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Magnetic field of a spherical capacitor

  1. Feb 20, 2014 #1

    ShayanJ

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    1. The problem statement, all variables and given/known data

    A spherical capacitor with inner and outer radii a and b, contains a dielectric material with small conductivity [itex] \sigma [/itex] between its spheres.Find the vector potential and magnetic field of this configuration.

    2. Relevant equations

    ce32c885c607fa9ba3fdce2fbf2e4e07.png

    3. The attempt at a solution

    [itex]
    \nabla^2\phi=0 \Rightarrow \frac{1}{r^2}(r^2\frac{d\phi}{dr})=0 \Rightarrow \phi=-\frac{p}{r}+q \\
    I=\frac{\Delta \phi}{R}=\frac{p(\frac 1 a -\frac 1 b)}{ \int_a^b \frac{dr}{4\pi \sigma r^2 }}\Rightarrow I=4\pi \sigma p \Rightarrow \vec{J}=-\frac{\sigma p}{r^2}\hat{r}
    [/itex]
    [itex]
    \vec{A}= - \frac{\mu_0}{4\pi} \int_a^b \int_0^\pi \int_0^{2\pi} \frac{\sigma p \hat{r}}{r^{'2}}\frac{r^{'2} \sin\theta d\varphi d\theta dr}{\sqrt{ r^2+r^{'2}-2r r^{'} \cos\gamma }}=
    - \frac{\sigma p \mu_0}{4\pi r} \int_a^b \int_0^\pi \int_0^{2\pi} \hat{r} \frac{\sin\theta d\varphi^{'} d\theta^{'} dr^{'}}{\sqrt{ 1+(\frac{r^{'}}{r})^2-2 \frac{r^{'}}{r} \cos\gamma }}=
    - \frac{\sigma p \mu_0}{4\pi r} \int_a^b \int_0^\pi \int_0^{2\pi} \hat{r} \sin\theta (\sum_{n=0}^\infty P_n(\cos\gamma) (\frac{r^{'}}{r})^n) d\varphi^{'} d\theta^{'} dr^{'}
    [/itex]
    And then,I expanded the Legendre functions using the following identity:
    [itex]
    P_n(\cos\gamma)=\frac{4\pi}{2n+1}\sum_{m=-n}^n Y^{m*}_n(\theta^{'},\varphi^{'})Y^m_n(\theta,\varphi)
    [/itex]
    After some calculations,I arrived at:
    [itex]
    \vec{A}=\frac{\sigma \mu_0 p}{12 r^2}(b^2-a^2)\sin\theta e^{i\varphi}(\hat{x}+\hat{y})
    [/itex]
    As you can see,its complex!!!
    Another issue is about the symmetry of the problem. The spherical symmetry of the configurations suggests that the magnetic field should be radial but a radial magnetic field violates Gauss's law in magnetism([itex] \oint \vec{B}\cdot \vec{dS}=0 [/itex]). On the other hand,any other form for the magnetic field, doesn't respect the symmetry of the configuration.
    I even thought that maybe the magnetic field is zero but as you can see easily,the vector potential isn't the gradient of a scalar field.
    Please help!
    Thanks
     
  2. jcsd
  3. Feb 20, 2014 #2

    TSny

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    I don't think you need to get into detailed integrations. From your general expression for ##\vec{A}## in terms of ##\vec{J}##, can you argue that ##\vec{A}## must be radial? What will that tell you about ##\vec{B}##?
     
  4. Feb 20, 2014 #3
    Hi.
    I don't think you can argue that [itex]\vec{A}[/itex] is radial only because we obtain it by integrating a radial function: in Cartesian coordinates, this logic would work because unit vectors [itex]\hat{x}[/itex] , [itex]\hat{y}[/itex] and [itex]\hat{z}[/itex] can be taken out of the integral but in spherical coordinates we can get any direction at the end of the integration...

    Also, a radial B-field seems perfectly good to me as long as [itex]\nabla[/itex][itex]\cdot[/itex][itex]\vec{B}[/itex] = 0, as for example: [itex]\vec{B}[/itex] = [itex]\frac{C}{r^{2}}[/itex][itex]\hat{r}[/itex]

    So i would start with Ampere's law in integral form (here in Gaussian units):
    [itex]\oint[/itex][itex]\vec{B}[/itex][itex]\cdot[/itex]d[itex]\vec{r}[/itex] = [itex]\frac{4\pi}{c}[/itex][itex]\int[/itex][itex]\vec{J}[/itex][itex]\cdot[/itex]d[itex]\vec{a}[/itex] + [itex]\frac{1}{c}[/itex][itex]\frac{d}{dt}[/itex][itex]\int[/itex][itex]\vec{E}[/itex][itex]\cdot[/itex]d[itex]\vec{a}[/itex],
    Around a circle centered at the origin on the left side, a hemispherical surface on the right and with the relation [itex]\vec{J}[/itex] = [itex]\sigma[/itex][itex]\vec{E}[/itex] (the electric field being easily found with Gauss' law and a charge Q(t) on the inner sphere). Then from the radial B-field, find a suitable vector potential...
     
  5. Feb 20, 2014 #4

    ShayanJ

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    As Goddar said,its not like that!
    I wrote the cartesian unit vectors in terms of spherical ones and now we have:
    [itex]
    \vec{A}=\frac{\sigma \mu_0 p}{12r^2}(b^2-a^2)\sin\theta e^{i\varphi} \left[ (\cos\varphi+\sin\varphi)(\sin\theta \hat{r}+\cos\theta \hat{\theta})+(\cos\varphi-\sin\varphi)\hat{\varphi} \right]
    [/itex]
    As you can see,all the components are present!

    You have two mistakes in your post.

    1-If [itex] \vec{B}=\frac{C}{r^2}\hat{r} [/itex],its divergence is not zero but [itex] 4\pi C \delta(r) [/itex]. Or, to put it in simpler terms, its surface integral on a spherical surface enclosing the origin isn't zero but [itex] 4\pi C [/itex]!

    2-In the integral form of Ampere's law,the line element isn't radial,but in the direction of the curving of the integration so in case of a radial B field,the line integral on the left vanishes!
     
    Last edited: Feb 20, 2014
  6. Feb 20, 2014 #5
    For your first point the tentative B is valid in the region b > r > a and B = 0 everywhere at r < a since E = 0, so the origin is not a problem and the divergence is zero everywhere. Anyway radial B-fields happen and i'm not saying this is the right one, you can Google some examples.

    I'll take your second point, i didn't pay enough attention.

    Then i guess you're right to go after the vector potential but it seems strange that in Cartesian coordinates you obtain A only in terms of x and y directions, given the spherical symmetry, so there might be a mistake there...
     
  7. Feb 20, 2014 #6

    ShayanJ

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    Yeah,that's right.
    Anyway.I think its not a problem for the vector potential because we can always add the gradient of a scalar field and so we can make the potential have a z component that way.But even if we don't do that,the B field is going to have a z component and because B is observable and not A,again there is no problem.

    Anyway,the curl of the vector potential I derived is:
    [itex]
    \vec{B}=\vec{\nabla}\times\vec{A}=-\frac{e^{i\varphi}\sigma mu_0 p (b^2-a^2)}{12r^3} \left\{
    \cos\theta\left[ -\cos\varphi+\sin\varphi+i(\cos\varphi+\sin\varphi) \right]\hat{r}-\sin\theta \left[ i(\cos\varphi+\sin\varphi)+2(\cos\varphi-\sin\varphi) \right]\hat{\theta}+3\cos\theta\sin\theta(\cos\varphi+\sin\varphi)\hat{\varphi}
    \right\}
    [/itex]
    Which is in no way radial!!!
    And is complex!!!
     
  8. Feb 21, 2014 #7

    ShayanJ

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    There is an alternative approach to the above one.The observation point can be arbitrarily assumed to be on the z axis(which doesn't harm the generality). The only change to the vector potential integral in the first post is that [itex] \gamma [/itex] becomes [itex] \theta [/itex] which means after expansion using Legendre functions, the addition theorem is not needed.The result I got with this method is:
    [itex]
    \vec{B}=-\frac{\sigma \mu_0 p \sin\theta \hat{\varphi}}{2r^3} \sum_{n=0}^\infty \frac{f(n)}{r^{2n}}
    [/itex]
    Where [itex] f(n)=\frac{b^{2(n+1)}-a^{2(n+1)}}{2^n} \sum_{k=0}^n \frac{(-1)^k [2(2n-k+1)]!}{k! (2n-k+1)![2(n-k)+1]![2(n-k)+3]!}[/itex]
    This solves the problem of magnetic field being complex but still doesn't seem right from the point of view of symmetry.This magnetic field is azimuthal but this can't be right because we can choose the z axis arbitrarily and the magnetic field can't be rotating around all directions of space!!!
    Damn it...Its too confusing.In fact in this case we have current wires in all directions of space (radii of the spherical capacitor) and each has a magnetic field circulating around it so the above field seems reasonable in some sense(or not???!!!)but what is it like?
    Too confusing!!!
     
    Last edited: Feb 21, 2014
  9. Feb 21, 2014 #8

    phyzguy

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    The Feynman lectures on physics, Vol 2 Chapter 18, has an analysis of just this situation. It is available for free online here. Understanding this problem will give you a key insight to electromagnetism.
     
  10. Feb 21, 2014 #9

    ShayanJ

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    Thank you very much.
    I should have known it,but...shame!
    It happens sometimes...you expect something so much that you can't think about it somehow else!
    I think one thing that should be solved for a novice physicist like me,is thinking out of structure and free from patterns.That takes time and practice though!
     
  11. Feb 21, 2014 #10

    phyzguy

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    Why should you have known it? It took the best minds of the human race more than a century to figure this all out. Don't feel bad if you didn't grasp it overnight.
     
  12. Feb 21, 2014 #11

    ShayanJ

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    Well, I knew that moving charges have changing electric field and I knew Maxwell's equation.But looks like I just was saying to myself "this is a problem which should be solved using the formula giving the vector potential of a current density".
    Anyway,I don't feel that bad,I'm just trying to remember it so I don't make the same mistake and think about problems using real physics, not that its this kind of problem or that kind of problem!
     
  13. Feb 21, 2014 #12

    ShayanJ

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    I still see another issue. In this problem,we have[itex] \vec{\nabla}\times\vec{B}=0 [/itex]. But we can write [itex] \vec{B}=\vec{\nabla}\times\vec{A} [/itex] which,along with the vanishing of curl gives us:
    [itex]
    \vec{\nabla}\times\vec{\nabla}\times \vec{A}=0 \Rightarrow \vec{\nabla}(\vec{\nabla}\cdot\vec{A})-\nabla^2\vec{A}=0
    [/itex]
    Using Coulomb gauge we'll have:
    [itex]
    \nabla^2\vec{A}=0
    [/itex]
    So we have the potential obeying Laplace equation which has non-trivial solution and its not at all obvious that the magnetic field coming out of such a potential is zero!
     
    Last edited: Feb 21, 2014
  14. Feb 21, 2014 #13
    Yes but boundary conditions are crucial in solving these equations and here they don't leave you much choice...
    Interesting problem
     
  15. Feb 21, 2014 #14

    ShayanJ

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    But what are the boundary conditions?
    Zero B-field at r=a and infinity?
     
  16. Feb 21, 2014 #15
    r = a and r = b
     
  17. Feb 21, 2014 #16

    ShayanJ

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    Ahh...yes...Charges on the conductors shouldn't accelerate and so there should be no Lorentz force parallel to the spheres' surfaces.But B is radial and [itex] \hat{r}\times\hat{\theta}=\hat{\varphi} [/itex] and [itex] \hat{r}\times\hat{\varphi}=-\hat{\theta} [/itex] and so B should be zero at the surfaces.
    Very beautiful!
     
  18. Feb 21, 2014 #17

    phyzguy

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    Since j has only a radial component, A will have only a radial component. Also, because of the spherical symmetry, the solution can only depend on r. So the solution for A must have the form:
    [tex] A = A(r) \overrightarrow{e_r}[/tex]

    this does have non-trivial solutions, but the curl of this is zero, so B = 0.
     
  19. Feb 22, 2014 #18

    vanhees71

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    I'm not so sure about that, because the magneto-static Biot-Savart law reads
    [tex]\vec{A}(\vec{x})=\int_{V} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.[/tex]
    Note that you have to evaluate the integral with cartesian components, i.e.,
    [tex]\vec{j}(\vec{x})=\frac{C}{\vec{x}^2} \vec{e}_r=\frac{C}{\vec{x}^2} [\cos \varphi \sin \vartheta \vec{e}_x + \sin \varphi \sin \vartheta \vec{e}_y + \cos \vartheta \vec{e}_z].[/tex] The volume element you can, of course write in spherical coordinates,
    [tex]\mathrm{d}^3 \vec{x}'=r'^2 \sin \vartheta' \mathrm{d} r' \mathrm{d} \vartheta' \mathrm{d} \varphi'.[/tex]
    However, I guess the integral is not so easily calculated. Perhaps, it's easier to find the answer by directly solving the local equation for [itex]\vec{A}[/itex] with the appropriate boundary conditions for [itex]\vec{B}[/itex].

    Note that in spherical coordinates, you cannot directly evaluate [itex]\Delta \vec{A}[/itex], which is only defined in Cartesian components. You have to use
    [tex]\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\frac{1}{c} \vec{j}.[/tex]
    Note that you can use one constraint on [itex]\vec{A}[/itex] due to ("static") gauge invariance, e.g., the Coulomb-gauge condition
    [tex]\vec{\nabla} \cdot \vec{A}=0.[/tex]
     
  20. Feb 22, 2014 #19

    phyzguy

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    Yes, clearly it's not correct to say that since j is radial, A must be radial. I apologize for muddying the water.
     
  21. Feb 22, 2014 #20

    TSny

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    I think this is correct. If ##\vec{J}## is radial of the form ##\vec{J}(\vec{r}) = J(r)\hat{r}## then you can show that the integral of ##\frac{\vec{J}(\vec{r'})}{|\vec{r}-\vec{r'}|}d\tau'## is also radial.

    Right.
    ---------------------------------------------------------------------------------------
    Is this because you are no longer assuming a gauge where A is the integral of J/|r-r'|?

    I think it would be odd to choose a gauge that would make A non-radial. It would disrespect the spherical symmetry of the problem. But, you could do that if you wanted.
     
    Last edited: Feb 22, 2014
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