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Motion of a charge due to induced electric field

  1. Dec 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Imagine a uniform magnetic field, pointing in the ##z## direction and filling all space (##\vec{B} = B_0\hat{z}##). A positive charge is at rest, at the origin. Now somebody turns off the magnetic field, thereby inducing an electric field. In what direction does the charge move?
    [There is a footnote for this problem: "This paradox was suggested by Tom Colbert...". The footnote than refers me to a problem about an electric field of the form ##\vec{E} = ax\hat{x}## that in spite of having a uniform charge density by its divergence, is not uniform itself and the problem asks to explain this. The answer is that the question is ill-defined and that the curl and divergence by themselves do not determine the field and appropriate boundary conditions are necessary.]

    2. Relevant equations
    A formula for the induced electric field in terms of the time derivative of the magnetic field:
    [tex]\vec{E} = -\frac{1}{4\pi}\frac{\partial}{\partial t}\int\frac{\vec{B}\times\hat{r}}{r^2}d\tau[/tex] where ##d\tau## is the infinitesimal volume element.

    3. The attempt at a solution
    Since the magnetic field is uniform and points in the ##z## direction, turning it to (abruptly) change in the ##-z## direction. The change in flux will also be in this direction. Thus, the electric field will have to create flux upwards in the ##z## direction by Lenz's law. In order to do so, it will have to circulate everywhere counterclockwise as viewed from above by the right-hand rule. Since the field was uniform everywhere, the circulations will also be uniform and it seems to me that by symmetry, the force they will exert on the particle will cancel out. Thus, I think that the particle won't move at all. However, this footnote makes me believe that this is a trick question and that I have missed something. Is my reasoning correct or have I missed a catch somewhere?

    Any comments will be greatly appreciated!
     
  2. jcsd
  3. Dec 23, 2014 #2

    BvU

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    There's nothing to get the charge moving (No Lorentz force).
    I wonder if your formula is applicable: it doesn't look as if it's invariant under translation of the volume element (or shifting the origin of the coordinate system). Was it derived with certain conditions attached ?
     
  4. Dec 23, 2014 #3
    Firstly, I believe that it is the induced electric field that is supposed to get the charge moving (##\vec{F} = q\vec{E}##), not the magnetic one.
    Secondly, I wonder about the formula too. I have attached its derivation from the book (Griffiths'). It is already the 3rd time he uses this kind of derivation and I wonder every time if it is legal...
     

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  5. Dec 23, 2014 #4

    vela

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    If there's no force, there's no electric field, right? So ##\nabla \times \vec{E} = 0## but ##\partial\vec{B}/\partial t \ne 0##.
     
  6. Dec 23, 2014 #5
    I'm not sure I understand what you mean. By Maxwell's equation: ##\nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial t}##. There is no force initially, but as the magnetic field vanishes, it is supposed to appear via the appearing electric field. The question is, what would that electric field be and what net force (if any) will it exert on the particle.
     
  7. Dec 23, 2014 #6

    BvU

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    Griffiths exploits the analogy with Biot-Savart. At this link we find back the formula 7.18 (we are completely enclosed by the non-zero ##-{\partial B\over \partial t}## everywhere). For any finite loop you get (Griffiths 7.19) a proper emf.

    But it looks as if ##-{\partial B\over \partial t} \times \hat r ## yields zero when you integrate over the azimuthal angle (##\phi##).

    I do find it a bit awkward: ##\nabla \cdot \vec E=0## and ##\nabla \times \vec E = - {\partial B\over \partial t}\;\ne 0## with ##- {\partial B\over \partial t} = (0,0,a) ## is satisfied (I think) with ##\vec E = (y,-x,0) \; a/2##, so zero at the origin all right, but there's nothing special about the origin in this situation. Perhaps we are overlooking some other (boundary or normalization) condition.

    I like the exercise -- must admit that I look at it just like you, as a (curious) student, non-expert. Hope some real expert listens in and helps out !
     
  8. Dec 23, 2014 #7
    Actually, you are right (I was so preoccupied with finding catches that I didn't think about trying to straight-forwardly integrate!): ##\vec{B}\times\hat{r} = B_0 (\hat{z}\times\hat{r}) = B_0 \left(-\sin\theta\sin\phi\hat{x} + \sin\theta\cos\phi\hat{y}\right) = B_0\sin\theta\hat{\phi}##. So when you integrate this over all space (and in particular ##\phi: 0→2\pi##) you get ##\vec{E} = 0## and the particle should not move. Then where is the paradox the footnote was talking about? Perhaps the "paradox" is that such a huge and abrupt change in ##\vec{B}## that gives rise, accordingly, to a big ##\nabla\times\vec{E}## leads to a zero ##\vec{E}##? I have to say that if it is so than it makes the connection with the previous problem I mentioned in the OP clear - it emphasizes, too, that the divergence and curl of the vector field are not enough to determine it.
     
  9. Dec 24, 2014 #8

    vela

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    If ##\vec{E}=0## as you argued, then how can ##\nabla \times \vec{E} \ne 0##?
     
  10. Dec 24, 2014 #9
    Thanks for the catch. After some thought, I figured that the formula I used for ##\vec{E}## was in fact a formula for ##\vec{E}## at the origin. The full formula should read:
    [tex]\vec{E} = -\frac{1}{4\pi}\frac{\partial}{\partial t}\int\frac{\vec{B}\times(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3}d\tau[/tex] where ##\vec{r}## is the position vector of the point where I am calculating the field and ##\vec{r}'## is the position vector of the source of the field (i.e. charges or, as in this case, a magnetic field). If I'm calculating the field at the origin, ##\vec{r}-\vec{r}' = -\vec{r}'## and, apart from a sign change, the formula turns to be as the one in the OP (##\frac{\hat{r}}{r^2} = \frac{\vec{r}}{r^3}##). Thus, ##\vec{E} = 0## only at the origin (the point of interest) and there is no contradiction with ##\nabla\times\vec{E} \ne 0##.
     
  11. Dec 24, 2014 #10

    BvU

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    So what's so special about the origin in this situation ? Would the charge experience a non-zero E field if it were placed at some distance away from the origin ?

    I'm just as unhappy as I was in posts #2 and #6 :(

    Merry XMas ! :)
     
  12. Dec 24, 2014 #11

    TSny

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    Maxwell’s equations do not have a unique solution for E for this problem where the B field extends to infinity. Knowing the divergence and curl of a vector field at every point does not necessarily allow you to uniquely determine the field. A nice discussion is given in the first section of the first chapter of Panofsky and Phillips, Classical Electricity and Magnetism. You can read most of their discussion here: https://www.amazon.com/Classical-El...&keywords=classical+electricity+and+magnetism.

    If the “sources” of the divergence and curl are nonzero only in a finite region of space, then there will be a unique solution. If the sources extend to infinity, then the solution does not need to be unique. In this problem, the source of the curl of E is - ∂B/∂t and this is nonzero throughout all of space.
     
    Last edited by a moderator: May 7, 2017
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