Griffiths 8.6

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Homework Statement


A charged parallel-plate capacitor (with uniform electric field [itex]\mathbf{E} = E\hat{\mathbf{z}} [/itex]) is placed in
an uniform magnetic field [itex]\mathbf{B} = B\hat{\mathbf{x}}[/itex]

(a) Find the electromagnetic momentum in the space between the plates.
(b) Now a resistive wire is connected between the plates, along the z axis,
so that the capacitor slowly discharges. The current through
the wire will experience a magnetic force; what is the total
impulse delivered to the system, during the discharge ?
(c) Instead of turning off the electric field, suppose we slowly reduce
the magnetic field. This will induce a Faraday electric field, which
in turn exerts a force on the plates. Find the total impulse.


I can do parts a) and b) so I do not think you need to read them. I need help with c).

Homework Equations





The Attempt at a Solution


I am stuck trying to find to find the electric field due to the changing B-field. What trick am I supposed to use? How can I figure out what direction it is in?
 

Answers and Replies

  • #2
olgranpappy
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can you tell us how you did part b? momentum conservation?
 
  • #3
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can you tell us how you did part b? momentum conservation?
No, I just used the Lorentz force law to calculate the force on the wire [itex]BdI=Bd\frac{dQ}{dt}=F[/itex] and then I integrated

[tex]Bd\int\frac{dQ}{dt}dt = Bd \int dQ =BdAE\epsilon_0[/tex]

where d is the distance between the plates and A is the area of a plate.
 
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  • #4
olgranpappy
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No, I just used the Lorentz force law
...okay, but that's the same thing you would have got if you had used momentum conservation, right? cuz the momentum in the fields initially was A*d*\epsilon_0*E*B

to calculate the force on the wire [itex]BdI=Bd\frac{dQ}{dt}=F[/itex] and then I integrated

[tex]Bd\int\frac{dQ}{dt}dt = Bd \int dQ =BdAE\epsilon_0[/tex]

where d is the distance between the plates and A is the area of a plate.
 
  • #5
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...okay, but that's the same thing you would have got if you had used momentum conservation, right? cuz the momentum in the fields initially was A*d*\epsilon_0*E*B
Yes, there are lots of ways to do this problem. I want to find the Faraday E-field even though it may not be the easiest way. Can you help me with that?
 
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  • #6
olgranpappy
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Yes, there are lots of ways to do this problem. I want to find the Faraday E-field even though it may not be the easiest way. Can you help me with that?

I think so. I think what you want to do is use

[tex]
\nabla \times \vec E_F = -\frac{\partial \vec B}{\partial t}
[/tex]
where E_F is the induced field due to the changing B field.

then convert to intergral form by considering the magnetic flux through the area between the plates (not A, rather d*L with L^2=A for a square plate, the surface element is pointing in the \hat x direction). Then you have
[tex]
\int d^2 \vec S \cdot \nabla \times \vec E_F = -\frac{d}{dt}\int |d^2 S||B|=-Ld\frac{dB}{dt}
[/tex]
and use stoke's on the LHS of the above
[tex]
=\oint d\vec \ell \cdot E_F = ({E_F}_y(z=0)-{E_F}_y(z=d))L
[/tex]

then you have an E_F field at the location of the plates and you can get a force by multiplying by the charge and rewrite in terms of dB/dt. and rewrite the (surface) charge in terms of your original "E"...

et cetera. give it a shot.
 
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  • #7
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[tex]
\oint d\vec \ell \cdot E_F = ({E_F}_y(z=0)-{E_F}_y(z=d))L
[/tex]
I am trying to figure out what happened to the other two sides of your integral. Did they cancel? What are you assuming about the direction of E_F?
 
  • #8
olgranpappy
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I am trying to figure out what happened to the other two sides of your integral. Did they cancel?
no. I have an equation
[tex]
L({E_F}^y(z=0)-{E_F}^y(z=d))=-Ld\frac{dB}{dt}
[/tex]
What are you assuming about the direction of E_F?
nothing.

But you know the direction from Lenz' Law or whatever.
 
  • #9
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no. I have an equation
[tex]
L({E_F}^y(z=0)-{E_F}^y(z=d))=-Ld\frac{dB}{dt}
[/tex]
I am so confused. But why is that equation true? This equation

[tex]

\int d^2 \vec S \cdot \nabla \times \vec E_F = -\frac{d}{dt}\int |d^2 S||B|=-Ld\frac{dB}{dt}

[/tex]

is certainly true and you can certainly apply Stokes Theorem to the LHS but I just don't see where this

[tex]L({E_F}^y(z=0)-{E_F}^y(z=d))[/tex]

comes in.

I'm sorry. I know I am probably missing something obvious.


But you know the direction from Lenz' Law or whatever.
I don't actually. I never learned how to use Lenz's law to calculate the Faraday induced by when a uniform B-field that fills all of space is reduced. I can use Lenz's law to calculate the emf induced in a loop! This is the heart of my problem here!
 
  • #10
Ben Niehoff
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I never learned how to use Lenz's law to calculate the Faraday induced by when a uniform B-field that fills all of space is reduced. I can use Lenz's law to calculate the emf induced in a loop! This is the heart of my problem here!
Ah, that's the problem!

First ask yourself the reverse question: if I have some current loop with a current I, it produces a B field. In what direction is this B field? This you should know from the right-hand rule. Suppose we have a current loop in the page, running counterclockwise. Which direction is B, in the interior of the loop? Into the page, or out of the page?

Now, let's go back to Faraday's Law. Suppose we have a loop of wire, in the page again, and a B field pointing out of the page. We know that if the B field is changing in time, then a current will be induced. We also know that the amount of current induced will be proportional to the total flux of B field through the interior of the loop. But which way will the current go? Here's a way to think about it:

In order to induce a current, you need to expend energy, to get the charge to flow against the resistance of the wire in question. You can't get energy from nothing; energy is always conserved.

Also, once we have a current flowing through the loop, we know (via Ampere's Law), that the current will produce its own B field! So now we have two possibilities. Either:

A. The B field produced by the loop will agree with dB_ext/dt, or

B. The B field produced by the loop will oppose with dB_ext/dt.

Note that the time derivative of B_ext must be used, because the current is proportional to this time derivative (and therefore the sign of the current must depend on the sign of the time derivative).

Now, suppose that possibility A happens. That is, if B_ext increases, then B_wire points in the same direction as B_ext, and vice versa. What happens to the energy of the system? What happens to the current? Note that the current induced in the wire is due to the TOTAL B; it reacts to its own B as well as the external B. This is called self-inductance. (Hint: you get runaway solutions in this case.)

Now, suppose that possibility B happens. That is, if B_ext increases, then B_wire opposes B_ext, etc. What happens to the energy? What happens to the current? In this case, you should have stable solutions.

The end result is that the B field set up by the wire loop MUST oppose the change in the external B field, or else the mathematics explodes. Knowing that, then, which way must the current go in order to set up this B field?
 
  • #11
Ben Niehoff
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So, the above post was about currents. Now, you have to translate that to what it means for E fields. Think about this. In order to induce a current in a wire, you have to cause charges to flow in a circuit. What causes charges to move? The E field, of course. Which way must it be pointing?
 
  • #12
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The end result is that the B field set up by the wire loop MUST oppose the change in the external B field, or else the mathematics explodes. Knowing that, then, which way must the current go in order to set up this B field?
The current always flows in the direction that opposes the change in flux.

So, the above post was about currents. Now, you have to translate that to what it means for E fields. Think about this. In order to induce a current in a wire, you have to cause charges to flow in a circuit. What causes charges to move? The E field, of course. Which way must it be pointing?
The E-field points in the same direction as the current.

But I don't see how this is getting us closer to finding the Faraday E-field IN PART C OF THE SPECIFIC PROBLEM I ASKED. Everything you talk about works nicely when things are circular. But in the problem I asked, it seems like things are rectangular. Apparently the B-field FILLS ALL OF SPACE and I am just not sure how to handle that. Of course in whatever direction the E-field is in, it will need to oppose the decreasing B-field. But there are an infinity number of ways in which it could do that. I am trying to figure out how the symmetry of the problem gives away the Faraday E-field in part c of the problem I posted.
 
  • #13
Ben Niehoff
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I believe that you are supposed to ignore fringing effects for this problem, correct? In that case, your fields must obey the same symmetries that they would have to obey if the two charged planes were infinite. In an infinite-plane geometry, which directions are the fields allowed to point?
 
  • #14
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I believe that you are supposed to ignore fringing effects for this problem, correct? In that case, your fields must obey the same symmetries that they would have to obey if the two charged planes were infinite. In an infinite-plane geometry, which directions are the fields allowed to point?
But shouldn't the Faraday E-field be independent of the geometry of the plates? The plates do not contribute to the "uniform magnetic field" in the problem so why does it matter whether they even exist when we calculate the Faraday E-field? Doesn't the Faraday E-field only depend on the symmetries of the B-field?
 
  • #15
Ben Niehoff
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I don't really know how better to explain it without simply telling you the answer. You keep asking questions about every single detail; perhaps you should slow down and think about the problem a bit more. I do notice you tend to rush through these things without gaining much understanding from them, and as you can see, it is hindering your ability to do the more difficult problems.

Whenever I get stuck on E&M problems, I find it helpful to look at Maxwell's equations. This seems obvious, but you'd be amazed how many false paths you can try to wander down when trying to reason about a problem. Always, always start with Maxwell's equations. Most of these things can be solved by taking the appropriate equations, doing some manipulations, and voila! an answer pops out.

Now, it should be obvious that you need Faraday's Law here. Is it more useful to use Faraday's Law in differential form, or in integral form? Do you actually care about the Faraday field specifically, or do you only need its curl integrated over some appropriate surface?

The very first thing you should do is write down the equations, and write down whatever integral you need to do in order to find the total impulse. Then, just do calculus.
 

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