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Today, I'm going to take a critical look at Ampere's Law.
I'm going to run a current through a wire, as we did before, but now I'm going to also put a capacitor in that line and so we are charging a capacitor.
Here is that capacitor.
And here is the wire.
We are running a current I.
And as we are running this current, clearly, we get a changing electric field inside the capacitor.
The electric field inside the capacitor, sigma free divided by kappa epsilon 0, which is also Q free divided by the area.
This is a circular plate capacitor.
Capital R, is the radius of this capacitor, so we get pi R squared kappa epsilon 0.
But since I run a current the Q free is building up all the time, and so the current per definition is dQ/dT, and so I now have ex- a changing electric field inside, dE/dt, which is the current I divided by pi R squared, kappa epsilon 0, because I simply take the derivative of this equation, I get dQ/dT, and dQ/dT is I.
And only if the current is 0 is there no changing electric field inside.
So how does this affect the magnetic field?
Well, if I take here a point P1 at a distance little r from the wire, if you're far away from this capacitor it's hard to believe that Ampere's Law would not give the right answer.
And we will apply that very shortly, Ampere's Law.
It's on the blackboard there.
Suppose you are at the same distance from this line here at point P2.
Well, yeah, you've got to admit there's an interruption of current now.
There is no current going through this space and so you expect that the magnetic field here would be a little lower perhaps than it is here.
But not very much.
So the question is, how can we now calculate the magnetic field here and there, now that we have this opening in the wire.
Well, Biot-Savart could handle it but I wouldn't know how to do it because if there's a current flowing like this there's also a current going up on these plates, and one like so, and I wouldn't know how to apply Biot-Savart.
In principle, yeah, but in practice, no.
How about Ampere's Law?
Well, let's give Ampere's Law a shot.
This is a cylindrical symmetric problem, so I choose a closed loop, which of course itself is a circle with radius R, and I apply -- I attach to this closed loop an open surface.
That's mandatory.
And I give myself an easy time, I make it a flat surface.
So now I apply Ampere's Law.
You see it there on the blackboard.
Anywhere on that closed loop, the magnetic field will have the same strength, for reasons of symmetry, and so we get B times 2 pi r equals mu 0 times I pen, and pen means the current that penetrates my open surface.
Well, that's I.
I goes right through that surface.
And so the magnetic field at that point, P1, mu 0 times I divided by 2 pi r.
We've seen this several times before.
Now I wonder about P2.
Can I apply Ampere's Law for point P2?
Well, yeah, you can try.
So now I attach a closed loop to this point.
Circle again, radius little r, and I use this flat surface and I apply Ampere's Law.
Well, I'm in for a shock, because B times 2 pi r is not changing but there is no current that penetrates that surface.
And so I is 0, and so I have to conclude that the magnetic field at point P2 is 0 which is absurd.
Couldn't be.
I can make the situation even worse.
I'm going to revisit point P1, and here is my capacitor, and here is my point P1.
My current is flowing like so.
Here's my closed loop.
According to Ampere's Law, int- closed loop integral B dot dL.
Why should I choose a flat surface?
I'm entitled to any surface! I like surfaces like this.
They are attached to a closed loop, so I will choose that kind of a surface.
The surface now goes like so.
[whistle] Right through the capacitor plates, and I apply Ampere's Law, it's open here.
B times 2 pi r, the radius is little r.
Mu 0 times I, but there is no I going through that surface.
Nowhere through this surface is a current poking, because there is no current going between the capacitor plates, so now I have to conclude that the magnetic field at P1, which we first concluded was this, is now also 0.
So something stinks.
So Ampere's Law is inadequate.
And so of course, Faraday and Ampere were both perfectly aware of this.
But yet it was Maxwell who zeroed in on this and he argued that any open surface that you attach to a closed loop should give you exactly the same result, same answer.
And so he suggested that we amend Ampere's Law, and so he asked himself the question, what is so special about in-between the capacitor plates?
Well, what is special there is in-between the capacitor plates there is a changing electric field.
And Maxwell reasoned, gee, Faraday's Law tells me that a changing magnetic flux gives rise to an electric field, so he says maybe a changing electric flux gives rise to a magnetic field.
And I want to remind you what an electric flux is.
Phi of E is the integral.
In this case it would an open surface of E dot dA.
That is an electric flux.
With Gauss's Law that you see on the blackboard there, we had a closed surface.
I'm talking now about an open surface.
That is an open surface.
This is an open surface, and this is an open surface.
And so Maxwell suggested that we have to add a term which contains the derivative of the electric flux.
And that's what I'm going to do there now, walking over to Ampere's Law.
I'm going to amend it in a way that Maxwell suggested.
He adds a term here, epsilon 0 kappa, d/dT, of the integral over an open surface attached to that closed loop of E dot dA.
This current, which is the one that penetrates, remember, through the surface is really a real current.
This term here, Maxwell called "displacement current.
I want to make sure that I have no slip of the pen, because I hate slips of the pen.
That is correct.
I have everything in place.
You may think now that we can start a party because all four Maxwell's equations are now in place.
Not quite.
We're going to make one small adjustment after spring break, and that adjustment is going to be made in this one, and then we'll have our party.
So now, I would like to use the new law and see whether we can clean up that mess.
So I'm going to revisit my point P1 and I'm going to apply the new law by first having a flat surface, that surface that we have here, and then trying this surface.
And I want to get the same answer.
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