the solenoid.
 
I'll be invoking this result
all the time.
It's one of the few results I
carry in my head.
Normally I encourage you not to
carry too much stuff in your
head, but this guy's pretty
helpful.
So here's another solenoid that
people have, which is better
than this one in the following
sense.
As long as you make a finite
solenoid--in real life,
everything is finite--you will
have this return flux.
There's a trick by which you
can keep all the flux hidden
inside the tube,
by wrapping the tube into a
doughnut.
 
So take the long tube and glue
the ends together,
then you have a solenoid which
is like a doughnut on which you
wrap the wire.
 
I've shown you a few turns.
 
So you wrap it all the way
round and you come back and you
drive the current from here.
 
Are you guys with me?
 
The cross section of the
doughnut can be circular,
it can be square.
 
It doesn't matter what it is.
 
Sometimes it even contains some
metal or iron,
which I'm not showing you.
 
It's fine.
 
Even if it's an empty tube,
you wrap the wire around it.
But now the flux lines go like
this along the tube.
So let me take a problem where
the cross section is actually a
square, in other words,
it looks like this.
I cut the doughnut right there
and I slice it and I look at it,
and you have the flux lines
doing that.
That's the cross section.
 
So which way is the flux going?
 
Well, you've got to figure out
again with the right hand rule,
the way I've shown the
currents, if you do this around
the wire,
the flux is actually going like
that here.
 
Our job is to find out what is
the magnetic field inside that
doughnut.
 
For that, I'm going to slice
the doughnut the way you
normally slice a doughnut.
 
You normally slice a doughnut
so it looks like this.
This is half the doughnut, okay?
 
You cut it, then let's say the
wire looks like that one.
That guy is coming in,
coming out of the board,
going into the board,
coming out of the board,
going into the board.
 
Yes? Can you see that?
 
If you take a doughnut with a
wire wrapped around it and you
cut it, it's going to look like
this.
Now the important thing to
notice is that we haven't really
cut that doughnut.
 
If you really cut the doughnut,
there's going to be no current.
This is a mental slice you make.
 
Then you ask yourself,
now what do you think is going
to be the Amperian loop?
 
Can you make a guess what the
loop will look like?
Yes?
 
Student:  It will look
like a piece of pie.
Prof: A piece of pie.
 
We're all getting hungry now.
 
I've talked about doughnuts.
 
Which pie?
 
Student:  It will have
it so that the sides
perpendicular to the--
Prof: Oh, okay.
Maybe you could do that,
but here is what the general
consensus, 9 out of 10
physicists use this contour.
This is going to be the
Amperian loop.
Then what can you say?
 
The loop has some radius
r,
so once again,
2Πr times B is
μ_0I times
the number of turns altogether
in the solenoid.
 
You understand?
 
Because this loop cuts through
every turn of wire.
As far as the direction is
concerned, it will go like this.
The field will go this way.
 
So the magnetic field here is
μ_0IN
divided by 2Πr.
 
So the field actually gets
weaker, so that if you cut this
doughnut, the field on the inner
rim will be stronger than the
field on the outer rim.
 
And if you go inside the inner
rim, you'll get 0;
outside the outer rim,
you'll get 0 because any loop
you draw on the contour will
intersect an equal number of
incoming and outgoing currents.
 
So B is going to vanish
outside the doughnut,
and it's going to be non-zero
right inside the heart of the
doughnut,
and the strength will be this.
Now if this doughnut had a
radius of say 1 light year,
you can ask what will happen to
the formula.
You should be ready for what
you expect.
Take a doughnut 10 miles long,
10 miles in circumference.
What do you hope the formula to
give?
Yes.
 
Student:  It should
approach the results of the
infinite solenoid.
 
Prof: For infinite
solenoid,
because the circle of radius 10
zillion miles,
for most purposes,
it's like a straight tube,
right?
 
You cannot tell it's going
around in a circle.
Let's see if that happens.
 
Well, you can see the field is
μ_0NI,
divided by 2Πr is
just the circumference of the
circle.
 
So it's going to be
μ_0 little n
times I,
because little n is big
N divided by
2Πr.
See that?
 
This 2Πr will-- in
other words,
you cut it here and you open it
into a solenoid,
its length will be
2Πr,
and the number of turns per
unit length will be big N
over 2Πr.
 
But that's an approximation,
because in a real finite
system, the magnetic field is
not uniform.
There's a slight variation from
this end to that end.
But if this thickness is 2
inches and that radius is 37
miles, that variation is
negligible and it becomes that
other easier result.
 
Student:  How did you
get the small n again?
Prof: We got the small n
because this guy.
This is the total number of
turns wrapped around the
solenoid and 2Πr is
the circumference,
right?
 
So number of turns divided by
the circumference will be the
number of turns per unit length.
 
In other words,
if it was a huge solenoid,
so huge that it looked like a
straight line to you from your
range of observation,
the number of turns it would
have per unit length would be
exactly this much.
So it will look like a straight
solenoid to you,
but it's curving around on a
big scale.
All right, so this marks the
end of one topic.
So I'm going to write down over
and over what we know,
because I think I talked to
some of you people,
and I asked you,
what's the part you find
difficult.
 
And one thing I heard,
which is very reasonable,
is the amount of stuff you're
learning every day.
This is just a lot of stuff.
 
"Drinking out of the fire
hose" was an expression
that came up.
 
You drink and you drink and you
drink and I keep throwing stuff
at you.
 
I would like it to end.
 
It's almost going to end,
but it's not over yet.
But I'll tell you where things
stand now.
All of electrostatics and
magnetostatics are summarized by
the following equations.
 
I'm going to write them again
and again.
E⋅dA
is the charge inside.
Line integral of
E⋅
dl.
 
or dr--I forgot what I
called it--that is 0.
Surface integral of the
magnetic field is 0 because
there are no magnetic charges.
 
The line integral of the
magnetic field is mu 0 times all
the current inside.
 
That's it.
 
And the force--after all,
who cares about E and
B,
except for this great equation,
which tells you that if a
charged particle goes into a
region with electric and
magnetic fields,
this will be the force acting
on it.
That's very important.
 
That's why we care about
E and B,
because they make things
happen.
So the cycle of physics will
proceed as follows - this is all
for statics.
 
That means J is constant
and rho is constant.
So somebody has to give you the
currents and somebody has to
give you the charges.
 
Then it's a purely mathematical
problem to solve for the
electric and magnetic fields.
 
And I was telling you yesterday
a fact that may not be obvious,
that these equations suffice to
determine electric and magnetic
fields,
given the cause of electric and
magnetic fields,
which is currents and charges.
Okay, so now we are going to
do--let me ask you something.
When do we keep on modifying
the equations?
I think I explained that to you.
 
Why don't we stop now?
 
Any idea why I'm going to
modify the stuff?
Do you know?
 
Do you know when we modify
equations?
What could be wrong with this
equation?
Who can convince me it's wrong?
 
Any idea?
 
I'm not talking about a
particular thing,
but I'm talking in general,
what makes people modify their
equations.
 
Pardon me?
 
Student:  New experiment.
 
Prof: New experiment.
 
That's the only reason you
modify your results.
If these things worked,
and it could have been a very
consistent world in which it
worked, but it is not the whole
story.
 
At least one reason it's not
the whole story is,
rho and J are not always
constant.
You can have charges shaking
around and you can have
currents.
 
You turn them on and off as you
wish.
One thing is to hope that this
continues to describe them even
in this situation.
 
The fact that we derive them
from steady currents and
densities doesn't mean it has to
fail when they are changing with
time,
but there is no guarantee that
they won't change.
 
So I'm going to now lead you
through the experiments that
forced us to change some of
these equations.
So these are called Maxwell's
equations, but they're not the
real Maxwell equations,
because they're not valid for
the time dependent case.
 
I'm going to start the
phenomenon, then we'll see where
the trouble begins.
 
So let's take a problem where I
have a uniform magnetic field
going into the board.
 
This goes on forever,
but it doesn't go beyond this
point.
 
I've only shown you that much.
 
Now I'm going to take a loop of
wire like this.
It's got some width w and maybe
some length l..
And I put a little light bulb
here.
Then I drag this to the right
with some speed V. And
this magnetic field
B going into the
blackboard.
 
Now if the loop is not moving,
nothing will happen.
If the loop is moving,
something does happen.
Now I'm going to ask you what
happens, because this is
something you must know.
 
What happens when I drag the
loop?
Yes?
 
Student:  The light bulb
will go on.
Prof: Good.
 
The light bulb will go on is
the answer that you've got to
give.
 
You may not know by how much
and whatnot.
But first of all,
why would I draw a light bulb
if it's not going to go on?
 
That's how you should reason.
 
That's how my kids did all
their SATs.
They asked extraneous questions.
 
One of them is,
you put the light bulb because
it's going to glow.
 
So you can ask yourself,
what's going on?
Why is the light bulb glowing?
 
Is it some new physics?
 
The answer is,
whenever youwant a light bulb
to glow, you're looking for a
battery.
There is no battery in the
circuit.
And yet there is an emf,
because every time a charge
makes one full turn and comes
around,
it has delivered some work,
because that's what makes this
thing glow.
 
You're constantly pumping
energy into the resistor inside
your bulb.
 
And who's providing the energy?
 
These charges are going round
and round doing some work.
That means something is pushing
them around this loop.
We defined the emf to be the
line integral of any force per
unit charge pushing them around
the loop.
This is per unit charge.
 
That's got to be not 0.
 
So what is that force,
and what is the emf is what I
want to ask.
 
Do you know what the force
might be?
And why does it kick in only
when I move the loop?
Yes?
 
Student:  The force is
the magnetic force that we
wrote.
 
Prof: The v x
B force.
That's because there's a
v here,
B into the blackboard.
 
v x B looks like
that.
The force is v x
B and it acts only here.
Let's understand why.
 
If you come to this portion,
if you take this v x
B,
this v is this way,
and B is into the board,
v x B is
perpendicular to the wire.
 
But charges cannot go--even if
there's a force perpendicular to
the wire, it doesn't contribute
to the emf, which I found by
going along the loop.
 
So dr is along the loop.
 
The only way to get that to
contribute is to have a force
along the loop.
 
That's only here and it's not
here.
So the whole force is non-zero
only in that segment.
So its line integral is equal
to simply the force,
which is Bv times
w.
That's the emf.
 
Any time a charge goes around
the loop once,
that amount of work is being
done.
Okay, so that explains it.
 
We're not in any kind of
trouble.
In fact, we understand this
problem completely without
bringing in any new stuff.
 
But there's one paradox.
 
We were told the magnetic field
doesn't do any work.
You remember that?
 
And yet here is a magnetic
field pushing these charges and
getting things done.
 
So what can be going on there?
 
Student:  It takes work
to move the whole loop.
Prof: It takes work to
move the loop.
That is certainly true,
but how about the fact that the
magnetic field manages to do
some work?
Is that a problem for you or
not?
Remember, the original argument
for why it doesn't do any work
is, the force is
v x B and the
power is the velocity times the
force and it's 0,
because v x B is
perpendicular to v.
But here, v x B
is this way, and the charge is
moving this way.
 
It looks like we got some work
done.
Okay, so the answer is that
this is not the full--there are
two kinds of velocities.
 
If you have a current in the
wire, the charges are also
moving in the wire at some speed
u.
So the real velocity has got
the velocity because the whole
loop is moving,
v, and the fact the
current is moving this way,
u, and w,
if you like,
is the true velocity.
And v x B is
actually in that direction.
That is v x B,
perpendicular to w.
So let me draw the picture for
you.
w and w x
B.
w is made up of two
parts.
It's got a part v here,
it's got a part u here.
So the magnetic force is not
just along the wire.
It's really at this strange
angle.
And we can easily find out how
much of the force is this part
and how much of the force is
this part.
The force here will be--let me
see.
It will depend on the other
component, so it will depend
on--yes, the force here is
B times v and the
power is that times u.
 
The power is Bvu,
due to this portion.
But in this side,
if you ask what's the power
delivered by that component,
it is B times u
times v,
but with a - sign.
In other words,
if you resolve this w x
B into two parts,
the vertical force is due to
the horizontal velocity,
and the power is that times the
vertical velocity u.
 
The horizontal force is due to
the vertical velocity.
That's how the cross product
works.
So when you're done,
both are equal to B
times u times v.
 
But this is and this is -.
 
In other words,
in the end, it would like to
move the charge this way,
because that's the force,
but it manages to move it along
the wire this way.
It would like to move it this
way, but it's force to go
oppositely.
 
So a force trying to move this
way, this part it is doing work,
but this part it is not doing
work.
It has work done on it,
because we forced the electrons
to move to the right.
 
Of course, that energy comes
from the person pulling the rod,
pulling this loop,
because every charge that has
this force,
it won't just move unless you
pull it,
and the force you apply is
really coming from you that
balances the magnetic force.
So there's yet another force.
 
That's the force due to me,
and I'm pulling this to balance
this part of the force.
 
So what is the role of the
magnetic field?
It doesn't do any work,
but you need it here,
because one component of the
magnetic field does work.
The other component has work
done on it.
So it takes with one hand and
gives with the other hand,
but you need that.
 
It's no used saying,
"I'm not very impressed
because I did all the hard
work."
You did all the hard work by
grabbing this loop and pulling
it,
but I challenge you to go and
push the individual electrons so
they can move down the wire.
You cannot even see those
little guys, right?
But the magnetic field does
that for you.
So it takes macroscopic
mechanical power,
turns into microscopic energy
supplied to the electrons.
Even though the field did not
profit in the end,
it just delivered to the
electrons what you gave them,
you need that field to do a
transfer from something you can
see and pull to something you
cannot see,
right?
 
I mean, I give you a light bulb
and you've got all the muscles
in your hand.
 
Let's see you make it glow.
 
You cannot say,
"I have the energy."
There's no way to transfer it.
 
That's the role of the magnetic
field, so you shouldn't say it's
no use.
 
Anyway, we got that answer.
 
I'm going to give you another
equivalent way to explain the
balance of energy.
 
We all know that the energy is
coming,
because once the loop carries a
current,
it's going to be hard for you
to pull the loop,
so you've got to do some work,
and that will explain the power
usage here.
 
So let's calculate the power
two ways.
The power in the resistor = emf
squared over R.
emf squared is
B^(2)v^(2)W^(2)/R.
Now how about the power that I
provide?
What's the power I provide?
 
I'm pulling this leg of the
wire here.
The force on anything,
you remember,
is BwI,
and I pull it with a force
v, with a velocity
v.
That's the power.
 
So it is Bwv times
I.
I = emf divided by
R.
But if you put the emf that we
just got, you will get this.
So they will match.
 
So do you understand how this
thing works?
This is a generator,
if you like.
If you want to light the bulb
in your house,
one option is to set up a
magnetic field,
take the light bulb and connect
it to a rectangle,
and grab it and keep running.
 
As long as you're running,
the light bulb will be glowing.
When do you think it will stop
glowing?
Yes?
 
Student:  I'm sorry,
what does that word say?
Prof: This one?
 
The power that I provide.
 
Okay, I'm just balancing the
power loss in the resistor.
Are you with me now?
 
You understand the two things?
 
I take the force I apply on the
wire, multiply by the velocity
with which I pull.
 
That's the power.
 
Likewise I take I^(2)R.
 
That's the electric power
dissipated.
And of course, they balance.
 
Notice that once the loop is
fully inside the magnetic field,
the light bulb will stop
glowing.
Do you know why?
 
Can you see why?
 
Yes?
 
Student:  Because the
flux in the loop isn't changing
any more.
 
Prof: No, no.
 
You're pulling new laws,
in terms of what I've told you.
Student:  The magnetic
field, there's no new field
coming through the wire.
 
The field is constant that's
coming through the wire.
Prof: Just use what
we've used so far.
Yes
Student:  The field is
acting on the left segment.
 
Prof: Right.
 
In other words,
without invoking any new ideas,
once this piece gets into the
loop, it's going to have a
v x B there and
a v x B there.
Which way should the current go?
 
In fact, it won't go anywhere,
because the line integral of
this will be 0.
 
You will get contribution from
there, and - the contribution
from there.
 
So we can demolish this problem.
 
You understand it completely.
 
It doesn't pose any
difficulties.
And it does not give any
grounds to monkey with those
equations.
 
And now comes the bad news.
 
Suppose I go to a frame of
reference moving with the loop.
It may well be that the loop
was just there,
and I hired a couple of guys to
carry a magnet and run under my
house, right?
 
And also, my light bulb will
glow, and I don't have to go
anywhere.
 
They will carry the magnet.
 
Now you realize that you're
certainly entitled to see it
from that frame of reference.
 
Now if you believe in
relativity, that means a person
should be able to say from the
point of view of the loop,
"I'm not moving.
 
Someone's carrying the magnet
and running with it."
But the light bulb must glow.
 
Lots of things are relative,
but whether a light bulb is
glowing or not is not relative.
 
You can go to any frame of
reference.
A glowing light bulb is a
glowing light bulb.
We don't know how much power it
consumes and so on,
but the fact that it's on and
glowing is undeniable.
So how is the person in that
fixed loop supposed to
understand the glowing of the
light bulb?
What do you do,
what do you say,
in terms of anything?
 
Why are charges going around?
 
I'm a static loop.
 
Okay, someone's moving with the
magnet, but there's no v
x B force because
nobody's moving in my loop.
The magnet's irrelevant.
 
Of course, there's one
difference.
This magnet is not like the
usual problem,
because it's a changing
magnetic field,
whereas previously you had a
static magnetic field.
So you might say, "Well,
maybe the force on a charge due
to moving magnetic field is
different from its force due to
changing magnetic field."
 
But that's not acceptable,
because if you believe in the
reality of the field,
what it does at a given point
depends on its value.
 
It doesn't depend on anything
else.
But I still haven't explained
why charges like to go around a
wire now.
 
So what force could be pushing
them?
What force do you think will
push them around a loop?
Yes?
 
Would you like to guess?
 
Yes, you.
 
Student: 
>
Prof: The magnetic field
is changing with time.
You agree that the loop is
going to be fixed,
and a little later,
it will look like this,
and so on, right?
 
But that's not going to move
the current around the wire,
because the magnetic field has
no cross product with anything.
So what's the only thing that
can make the charges move around
the wire now?
 
What made them move in the DC
circuit?
Student:  Electric field.
 
Prof: Electric field.
 
Look, I want you to understand
that these equations carry a
large amount of information,
and they in fact carry
everything as far as you need to
know.
The only force on a charge is
either E or v x
B.
 
That's it.
 
If v x B is out,
it's E.
I mean, there may be brand new
phenomena.
It turns out that we don't need
anything else.
E and B will do.
 
In which case,
what are we saying?
We are saying that in the other
frame of reference,
where the loop is fixed and the
magnet is moving,
there is an electric field.
 
Not only that,
it's an electric field whose
line integral is not 0,
because it is still true that
charges in the loop are going
round and round and doing work,
and someone's providing work
for every charge that goes
around the loop.
 
And the emf is the work done on
a unit charge.
So this makes us believe now
that whenever there's a changing
magnetic field,
there is an electric field.
This is an electric field of
brand new origin.
All the electric fields you
studied before were produced by
charges, by coulombs force.
 
This says, without the help of
any charge.
Charges are nowhere to be found.
 
They could be infinitely far.
 
If I change the magnetic field,
I'm going to get an electric
field.
 
Of course, we have to find out
what is the connection.
How much electric field do you
get?
How is it connected to the
magnetic field?
These are things you have to
deduce and you cannot deduce
them from logic.
 
You can deduce them only from
experiment.
Certain things come from logic.
 
If you have an equation,
you solve it using mathematics.
If you like,
that's a kind of logic.
Or you use symmetry and say,
"If I turn the wire and it
looks the same,
the answer should look the
same" and so on.
 
But what electric field should
be produced by changing magnetic
field you cannot derive from
pure logic.
If you combine it with other
principles like relativity,
you could try to get them,
but right now,
with what we have,
we just know that in the moving
frame--
look, you don't need Einstein
to tell you this.
 
Even before Einstein,
you knew that you're certainly
free to imagine a loop being
fixed and somebody moving the
magnet.
 
And everything in you tells you
the answer's got to be the same,
because in the end,
it's the relative motion
between the loop and the magnet
that decides.
And if you provide the same
relative motion,
you've got to get the same
answer.
And the only way to get that is
to say when there's a changing
magnetic field,
I have an electric field,
and furthermore,
electric field which is no
longer conservative because its
line integral is not 0.
So we've got to find out what
it is.
So I'm going to tell you now an
equation that really answers
that question,
and we cannot derive it.
It's a summary of a lot of
experiment.
And that says that the
electromotive force--
in fact, due to electric and
magnetic origin,
em = -d
Φ/dt,
and I'll tell you what this
equation means.
So you take a loop,
some contour c,
and you define Φ
to be equal to the integral of
the magnetic field on a surface
whose boundary is this loop.
That's called the magnetic flux.
 
So you know operationally how
to calculate it.
Take any surface you want.
 
It may be a flat one,
a bulging one,
doesn't matter.
 
Take the flux going through any
surface with that as the
boundary, and find its rate of
change, and that will be equal
to the electromotive force.
 
In fact, you just write
E electromotive force,
= -d
Φ/dt.
So you know operationally how
to find this thing,
right?
 
Now again, you may ask,
"Which surface do you want
to take for a given loop?"
 
This flat one or the curvy one
or the one with more bumps in
it?
 
The answer is,
again, it does not matter.
The reason is the same.
 
If you've got some magnetic
lines coming through this guy,
they also have to come out
through that guy because they
don't stop.
 
Magnetic field lines that enter
one surface have to also enter
another one with the same edge,
because they cannot escape.
So you can take any surface you
like.
So whenever you have these laws
where a surface is invoked,
and it's ambiguous and not
defined by the perimeter,
the answer is,
it doesn't matter which
surface.
 
The reason is that field lines
don't stop and end,
therefore whatever enters this
surface has to leave this.
If you had magnetic monopoles,
the lines can enter this and
terminate on a negative
monopole,
then of course it won't be
true, but you don't have that.
Now this - sign has got the
name Lenz attached to it,
and this thing is called
Faraday.
Faraday and Lenz.
 
And I will tell you what the -
sign due to Lenz means.
Lenz's law tells you that if
you change the magnetic flux in
a system, there's going to be a
very long sentence.
If you take the magnetic flux
in a system and it changes,
it will induce an emf which,
if it could drive a current,
will do so in order to fight
that change.
That's why it's a long
statement.
In other words,
if you had a real circuit and
you shove it into region of
changing field,
an emf will start driving a
current.
That current itself will
produce its own magnetic field.
Which way will it point?
 
The answer is,
it will point it such a way as
to neutralize the change.
 
For example,
if you took a loop and it had
some flux going through it,
if you tried to increase the
flux going through it,
it will try to decrease it,
so the current will flow like
that.
If you try to decrease the flux
through it, it will try to prop
it up to its old value.
 
The current will go like that.
 
That helps you find the
direction of the current.
The emf will be such as to
neutralize the change.
It does not necessarily oppose
the external field,
the magnetic field.
 
It opposes the change in the
magnetic field.
That's what I want you to
understand.
For example,
suppose I've got lines coming
out here.
 
And they get weaker,
namely some source of magnetism
makes the lines weaker.
 
Then the current will start
going this way to prop up the
thing back to its old value.
 
That's what you have to
understand.
That's the meaning of the -
sign and that's going to guide
us much better than all the
cross products.
Now I will show you that with
this one law,
you can explain everything.
 
In other words,
you can explain the wire,
the current loop moving in the
magnetic field,
in the laboratory frame,
in which the magnet was at rest
and the loop was moving,
or in the loop frame,
where the loop was at rest and
the magnet was moving.
You can explain everything with
this one great law.
It's a remarkable law.
 
And you just have to say,
we got it from experiment.
I'm going to tell you how it
explains everything.
So let's go back to this loop.
 
By the way, some of you gave
that answer, so it was not the
wrong answer.
 
I mean, I think it's perfectly
valid,
except that I did not want you
to use something we had not done
yet,
but from this moment on,
it's a valid answer.
 
So I'm going to give you the
other explanation.
So here is my field.
 
I think I had everybody going
in, right?
Here is the thing-- so I want
you to know what I'm trying to
do now.
 
I'm trying to tell you that
this one law will explain the
loop going through the field in
both frames of reference.
So first I'm going to do the
easy part.
When there's a fixed magnetic
field and I'm dragging the loop.
I'm not going to worry about
and - signs, because I will get
that in the end.
 
Let us ask, what is the flux
penetrating this loop?
Can you see that it's just the
magnetic field times the width
of the loop times the length of
the loop that is inside?
Not the whole length,
but the length of the loop that
is still inside the magnetic
field.
That's a very simple result.
 
B times A is the
flux.
Now let's take the time
derivative of this,
dΦ/dt.
 
B is not changing,
w is not changing,
but l is changing,
because when I move the loop,
the rate of change of l
is just the velocity of the
loop.
 
But that's exactly the emf we
got before.
Now the only thing is with the
- sign, let's figure out if the
sign is correct.
 
This has got magnetic flux
going into the board.
When you pull the loop to the
right, you've got less magnetic
field going into the board.
 
Therefore the current should
flow in such a way as to
continue the flux.
 
So which way should the current
flow?
So if I'm pulling the loop--I'm
sorry.
I've got the loop drawn
backwards.
This is okay too.
 
Let's take this loop and pull
it to the right with a velocity
v.
 
So the flux into the board is
increasing, right?
So it should produce flux
coming out of the board now to
compensate that.
 
That means the current has to
flow like this.
See that?
 
When the current flows like
this, you'll start producing
upward flux.
 
In other words,
you're getting more and more
flux into the loop going into
the board.
To cancel it,
I will produce flux coming out
of the board.
 
That's what the loop tells you.
 
That way you don't have to
worry about all the cross
products.
 
At the very end,
you ask what's happening to the
flux?
 
How do I keep it from changing?
 
And the answer is,
you keep it,
in this case,
from changing by producing flux
coming out of the blackboard.
 
And finally,
remember when the loop is
entirely inside,
there's going to be no more
emf, because the flux through it
is not changing.
Because in the beginning,
as it moves to the right,
it's gaining more and more flux
from the back end of it,
but once it's fully in,
it's not getting any more flux,
so the current will stop
flowing.
That comes from this argument,
from this way of thinking,
that
dΦ/dt
is 0.
 
So in other words,
I've given you something that I
say is a new law.
 
The law says the rate of change
of flux = emf,
but so far, it doesn't look
very new,
because everything it gave you,
you were able to get without
this law,
right?
We managed to understand the
loop completely.
So is there any new information
in this?
And that's what I'm going to
talk about.
The new information's going to
come in when I eventually use it
to understand the answer in the
moving frame of reference,
namely, what's the electric
field produced by changing
magnetic field?
 
That's also contained in this
one.
But I'm just telling you,
this part of it is a letdown,
because it's not very
impressive.
It gave you an emf you can get
from the v x B.
It is just v x B
all over again.
But let's see what is new.
 
To really find what is new,
we have to take this problem
and apply it to the very general
case of a loop.
Here is some loop.
 
I'm going to have the loop
actually move in time.
Take a loop in one shape.
 
A little later,
it's got another shape.
And the magnetic field is free
to change.
I'm going to take a very
general situation.
The loop is moving,
field is changing.
So let's say this surface is
S_1 and a
little later there's a surface
S_2.
You should imagine this is
stacked on top of this.
Maybe I should hide that part.
 
Here is this surface,
and it's moved to a new
location.
 
Now this surface,
let me call S and let me
call S Δs and you
will see what I mean by S
ΔS.
 
You can build up that surface
by taking this surface and
gluing to it this edge.
 
Is that clear or not clear?
 
Not clear.
 
Let me take a simpler surface
than this one.
Here is a surface,
here is another surface.
So one is a cylinder with the
round sides on the bottom.
The other is the top.
 
Do you agree,
they are both surfaces with the
same boundary?
 
Are you with me there?
 
This top loop can either take
the top of the cylinder as its
boundary,
or it's a hollow cylinder,
the curvy side and the flat
bottom,
are also another surface with
the same boundary.
But I can think of the second
surface as essentially the old
surface the sides.
 
So you agree that S' = S
ΔS,
where delta S is the
stuff you glue on the side to
the old surface to make up the
new surface.
Yes or no?
 
I think the cylinder is the
easiest one.
I tried more ambitiously to
draw this, but here is the
simpler case.
 
So now the question is,
what is the rate of change of
flux in this problem?
 
So the rate of change of flux
I'm going to find as follows.
I'm going to find the flux at
time t Δt - phi at
time t.
 
That's what I want.
 
Now at time t Δt,
my loop is here,
but I'm going to use,
rather than that surface,
I'm going to use the old
surface the patch job to make it
into the new surface.
 
So therefore,
I want the magnetic field,
B at t Δt
on the surface S ΔS
- the integral of the magnetic
field,
dA on the old surface at
time t. This part is
quite subtle and it took me a
while,
when I first learned it,
to understand what the point
was.
 
See,the surface used to be the
bottom half of the cylinder,
the bottom plate.
 
The new surface is the top
plate.
So natural instinct will be,
let's find
B⋅v as
on the top face,
but we don't want to do it that
way.
We purposely want to take
another surface with the same
boundary, because we're allowed
to.
That is the old boundary the
vertical sides.
That's the main point.
 
You take the old boundary
because it's going to make it
easy for you to cancel some of
the stuff with some of this.
That's why.
 
So B(t Δt)
⋅dA,
you can write it as B(t
Δt) ⋅
dA on S integral
B(t)⋅
dA on ΔS.
 
In other words,
the surface integral is
changing for two reasons.
 
The surface itself,
the loop itself around which
you take the line integral is
changing, and the fields
themselves may be changing with
time.
So they can both contribute to
the change.
And I've broken up into two
parts.
One says take B at time
t Δt on the same
old surface,
and then take
B⋅dA
on the incremental surface,
and subtract from it
B(t)⋅dA
on the old surface.
 
So do you guys see that?
 
Phi of t Δt
I've written as two parts.
One is the flux on the lower
face, but at a later time,
and the flux on the curvy
sides.
Now on the curvy side,
you don't have to worry about
whether it's at a later time or
at an earlier time,
because ΔS is
first ordered in time.
It's proportional to
Δt,
and this difference is going to
be proportional to
Δt,
so you don't have to worry
about the change in time over an
infinitesimal surface,
because that's second ordered
in time.
So now I have to bring all that
stuff over here.
So I'm going to say the change
in Φ = dB/dt times
dA,
Δt on the old
surface,
here is the stuff we have to do
that's extra.
 
So do you understand how I
cancel B(t Δt) −
B(t),
that's given by the rate of
change of B with time,
times the change in time?
You get their contribution only
because B is changing
with time,
because you're taking the very
same surface,
as you did the first time,
but at a later time.
 
Therefore the difference is due
to the rate of change.
But now we have to do the
second part, which is,
on the surface Δs,
we want to take
B⋅dA.
 
Now what is ΔS?
 
Let me see.
 
 
 
So I will have to draw the
picture like this.
So here's one surface,
here's another surface,
and I've gone from here to
here.
I've gone from this loop to
this loop in a time
Δt.
 
This is my segment of the loop,
and that's the distance,
vΔt.
 
So the area vector associated
with this section is just
vΔt x
dl.
That is the surface area.
 
You see that?
 
A parallelogram formed with the
loop size dl here
and the distance traveled in
dt.
Area of that is just the cross
product.
So this = 1 v (I'm going
to put the Δv outside)
v x
dl⋅B.
So I'm not going to touch this
expression.
I'm going to fiddle with this
for a while.
Cancel dt everywhere if
you like, divide by dt.
Forget that and forget that.
 
Then I get
dΦ/dt =
dB/dt
⋅dA on the
surface,
the line integral of v x
dl⋅B.
 
Do you understand the last part
what I did?
The extra surface that you got
here is made up of little
rectangular tiles with the one
end being dl, other end
being Bdt.
 
The area vector comes like
that, the magnetic field may
look like that,
and the dot product of the two
is what I have as the extra flux
coming out of the curvy side of
the surface.
 
Now you should know enough
about cross products to know
that this cross product is the
same as
dl⋅B x
v,
because you can rotate the
factors in an A x
B⋅C is
B⋅C x
A,
is C⋅A
x B.
 
I've done that once.
 
But then this = - the integral
dl
⋅v x
B.
So I'm going to write down what
I have.
What I have is that the rate of
change of flux,
dΦ/dt
= -dB/dt times the area -
integral v x
B⋅dl.
I'm sorry, it's d/dt
that.
Therefore −d
Φ/dt,
we'll have a - sign here and a
sign here.
So just write the formula for
−d
Φ/dt.
 
Put a - on everything here.
 
This is what I want you to know.
 
Now you've got to go home and
you have to think about this.
I don't claim it's easy.
 
I will also tell you something.
 
If you really don't want to
know everything,
you don't have to follow this
particular detail.
I'm going to tell you what I
get out of this.
That's what you have to know.
 
I want to make sure that if you
want to follow everything,
you're given a chance.
 
So here's what I get from the
old Faraday and Lenz law.
This says that rate of change
of flux has two parts.
You take a contour and you find
the flux changing through it,
it's got two parts.
 
One, because the field itself
is changing over the region.
Second, because the loop itself
is changing.
And together,
they make these two
contributions.
 
But emf = - integral of all the
electric forces and the magnetic
forces.
 
That's =
-(dB/dt)dA
integral v x
B⋅dl.
Now here's something very
beautiful to look at.
It says the rate of change of
flux by miracle manages to
measure the line integral of the
electric field and the magnetic
field.
 
And the rate of change of flux
contains two parts.
The second part is really the
v x B force that
comes because the contour itself
is moving, the wire itself is
moving.
 
So you balance this term with
this term.
If this is all you were looking
at, there's nothing new in the
equation.
 
That's why I said we're not
impressed, because it's just the
v x B.
 
But the real beauty is,
when you cancel these terms and
you look at what it says here,
that says line integral of
E⋅dl =
- rate of change of
B⋅dA.
 
The meaning of the partial
derivative is that B can
depend on x,
y, z, and t, but
this is only due to change in
t.
Only due to change in t
because I did the integral on
the same surface,
only a little later.
This is called Faraday's law.
 
You can see that because of
Faraday's law,
one of the four equations I
wrote hidden somewhere here,
in which the line integral of
E is 0 is no longer 0.
So this finally tells you that
in general,
if you take the line integral
of the electric field around a
closed loop,
you will not get 0 if the
magnetic flux through that loop
is changing.
But this has nothing to do
whatsoever with a real
conducting loop any more.
 
See, this formula has a lot to
do with the conducting loop,
because v is the
velocity of the portion
dl,
as the loop moves through space.
But the loop has been banished
from the two sides.
These equations have nothing to
do with any loop.
They just say take a fixed
contour in space and the line
integral of E on that
contour is the rate of change of
flux.
 
It's an intrinsic relation
between electric and magnetic
fields, not having anything to
do with conductors.
It's a property of E and
B, namely that a changing
B can produce a
circulating E.
That's the content of this
equation.
