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Moving sphere in magnetic field

  1. Jul 29, 2015 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I got no credit in an exam for the following exercise and I've been told what's wrong but even then I am unable to solve it correctly.
    A conducting sphere with radius R moves with constant velocity ##\vec v =v \hat x## (v <<c) inside a constant magnetic field ##\vec B =B \hat y##. Find the induced charge distribution on the sphere to 1st order in v/c in the laboratory inertial reference frame.
    2. Relevant equations
    Fields transformations: ##\vec E '=\gamma (\vec E +\vec \beta \times \vec B ) - \frac{\gamma^2}{\gamma +1} \vec \beta (\vec \beta \cdot \vec E )##
    ##\vec B' =\gamma (\vec B -\vec \beta \times \vec E ) - \frac{\gamma ^2}{\gamma +1} \vec \beta (\vec \beta \cdot \vec B )##.
    Lorentz boost.
    ##\rho = \gamma \rho'##. Where rho is the charge distribution in the laboratory K and rho' the charge distribution in another inertial reference frame moving with constant velocity with respect to K.
    3. The attempt at a solution
    My idea was to consider K', an inertial reference frame that would move alongside the conducting sphere. From there, calculate the E' and B' fields "seen" by the sphere. From there, calculate the induced surface charge density ##\sigma '##. And from there simply convert using the fact that ##\sigma = \gamma \sigma '##.
    So here I go:
    Using the formulae for fields transformations, ##\vec E'=\frac{\gamma v}{c}\hat z## and ##\vec B'=\gamma B \hat y##. Thus for the reference frame K', there's a static EM field. Now the sphere is a perfect conductor so the E' field inside of it must vanish. So that there's a discontinuity in the E' field that is proportional to the surface charge density sigma'. Mathematically, ##\hat n \cdot (\vec E' - \vec E'_{\text{inside}})=\sigma ' /\varepsilon _0##. Where ##\hat n = \hat r'## and as I said, ##\vec E'_{\text{inside}}=\vec 0##.
    Furthermore, ##\hat z' \cdot \hat r' = \cos \theta '##.
    Therefore ##\sigma '(\theta ') = \varepsilon_0 \frac{\gamma v}{c}\cos (\theta ')##. Now ##\sigma = \varepsilon_0 \frac{\gamma ^2 v}{c}\cos (\theta ')## but I must transform back ##\theta '## into non primed coordinates. I won't do it though, because this answer is completely wrong according to my professors and I don't deserve any credit.
    Instead, according to them, what I should have done is to calculate the electric field generated by the surface charge density. The total electric field seen in K' will be the sum of the field I had calculated plus the field due to this charge density.

    That's basically what I have been told. But then what? I wasn't asked to calculate the fields... I was asked the charge density. So if I understand them well, I must calculate this new E' field, and from it recalculate the new charge density? This doesn't make any sense to me, I could do this process forever, i.e. calculate the E field generated by a charge density, then calculate the new charge density, then calculate the new E field, then calculate the new charge density, etc. So why should I stop at step 1 or 2? Is it because they asked "to 1st order in v/c" in the problem statement?
     
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  3. Jul 29, 2015 #2

    Orodruin

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    You can solve the problem exactly for a constant external field at infinity and a fixed potential on the sphere. From there you can deduce the charge.

    Since you are only looking at leading order terms in v/c, you may ignore any gamma factors. However, note that when you cannot, surface charge densities are not related in the same way as volume charge densities, but also depend on the orientation of the surface.
     
  4. Jul 31, 2015 #3

    fluidistic

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    Thank you very much for the explanation/clarification. I'll keep using gamma up until the very end then.
    So I solved for the potential outiside of the sphere, considering azimuthal symmetry (so ##\Phi = \Phi (r, \theta)## and doesn't depend on phi, the azimuthal angle). I solved Laplace equation ##\nabla ^2 \Phi =0## in spherical coordinates. I applied 2 boundary conditions, namely that ##\Phi (r=R, \theta ) =V_0## and ##\Phi (r \to \infty , \theta )=-r \gamma \frac{v}{c} \cos \theta##.
    So that the general form for the potential ##\Phi (r, \theta )=\sum _{l=1}^\infty (A_l r^l +B_l r^{-l-1})P_l (\cos \theta)## becomes ##\Phi (r,\theta)=\underbrace{-\frac{\gamma v}{c}r \cos \theta }_{A_1rP_1}+\underbrace{\frac{RV_0}{r}}_{B_0P_0/r} + \underbrace{\frac{\gamma vR^3}{cr^2}}_{B_1P_1/r^2}##.
    Now I don't really know how to get the surface charge density from it. Some books give the formula ##-\sigma /\varepsilon_0 =\frac{\partial \Phi}{\partial r} \big | _{r=R}##. If so, then I'd get ##\sigma(\theta )=\frac{3\gamma v}{c} \varepsilon _0 \cos \theta + \frac{V_0 \varepsilon _0}{R}## However I don't trust that formula, because I don't see what's wrong with the following:
    Matching condition ##\hat n \cdot \vec E=4\pi \sigma(=\sigma /\varepsilon _0)=E## where ##\vec E = - \nabla \Phi##. But if I take the gradient in spherical coordinates, I get a much more complicated relationship between sigma and Phi than the one given in Griffiths for example (the one I posted above).
    Therefore I'm not sure which way is correct from there.
     
  5. Jul 31, 2015 #4

    TSny

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    Shouldn't the magnetic field B be a part of your expressions for ##\Phi##?
    The last term of your equation for ##\Phi (r,\theta)## is missing the appropriate Legendre polynomial factor.

    OK. You should be able to determine the value of ##V_0## by thinking about what the total induced charge must be.

    You should get the same result this way. Remember that you are evaluating the gradient at the surface of the sphere where ##r=R##.
     
  6. Aug 1, 2015 #5

    fluidistic

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    Hmm I don't think so. Do you mean that the B field has an effect on the charge distribution?
    Indeed, I made a typo there; on my draft I have the cos theta term.


    The total induced charge must be 0 right? So that the integral of the surface charge density over all theta should be 0. Correct?
    And ok, thank you very much for the comment about the gradient and that both formula should yield the same result.
     
  7. Aug 1, 2015 #6

    TSny

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    In the frame of reference of the sphere, there is an electric field E' which is proportional to B. So, the induced charge should depend on the magnitude of B. Note that your expressions for ##\Phi## do not have the correct dimensions for electric potential.

    Yes
     
  8. Aug 2, 2015 #7

    fluidistic

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    Oops right, I made a mistake there...
    I get [tex]\Phi (r, \theta) =-\frac{\gamma vB}{c}r\cos \theta + \frac{RV_0}{r} +\frac{\gamma v B R^3}{cr^2}\cos \theta[/tex]
    If I use a unit system such that ##4\pi \sigma (\theta ) = - \frac{\partial \Phi}{\partial r} \big | _{r=R}##, then I get that ##\sigma (\theta )= \frac{3\gamma vB \cos \theta}{4\pi c}+\frac{V_0}{4\pi R}##.
    I am unsure about how to integrate sigma over all angles. Is it just ##\int_0^\pi \sigma (\theta ) d\theta =0##? If so, then I get ##V_0=0##.
    But maybe it's ##\int_0^\pi \sigma (\theta ) \sin \theta d\theta =0##? I am really not sure.
     
  9. Aug 2, 2015 #8

    TSny

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    ##\sigma## is charge per unit area. So the amount of charge in a patch of area ##dA## of the spherical surface is ##dq = \sigma dA##.
     
  10. Aug 2, 2015 #9

    fluidistic

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    I see. Then I'm probably doing some mistake. If ##dA=R^2 \sin \theta d\theta d\phi##, I get ##V_0=0##.
     
  11. Aug 2, 2015 #10

    TSny

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    Why do you think that answer is wrong?
     
  12. Aug 2, 2015 #11

    TSny

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    I think I see what might be causing confusion. In your summation over ##l## in the expression for ##\Phi(r, \theta)##, the sum should start at ##l = 0##. So, there is an additional ##A_0## term that you left out.

    [EDIT: So, you cannot assume that ##B_0 = RV_0##].
     
    Last edited: Aug 2, 2015
  13. Aug 2, 2015 #12

    fluidistic

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    I thought that this would have caused problem if I had assumed that ##V_0=0## when I applied the boundary conditions but that would have been fine I guess, so ok, I accept this as answer.
    So overall I get a charge density 3 times greater than what I did in my original attempt. I get that ##\sigma ' (\theta ')=\frac{3\gamma vB \cos \theta}{4\pi c}##. Where the primes mean as seen from the system K'. Now I need to transform this for K. I can write ##z'=r' \cos \theta '##. So that ##\cos \theta '=z'/r' =z/r'=\frac{z}{\sqrt{x'^2+y^2+z^2}} = \frac{z}{\sqrt{\gamma ^2 (x-vt)^2 +y^2 +z^2}}##.
    Thus [tex]\sigma (x,y,z)=\frac{3\gamma vBz}{4\pi c \sqrt{\gamma^2 (x-vt)^2 +y^2+z^2}}[/tex]. From here I could approximate gamma by something with v/c (I haven't seen the approximation yet).
    Is this correct?
     
  14. Aug 2, 2015 #13

    fluidistic

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    Hmm but from ##\Phi (r \to \infty, \theta)=-r \cos \theta \frac{\gamma vB}{c}## this seems to imply that ##A_0=0## to me.
     
  15. Aug 2, 2015 #14

    TSny

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    For ##r \gg R##, ##\Phi \rightarrow -E' r \cos \theta +C##
    where ##C## is a constant to be determined. For general boundary conditions you cannot assume ##C = 0##.

    [EDIT: You should be able to show that ##C=0## only when ##V_0 = \frac{Q}{4 \pi \epsilon_0 R}##, where ##Q## is the net charge of the sphere.]
     
    Last edited: Aug 2, 2015
  16. Aug 2, 2015 #15

    TSny

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    I suspect that you are just supposed to describe how the charge is distributed over the sphere in the K frame. I don't think you need to express ##\sigma## as a function of ##x, y, z, t## in the K frame. Just express ##\sigma## as a function of spherical coordinates ##\theta, \phi## on the sphere from the point of view of the K frame.

    Since you are only required to express the answer to first order in ##v/c##, you don't need to worry about effects of the Lorentz transformation from the K' frame back to the K frame. In the K' frame, you found that ##\sigma## is already proportional to ##v/c##. Any relativistic effects of transforming back to the K frame from the K' frame would just introduce higher order terms in ##v/c##.

    So, unless I'm overlooking something, I think you are done. :smile:

    But this exercise is showing how relativity provides a nice way to solve the problem by going to the frame of reference of the moving sphere and using the fact that in this frame there will be a uniform E' field that induces the charge distribution on the sphere. If you stayed in the K frame, you would have to try to find the induced charge distribution by finding a distribution of charge where the magnetic force on the (moving) charge distribution is balanced by the electric force due to the electric field produced by the induced charge. (Yuck!)
     
  17. Aug 2, 2015 #16

    fluidistic

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    Thanks for all the comments.
    Hmm I really don't see how to do this. I don't see how to avoid the r dependency either.
     
  18. Aug 2, 2015 #17

    TSny

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    Let the center of the sphere in the K frame be at the origin of coordinates at t = 0. At this instant of time, you can use spherical coordinates to describe the charge distribution. I would think that's all that is asked for. The distribution of charge over the surface of the sphere does not change with time. You could go further and write the distribution as a function of x, y, z, and t for any time t, as you essentially did. But you would need to restrict the values of x, y, and z so that the point (x,y,z) lies on the surface of the sphere at time t. But I don't see the point of doing all that.

    Note that to first order in v/c, ##\gamma## can be approximated by 1: ##\frac{1}{\sqrt{1-v^2/c^2}} \approx 1 + \frac{v^2}{2c^2}##.
    So you can set all gamma factors to 1 as pointed out by Orodruin in post #2.

    I don't know how your professor wanted you to approach the problem.
     
  19. Aug 2, 2015 #18

    fluidistic

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    Ok thank you very much for everything. So I'd give as answer ##\sigma (\theta) =\frac{3vB \cos \theta}{4\pi c }## neglecting the fact that the sphere doesn't appear spherical. Yeah I don't know either what my professor was expecting from me. Next time I'll ask him if I have doubts on what he's expecting.

    Edit: In post #15 you wrote sigma in terms of theta and phi, what did you have in mind about phi?
     
  20. Aug 2, 2015 #19

    TSny

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    Sorry, I didn't mean to imply that there would be phi dependence. Sigma will just depend on theta, at least to order v/c.
     
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