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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.