Prof: Okay,
this is the final lecture on
electromagnetism.
 
When you come back from the
break we will start doing
optics.
 
Of course you know by now
that's really electromagnetism,
but it's going to be somewhat
easier on your imagination
because it doesn't involve many
abstract things,
pretty concrete things.
 
Then it'll take a turn for the
worse when we do quantum
mechanics, but I think the
mathematics of that should be
quite simple.
 
Before you do that you should
make sure you know all about
elementary complex numbers like
what is e^(i)
^(θ).
 
You should get that completely
into your system.
It's going to be all about
that, then very elementary
calculus, just one variable.
 
So last time I told you that if
you want to solve Maxwell's
equation in free space you can
write down any function you want
for
E and B that
satisfy the four Maxwell
equations.
Two involve surface integrals,
and two involve line integrals,
and the answer we got looked
like this.
This is x,
y and z.
The electric field is going to
point that way,
magnetic field is going to
point that way,
and the functional form E
was k (k is
unit vector in the z
direction)
E_0sin(ky -
ωt),
and B was
iB_0sin(ky -
ωt).
 
When you take this assumed form
with this y and this
t dependence you will
find,
as written, they automatically
satisfy the surface integral,
and the surface integral of
these fields on any tiny cube
will be 0.
 
When you do line integrals you
can take a loop in this plane,
and that plane,
and the third plane,
and you fool around you get
some additional constraints
which I derived,
but I don't want to do that
again.
 
But the conditions you derive
are that ω should
be equal to k times
c,
and remember the velocity of
light emerged at this following
combination.
 
Then you will find the magnetic
field should be electric field
divided by c.
 
These are the two conditions
you get from writing the two
other.
 
There are three equations for
loops.
One of them is trivially
satisfied 0 = 0,
the others give you this.
 
Now I want you to be able to at
least visualize these fields.
You can write them on a piece
of paper, but E is not
the letter E.
 
E stands for something
physical.
So what do these things do?
 
First of all what is k
and what is ω?
I realize that some of you are
not familiar with what k
and ω mean.
 
First of all you know it's a
periodic function in y at
fixed t,
if you like,
because sine of anything is
periodic.
That means at a given time if
you take a snapshot of this
E it'll be oscillating as
you vary y.
And at a given point in space,
if you study it as a function
of time,
it'll be also oscillating in
time with an angular frequency
ω.
So let me relate these
quantities to more familiar
things.
 
If the wave looks like this,
and E is going to point
in that direction,
you know that if you go a full
wavelength you come back to
where you are.
So let's call the wavelength
λ.
It's clear then whatever
happens at 0 happens at
λ.
 
At zero, ky is 0,
at λ,
ky should be 2Π.
 
When you go a distance lambda
it should be equal to a 2Π
change.
 
So λ,
which is more familiar to you
as the wavelength,
is the reciprocal up to the
factor of 2Π
of this k.
k is called the wave
number.
The bigger the k the
more rapidly the wave
oscillates.
 
That is bigger the λ
the long wavelength means more
slowly the wave oscillates.
 
Similarly you can come to the
second guy and you can ask
yourself--let me continue.
 
So that's what E is.
 
B, again,
is a function of space at a
given time if you draw it,
but the B vector's
pointing this way,
so it'll look like this,
and you can keep repeating
this.
So E is going back.
 
So now this is the thing to
understand with this graph.
You can see this graph,
but you should know what it
means.
 
This whole graph describes the
situation at one given time only
at points on the y-axis.
 
I've drawn it only at points on
the y-axis.
So this arrow,
even though it sticks out here
and there, is not talking about
a whole region.
It's talking about that one
point.
For example,
at this one point the electric
field points this way and has
that magnitude.
At this point the electric
field points down and has that
magnitude,
and the magnetic field points
away from the board,
here the magnetic field into
the board and away from the
board,
okay?
 
That's the situation at one
time.
Then you can ask yourself what
happens later.
You can do it in two ways.
 
First of all let me tell you
the relation between
ω and the time
period.
If you wait a full time period
everything should come back to
where it was,
so the time period T by
definition will be this
condition.
Therefore, the time period that
you guys are all familiar with
is also related to
ω by a similar
formula.
 
So one way to write this wave
that is in terms of things
you're more familiar with would
be to say it looks like
E_0
sin(2Πy/λ
- 2Πt/T).
 
This is completely equivalent
to what I wrote because these
are the definitions of
ω and k in
terms of λ and
T.
Maybe writing it this way is
more transparent - that if you
add a λ
to y on the top,
if you say y goes to
y λ
you're adding a 2Π.
 
If you add capital T to
small t upstairs you're
subtracting the 2Π.
 
That doesn't matter.
 
This shows you the periodicity
in space of λ and
in time of capital T.
 
You can also write
ω as
2Π/T,
but 1/T is the
frequency.
 
This is what we normally think
of as frequency,
so many hertz,
megahertz or whatever.
That's the frequency.
 
The angular frequency omega we
use is 2Π times that,
because every time the thing
completes a cycle it goes by
2Π,
and if it does f
revolutions per second the
angular frequency is
2Πf radians per
second.
That's the difference between
revolutions per second and
radians per second.
 
Okay, so this is another way to
write it, and B will be
the same thing.
 
So you should understand what
this wave does.
So imagine the plane wave is
coming from the blackboard
towards you.
 
If it's polarized this way--now
I should tell you what
polarization is.
 
Polarization is the direction
of E.
B is understood to be
perpendicular to it,
and the direction of
propagation is perpendicular to
both of them.
 
You can see E x B
is the direction of propagation
of the wave.
 
You do the cross product from
E to B and that's
the way it advances.
 
So what you will see is that if
these rows in this classroom
were not curved,
but straight,
then all the people in the
first row will see exactly the
same electric field.
 
So maybe it's pointing up.
 
Then people in the second row
may see no electric field
because they happen to be
sitting at that distance.
People in the third row may see
a negative electric field.
So every half-wave length it'll
go from 0 to maximum to 0 to
minimum back to 0,
and so it oscillates at one
time.
 
If you wait a little bit the
whole pattern will shift towards
the back of the class.
 
So what these guys saw now you
will saw a little later.
How much later?
 
It's the time light takes to go
from this row to your row.
That's the delay,
ωt,
in fact I told you,
you can write ky - ωt
as--
let's see, as k times
y - ct because
ω = kc.
And I told you very early on
that any function of y - ct
is a solution to wave
equation,
and this tells you there's a
wave propagating to the right at
speed c,
because whatever it does at one
time,
say at t = 0,
at later time it's shifted by
an amount ct.
Now if I send an
electromagnetic wave towards
you,
first thing you should notice
is the electric field is
perpendicular to the plane,
in which things are happening
are normal to the direction of
propagation,
so the field vector is
perpendicular.
 
Both E and B are
in this plane and the wave is
going that way.
 
For this reason it's called a
transverse wave.
So transverse means all the
action is in a plane
perpendicular to the motion.
 
So one example of transverse
waves is if you hold a string;
I give it to the people in the
last row and I hold it and then
I shake it.
 
These ripples will go from me
towards the end of the room.
The ripples are going like this
but the wave is going that way.
That's a transverse wave.
 
Or if I shake it this way,
same thing, it's going from
side to side.
 
The wave is going from me to
you.
A longitudinal wave is a sound
wave.
For example,
when I talk now some diaphragm
in the throat pushes the air out
and compresses it,
and maybe when it goes the
other way decompress it,
and the motion of the air is
back and forth,
namely in the same direction as
the propagation.
So sound waves are
longitudinal,
light waves are transverse.
 
Yes?
 
Student:  You said the
direction of
E x B gives the
direction of propagation.
Is there a significance to the
magnitude of that cross product?
Prof: Yes.
 
We'll come to that later today.
 
That's correct.
 
So E x B not only
tells you the direction it's
reasonable to ask "If it is
a physical quantity,
what does it mean?
 
And we'll find out shortly what
it means.
Now polarization is the
direction of the electric field
in the solution,
but we've got to realize that
not every electric field has to
look like this.
I can make a new problem where
I have solution looking like
this.
 
What is this guy doing?
 
Can you tell it's going the
other way, because I just
changed that to that.
 
So that's the plane wave going
to the left.
Now what should I do with the
B field?
Its magnitude should be
E/c.
I can also change this to going
backward, otherwise they don't
form a pair.
 
E and B have to
form a pair.
But I claim this is not enough.
 
If you took this pair it'll,
of course,
satisfy the wave equation
because any function of y
ct will satisfy the wave
equation,
but I want you to remember
there are four Maxwell
equations.
 
You jiggle them all you get the
wave equation.
It's just one consequence of
the four.
You've got to satisfy all the
Maxwell equations,
and some of them won't be happy
with this choice in which I
reverse just the velocity.
 
Can you think about what's the
way to fix that or what's wrong
with that?
 
I'm just saying that if the
wave traveled to the left
instead of to the right that's
what I've done here.
I would just reverse the
direction of propagation.
That's not a good answer.
 
Yes, you've got an answer?
 
Student:  Does it have
to do with the fact that they're
a cross product?
 
Prof: Yeah,
so what's about the cross
product?
 
Student:  To make an
inverse cross product you have
to make it negative.
 
Prof: You want the cross
product to dictate the direction
of motion.
 
So if E and B are
like this the cross product
comes this way.
 
If you want them to go the
other way you've got to reverse
the direction of B so it
points that way.
That is another pair.
 
So this, in other words,
you change the velocity here,
but you forgot to change
it--not you,
I mean, I did not change it
because I wanted to show you
that you just cannot put
together any E and
B that you like.
 
You've got to make sure that
these form a pair.
This is the pair that obeys
Maxwell equation.
Now I know this will work for
another reason.
If you took this long wave,
okay?
I want you to think in your
head.
Rotate it around this axis by
180 degrees.
Can you do that in your head?
 
Take that axis turn the whole
pattern and it's going the
opposite way,
but in the process you can see
this B will rotate from
pointing this way to pointing
that way.
 
And one of the properties of
nature is that if something is a
solution, then you rotate the
whole thing, that's also a
solution.
 
You've got to be careful when
you say rotate.
For example,
if you have a grandfather clock
and you rotate only the clock it
won't work the same way.
That's because the clock is
very sensitive to the earth,
but if you took the earth and
the clock, you rotate both of
them it doesn't matter.
 
In fact, that's happening all
the time.
So all the things relevant
should be rotated,
but for electromagnetic field
in vacuum there's nothing else
to rotate that matters so you
rotate it.
Now another thing you can do
instead of rotating around that
axis you can rotate it around
this axis, around the axis of
propagation.
 
You can turn E to that
or you can turn B to
that.
 
I'm not good at drawing
pictures, but in the E-B
plane you had them like
this, E and B,
the thing is coming towards
you.
I'm saying you can turn this by
some angle and turn that by the
same angle.
 
That'll also form a good
solution.
It must.
 
I mean, what's wrong?
 
If you look at a wave this way
it should still obey Maxwell
equations, right?
 
The titled frame of reference
is as good as the original one.
All right, so now we've got a
few properties of waves,
and one thing maybe worth
mentioning is the light from the
light bulbs in this room is a
chaotic set of waves being sent
to you.
 
Each atom emits light at a
different polarization not in
sync with the other atom,
so it doesn't have a definite
frequency.
 
It's a big jumble,
a big mess, but it's a
superposition of elementary
plane waves.
Actually, I should amend myself.
 
Plane waves are an idealization.
 
Every wave you have,
like if you turn on the light
bulb the waves go out
spherically from the center,
but far from the center when
this sphere is 10,000 miles in
radius if you're a little
creature at the end it will look
like a plane wave,
just like a sphere would look
like a sheet to somebody who is
very near like the earth does to
us.
 
Yep?
 
Student:  Professor,
could you just say again what
you said about switching the
cross product or just ________?
Prof: Yeah,
I was saying that the property
of the electromagnetic field
that you saw here is that if you
take the cross product,
E x B...
Student:  You get the
direction.
Prof: You get the
direction.
I then took an opposite
direction of propagation,
but took the same E and
B.
They don't fit.
 
If you reverse the velocity
here you've got to change this
so that the cross product of
E and B vectors is
pointing the opposite.
 
Student:  Okay,
thank you.
Prof: Now when you buy
Polaroid glasses I think you
know what they do,
right?
The polarizers in the glasses
will allow light to travel in
only one polarization,
so even though the stuff coming
in can be randomly polarized,
what comes out of the other
side towards your eye is
pointing only one way,
so you cut down about fifty
percent of the light if you
polarize it.
 
In fact, when light reflects
off a shiny surface and comes to
your eyes it tends to be
polarized horizontally.
Therefore, if you choose your
Polaroid to be vertical in your
lens, that's the most effective
way to cut the glare.
Also, you can take two lenses,
Polaroid lenses,
and you turn them one relative
to the other you can cancel the
light completely,
because the first one may let
it go this way,
the second one will only let it
go that way.
 
So if you chop it like this
first then it won't go through
the other way.
 
Anyway, light is a very
extensive and interesting
subject which probably is more
important to you than some of
these fields,
but we don't have time to go
into all the aspects of that.
 
I do want to mention to you
that the light that you and I
see has a very limited range of
the possible wave lengths like
400 nanometers or 4,000
angstroms,
namely 400 times 10 to the
minus 9 meters.
To roughly double that is all
you can see.
Stuff on the other side is
ultra violet,
then you've x-rays,
on the other side you've got
infrared,
you've got radio waves,
but they're all electromagnetic
waves.
All you're doing is varying
ω,
but nature designed our eyes to
respond only to a range of
ω's,
because that's where our
enemies were.
 
If you've got different enemies
maybe you will have different
eyesight.
 
Maybe if you've got a lot of
enemies you'll have eyes all
over your head like some animals
do, so we got two.
Anyway, that's another
interesting thing on how nature
adapts to properties of light.
 
All right, so let me tell you
now about the energy contained
in this wave.
 
You've got to agree that when
you have an electromagnetic wave
you have energy that you did not
have before.
Let me ask you a first question.
 
If I'm sending an
electromagnetic wave towards
you, and I ask you,
which way is it polarized,
what will you need to check
that, any idea?
Look, I've told you.
 
If you cannot measure something
or tell me in principle how
we'll measure it you don't know
what you're saying.
Yes?
 
You have an idea?
 
Anybody?
 
Yeah, go ahead.
 
Student:  Was there
anything lost maybe?
Prof: Pardon me?
 
Student:  Was there
anything lost?
Prof: Yeah,
but forget the polarizing.
Go back to basic definition.
 
How will you know there's an
electric field somewhere?
You think you'd see a little
arrow sticking out in space?
Yes?
 
Student:  You take a
test charge.
Prof: Take a charge and
you put it there and you see
which way it moves.
 
If it moves like this that's
the polarization,
if it moves like that,
that's the polarization.
In fact, the antenna on your
radio has got a wire and the
electric field from the radio
station comes and starts moving
the charges in the direction of
the polarization.
So you guys should know this.
 
I'm going to assume that you
didn't answer me because you're
unusually modest,
because you must know how to
measure electric field.
 
I could even ask you,
how would you find the magnetic
field," But I don't want to
do that now because we should
always know how to measure these
things.
So anyway, we know that when
you have an electric field it
has got energy.
 
For example,
when you took a capacitor,
it took some energy to charge
it, to rip the charges from one
plate and ram them in the other
plate.
We saw that as an energy stored
between the plates in the form
of an electric field,
and the energy of the electric
field was
ε_0
E^(2)/2.
 
Now you may ask me,
you found the energy density in
a capacitor,
now we're talking about the
electric field in vacuum
traveling from some radio
station,
how do I know this formula is
good?
 
And the remarkable thing about
the field is that any expression
you derive for it is a local
expression.
It only cares about what the
field is at this point.
It doesn't care what the origin
is.
It doesn't matter if this is
produced by static charges or
maybe it's an electric field
produce by changing magnetic
field.
 
It does not matter.
 
The answer does not depend on
the context.
This is the energy density,
energy per unit volume.
For a magnetic field it looks
like
B^(2)/2μ_0,
and you all know by now that
whatever ε_0
does μ_0 does in
the other place.
 
If this guy's up that guy's
down, or in the field laws
μ_0/2Π's up and
1/4Πε_0
epsilon is down.
 
Therefore, if you have a region
where there is,
let's say, no electromagnetic
field and suddenly a wave goes
by that region has now got
energy,
and how much energy do we have?
 
The electrical energy,
you can see,
is ε_0 over
2E_0^(2)
sin^(2)(ky -
ωt).
And the magnetic energy is 1
over 2μ_0
B_0^(2)
sin^(2)(ky -
ωt).
 
I will now show you that these
energy densities are actually
equal.
 
They're equal because
½μ_0--
B_0 is
E/c,
so let's write it as
E_0^(2)/c^(2).
From the definition of the
velocity of light and
μ_0 and
ε_0 this is
simply ε_0
over 2E_0^(2).
Sorry, E_0^(2)
plus all the sine squares,
which I forgot.
 
Therefore, the total energy is
equal to E_0^(2)
ε_0
sin^(2)(ky -
ωt).
 
You got a half from this and
you got a half from that.
Even though the formula looks
different by the time you put in
the relation between
μ_0 and
ε_0 and
B_0 and
E_0 it turns
out to be equal.
So even though the magnetic
field is weaker than the
electric field by a factor of
1/c you might think it's
negligible in terms of energy,
but it's got same energy
density, and you add them up you
get that.
Now you can see this energy
density is time dependent and
space dependent because it's
oscillating with time at a given
point,
and oscillating with space at a
given time.
 
So you can ask yourself,
let me sit at one place and ask
myself,
"What does the average
energy density average over a
full cycle?"
If something is up sometime and
down sometime what's the
average?
 
The average will be
ε_0
E_0^(2)/2,
because the average value of
sin^(2)θ over a
full cycle is 0 to 2Π
divided by 2Π
which is ½.
This is something we have done
before in circuits.
You should know that.
 
Average of sine squared is
half, average of cosine squared
is half,
and the check is that sine
squared plus cosine squared
average is 1 because that's 1
identically.
 
So try to remember this.
 
So this is the average energy
density.
Next I want to ask the
following question.
What is the rate at which
energy is coming at me?
And that is called intensity,
I, and it is equal to the watts
per meter squared,
just like a flow of liquid
except this is the flow of
energy.
So what I want to do is take a
square meter,
stand in the way of the beam
and ask how many joules cross
per second,
or if you want,
how many watts per square
meter.
That's easily calculated from
the energy density and I'll give
you the reason.
 
It's identical to the reasoning
for currents traveling in a
wire, or fluid flowing in a
tube.
Suppose you take a region,
a cylindrical cross section
through which electromagnetic
waves are traveling?
This is cross section A and I
wait 1 second it'll go a
distance c.
 
So all the stuff in a cylinder
of base A and length
c that energy will go
past this checkpoint.
Can you see that?
 
It's like toothpaste.
 
You squeeze the toothpaste
these guys get passed in 1
second.
 
If I want per square meter
we'll just call that 1.
Therefore the intensity is
simply the energy density
multiplied by the velocity of
light.
That's the rate at which the
energy flows.
So let us now calculate that
and you'll get a very
interesting result.
 
The intensity is equal to the
energy density times c.
The energy density was
ε_0
E_0^(2)
sin^(2)(ky - ωt).
This is not the average energy.
 
This is instantaneous.
 
I'll average it in a moment.
 
Are you guys with me now
somewhere here,
here?
 
Now I'm going to write it as
follows, ε_0
and E_0
then at B_0--I
forgot a c here.
 
You realize that?
 
That's another c.
 
Then E_0 is
B_0 times another c
times sin^(2)(ky -
ωt).
And what is that?
 
c^(2)
ε_0 is
1/μ_0.
 
So E_0
B_0 over
μ_0 sin^(2)(ky -
ωt).
But if you now define a vector
S, it's called a
Poynting vector,
which is spelled with a p,
to be E x B over
μ_0.
The magnitude of that vector is
precisely the intensity.
The magnitude of S,
which you can either call
S or you can call it
I, is exactly this.
So E x B except
for a factor 1/μ_0
not only gives you the direction
of the way of propagation it
tells you how many watts are
going to cross a square meter,
or how many joules are going to
cross square meter in 1 second.
Now these are things you'll
find in all the textbooks,
so.
 
All I've done is take this u c,
put it in the formulas with
E and bring B back
into the picture so that it's
symmetric between E and
B,
because this gives the
impression it's all electric.
Remember, this is electric and
magnetic.
They just happen to be equal.
 
This way you can see the role
played by E and B.
The Poynting vector tells you
the flux of energy.
So here's one example.
 
At the surface of the earth if
you take a square meter,
and here's the sun,
emitting light.
You can ask,
"What's the intensity of
sunlight?"
 
Anybody have an idea how many
watts per square meter from the
sun.
 
Student:  A thousand?
 
Prof: Yes,
very close to a thousand.
I think there's some other
numbers, but for our purposes
it's very close to 1,000 watts
per meter squared.
That's pretty amazing over the
entire surface of the earth
every second the sun is pumping
in 1,000 joules.
And you've got to remember the
context of the sun.
I mean, here is the sun and
here we are ninety-three million
miles away, and the light energy
is going and this is our share.
A tiny circle like 7,000 miles
in diameter you're intercepting
the light, and every square
meter of it gets a thousand
watts.
 
You can see the stuff coming
out of the sun,
its a prodigious amount of
light.
Anyway, that's the electric
field.
I mean, that's the intensity,
so let me write the following
thing.
 
This is oscillating rapidly
with time, so let's define an
average intensity as the average
of the sine squared.
That's E_0B
_0
/2μ_0.
 
All these averages are simply
half, anything involving in sine
squared.
 
I'm not going to worry about
the half,
but if you took this to be the
average intensity you can ask,
"How big is the electric
field that comes with it,"
because that light is going to
be this electric and magnetic
field.
 
That's all light is.
 
So the sunlight produces
electromagnetic field,
is electromagnetic waves and
I'm asking, "How big is the
E vector?"
 
And all you have to do is stick
that into this number.
If you want you can get rid of
B and go back to E
because E is just
B times c.
You'll find a pretty surprising
amount.
It's roughly 1,000 volts per
meter.
Remember electric field is
measured in volts per meter.
What that means is if you took
the field,
and it's uniform in space,
between one place and another
place there's a potential
difference of 1,000 volts,
or it'll take 1,000 joules to
shove a coulomb from lower
potential to higher potential.
 
That's a pretty strong field,
but it's very incoherent in
direction.
 
It's doing this for a while and
doing that for a while,
but if you roughly approximate
it and just ask,
"What is the average field
strength,"
this is the number you get.
 
So let's talk about one thing
which I've not discussed at all,
which is, where are these
electromagnetic fields coming
from?
 
I said you don't need ρ,
you don't need I,
you don't need the current,
you don't need the charge,
these can exist in free space.
 
But what's the origin of the
electromagnetic waves?
Anybody know?
 
When can you get them?
 
It didn't happen in anything we
studied.
Yes?
 
Student:  Electron
changes energy level?
Prof: Well,
you can say in the atom the
electron changes energy level,
but before you do the quantum
mechanics.
 
In classical theory one can
ask, what is the electric field?
When does the electric field
produce these waves?
You take a static charge it has
a 1/r^(2) field which is
pinned to the charge.
 
If you go too far from the
charge you don't see it,
but these guys can go off into
space without being anywhere
near a charge or current,
but what produces them?
Yes?
 
You don't know?
 
Yep?
 
Student:  When it
oscillates.
Prof: Pardon me?
 
Student:  When is
oscillates.
Prof: Which oscillates?
 
Student:  The electric
field.
Prof: The electric field
is itself oscillating,
but what's producing the
electric field?
What's the cause of the
electric field?
Electric field fairy?
 
Yes?
 
Student:  When a charge
oscillates?
Prof: Yes.
 
In other words,
rather than oscillates the more
general answer is whenever a
charge accelerates.
This is a very important result.
 
Waves are produced by
accelerating charges.
If they travel at uniform
velocity like in a wire they
don't produce oscillations,
electromagnetic waves,
or if they're going in a
circuit at constant rate that
doesn't produce.
 
But if you have charges which
are say oscillating is one
example when things are
accelerating,
right, because you're going
back and forth when you radiate
light.
 
So every single source of
electromagnetic wave is
oscillating charges.
 
You can sort of imagine why
that will happen.
I mean, if you took a capacitor
plate,
and you connect it to an AC
source,
let's say, there may be other
things to keep it from burning
out,
then what will happen is
charges will be like this for a
while then they've got to go
back and forth in order to
change polarity with the
alternating field,
therefore the charges are going
back and forth,
and you have a time dependent
electric field here.
 
When you have a time dependent
electric field you'll have a
magnetic field going around it
because the line integral of B
will involve the rate of change
of electric flux.
And that will also be time
dependent, but if that gets time
dependent there'll be an
electric field going around
that.
 
So basically these will curl
around each other whenever
they're dependent on time,
and they can then free
themselves loose from the
capacitor and take off.
All you need is two plates,
and an AC source,
and you connect them,
you will make electromagnetic
waves.
 
You'll make them at the
frequency of the source,
so you won't be able to see it.
 
Your dog won't be able to see,
but some gadget will be able to
pick it up.
 
That's all you need,
oscillating charges.
So what happens in a radio
station is you could imagine a
simple radio station with an LC
circuit,
the current is oscillating at
some rate,
and part of the circuit is in
the antenna,
and the charges are going up
and down as the current goes
back and forth that sends out
the waves,
and the waves come to your
house.
So here's the picture.
 
Here's the radio station,
and the waves are emitted in
big circles, and this is your
house, and here's your little
antenna.
 
It's a piece of wire,
and the electric field,
if it's polarized this way,
will move the charges up and
down, and charges can be part of
an LC circuit.
So here are the antennas,
if you like;
part of the circuit.
 
As the charge goes up and down
the AC current will try to flow
here.
 
And if you tune this capacitor
so that it resonates with the
frequency you'll get a hefty
signal from the radio station.
So in the end it's all charges.
 
Charges produce the field.
 
Charges respond to the field.
 
That was true in the static
case.
That's true in the time
dependent case.
Yes?
 
Student:  When you say
that the field if it's loose
from the capacitor,
what do you mean free from the
capacitor?
 
Prof: It means it can go
thousands of miles from the
capacitor.
 
And if you simply look at the
Coulomb force due to the
charges, the plus and minus
charges, they die like
1/r^(2).
 
So if you calculate
1/r^(2) you will get a
negligible number compared to
the actual electric field.
So it's really like you and
your parents.
I mean, at some point you are
free from your parents.
You are able to manage on your
own, but you had parents
somewhere sometime,
right?
That's what it is.
 
The electromagnetic waves can
go on their own,
but they are not produced on
their own.
They're produced by charges.
 
It is just that unlike the
static fields,
which are very near the
currents and charges that
produce them,
the time dependent fields
propagate on their own.
 
E keeps B alive
and B keeps E
alive.
 
If E tries to die
there's a
dE/dT that
produces a B.
If the B tries to go
down it produces at
dB/dT that
produces an E,
so they go back and forth.
 
It's really like oscillations
in which you have kinetic to
potential transfer.
 
You can have energy transfer.
 
So the fields cannot die.
 
They are self-sustaining,
but to get all of that physics
you had to put the term that
Mr. Maxwell put in.
Without that term you don't
have this phenomenon.
You don't get it from statics.
 
So one typical problem you have
is this radio station is,
let's say, 100 kilowatts and
you're sitting here at some
distance r from the radio
station.
Then the intensity in your
house will be 100 kilowatts
spread over a sphere of radius
R.
That will be your intensity.
 
Then you can go from the
intensity and translate it to an
electric field and say,
"The electric field
produced by the radio station
oscillates with the following
amplitude.
 
I've got to build a circuit
that's smart enough to pick up
that tiny field."
 
Okay, so I want to switch now
to my favorite theme.
The remarkable thing about
electromagnetism is that you can
ask what happened when physics
went through the Einstein's
revolution with special
relativity.
We know everything changed
after Einstein,
and all the Newtonian mechanics
had to be modified.
And so far I never mentioned
the word relativity,
so you can ask yourself,
"How are these modified by
Einstein's work?"
 
So first let me tell you,
remind you, now you guys did
relativity last term,
right?
Is there anybody who's never
seen it before?
Okay, well you don't have to
know a whole lot,
but let me just say the
following.
In the Newtonian world there
was a principle of relativity
according to which the equations
like F = ma you can ask,
"Who is allowed to use
this equation?"
It says by definition an
inertial observer can use this
equation.
 
And you say,
"Who is an inertial
observer?"
 
You say, "Inertial
observer is anybody who can use
this equation."
 
Sort of seems to be meaningless
tautology.
What makes it meaningful is the
following.
There are some people,
at least, for whom this
equation works.
 
So there are at least some
inertial observers.
For example,
we are an inertial observer
because if you want to test this
equation you can say,
"I leave a piece of chalk
here.
I don't apply force.
 
Does it have an
acceleration?"
It doesn't.
 
Okay, so I'm obeying at least
the first of the three laws of
Newton.
 
On the other hand,
if I leave my iPod in Grand
Central Station and I come back
it's gone, that's not a
violation of Newton's Laws.
 
That just means I'm stupid
because there are other forces
acting on that iPod,
and those forces moved it,
so I can understand that.
 
So I won't be that traumatized
by the loss of the iPod because
I feel as I know there's an
explanation.
But if you go to a plane and
the plane is about to take off
you leave stuff on the floor it
will slide to the rear end of
the plane.
 
You have no F,
but you have an a.
That means an accelerating
plane is a reference frame in
which people cannot use F =
ma.
Stuff will accelerate for no
apparent reason,
so not everybody's inertial.
 
So you can ask,
"If I have at least one
inertial observer in the
universe does it imply
others?,"
and the answer is yes.
If I'm inertial and you move
relative to me at constant
velocity you're also inertial.
 
There's a large number of
people in the universe all
allowed to use Newton's Laws if
you've got one of them,
and they differ by velocity,
constant velocity,
and we can understand that from
Newton's Laws.
Here is a mass and spring
system.
Newton's Law takes the form
-kx = ma,
let's say, or let's write
mdv/dt.
Now if you go to a train or you
carry this on a train and you go
at constant velocity I know it
will obey this equation because
I've not done anything.
 
It's not my fault you're going
on a train and this is riding
with you on the train.
 
It will obey this equation.
 
By the way, this x is really
x - x_0,
where x_0 is
the rest length of the spring
and x - x_0 is
the deviation from that.
That's the equation.
 
Now you and I differ by what?
 
You and I differ by a constant
velocity.
The constant velocity just
means x prime is equal to x -
ut.
 
Our origins are differing by an
amount ut after time
T,
so if an event occurs here for
me, you have moved a distance
ut and it occurs at a
distance x' which is x
- ut.
Then you can see the laws are
going to work for you also
because even though we disagree
on velocity we don't disagree on
the rate of change of velocity.
 
Your velocity and mine differ
by a constant,
which has no derivative,
so they have the same
acceleration.
 
As far as the spring is
concerned if I think it's
stretched by two inches you will
also think it's stretched by two
inches.
 
I mean, your x is
zooming to the right according
to me and your
x_0 is also
zooming to the right,
but the extension of the spring
is the same.
 
So F = ma doesn't have
to be modified by going to a
moving frame.
 
And that is the relativity of
Newtonian mechanics.
If these are the laws of motion
we can understand why if you're
inside a train,
which is completely closed,
you cannot look outside and
it's going at uniform velocity
with respect to the ground,
you cannot tell.
You cannot tell because nothing
you do will be different.
Because everything you observe
in the world is controlled by
Newton's Laws,
and Newton's Laws are
unaffected by adding a constant
velocity to everything.
So when you wake up and I say,
"Is the train
moving,"
you cannot tell.
Okay, so that's by doing
physics experiments.
You cannot say,
"It says Amtrak so I know
it's not moving."
 
That kind of argument is based
on sociological axioms.
I'm just saying can you--after
all it's possible,
theoretically,
Amtrak trains can move,
so you must admit the
possibility.
Okay, now you come to Maxwell's
equations and electromagnetic
theory.
 
Let me write down what we have.
 
The first thing we have is
q equals--I'm sorry,
q times E
v x B is the
force.
 
Then let me write down one
other consequence,
d^(2)E/dx^(2) −
(1/c^(2))d^(2)E
/dt^(2) is 0.
 
These are some of the results
we got from electromagnetic
theory.
 
Now comes the important
question.
This is the velocity of a
particle according to whom?
That question came up earlier
in the class.
Who is supposed to use it?
 
I may assume it worked for me,
but how do I know then when you
see it it's got a different
velocity will you get the same
physical world that I get.
 
Now let's look at this
equation, this also came from
Maxwell's equations,
and compare the equation for a
string.
 
Let me call ψ
as the displacement of the
string rather than y,
d^(2)ψ/dx^(2) is
(1/v^(2))d^(2
)ψ/dt^(2).
I just wrote the equation
shifting to the other side.
They look very similar.
 
Here v is the velocity of the
waves according to a person for
whom the string is at rest.
 
Okay, according to whom the
string is at rest.
You understand that?
 
This velocity,
because the waves are traveling
in the string,
a at speed v.
So this equation is to be used
in its present form only by a
person for whom the string is at
rest.
If you want to see the string
from a moving frame then
x' is x - ut,
and in classical mechanics
t' is equal to t.
 
You can do the change of
variables with these partial
derivatives.
 
I don't want to do that,
but you can say d/dx of
ψ is equal to d/dx'
of ψ times dx'
over dx,
then dψ/dt',
then dt' over dx.
 
Now t' is the same as
t, but formally we can
change variables,
and we can take this equation
and rewrite it in a frame moving
to the right at speed u,
and I promise you it won't look
like this.
It will look very different.
 
More importantly at least
understand conceptually that the
velocity u of the moving
observer will appear in the
final equations,
because these derivatives,
x' over x and so
on contain the velocity
u.
 
Can you see that,
dx'/dt has the velocity
u?
 
So when you make all these
changes and put them in you'll
get a new equation involving
x' and t' in which
the velocity u will
appear.
So if you are that person you
have to decide what's the
velocity to use for you,
and the answer is unique.
It's your velocity relative to
the string.
The string is anchored in the
lab.
If you happen to have a speed
u relative to that that's
the speed you should put in.
 
Therefore, the equation is not
the same for everybody.
There is a special observer for
whom the string is at rest,
namely in the laboratory frame
for whom this equation works,
and v is the velocity
for that person.
Now we come to this equation.
 
It has no reference to the
velocity of the observer.
It's got a velocity of light,
and you can ask who is supposed
to use it.
 
Whereas in the string we know
the privileged frame of
reference is where the string is
nailed down, but the light is
traveling in vacuum.
 
There is no frame of reference.
 
People thought maybe even the
vacuum contains a medium called
ether.
 
Because everything needs a
medium to travel they said there
is an ether.
 
Then, of course,
this is to be used only by
people who live in that ether at
rest,
but we are moving relative to
the ether because we are on the
earth which is going around the
sun.
You may say,
"Well, maybe today I just
happen to be at rest relative to
the ether.
That's possible,
but then tomorrow I cannot be
because I'm going around the
sun.
Six months from now I'm going
the opposite way around the sun
at a huge speed,
but what I find is every single
day of the year I'm able to use
these equations.
That means they apply to me no
matter what my velocity is.
So these equations,
it turns out,
are valid for any observer who
is inertial, namely one for whom
at low velocities Newton's Laws
apply.
Now what people were worried
about in the old days they said,
"Look, let's take x'x -
ut and t' equal to
t and put it into this
equation."
Then they found the equation
changed their form because it's
just like this wave equation
here.
Then they said,
"We've got to change the
equation because it depends on
our speed u,
but there is no valid choice
for what our u is.
What is our speed relative to
this magical ether?"
That was what they were worried
about until Einstein came and
said,
"There is no either and
this is the wrong set of
transformations."
If you use x' equal to
x - ut divided by this 1
− u^(2) over
c^(2) and t' is
t - ux over c^(2)
divided by the same square root,
if you change coordinates this
way you'll find amazingly,
if you change all the
d/dx's to d/dx'
and did the whole partial
derivatives and chain rule and
so on you will find that in the
new frame of reference you'll
find the equation will look like
this.
It will look the same for
anybody moving relative to me at
any speed.
 
That's why it's not clear that
I'm the privileged user.
All people at uniform relative
motion can use the very same
equation with the very same
number c entering.
So this was the great triumph
of the Maxwell theory.
It was that it was already
consist with the relativity.
In fact it is what led to
Einstein's revolution because
this equation said no matter who
you are a light pulse is going
to travel at a speed c
for you,
no matter who you are.
 
That's very strange because
every signal we know has a
property that if you move along
the signal its speed is reduced,
right?
 
If you've got a bullet going at
700 miles per second if you
travel at 400 you will think it
is going at 300,
but if it's a beam of light
it's supposed to have the same
velocity for everybody,
even those moving in the same
direction and the opposite
direction.
It doesn't matter.
 
Therefore, something had to
change with our definition of
space, and time,
and velocities,
and Einstein replaced it with
these new equations.
And one of the consequences of
that equation is that if I have
an object that is going at a
speed v and they're moving the
same direction at the speed u in
the old days you will subtract
your speed in the same
direction,
but the correct answer is--this
is 1 - uv over
c^(2).
 
This is the relativistic
equation about how to change
velocities.
 
v is the speed of an
object according to me.
You are moving in the same
direction at speed u.
You will measure the speed w
given by this.
If u and v are
much smaller than c you
can forget this and it looks
like the good old Newtonian
days,
but if u and v
are comparable to c then
the denominator is one less
than,
one minus something,
so the whole thing will be a
little bigger than what you
thought because you're dividing
by 1 minus something.
And finally,
if what I was looking at was a
light pulse,
that means v is equal to
c,
I get c - u divided by 1
- uc over c^(2).
 
And if you fiddle with that
you'll find it's c.
So this law of transformation
of velocity has the amazing
property that if I'm observing a
light pulse that's got a speed
c you will also get a
speed c.
So it's a very beautiful way in
which the mystery was resolved.
So Maxwell gave these
equations, did not give a
preferred frame of reference.
 
Then you've got to ask
yourself, "What coordinate
transformation should exist so
that the equation has the same
form for everybody,"
because it doesn't tell you who
supposed to use it.
 
Then you get this equation.
 
You can get to this equation
simply by demanding that two
observers looking at a light
pulse somehow get the same
speed.
 
If you fiddle with that and the
symmetry between the two
observers you will get this.
 
Anyway, I just wanted to tell
you that there are many things
you have to change,
but you don't have to change
any of electromagnetic theory.
 
The equations I wrote down are
correct.
Okay, the last thing I want to
do is something I promised long
ago, which is the following
thing.
If you believe in relativity,
namely if you believe that the
laws of physics should have the
same form for people in uniform
relative motion,
you can deduce the presence of
magnetism given just the
presence of electrostatics.
In other words,
suppose you never heard of
magnetism.
 
I can show you that it must
exist.
magnetic forces must exist,
and I show that as follows.
Before I show that you need a
couple of results that you guys
may not remember all the time.
 
First result is if you've got a
wire and it's got n is
the number of carriers,
charge carriers per unit
volume, and A is the
cross section of the wire,
and e is the charge of
the carrier,
and v is the velocity of
the carrier,
then the current is equal to
n,
sorry, nAve.
 
Let me tell you why that is
true.
Look at the wire.
 
I'm going to stretch it a
little bit so it looks like a
thing with a finite cross
section.
It's just like what I did
earlier on.
If you wait 1 second the
carriers in that cylinder will
have crossed the checkpoint,
and the volume of that region
is A times v.
 
This is the number of carriers
per unit volume,
and each one carries charge
e,
that many coulombs would have
gone past this point and that's
the meaning of current,
first thing to know.
Second thing to know,
if I took a rod of length
L and I put some charges
on it,
and they had a certain density
n_0,
if the rod moves at the
velocity v you know it
will shrink,
therefore these plus signs will
be compressed,
and to a person seeing the
moving rod the density n
will be n_0
(the density at rest) divided by
1 − v^(2) over
c^(2).
 
This is a relativistic effect -
that the number of charges get
squeezed because the rod itself
gets squeezed.
So a moving rod which is
charged will appear to have a
higher charge density.
 
And the final result I'm going
to invoke is the following.
What is the charge per unit
length?
If the charge per unit volume
is n then I claim the
answer is n times
a.
That's also easy to understand.
 
If you took a unit length of
this wire,
unit length,
the volume of that is just
A times 1 and that times
the density is the amount of
charge that's there.
 
So if you have a very thin wire
you may like to think about
charge per unit length rather
than charge per unit volume,
and this is the way to go from
one to the other.
Now we are ready.
 
Now I'm ready to show you how
just by thinking you can deduce
the presence of magnetism.
 
And I like this argument
because quite often this is how
people make, theorists make
discoveries.
They will take something that's
known.
They'll appeal to a principle
like symmetry,
or relativity or whatever.
 
Then they will say,
"This implies that there
is a new force,
and I'm going to tell you what
the new force is."
 
So here's what I want you to
imagine.
There's a very long infinite
line of charge and right on top
of it there's another infinite
line of negative charge.
If they're just sitting there
there's nothing interesting.
Now what I want to do is I want
to have the upper thing going at
a speed v.
 
Now n_0^( )^(
)is charge density of plus
charges in the rest frame of the
rod, of the plus rod,
and likewise minus.
 
Are you with me?
 
Each rod has a certain density
when it's at rest.
I'm going to call that with a
subscript zero.
Now I want to arrange this wire
to be electrically neutral.
I want it to be neutral,
and I'm producing the current
by moving the plus charges to
the right.
I'm dragging them.
 
The minus charges are not doing
anything.
So what is the current?
 
The current is going to be the
density of plus charges,
times the velocity,
times area, times the value of
each charge.
 
But for the wire to be
neutral--this is where I want
you to follow me closely.
 
The two of them cannot have the
same densities at rest because
if the plus charge had the same
density at rest as the minus
charge when it starts moving and
it starts compressing the plus
density will exceed the minus
density and the wire will be
neutral.
 
So I cook it up so that
n_0^( ) with
this factor is the density of
minus charges at rest.
Are you with me?
 
I take a rod with positive
charge somewhat less than the
density that the negative
charges are going to have,
but by moving it at the
suitable speed I bring these two
densities to the same value.
 
So I want you to follow this in
some detail because it's not so
difficult.
 
This is the neutrality
condition for the wire.
I want the wire to be neutral.
 
See normally when you don't
think in relativist terms you'll
say,
"Take a plus charge rod
and a minus charge rod and make
sure the charges are equal per
length,
and drag one of them,
or you've got a current."
 
That works in a
non-relativistic limit,
but if you take into account
length contraction you won't get
neutrality unless the plus
charge had a slightly smaller
density in its own rest frame
that got boosted by this factor
to equal the density of negative
charge.
Okay, now imagine a particle at
rest here, and you know all
about electrostatics.
 
What do you think it will do?
 
You've never heard of magnetism.
 
You've heard of electrostatics.
 
What will this charge do?
 
Yep?
 
Student:  It is positive.
 
It'd probably be attracted.
 
Prof: Why would it be
attracted?
Student:  Because the
negative wire has more density
and...
 
Prof: No, no.
 
That's what I said.
 
No.
 
I arranged it so that the
positive wire,
the moving positive wire has
the same density as the static
negative wire.
 
You understand?
 
I cooked it up so that the wire
is neutral.
All I'm telling you is,
if one set of charges are
sliding to the right at a
certain speed that rod must have
a density that's somewhat lower
in its own rest frame,
but by the time you translate
it to the laboratory frame it's
electrically neutral.
 
So what'll this charge do is my
question.
Student: 
>
Prof: Pardon me?
 
It should not experience any
electrical force,
you understand that?
 
Our prediction is that it will
just stay there because it's
neutral and all we know is
electrostatics.
It can have any velocity,
but I want it to have a
velocity v.
 
Electrostatics doesn't care
what the velocity of the charge
is.
 
It's still supposed to go in a
straight line because the wire
is neutral.
 
But now I say I'm allowed to
view the physics from any frame
of reference I want.
 
Let me go to the frame of
reference of this charge.
I move to the right.
 
So in general,
if you want to analyze the
problem carefully,
you can have this wire moving
at one speed and this charge
moving at a different speed,
but I want to do the easier
case where they're both the
same.
 
It makes life easier.
 
So go to the rest frame of
charge.
You can already see what's
going to happen in the rest
frame of the charge.
 
If you go to the frame moving
with the particle the plus wire
comes to rest,
so the density of the plus
charges will just become
n_0^( )^( ),
the rest density of the plus
charge,
but the minus charges now they
will be boosted by this factor
because they're going the
opposite way,
and it doesn't matter which way
they're moving.
Things depend on v squared.
 
The minus charges will appear
compressed.
So if previously they're
balanced now they won't balance.
In fact, n_-
is n_0^(- )
over this,
which is n_0^(
) divided by 1 -
v^(2)/c^(2).
Therefore the wire will appear
to be negatively charged because
the negative charges are that
value,
the positive charges are that
value,
so net charge on the wire,
net charge density will be
equal to -n_0^(
)^( )over 1 -
v^(2)/c^(2)
n_0^( ).
If you like,
this is the plus charge and
that's the minus charge times
e, if you like.
And I'm going to make an
approximation.
I'm going to approximate this
is n_0^( )^(
)times 1 v^(2)/c^(2).
 
There are more terms,
but in the binomial expansion
I'm going to stop with the first
term.
So this is the result that's
very good in the limit of small
v/c, but v/c is
not set equal to 0.
It is set equal to a finite but
small amount.
So if you compare those two you
will find the net charge is -
n_0^( )
v^(2)/c^(2).
That's going to be the charge
on the wire.
So to the person moving with
this charge the wire looks
charged.
 
In fact it looks negatively
charged.
So you know,
knowing just electrostatics,
what the charge will do.
 
The charge will be drawn to the
wire, start moving towards the
wire.
 
That's a fact.
 
Well, if it's moving towards
the wire for me it better be
moving towards the wire for the
original person also because
even if you and I move
horizontally the fact that your
moving transverse to a velocity
is going to be a true statement
in both frames.
 
From that you will conclude
that even in the original
laboratory frame this moving
charge would have been attracted
to that wire,
and it's attracted by virtue of
its velocity.
 
So that means there is a new
force in which a charge is
attracted to the current in the
same direction,
right?
 
So let's ask,
"What is the force of
attraction?"
 
In Newtonian mechanics--I mean,
at this point once you've got
this length contraction you can
just think in terms of Newtonian
formulas because the errors you
make involve higher powers of
v^(2)/c^(2).
 
The leading term is this.
 
So look at that.
 
You have a net charge of
n_ ^(0 )v^(2)
/c^(2).
 
You have a charge per unit
length which is the area of the
wire times that.
 
I sorry, that's charge e.
 
That's the density.
 
That's the actual charge.
 
That's the λ.
 
So what's the electric field of
a wire with charge per unit
λ?
 
I remind you it's λ
/2Πε_0
r.
 
So for this particular lambda
it's n_ ^(0 )v^(2)
over c^(2) eA times
½Πε
_0r.
 
In other words in its particle
rest frame there'll be an
electrical field because there's
a net charge,
negative charge,
of that strength on the wire.
Once you've got a charge per
unit length
λ/2Πε
_0r is the
field attracting it toward the
center,
so that's the force.
 
So I'm going to write it as
follows.
I'm going to write it as
n_
^(0)veA_1
over 2Πε_0
c^(2) times another
e and another v.
This is the force,
this e times E.
In Newtonian approximation the
force perpendicular to the
motion is the same for all
observers,
therefore I ask you--and this
force must be present even in
the lab,
but look at this force.
This guy is the current,
nevA is the current.
This guy
1/ε_0c^(2) is
μ_0/2Π.
 
This is ev.
 
You see, this is the exactly
the magnetic force we got,
μ_0
I/2Πr.
I forgot a 1/r here.
 
That is the azimuthal magnetic
field around the wire that times
ev is the force.
 
Because you can tell that no
matter where the charge was,
whether it was here,
or whether it was there or
anywhere around the wire it'll
have the same attraction.
So you can deduce in the
laboratory frame there is an
attractive force towards the
wire,
and that agrees with the exact
formula you got for the
magnetism.
 
If you don't ignore the higher
powers of v^(2)/c^(2) you
will find that the force in the
lab frame and the force in the
moving frame are slightly
different.
That's because in relativistic
theories the force is not the
same for everybody.
 
The force for me and the force
for you do not actually agree
because time is not the same for
everybody.
But the leading order in
v^(2)/c^(2) you can see
this.
 
There's only one thing I did
which is a little fudge,
which you guys may not have
noticed,
is that the density here is
n_ ^(0)^(
)whereas the actual current in
the lab is n_
_ and the
two of them differ by this
factor.
 
However, if you have a
v^(2)/c^(2) in front of
your expression and you've
ignored v^(4) or
c^(4) there is no point
in keeping this guy here because
that makes an error of order
v^(4)/c^(4).
If I'm going to keep that I
should go back right here and
keep such terms.
 
So the leading order you don't
have to worry about the
difference in density.
 
This is a very subtle
calculation because sometimes
you worry about the difference
and sometimes you don't.
So the rule always is if you're
trying to do calculations to a
certain order,
namely v/c the whole
thing squared then things that
make corrections of order
v/c to the fourth can be
dropped.
Anyway, I'm not that worried
about the details,
but I want you to understand at
least what the logic is.
That's more important.
 
I know about electrostatics.
 
I cleared the neutral current
carrying wire.
It's neutral because the
negative charges are at rest at
some density.
 
The positive charges in the
moving rod are also at the same
density, but the rest density is
somewhat lower.
Then I predicted the particle
won't move because it's got no
electric attraction.
 
Then I go to the rest frame of
the particle,
then I find that the positive
charges have come to rest,
and therefore to a lower
density.
Negative charges are moving the
other way, therefore at the
higher density.
 
They no longer cancel.
 
The wire is charged.
 
I expect this charge to be
attracted to the wire.
That means going back to the
lab I expect the moving charge
also to be attracted to the wire
because when something goes
towards the wire that goes
towards the wire according to
all people.
 
But you can go beyond and
actually compute the force and
equate the force and deduce that
that is a new force now.
Whenever you have a current
carrying wire with current
I,
and that's a particle of charge
e moving at speed v it
will be attracted to the wire by
this amount.
 
Okay, so I'll see you guys
after the break,
and my suggestion to you on
what to do with the break is to
try to carry your textbook with
you,
and the textbook has got lots
of problems,
some of which are a lot simpler
than the one I gave in the early
days,
but roughly the same level of
difficulty as the midterm.
 
And do as many problems as you
can.
Look at the worked examples and
try to solve them.
 
 
