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Direction of induced electric field due to uniform B field decreasing

  1. Sep 14, 2013 #1
    As part of a Griffiths problem, I have a charged capacitor placed in a uniform magnetic field which points in +x direction. the capacitor plates are parallel to xy planes. Suddenly, the B field starts decreasing in magnitude...what is the direction of the induced E field?

    It makes most sense to me that the induced E field is parallel to y axis, but I do not know which direction. If I imagine just a single layer of E field lines parallel to xy plane and pointing in -y direction, then below the layer, the induced B field will point in +x direction, which is what we want. However, above the layer, the B field will point in -x direction....which is not what we want.

    Furthermore, if we stack these layers, the resultant B field induced should just be zero...

    Any help in improving my understanding is appreciated!
     
  2. jcsd
  3. Sep 14, 2013 #2
    Actually now that I think more about this, E-field layers parallel to xz plane, with the E vectors pointing toward +/-x direction, will also produce the desired induced B field (but again only on one side of the layer, and if the layers are not stacked)....now i am completely confused lol
     
  4. Sep 15, 2013 #3

    vanhees71

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    You always have
    [tex]\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}.[/tex]
    So if [itex]\vec{B}=\vec{\beta} t[/itex], you have
    [tex]\vec{\nabla} \times \vec{E}=-\vec{\beta}.[/tex]
    Now you can guess the general form of the solution (which of course is only given up to a gradient field) by the fact that obviously [itex]\vec{E}[/itex] must be linear in [itex]x,y,z[/itex] and in compontents of [itex]\vec{\beta}[/itex]. Further it must be a vector. Just think, how you can form a vector field out of [itex]\vec{\beta}[/itex] and [itex]\vec{x}[/itex] and then plug the corresponding ansatz into the equation.
     
  5. Sep 15, 2013 #4
    I have no clue how to guess the solution, and why would E have to be linear in x y and z?

    Anyways tho I dont think solving a differential equation is necessary for this part, I really just need the direction of the induced E field, and once I have that I can find the magnitude using integral form of faradays law. Before I was always able to get the direction by just using lenz's law..but this doesnt seem to work for this case
     
    Last edited: Sep 15, 2013
  6. Sep 15, 2013 #5

    TSny

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    A similar question came up recently here: https://www.physicsforums.com/showthread.php?t=709452

    Looks like you might be working on the same problem from Griffiths. Not sure.

    Does Griffiths give any other information about the B field other than it is uniform and points in the x direction? Does he state if the B field extends to infinity in the y and z directions, or does it occupy a finite region in the y and/or z directions. It seems to me that without this type of information, it will be impossible to deduce the direction of the induced E field when B decreases.

    Note that at the above link, I give a reference to a 2009 paper of Griffiths and co-authors that discusses some subtle issues associated with the capacitor in a B field. In this paper Griffiths points out that the solution (in the solution manual, I guess) for this problem in the 3rd edition of his text is incorrect.
     
  7. Sep 16, 2013 #6
    according to my professor, the way to solve this is to imagine a current induced at the capacitor surface, in a direction such that it would satisfy Lenz's law for this case. So I think that on the top plate, the current would be moving purely to the -y direction, and on the bottom plate, it would be moving purely in +y direction. The induced E field is then in the direction of this induced current.


    This still however does not give me the right answer. According to solutions manual, we cannot know the field exactly on the bottom and top plate, so they just leave it at E(at z=0) and E(at z=d), but apparently the field should be in +y direction for both top and bottom plate...but if I follow my professor's logic, then we do know the induced field at the top and bottom (it should be uniform everywhere on the capacitor, since the induced current must be uniform), and they are definately not in the same direction...

    and I dont see how fringing plays a role in this case...the induced field should be parallel to the plates..with such a configuration i cannot picture where the fringing occurs..
     
  8. Sep 16, 2013 #7

    TSny

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    You might ask your professor to take a look at Griffiths' paper:

    "Hidden momentum, field momentum, and electromagnetic impulse" by
    David Babson, Stephen P. Reynolds, Robin Bjorkquist and David J. Griffiths
    Am. J. Phys. 77, 826 (2009)

    The fringing of the E field of the capacitor comes into play in part (a) of question 8.6 on page 358 in the 3rd edition of Griffiths. The E field of the capacitor is not just between the plates, it fringes out. (Also, the field is not actually uniform between the plates - it gets weaker as you approach the edges of the capacitor.) The total momentum associated with the fields must include the contributions from the entire space where E and B exist, not just between the plates.

    For example, in the figure shown, the Poynting vector ##\vec{S}## is shown at a particular point of the fringe field assuming B is out of the page. Griffiths' paper does not show the calculation that includes the fringing field, but he does give a reference to another paper.
     

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  9. Sep 16, 2013 #8
    Yes I read these posts explaining that already, but my question is not about part (a) but about part (c). The picture you attached shows the E field distribution for a regular charged capacitor. I know what fringing looks like for a regular charged capacitor. However, this is not what the induced field looks like...the induced field is in the direction of the induced current, which is along the capacitor plane...it is not between the capacitor. So I cannot visualize this kind of E-field fringing, since it seems to me that the induced current should be uniform. Even if I ignore this for the moment however, I especially cannot see how the induced field would be in the same direction for both the top and bottom plane.,
     
  10. Sep 16, 2013 #9

    TSny

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    The nature of the induced E field caused by the changing B field will depend on the geometry of whatever is producing the magnetic field B. If the B field is produced by a large solenoid, the the induced E-field lines will be circular.

    For this geometry, you can work out the magnitude of the E field as a function of distance from the axis of the solenoid. You can then integrate ##dF_y = \sigma E_y dA## over one of the plates of the capacitor to get the force in the y direction on that plate caused by the induced E field. Here, ##\sigma## is the charge per unit area on a plate and ##dA## is an element of area of the plate.
     

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  11. Sep 16, 2013 #10

    rude man

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    Imagine a closed circular path in the yz plane, centered at the origin.

    Then use Faraday's law to give you the direction of the emf around the loop, which is the direction of the E field around the loop.

    If you need extra help determining direction, invoke Stokes' theorem with ∫∫curl E * dA = ∫E * ds and Maxwell relating curl E to ∂B/∂t.

    This assumes an infinite pair of plates (in both dimensions). If the plate is of finite dimensions I pass!
     
  12. Sep 16, 2013 #11
    Yes in the case of a solenoid producing this constant B field, I can understand that the induced field goes around in loops....but the problem does not say what is causing this B field, just that it is constant, and the figure corresponding to the problem does not imply a B field due to a solenoid at all, so I just imagined that the B field fills all of space, is uniform, and points in +x direction.

    Either way, if we model this problem as a solenoid being the cause of this B field, there is not reason for the induced E field to flow in the same direction on both the top and bottom plate..
     
  13. Sep 16, 2013 #12

    TSny

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    Yes, that's right. (Of course the E field doesn't really "flow", it just "points" :smile:)

    Anyway, I don't think I understand what your specific question is. You have referred to "induced current", but I don't see how that's coming into the picture. The force on the capacitor plates that gives an impulse to the capacitor is coming from the force of the induced E field on the static charge of the plates.
     
  14. Sep 16, 2013 #13
    No it is not right since I am wrong lol. Somehow the induced E field is supposed to point in the same direction everywhere.

    By induced current I mean the current that is induced on the capacitor plates due to the changing magnetic flux in between the capacitor. I do not know what you mean by static charge...the capacitor is a conductor, so if there is an E field along the capacitor plane, current will flow (induced current)
     
  15. Sep 16, 2013 #14
    nvm, i found a great solution here: http://lief.if.ufrgs.br/~ambusher/griffiths/Pace_EM13.pdf in case anyone is interested. So the E field does point in opposite directions for the plates, and if we suppose that the B field is caused by a large solenoid then it is easy to see why there is fringing (although I hate making assumptions in a problem). The only thing left to solve the problem is to notice that the force is the same direction on both plates, since one is positively charged and the other negatively.
     
  16. Sep 16, 2013 #15

    rude man

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    The question did not ask for the force on the plates. It asked for the direction of the induced E field.
     
  17. Sep 17, 2013 #16

    TSny

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    But, the solution you linked to does not explain why the induced E field should be parallel to the plates of the capacitor. Why should the induced E field choose to orient itself parallel to the capacitor plates? We know, for example, that if the B field is produced by a solenoid, then the induced E field will be circular rather than parallel to the plates.

    The fringing of the E-field of the capacitor is not due to the B field. The capacitor is charged and there is an electrostatic field between the plates of the capacitor. This electrostatic field "fringes out" from between the plates as shown in post #7. This is relevant to part (a) of Griffiths' question as discussed in the paper by Griffiths et al.
     
  18. Sep 17, 2013 #17

    TSny

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    For the record, here is the complete statement of Problem 8.6 in the 3rd edition of Griffiths' Introduction to Electromagnetism.

    ----------------------------------------------------------------------------------------
    A charged parallel-plate capacitor (with uniform electric field ##\textbf{E} = E \;\bf{\hat{z}}##) is placed in a uniform magnetic field ##\textbf{B} = B \;\bf{\hat{x}}##, as shown in Fig. 8.6.

    (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 (as in part (b)), suppose we slowly reduce the magnetic field. This will induce a Faraday electric field, which in turn exerts a force on the plates. Show that the total impulse is (again) equal to the momentum originally stored in the fields.
    ---------------------------------------------------------------------------------------

    Comments. If the electric field of the capacitor is assumed to be uniform between the plates and zero everywhere else, then it is easy to calculate the answer for (a). However, this distribution of E field violates Maxwell's equations. The E field cannot suddenly go to zero at the edges of the plates. The actual field must be non-uniform between the plates and include fringing. In the AJP paper mentioned in post #7, Griffiths et al. note that including the fringing of the field reduces the answer for (a) by a factor of 1/2.

    If you work out (b), it appears that you get an answer equal to that of (a) for the case of no fringing of the E field of the capacitor!

    For part (c), Griffiths does not give enough information in the problem to deduce the pattern of the induced E-field. If you assume that the induced E-field is parallel to the plates of the capacitor, then you appear to get an answer that is equal to that of (a) for no fringing of the capacitor's E field. If you assume the induced E field is circular, as produced by the changing B field of a solenoid, then you get an answer for (c) that equals the answer for (a) with fringing taken into account!

    So, there appear to be inconsistencies in this problem which make it very interesting. These are all discussed and reconciled (apparently) in the paper of Griffiths et al.
     

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    Last edited: Sep 17, 2013
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