Prof: Okay, let's start.
I'm going to do some leftover
stuff from circuits, because I didn't finish it last
time. So I want you to remember the
following things about a circuit.
I'm always thinking of
different ways to explain it. So the things we normally learn
is, if you've got some battery
here, some number of volts, let's give it the symbol
E, and you have a resistor,
we just say there's a 1.5 volt difference.
So you can say this is 0 this
is 1.5. Current goes around and comes
back. And we say the work is done,
because the battery is somehow making the current go around.
But we look under the hood and
found it's somewhat more complex.
It is not done entirely with
the electrostatic force. The electrostatic force is due
to these charges here, sitting on the plate,
but they cannot do work for a charge that goes round and
round. Because if you go round in a
full loop in an electrostatic field, the net work done by the
field is 0. That's because the net work
done is the integral of the electrostatic field and because
it's conservative, you'll get 0.
Yet when charges go round and
round the circuit, they are doing some work,
because the resistor is heating up, delivering heat.
And the hidden secret of course
is not the electrostatic force, but another force inside the
battery that's doing the job. So I gave an analogy which I
think is worth repeating. So here is a ski slope and you
start here and you come down. Let's imagine on the way down
you bump into a lot of trees, that you pretty much come down
at 0 speed. Then you stagger back to the
foot of this ski lift. Now that's the end of your
skiing. There's nothing else,
because gravity is not going to help you anymore,
because gravity is not going to help you go from here to there.
So it's not obviously run by
just gravity. That's when you have the ski
lift, which I'm going to portray as a vertical thing,
where you've got a lot of chairs, so you can sit on one of
them and you can ride to the top.
So that ski lift exerts a force.
Gravity, which is acting down
here, is also acting down here,
and the ski lift, F_ski and
F_gravity, ski lift cancels gravity here and
takes you to the top. So if you define the analog of
an emf, there is no "electro"
about this one. Let's just call it emf.
As the integral of the forces
acting on this one skier, as he or she goes around the
loop, you can write that as the force due to gravity ⋅
dr and the force due to the lift--
or what did I call it? Hey guys, you know what this
was called? Student: Ski.
Prof: Ski.
Okay, ski dot dr.
this is no good.
Gravity does something to you
when you come down and takes it all back when you go up.
This guy, the force inside the
ski, has a non-zero line integral, because it's a
non-zero force only in here. The line integral's basically
the ski force times the distance.
The ski force is just
mg, the magnitude. That times the distance is
mgh. That's why you can say this
point is a potential mgh that lifts up to that.
But inside this region,
gravity is pointing down and the machine takes you against
gravity and puts you on the top. And that's the non-conservative
force that does the work. So the main point was,
electrostatics is conservative, and yet when a charge goes
around and around the loop, the line integral of the total
force is not 0, so it's not entirely driven by
electrostatics, but electrostatic forces are
important. Just like gravity's important
here, because gravity's what helps you come down here.
The ski lift is all hidden
inside, and then it takes you back up.
So I'm going to draw the
circuit in a way that really indicates the gravitational
analogy. So I'm going to draw it not
this way but this way. Here's the top plate and here's
the bottom plate and you may directly visualize the height
here to be the potential. Then there is a wire that has
no resistance. It's going to be flat.
Potential doesn't have to be up
or down for current to flow down this.
Then here is a resistor.
Then there's another little
wire, which is pretty much flat and joins at the bottom.
There are charges piled up
here, there are charged piled up here,
and they will produce an electrostatic field that will do
something there, but inside, they're bad.
Electrostatic field goes down
here. So if this is the whole story,
you cannot do anything, because maybe if you charged it
once, it will go round and come back
here and it's stuck, because it cannot go up.
That's when the tiny little
hands of the chemical forces will come and drag the charge,
against this field. That's the F_chemical
force. And it's the line integral of
the chemical force around a closed loop that's called emf.
If you write the chemical force
per unit charge, integrate over a closed loop,
that's what we call the emf. But that's very easily computed.
That's what I didn't realize,
a much easier way to do the computation.
This thing is just a product of
that force times the distance, but in magnitude,
that force precisely cancels the downward electric force,
just like the lift cancels the force of gravity.
So the line integral in
magnitude is just the electric field times the distance between
the two plates, and that is just the voltage.
Therefore the voltage
difference between this point and this point will be the emf.
Now if you don't want to know
the whole details, you can go back to drawing
pictures this way, and if you have a circuit like
this, and maybe there's a switch
here, you can be assured that between this point and this
point, if you go around and measure
the voltage difference, that will be the emf of the
battery. That's all you really need to
know. The way it comes out is that
this battery is like a pump that lifts you up by an electrical
height equal to 1.5 volts. The emf is literally the
voltage difference. Yes?
Student: What's the
symbol under F_c in the
first? Prof: Charge.
Because electric field is the
force per unit charge. We should take whatever force
there is and write it as a force per unit charge.
All right, so I wanted to do
some simple problems. This stuff you've done already
in high school, but I want to go over it in
view of everything we have learned.
So here's the simplest circuit
in the world. You have an emf.
There is a resistor and it
comes right back. You want to know what current
is flowing in the circuit. So the logic,
there are two equations you write now.
One equation is,
if you start at any point on the circuit and you go around a
path and you come back to the same point,
the change in voltage should be zero,
because voltage is like height. You can go anywhere you want
and you can come back, then the change in height
should add up to zero. If you want,
sum of all the change in voltage should be zero around a
closed loop. Second thing is,
current is conserved. These are the only equations
you need. Current doesn't disappear.
It's uniform here.
So current I comes here,
drops through the resistor and comes out of the other side.
So I'm going to write the
following equation. I'm going to start at this
point and keep track of all the changes in voltage.
First when I go around this
battery, I told you many times, the change in voltage will be
just E. When I come here,
current flows downhill. I know the voltage drop here is
RI, but if I follow it, I'm dropping in potential.
It's a skier;
it's going downhill. It's a ski lift where it's
going up. And then you come back here to
where you started, the total change must be 0.
Now this is pretty trivial.
You probably wrote this without
any further thought, but we then just say we get
from that E = IR. Then we're going to do a more
complicated problem, one resistor,
second resistor, R_1,
R_2. Again, if you apply an emf
here, that current is going to go through this guy and go
through that guy, and it's the same current
because it has nowhere to go. So the equation now will be
E, which is increased when you go up here.
Then you drop
R_1 times I R_2
times I and that equals 0,
so current is E/(R_1
R_2). That means if you put two
resistors in series, it's equal to a single resistor
equal to their sum. In other words,
if these two were enclosed in a black box,
you didn't know what's inside, you've got two leads coming out
and someone says, "Hey, what's inside the
black box? What's the resistor
inside?" you will apply voltage,
you will measure the current and that ratio you will call
R. E/I = R,
but here, you can see E/I =
R_1 R_2.
So that's the value you will
ascribe to the two guys in series.
Now if you put them in
parallel--I think again I'm repeating it,
but maybe it's harmless right now--so here is one resistor,
here's the second resistor. This is R_1,
this is R_2. This is some E.
If you put this inside a black
box and someone says, "Hey, tell me what's
inside," you don't know.
It may not be this,
but there'll be some effective resistance.
And we compute that by saying
let a current I leave this battery here,
do whatever it wants and come out of the other side.
Inside the box,
this current will split into an I_1.
I can call it
I_2, but I'm just going to call it
I - I_1, because that's what
I_2 should be. And it comes out of this side.
So let me say the total current
= I_1--I changed my mind,
guys. Sorry.
Let me call this
I_2. The total current is
I_1 I_2.
But I_1 is
V/R_1 because the voltage V is acting
across these two ends. The voltage V is also
acting across those two ends, so it will be
V/R_2, and that's going to be equal to
V divided by whatever the resistance inside the box is,
by definition. If you compare the two,
you find 1/R is 1/R_1
1/R_2. I mention this just to tell you
that it comes from current conservation,
where I used I, is I_1
I_2. We might as well deal with the
two others, one other circuit element, which is the capacitor.
So suppose someone takes two
capacitors. This guy is
C_1, this guy is
C_2, puts them in a black box and
says, "Tell me what's inside."
So you will take your battery,
apply voltage E, then see how much charge leaves
the battery. Then the capacitance will be
the charge that leaves the battery divided by the emf or
the voltage. If a charge comes from here,
a charge Q_1 will go from there and
-Q_1 will go there.
Q_2 will go
there and -Q_2 will go there.
So the Q that's going in
will be Q_1 Q_2...
over E.
But Q_1 = C_1
times E. Q_2 =
C_2 times E.
So divided by E, you'll
get C_1 C_2.
This will tell you that the two
capacitors in parallel will act like a single capacitor whose
value is the sum of the two. That's actually very easy to
understand because here's one capacitor.
Here's another guy.
Let me just glue those plates
together. It doesn't matter because they
are the same voltage. And you don't need these two
wires. You can just dump that and dump
that. Current can just come in this
way. You've got a single capacitor
with the area equal to the sum of the two areas.
Capacitance was
ε_0 A/d.
They have the same d,
but the areas add up so capacitances add up.
And the final thing which I
don't want to do, because I don't want to move
onto more interesting things, is when you put two capacitors
in series, what happens is,
you can take it as a homework problem,
that if you put a charge Q on this guy,
and a -Q comes out of the other terminal,
these two guys have no choice but to take -Q and
Q, because these two plates are
isolated from the world. The charge they can have has to
be equal and opposite, but -Q has to balance
this, and Q has to balance that.
Now if you use the fact that
the voltage drop on this one the voltage drop on that one is the
total emf, you will find 1/C =
1/C_1_ 1/C_2.
So capacitors and resistors
combine in exactly the opposite way.
This is just in series,
they're additive; capacitors in parallel are
additive. Whereas for resistance in
parallel, you add their inverse to get
the inverse of the total, and capacitors in series,
you add the inverses to get the inverse of the total.
So now I want to do one
slightly more interesting problem.
It looks like this.
You have a voltage E
here. Then you have a capacitor C
and a resistor R, and there's a little switch
which we will close at some point.
So we want to know what will
happen here. Initially you are told the
capacitor's empty, there's no charge on it.
Once you close the switch,
you should sort of imagine what will happen.
This plate is full of positive
charges. They don't want to be there.
And this is full of negative
charges. They don't want to be there.
So positive charges would like
to go there, negative would like to come around,
but the switch did not allow that.
But now that it's closed,
some charges will leave this and some negative charges will
be formed on the other side and they'll come back here.
It's very interesting to note
that current doesn't really flow through the capacitor.
Instead what happens is,
positive charges come here, and positive charges leave that
plate, leaving behind negative charge.
And that current will come
here, that's the current I, and goes back there.
But once the capacitor begins
to charge, it bites the hand that feeds it,
because it's then trying to drive its own current.
If you ask this guy,
"What do you want to do?"
he'll say, "I want to
drive the current that way."
The capacitor will start
opposing the external voltage, and as it charges up more and
more and more, eventually, its voltage will
equal the applied voltage and then the current will stop.
And when that's happened,
the charge on this capacitor at the end,
I'm going to call it Q_infinity,
you'll see why, will be C times
E. Then the current would stop.
So let's understand in detail
what happens to current from the time you close the switch.
So you write down the basic
equation which says start here, go around and keep track of the
change in voltage and equate it to 0.
So I get a E when I go
through this battery. When I come from here to here,
I drop by an amount Q/C.
Then I drop by an amount
RI. Then I'm back to where I
started, which was my 0. That's the equation.
So now you can write this
equation as--now what's the relation between I and
Q here? You should think about it.
When we drain the capacitor,
remember, I did a problem earlier on where I only had a
capacitor and a resistor. There the current was
−dQ/dt because as the current flows,
the capacitor's getting drained.
Here actually the current is
dQ/dt, so if you're not careful,
you'll get in trouble. It's dQ/dt,
because any charge that piles up on the capacitor leaves the
other plate and goes through this.
So the current flow is the
charging of the capacitor so I is in fact
dQ/dt. Otherwise if I is
positive and the current is going this way,
charge is building up on the capacitor.
So if you remember that,
you get the equation E = Q/C R dQ/dt.
So that's the equation you want
to solve. So how do we think of this
equation? We tell ourselves if we wait a
long, long time, until the capacitor
is completely loaded, so that it doesn't allow any
more current to flow, its voltage will equal this one
and at that point, dQ/dt will vanish.
And let me call that final
charge Q_infinity,
because you will find out it happens only after infinite
time. That
Q_infinity/C = emf.
So the capacitor is empty to
begin with. After infinite time,
it's fully charged, meaning this battery cannot
charge it to any higher voltage, because its voltage is
precisely balancing that of this.
So let me solve--our goal is
still to find Q as the function of t,
starting with this equation. So we are going to write
Q as a function of t, as
Q_infinity, which is the asymptotic value
some Q twiddle. You can always write your
answer as 96 something else, and something else will adjust
itself to make this true. But let's find the equation for
Q twiddle. So take this Q,
put that into the equation and see what you get.
Left hand side is E.
Right hand side is
Q/C, Q_infinity
/C Q twiddle/C.
And when you take dQ/dt,
this is a constant. So this is just dQ
twiddle/dt. But Q_inf
inity/C is exactly equal to E,
so these terms cancel, and the equation I get is this
combination = 0. Right?
This says dQ/dt,
dQ twiddle/dt = -RC--I'm sorry,
-Q/RC. Q twiddle/RC.
And we know how to integrate
this. We've done it many times.
With Q twiddle =
Q twiddle at time 0 times e to the
−t/RC. Now Q twiddle at time 0
has to be chosen to satisfy the following condition:
what is Q at time 0? You guys know what charge is on
the capacitor at time 0? Do you know?
Student: Sorry,
my mind is elsewhere. Prof: Okay.
Welcome back here.
So what is the charge on the
capacitor when I started everything?
Student: Charge on the
capacitor. Prof: At the beginning
of the experiment. Student: The charge on
the capacitor should equal the 0.
Prof: No,
where I just started. Anybody know?
Yes?
Student: 0?
Prof: 0,
because I told you the capacitor started out with
nothing on it. That's not a mathematical
result. It's a historical fact.
You can't begin an experiment
with charge on the capacitor. I told you today that the
charge on the capacitor is 0. So I've got to get 0 at t =
0 and this guy is E/C Q twiddle, that's Q
twiddle of 0 over C. I'm sorry.
Q_infinity is
EC. Is that right?
Yes, Q_infinity
= EC Q twiddle of 0 and that's got to be 0.
Therefore Q twiddle of 0
= -EC. That means Q(t) =
Q_infinity Q twiddle.
That becomes EC times 1
- e^(-t/RC). So if you draw the graph of
this, it will satisfy all the expectations we have.
If you draw Q as a
function of time, at t = 0,
e to the -0 is 1. 1 cancels the 1,
you start with 0 charge. At Q = infinity,
this guy is gone. It is EC, which is just
that value. So the charge builds up to that
value, but it never quite reaches the value EC.
In other words,
the voltage in the capacitor is never quite equal to that of the
battery. There's always some left over.
Yes?
Student: Where does the
negative come from? Prof: Here?
Student: Yes.
Prof: Okay.
I said Q at time 0 is
this guy at times Q_infinity,
which is EC Q twiddle at time 0.
But that had to be equal to 0.
Student: Oh, okay, yeah.
Prof: So that's how I
got that. This is a trick to solve the
equation. See, we all sort of know how to
solve this equation. You may not know how to solve
an equation with an extra term here, so this is a trick for
solving that. So this is the part that's
interesting, in the sense that once you write the basic rules
of physics, you get some equations.
You've got to solve the
equations and it's no longer up to you to see what happens.
The mathematics rules after
that point. And whatever it tells you,
you rush out to the lab and see if it's true.
And with the capacitor,
whose capacitance you have measured carefully,
and a resistor which you have measured carefully,
you put them together, here's a nice prediction on
what will happen as it charges up.
For example,
you may want your capacitor to hold 80 percent of its maximum
charge and you may like to know, "How long should I
wait?" If you want it to hold 100
percent of the maximum charge, it's never going to get done.
So pick some number.
They'll tell you,
if you want 75 percent, that's how many seconds you
wait. So this is how you're supposed
to have interplay between mathematics and physics,
because what happened was, you got into a situation where
you have to solve an equation. So this is a differential
equation, but differential equations are just questions
which are opposite of derivatives.
You're trying to guess a
function about whose derivative you know something,
namely this one. And it's all guess work.
You keep on guessing and keep
on guessing and you make a table of integrals where people tell
you what they guessed, and that big fat table of
integrals, you can turn to.
But in easy cases,
you can solve it yourself. Okay, now there's a homework
problem I'm going to give you guys, but I'll tell you what's
in store for you. We want to ask ourselves,
what happens to the energetics in this problem.
When I started,
how much energy did I have, and where was it?
Anybody know?
Was there any charge in the
capacitor? Nothing.
There's nothing to begin with.
Yes.
Student: It's all in
your battery. Prof: Yes,
the battery had some internal energy.
That's correct.
But now, during this
experiment, once I close the switch, I think of the battery
as outside my universe so it gives me some energy.
How much energy did the battery
give me during the whole process?
Yes?
Student: Equal to the
potential energy stored in the capacitor the energy dissipated
by the resistor. Prof: Right.
So let me repeat what he said.
He said it's got to be = to the
energy stored in the capacitor the energy dissipated in the
resistor. And that's got to be equal to
the work done by the battery. It's like saying,
how much work did the ski lift do?
Well, the work done by the ski
lift is the work it takes to carry each person from the
bottom to the top, multiplied by the number of
people. That's the work done.
In electrical language,
that just means the charge transported from the bottom to
the top times the voltage. That means the total work done
by the battery = the emf times I(t) dt
from 0 to infinity. I(t)dt is
a charge that flows in a small time dt,
and I is not a constant. In fact, we can find
I(t) by the formula of Q(t).
Remember, Q(t) =
EC times 1 - E to the −t/RC.
Then I, which is dQ/dt,
you can calculate is EC divided by RC times
e^(−t/RC). It becomes E/R times
e^(−t/RC). So the current falls
exponentially in this form. At t = 0 the current is
as if the capacitor were not even there, because the
capacitor's inert at t = 0.
It has got no charge.
It's not opposing you.
And eventually,
the current goes to 0. So take that formula for the
current, put it here and do some integral.
You get some answer.
That's the energy given by the
battery to us. And what have you done with it?
You charge the capacitor to
½Q^(2)/2C, where Q in the end
should be Q_infinity.
The resistor burns power at the
rate VI, which is I^(2)R.
So you take this I^(2),
take this I, square it, do the integral from
0 to infinity, multiply by R.
That is the work done in the
resistor, or work dissipated in the resistor.
You should make sure that this
energy delivered by the battery = what is stored by the
capacitor what is dissipated in the resistor.
That's your homework.
That just means doing these
integrals and making sure everything works out.
Yes?
Student: Shouldn't it
be negative, negative t/RC?
Prof: Here?
Student: Yes.
Prof: Okay,
finally I have a chance. This negative sign is going to
cancel that. Student: Oh, okay.
Prof: Look,
don't give up, okay, because just like you
thought, I make mistakes too. So I never mind it when you
guys do that. In this case,
I'll tell you why I sort of knew what the answer was.
I know the answer's positive
because the current's going to be positive.
So if I'd kept the extra -
sign, it will mean the current's going against the battery and I
would know that doesn't work. Quite often,
we're allowed to use the sign of the answer,
we're allowed to guess the sign.
For example,
what's the height of a person? You said the answer was 6 feet,
but I don't know if it's or -. Well, you should know it's .
You're talking about the height
of a person. What's the depth to which he
sank? Well, there it can be negative.
So you should know by context
whether the answer is positive or negative.
That's a very useful check as
far as signs go. Now as far as overall formulas
go, you can always take extreme limits of every answer and see
if it's correct. In the beginning,
just when you close the switch, there's the battery,
there's the resistor, the capacitor doesn't oppose
you, so you get e/R. At the end of the day,
the capacitor's fully charged, it's neutralizing the battery
by driving the current in the opposite way with equal voltage,
so the current should vanish. These are some tests.
All right, so let's continue
now to the last thing in circuits,
which I'm not going to do any more,
but I'll tell you the kind of stuff people throw at you.
So here is some guy,
some resistor, branches into two other
branches, some other voltage, join it here.
This is E,
this is R_1, R_2,
R_3. This is E_2,
let's say. You've done stuff like this
before, but let me remind you the trick.
The trick is very ancient.
We all know what the trick is.
First is, you've got to know
how many unknowns there are. If you can say the current here
is I_1. I don't know what
I_1 is; that's an unknown.
The current flowing through
this guy is I_2.
The current flowing through
this guy is I_3,
but I know that I_3 = I_1
- I_2. Do you agree?
I_1 comes here.
If I_2 goes
there, I_1 - I_2 goes there.
So I'm actually solving one of
the equations, which is that I_1
= I_2 I_3.
Let's put that into the
equation. Then you have to write,
so how many unknowns do I have? I have two unknowns,
I_1 and I_2. So I have
to get two equations for two unknowns,
and one equation will be--you can do many things.
You can start here,
go around like this and say the change in voltage is 0.
That would tell me
E_1 - R_1 I_1 - R_2
I_2 = 0. Are you guys with me now?
You drop here,
you drop here, then you come back.
Then you can take another loop
that goes like this. Then you will say
E_1 - R_1 I_1 - R_3
times I_1 - I_2.
Now let's look here.
You keep going the same way,
but here you drop by an amount E_2.
Then you've come over here and
you come to the other end and you get 0.
Now somebody can say,
"Hey, why don't I start here, go around that loop and
say that voltage difference is 0?"
I'm getting the third equation,
but there are only two unknowns.
So you math-minded people
should sort of know what will happen if I write a third
equation. What do you think will happen?
Student: Nothing.
Prof: What do you mean,
nothing will happen? Student: It's not going
to help you get any more. Prof: In what manner
will it prove to be useless? Student: It will be a
combination of the other two. Prof: Right.
You can deduce the third
equation by fiddling with the other two.
Maybe 9 times one equation - 6
times the second equation will be the third equation.
Therefore it's not an
independent equation. It will always turn out that if
you've got two unknowns, you need two independent
equations. If you can get three and four
and five, you will find out they're not telling you anything
new, so that's when you stop. If you don't know that,
you may start writing all kinds of equations.
You may think,
here's another loop. Here's another guy who wants to
do this. You can do all that,
but you'll keep getting the same stuff.
Okay, now we will really start
new topic: magnetism. Well, you start this because
every time you think you're done with physics,
somebody does some experiment and it doesn't fit what you
know, so you've got to make up new
stuff. So magnetism you know was
discovered in Ancient Greece, when parents noticed kids are
sticking stuff on the refrigerator using some little
black things. So more experiments.
I obviously don't know all the
details. But it was discovered in many
ways. One was in little compass
needles which told you which way was north.
So I'm going to give you a
string of things that happened that tell you there is something
going on that is not covered by anything I've written down so
far in this course, new phenomena that don't make
sense. Here's the simplest one.
There's a wire carrying some
current. There's a little charge sitting
here. Nothing happens.
Because the wire is
electrically neutral, the charge Q doesn't do
anything. Now the charge begins to move
at velocity v. Then you suddenly find the
charge attracted to the wire. It starts bending in.
That is not the electrical
force, because the electrical force is 0.
The wire is electrically
neutral. It doesn't care if the charge
is moving or not, so it is a new force.
And if the charge goes this
way, the force is repulsive. That's just one phenomenon you
notice. Another thing you notice is
there are little things called bar magnets.
They seem to have a north and a
south. And if you bring it next to
another bar magnet, which has got a north and
south, you find they repel and south and north attract.
Then you take this compass
needle that people use to find the north direction for the
earth, and you put it somewhere here.
You find that it swings,
if it's free to pivot, and points in a certain
direction. And if you use that as a
direction of a certain field, you can draw these pictures and
you sort of know what it's going to look like.
They look like this.
So this means if you put a
compass needle here, the north will point like that.
That's what I mean.
That's another phenomenon.
Then once you found out about
electric currents, you also found that if you took
a coil of wire, then a compass needle somewhere
here began to respond to a magnetic field,
which if you plot, seemed to look like this.
It seemed to look just like a
magnet with a north pole here and a south pole here.
There are no magnets at all.
You just had a coil of wire
carrying the current in a certain direction.
It goes like this.
Then the magnetic field lines,
which you plot by moving the compass needle,
they do this. So all I'm trying to say is
there are several phenomena going on,
and the clue that you get is, why didn't I need this before,
and why do I need it now? What is new in this problem
compared to the problems I've solved?
Can you see what's making this
problem outside the realm of what we studied?
What's the one feature you
notice from electrostatics? Yes?
Student: The charges
are moving. Prof: Everything is
moving. The wire carries charges which
are moving and this little guy who got attracted or repelled,
he's also moving. So magnetism is caused by
moving charges and it's felt by moving charges.
They've got to move to play
this game. So magnetism is caused by
moving charges and it's felt by moving charges,
whereas in electrostatics, we didn't have that motion.
Yes?
Student: What about
reference frames? Prof: Yes,
her question was, how about reference frames,
namely moving according to whom, right?
So I will come to that near the
end when we do relativity, when I remind you of some ideas
of relativity and see what that has to say about electricity and
magnetism. But you are free,
even in the relativistic theory, to take the view that
you are not moving. As long you're an inertial
frame, that means a frame in which Newton's laws are valid,
you can apply all the laws of physics as if you were not
moving. I'm just saying for that
observer, who's inertial, it's found that when charges
are moving according to him or her,
they produce currents which produce a field,
and the charges in that field will also respond.
Now you can say,
"What's so special about you?
I will go to a new frame of
reference,right?" I'll come back to that later,
but I cannot resist telling you at least the answer to some of
the questions. For example,
if this charge is moving at a speed v and I'm
completely stymied by this velocity, because I don't know
how to deal with it. There's one way to deal with
it, which is to get on a train that goes at the same velocity
as this charge. Then this charge is at rest and
if it still bends towards the wire,
which it will--you agree that if you go on a moving train,
a charge attracted to the wire will continue to be attracted to
the wire. You've got to say,
"How come? How do you explain that?"
Neutral wire attracting a
charge. Either you can say it's all
happening because you're in a moving train,
but relativity tells you, people in a moving train are
entitled to the same laws of physics as people in a
non-moving train. Now it's true for Amtrak.
As long as you can get the
train to move, you can make the same
statement. So how does a person in a
moving train explain it? What do you think happens?
Yes?
Student: The wire looks
like it's moving back in that direction.
Prof: The wire may look
as though it's moving backwards, but still, a neutral thing
moving forwards or backwards shouldn't matter,
right? A neutral rod, let it move.
Why does it matter?
So what do you think happens?
First of all,
do you think it will be attracted to the wire,
even in the moving train? Yes, no?
It will be attracted,
because if I see it moving towards the wire,
you can go on a train, you can go on a plane,
you will also see it's moving towards it.
Maybe the rate will be
different and so on, but the fact that it's getting
closer to the wire is undeniable.
And there's going to be no new
physics. You're just going to rely on
good old electromagnetism, because electromagnetism is
supposed to work for everybody. In fact, that's how the
relativity-- even though electromagnetism
was discovered before relativity,
it obeys all the principles demanded by relativistic theory.
So everybody should have the
same laws of motion. So have you found a way out now?
Yes?
Student: Would the wire
look like a bar magnet? Prof: No,
because bar magnet means you're going to magnetism,
right? So let's say I don't know any
magnetism, because the charge is at rest.
A charge at rest doesn't care
if it's near a bar magnet. Yes?
Student: ________ lines.
Prof: So what should be
happening to the wire, you guys?
Tell me?
Yes?
Student: If you're
moving and then the charges in the wire are moving,
say you're moving at the same speed as the charges on the
wire, it would seem stationary.
So it's just like the point
charge being attracted to a bunch of--
Prof: Right. Why is the point charge
attracted to a neutral wire? That is my question.
Yes?
Student: The wire isn't
neutral, because you're moving along
with the negative charges, so they stay there,
but then the positive charges keep--
Prof: That's correct. So let me explain what he said.
In fact, the answer is,
the wire will not be neutral. So how did that happen?
You might say,
"Oh, there's a Lorentz contraction,"
meaning the lengths get contracted,
but if the wire is neutral, the positives and negatives all
get compressed, it still should look neutral,
but that's not how it is. Remember, in a real conductor
you've got positive charges, which are at rest and negative
charges which are moving in the opposite direction,
right? So I do that by taking a rod,
which is positively charged, and I take another rod,
which is negatively charged, and I simply move this rod.
That will produce a current.
Now if the wire is neutral,
I think you guys can understand that the density on this,
in its own rest frame, will be different from the
density of charges here, because lengths get contracted.
It is a contracted wire that
should have the same density as the positive wire,
therefore in reality, the contracted wire,
length contracted with increased density,
should match the density of this.
But if I go to the moving
frame, what will happen is this guy will freeze,
and that guy will move the opposite way.
This guy will stop moving;
the other will move the opposite way.
But then you can show--I will
do that calculation for you later--
that the densities that previously canceled will no
longer cancel, and the wire will have a net
positive charge. The point is,
there's a lack of symmetry between the positive and
negative charges, because one of them is moving.
In fact, if you took the simple
example where the charges in the wire are moving at the same
speed as this guy-- let's imagine positive charges
can move-- and if they're moving at the
same speed as this guy, if you stop this,
you will also stop those, and they will go the opposite
way. But if they go the opposite
way, they will contract, and when they contract,
they will no longer be balanced by this,
and the positive charge will be attracted to the negative wire,
and not only attracted. It will be the same force that
you will get when you learn about magnetism.
So I'll do this in detail,
but I'm just telling you that you don't have to worry about
which frame of reference you're in.
The laws of physics are
guaranteed to work for all frames of reference in uniform
relative motion. So when I say velocity is
v, I mean according to any one observer.
That could be you.
So you've got all this
phenomena. Now I'm going to give you the
fundamental equations of magnetostatics,
that will explain to you everything I've described so
far. So it will have two parts.
The first part will be,
what is the force felt by a charge that's in motion in a
magnetic field? The next thing is,
how do you produce a magnetic field?
Well, the answer is electric
currents. Then you can ask,
how does a current here produce a magnetic field there?
What's the analog of Coulomb's
law? Coulomb's law is charges here
producing the electric field there.
Magnetism is a current here
producing a magnetic field there, so you will have the
second part. So I'm saying that this
formula, q times E is the electric force.
Then there's the fact that the
E here due to this guy is 1/r^(2) times the
q here, etc.
You need both parts.
I'm going to give the two parts.
The second part will come
later, but first is, if you have a charged particle
in a magnetic field, what's the force on it?
So here's the answer.
The total force of a charge
particle q is this guy that we already know,
the cross product of the velocity with the magnetic
field. The direction of the field is
determined by the compass needle.
That's how the B is
directed. But if you want,
this force is called the Lorentz force.
It is not invented by Lorentz,
but he did so many things in this field, it's named after
him. You can take this to be a
postulate, the summary of years of
experiment, and you can say,
"I want to begin here," you can begin here.
You're never going to derive
this, but this is an experimental fact.
So if I go to a part of the
room and say, "Find the electric field
here," I think you all know what to do.
We've done it many times.
Take a coulomb and put it there.
Find the force on it,
and that's actually the electric field.
If you put 5 coulombs,
you divide the force by 5 and that's the electric field.
If I tell you,
what's a magnetic field, you can take a coulomb there,
but you still have to find out--
by the way, the electric field problem's very easy,
because the direction of the electric field is simply the
motion of the charge. In the magnetic problem,
there are lots of directions involved.
There's the velocity of the
charge. There's a magnetic field,
there's a magnetic force, and they form a cross product.
v x B is the
force. So if someone says,
"Which way is B pointing?"
you'll have to shoot a few
particles and find out how they bend.
If you shot it exactly parallel
to B it won't bend at all.
That's the direction of
B. If you sent it perpendicular to
B, it will bend the most, because the cross product will
be the biggest. Then you can slowly determine
by some experiments what the value of B is.
So let's do a couple of simple
problems where we just use this part, v x B.
But there's a very important
aspect of v x B. Wherever a force acts on a
body, you know force dot velocity is the power or the
rate at which work is done. If you take the electromagnetic
force dot velocity, you get q times v ⋅
E q times v ⋅ v x B and
that is 0. So you know why v ⋅
v x B is 0? It's again a purely
mathematical result. Yes?
Student: Because
v cross B is going to be perpendicular to--
Prof: v x B is orthogonal to both
v and to B, so it's coming out--in this
case, if B and v are in the blackboard,
v x B is outside the blackboard.
Its dot product with anything
in the blackboard is 0. So whenever you have a vector
dotting itself cross something else, the answer is 0.
That means the magnetic field
is always perpendicular to the motion of the particle.
That means it doesn't do any
work, never. Not just weekends.
The magnetic field doesn't do
any work. So you can say,
"Who cares about such a thing?"
Electric fields do a lot of
work. They speed up particles,
they slow down. The kinetic energy of a
particle will never change due to the magnetic field.
And yet you will see,
even though it's not able to do anything by itself,
it's extremely useful in getting a lot of things done,
like generators and so on. They rely on the magnetic field.
So we'll see that,
but at the moment, it's fact.
The force is always
perpendicular to the velocity for a magnetic field.
So now we are going to do one
or two simple problems. First problem,
let me take the easiest one. I want to find a way to select
from a beam of particles, which are all going from left
to right, those which have a certain velocity.
I want a velocity filter,
and I cannot seem them microscopically;
they are little guys. But they're moving,
they're all moving like this. Here is how we can make a
velocity filter. You take two parallel plates,
charge them up so the electric field looks like this.
This particle q will then bend
like this in the electric field. Now put a magnetic field into
the paper. Let me make sure I got this
right. Yes.
Put a magnetic field into the
board that's shown by this symbol.
Magnetic field coming out
towards you is shown by dots, and coming away from
you--coming towards you is the dot, and going away from you is
the cross. That comes from the arrow.
If you have an arrow,
if you look at the arrow from behind, it looks like this,
and if it looks like this, you run like hell.
These are the two things.
So I want to show you that
B is going into the blackboard.
Now what's the force due to
B? Take the cross product--that's
what I was trying to do in my head,
make sure I got it right--v x B is
turning a screwdriver from v to B.
Let's see, B is into the
board, v x B will be like this.
Therefore the total force on a
charge will be q times E,
which is downstairs--I mean, pointing down,
and v x B which is pointing up.
You can see then for a given
value of E and B, for any randomly chosen
velocity, these forces will not cancel.
So some guys may bend like
this, some guys may bend like this, or some guys will go
straight through. They are the ones for whom
E = v x B or the magnitude of the velocity
is E divided by B. So only those particles with
that velocity will make it. In other words,
imagine this thing is very, very long.
Some things will go hit it
there, others may hit it here. If you are just right,
you will make it to the other side, so it's a velocity filter.
Now you should be able to
answer the following question without doing long calculations.
The guys hitting the top,
are they faster than the desired speed or slower?
Student: Faster.
Prof: Because what?
You understand that?
If they're going up,
the magnetic force is winning. The magnetic force is the only
one who cares about your velocity, so the velocity is too
big. If they're falling down,
electric force is winning. Electric force is a constant,
magnetic force is velocity dependent.
So the fast guys go here,
slow and the right one comes here.
Here's another exercise you can
do. There is a magnetic field
coming out of the blackboard and it's uniform in density.
And I send a particle here with
the velocity like this. What will it do?
So B is coming towards
me. B is here.
v x B is a force
in this direction. It will bend;
it will go there. So it's constantly being
applied a force perpendicular to its velocity,
so it's like planetary motion. It will go in a circle.
Never speeding up.
It's not speeding up,
because the force is always perpendicular to velocity.
You don't change the kinetic
energy, but you change the actual direction of motion.
So this is the way to trap a
particle. If you want to trap particles,
you put them in a magnetic field, you don't have to touch
them. They will not go anywhere.
They'll form an orbit.
So let's find out what we can
say about this orbit. If that distance is R,
the dynamical equation is mv^(2)/R,
is the force you need to bend something into a circle of
radius R, if it's going at speed v.
And we can take V to be
constant, because v is not going to change.
And that's going to be equal to
the magnetic force, which is v times
B times q. I didn't write the cross
product, because everything is perpendicular to everything.
The magnetic field and v are
perpendicular, so the cross product and
magnitude is just v times B.
The direction of the force is
of course towards the center. If you balance these two
equations, you find v/R is
qB/m. So what is v/R?
Let's look at what's v
over R. If the particle has a velocity
v, the time period due in orbit will be
2ΠR/v. The distance divided by speed
is the time period. So v/R = 2Π/T
or 2Πf. That's called the omega or the
angular speed. It goes around and around the
circle with an angular speed, ω.
So v/R = ω.
So this fact was known for some
time. It's not very hard to find.
But do you know who made a
living out of this? Notice something very
interesting. The angular frequency does not
depend on the speed or on the radius.
It just depends on the charge
to mass ratio of the particle in a given magnetic field.
So that circle,
that circle, they all go around at the same
rate. As long as the particle has a
given charge, a given mass,
it's in a given magnetic field, you can launch them in orbits
of various velocities, they will all be in synch.
Now do you know who used this?
Does it look like anything,
these guys? That person was actually
associated with Yale. That tell you anything?
Okay, his name was Lawrence.
Do you know what Lawrence
invented? No?
Pardon me?
He invented the cyclotron.
I'll tell you what--in fact,
he had the ideas when he was a young faculty member here,
but he didn't build it here. He went off to Berkeley and he
built the cyclotron there. Then he built bigger and bigger
cyclotrons. And I want to tell you what the
idea is behind a cyclotron. Here's what you want to do.
You want to accelerate
particles so you can smash them against other things.
So one thing you can do is you
can take a little battery here, connect it to two plates.
This is positive,
this is negative. If you release a proton here,
for example, it will fall down the
potential. So it will lose a potential
energy, q times V. That will turn into
½Mv^(2). And just when it hits the
bottom plate, if you have a hole there,
it'll go through the hole and you have a particle accelerator
whose output velocity will be v,
given by this equation. But now if you want to
accelerate it to a higher and higher energies,
you've got to get bigger and bigger voltage.
So how are we going to do that?
Here's what Lawrence did.
So take a magnetic field.
In fact, there are two metallic
halves called Ds, because of the way they look.
You send a particle here and at
the instant you release it, imagine that this plate is
positive and this plate is negative,
and the particle is positively charged.
That means the whole plate has
one potential, and the other plate has a
negative potential. So this particle will speed up
when it comes here. Now the whole thing is immersed
in a magnetic field, which I'm not showing you.
So this guy will bend and come
out of this side. Now what will happen to it?
What will it do now?
If you don't do anything,
what will happen to the particle now?
Student: It will slow
down. Prof: It will slow down,
so that's not the accelerator. You've got one gain and
immediately lost it. But suppose when it's here,
you very cleverly change the polarity of this?
Just when it's leaving this
hole and going to the other D, you change the
polarity, so it's falling again. So it picks up a lot of speed
and goes on a bigger spiral. Here is doesn't experience any
field, goes around and comes here.
When it comes there,
you quietly change it one more time, so it's always falling
downhill. So every time it's ready to
cross the gap, you flip the voltage.
How do you do that?
You're not going to stand there
with the battery and keep switching the wires,
right? So what do you think you do?
Student: Alternating
current? Prof: Put an alternating
voltage. Take the power supply from your
house, 60 cycle a second, denoted by this symbol,
AC voltage, will reverse its polarity automatically.
And the beauty of this process
is the following: once it's got a certain
frequency, it will produce a cyclotron
orbit whose omega matches that frequency.
Even though the particle is
speeding up-- here's the whole point--even
though the particle is speeding up and going in bigger circles,
the ratio of the speed to the radius is such that it takes the
same time to go around these semicircles,
no matter how big they are, no matter how fast the
particle's moving. It's the fact that v
over R does not depend on R, R on v.
So that's the omega that you
will take for this guy. Then it will keep on picking up
speed and at the very end, it will do a few more such
things, then it will come shooting out of here.
So you don't need--in the end,
you can get 1 million volts, but you don't have a 1 million
volt battery. You have a 1 volt battery.
It gives it a kick a million
times. A million times you cross the
midpoint. It's a very clever way to make
the particle accelerate. The Stanford linear accelerator
is somewhat different. There you have a lot of
cylinders and charged particles go from one to the other.
But once again,
the polarity is reversed, so every time it goes there,
it's constantly falling downhill.
It's just this thing taken out
and made into a straight line, that's about two miles long,
I think, in Palo Alto. And that was the machine that
discovered a lot of great things, including what's called
the J particle. So the linear accelerator's
one, but this is the cyclotron. So the next question you want
to ask is--these are all examples of forces on a single
charge--what's the force on a wire?
Sometimes you do microscopic
experiments with tiny particles. Other times you do macroscopic
experiments with wires carrying current.
So I want to look at that
problem. So let's look at the wire
carrying current. So here's a piece of wire,
and I want to find just a little segment,
which I write as vector dl.
There's some current going here
and the whole thing is bathed in a magnetic field B,
which is constant over the tiny segment.
We want to know what's the
force on the little segment. There's going to be a force
because there are guys moving in the wire.
That's it.
Each one feels a v x
B force. You've got to add it up.
So if this has got length
dl, how many charges am I talking about?
The number of charges is the
density of carriers times the charge of each one.
That's the number of charges,
per unit volume. If the cross section of the
wire is A, the length is dl,
that's how many charges are involved here.
Do you understand?
This cross section is A.
A times dl is the
volume of the cylinder. That times the number per
volume times the charge for each guy is the total charge.
And each one of them
experiences the force v x B.
Now I'm going to do a little
switch here. I'm going to write it as
Anev dl x B.
So you see if you follow that.
The velocity of the carriers is
along the wire. dl is also along the
wire. Therefore either you can take
the magnitude of this vector times the vector v,
or the magnitude of vector v times the dl.
It's the same thing.
They're both the vector
parallel to either one. But if you write it this way,
if you go back to what we did earlier, this is just the
current in the wire. So we have a nice formula that
says the force on a segment of wire carrying current I
is I times dl x B.
So here's an example.
Suppose you've got a magnetic
field like this. And I've got a wire.
I don't know where it's coming
from, I don't know where it's going.
Let's say it's just carrying a
current. And imagine this is in the real
vertical plane, so that there are some weights
hanging from it. It's trying to fall down.
You can balance that by
applying a force of magnetism to that wire, because the magnetic
force will be dl x B.
Let's see, so dl is this
way, B is out of the board, so let me set the current
like this. Then dl is this way,
cross B will be a force upwards.
There'll be an upward force of
B times the length of the wire times the current.
And if that's equal to the mass
of the wire times g, then the wire won't fall up or
down; it will just stay there.
So you can hold up a piece of
wire by driving a current through it and putting in a
magnetic field. But here's another simple
homework problem, or some problem you can find in
a textbook. This is a semicircular wire.
Now this wire cannot just start
and end here because the current has to come from somewhere and
go from somewhere, so maybe it's doing this.
But I'm just focusing on this
section. And imagine there is a magnetic
field going like this. I want to find the force on the
semicircle. This is supposed to be a
semicircle. What's the force on it?
So take a portion of the wire
here. This is my dl.
This is my B.
Take the cross product of dl
with B. Can you see which way it will
go? Do the old screwdriver,
your favorite thing. Do this.
It's coming out of the
blackboard. And what's the force that's
coming out of the blackboard due to the little section = I
times dl times B times the angle between these
two. And that will turn out to be a
sine theta. If you want,
you can measure theta from here and you can see that dl
is like that, B is like that,
and the angle between the horizontal and the radial is the
same as the angle between the perpendicular to the radial and
the perpendicular to the horizontal.
So this theta is the same as
that theta. Then you integrate the force,
you get I. dl will be just
2ΠR, times B.
Sorry, dl will be
ΠR time B, times integral of
sinθ dθ,
which is -cosθ from 0 to Π,
that will give you a 2. So that will be the force
coming out of the blackboard on this section.
I should say theta from 0 to
Π. I think there's something wrong
with this formula. Hold on for a second, guys.
It should not be done this way.
dl = Rdθ.
There's I times B
and Rdθ sinθ from 0 to
Π, and that will give you 2IBR.
Do you understand that the
length of the wire, this segment dl,
is R times dθ.
Theta is measured in radians.
Anyway that is a force and this
section tells you how to calculate that force.
All right, the last thing which
I will start but which I will not finish today,
and you can try to look at some good pictures before you come to
class next time. There is just no way I can get
this right. So here is a current loop.
And the current goes like this.
And it's in a magnetic field
like this. The question is,
what happens to it? So you've got to go to the four
sides of the loop, in this side the v x
B, your current is going this way.
B is going straight up.
v x B will act
like this. There the v x B
will act like this. Here it will act like this and
here it will act like this. And this is the normal to the
plane, it's the area vector. So let me draw you a side view,
because that's the only thing I can even try to do.
In the side view,
it looks like this. These two edges are getting
pulled. This edge that you can see here
is getting pulled out of the board.
The other one's getting pulled
away from the board. They just try to distort the
loop, but they cancel each other and they don't do anything.
But these two are equal as
forces, but as a torque, they are not equal and
opposite, but they're additive. So they together produce a
torque and let's find out what the torque is.
The force on this segment is
B times l times I, where l is
this. Let the other dimension be
w, which stands for width.
This is w here.
Then the torque will be that
times wsinθ. That is the torque.
But l times w is
the area of the loop. B times A times
sinθ. If I represent the area as a
vector, pointing on this direction,
then the torque is simply given by μ x B,
where μ is a vector whose magnitude is equal to the
current in the loop times area. I forgot the I.
And this is called the magnetic
moment of the loop. You call it a magnetic moment,
because it's just like a dipole moment.
You might remember,
if I take a dipole, put it in a field,
the torque on it is p x E and the energy of it is
-p⋅E. So magnetic loop looks like a
dipole in a magnetic field. In other words,
if there were magnetic charges in nature,
which there aren't, then this guy behaves exactly
like a magnetic charge and a - on the other side of the loop.
So if magnetic charges existed,
this would be simply a magnetic dipole,
aligning itself with the magnetic field,
but in reality, there are no magnetic poles.
But there is just the loop that
behaves like a dipole. And the dipole moment of the
loop, which is the analog of distance times the charge is the
current times the area. So we'll take it from here next
time.
