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PROFESSOR: Problem 4 is 30-21.
Here, linear axis: 0, 'L', '2L', '-L', '-2L'.
And here in space, perhaps a little bit artificial, I have
an area '2L' by '2L', which has a magnetic field pointing
in the paper, and the magnetic field is
constant in the paper.
Here I have a loop, which has sides, capital 'L', and that
loop is being moved by me with a certain velocity, 'v',
dragged into this magnetic field.
And I'm being asked now, what is the induced current in this
loop as a function of position, and what is the
force of this loop as a function of position.
I, Walter Lewin, have to drag it in.
Is it positive force to the right or is it a
force to the left?
Do I have to push it?
Maybe it sucks in itself.
I will use this point 'P' as my reference point
relative to this axis.
Follow me closely.
Before 'P' reaches this point, there is no magnetic flux
change; it's 0, so there cannot be any induced current.
The moment that 'P' hits '-L' there is going to be an
increased magnetic flux through this surface.
This is the surface attached to my loop, it's an open
surface; and therefore, there will be an induced current.
The magnetic flux is increasing, and therefore,
there will be an induced current.
If the magnetic flux is increasing in this direction,
the current that will flow will try to oppose that
increase, and so it will create an induced magnetic
field in this direction, so that the flux in the paper
doesn't grow so fast. Once the loop is completely inside,
that means once point 'P' reaches 0, there is no longer
any flux change, and so there is no longer any induced
current, and the current will stop once it
reaches this point.
And all the way from here to here, 'P' being here to here,
the magnetic flux will not change.
But the moment that 'P' emerges here, the magnetic
flux, which is in this direction, is changing again.
It's losing magnetic flux, so something is
going to happen again.
I will do the early part of the problem, and you will then
do the second part of the problem.
Well, the magnetic flux change, 'd phi dt', when 'P'
begins to move into the magnetic field equals 'vBL'.
Every second the magnetic flux increases with a value, which
is 'B' times the increase in surface area, and the increase
in surface area is 'vL', and so the closed loop integral of
'E dot dl' along the loop that I have chosen--
and let's not worry about signs, because we know the
direction anyhow--
equals 'vBL', and we already determined that in the loop
when it starts to enter the magnetic field the induced
current will be in this direction, so the induced
electric field will be in this direction.
That's the 'E' field in this loop, and if you want to know
what the induced current is, well, that is the EMF value,
which is 'vBL' divided by the resistance of this circuit, if
'R', capital 'R' is the resistance of that circuit.
What is the force on this loop?
Well, there are two ways that you can do this.
One way is to say, aha, I have this wire here, which has
length 'L', I'll make a new drawing.
This wire, which has length 'L', it has a current flowing
through it, which is 'i induced', which we just
calculated.
It has a magnetic field, 'B'--
I should really make it black--
'B', in the paper, which is this external field, and so
the force on the wire equals 'i' times 'L' times 'B'.
This is part of Assignment 5, and that
force is in this direction.
Remember, it is the cross-product between 'i',
'L', and 'B', and if you take that into account, you'll see
that it is in this direction.
But I, Walter Lewin, have to push this in, have to pull it
in, so my force, 'F', Walter Lewin, is in magnitude the
same, but it is to the right.
And therefore, notice that I do positive work.
I do positive work, because the force is to the right, and
I'm dragging it to the right.
That's one way of doing it.
There's another way, which I like, and that is: as the
current is flowing in here, there is power dissipated in
that loop, which has a resistance 'R', and that power
is 'i squared R'.
That's the amount of work per second that is released in the
form of heat in this whole loop.
Now, who is going to provide that work per second?
That's Walter Lewin; so this must be the force by Walter
Lewin by the velocity with which I drag it through.
Remember this is also power, and if you solve this
equation, you will find exactly the same
value for my force.
It must give the same answer, and I think you should
convince yourself, both methods give the same answer.
If I make a drawing, a sketch, of the induced current as a
function of this linear dimension: 'L', '2L', 0, '-L',
'-2L', and if I draw here, 'i induced', then we would agree
that when point 'P'-- this is all relative to point 'P'--
when 'P' is outside capital 'L', there
is no induced current.
Instantaneously, when 'P' penetrates a magnetic field,
it goes to a value 'BvL' divided by 'R', and I called
this negative, but it's really a counter-clockwise current,
and I would like to call a clockwise current positive.
Then while it's inside all the way from 0 to 'L', there is,
again, no flux change--
I stress that point, here it is, again--
when the loop is in here, all the way marching up to here
there is no flux change, so the induced current goes to 0,
and this, now, is the part all the way on the right side,
here, you do.
I think you have all the ingredients, now.
You should have no problem.