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Current distribution, magnetostatics

  1. Feb 8, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    If we want to obtain a magnetic dipole in the interior of a sphere of a radius R, what should be the current distribution over the surface of the sphere? Note that its permeability is the one of the vacuum. Determine the magnetic field outside the sphere.


    2. Relevant equations
    This is where I'm lost. I have Jackson's book opened on chapter 5 but I don't know where to look at.


    3. The attempt at a solution
    So they ask me for ##J(\vec x )##. I know it will involve the Dirac's delta in spherical coordinates. In other words ##J(\vec x ) = \frac{\delta (r-R)}{R}f(\theta ) f (\phi )##. So I must find ##f(\theta )## and ##f(\phi )##. In fact that would be ##J(\vec x )## in all the space, not only the surface of the sphere. For the surface of the sphere only, I remove the Dirac's delta.
    Maybe I must use Ampere's law under the differential form, namely ##\nabla \times \vec B = \mu _0 \vec J## but although they tell me more or less how the magnetic field is, I don't have it under any mathematical form and I'm not even sure there's only 1 possibilty for such a B field given the problem statement.
    Any tip will be appreciated.
     
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  3. Feb 9, 2013 #2

    BruceW

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    hmm. I haven't done this question before, so I am not certain. But solving
    [tex]\nabla \times \vec{B} = \mu_0 \vec{J} [/tex]
    seems like the right thing to do (as you said). The magnetic field for a magnetic dipole does have a specific form (assuming that you are not too close to the dipole). So the answer to this problem should be fairly easy to solve, once you know the form. I am surprised that the book does not give you the form. I guess they assume you can either derive it from first principles, or look it up on wikipedia... Unless there is some cheap & easy way to work out the current, by using symmetry, and without using the form of the magnetic dipole. But I cannot think of any such shortcut.
     
  4. Feb 12, 2013 #3

    fluidistic

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    Ok thank you very much. The problem is not taken out of Jackson's book, what I meant is that I used that book in order to tackle this problem. It was given by my professor so I'll ask him exacty how I should tackle it.
     
  5. Feb 12, 2013 #4

    marcusl

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    I suggest that you consider an analogous system that has the fields you want. A uniformly magnetized sphere is a dipole--it has uniform B inside and a classic dipolar field outside. Think about how to find the equivalent surface current density K.
     
  6. Feb 15, 2013 #5

    fluidistic

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    Good idea.
    Ok I will think on how to find the equivalent surface current density, K.
    As of now I don't come up with any idea, though I know that if I use spherical coordinates and the lines of the B (or H) fields leaving the magnetized sphere are along the z axis, then ##\vec K## will not depend on the coordinate ##\phi##, nor r for that matter. So that ##\vec K = K( \theta )## where theta is the zenithal angle. I hope this reasoning is right.
     
  7. Feb 15, 2013 #6

    marcusl

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    You're good so far.
     
  8. Feb 16, 2013 #7

    fluidistic

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    Ok thanks.
    I've been thinking about this problem but I'm stuck. Should I look up for the B field (or calculate it) of a magnetic dipole and then use some kind of Ampere's law to get the current density K?
     
  9. Feb 16, 2013 #8

    marcusl

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    Look at the boundary condition that relates a discontinuity in M to an equivalent surface current K.
     
  10. Feb 16, 2013 #9

    BruceW

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    you can either do that (which is the way I did it). Or you can do it marcusl's way. Both methods should give same answer.
     
  11. Feb 16, 2013 #10

    fluidistic

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    I notice that M and H are very much related, they differ by a unitless constant.
    An equation I found in Jackson's book is ##\hat n \times (\vec H_2 - \vec H_1 ) = \vec K##. Where ##\hat n## is a unit vector pointing from region 1 to region 2.
    My problem is that I don't have any mathematical expression for ##H_2##, the magnetic field outside the sphere. I know that for both poles, ##H_2=H_1## where H_1 is the magnetic field inside the sphere (and is a constant so I'll assume that ##H_1## is known).
    It means that at both poles there's no current because ##\vec H_2= \vec H_1 \Rightarrow \vec K =\vec 0##. But as soon as I leave the poles, I'm lost about ##\vec H_2##.
    Ok thanks! I'm going to try marcusl's method first.
     
  12. Feb 16, 2013 #11

    marcusl

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    There should be another form that relates to M which is valid in the absence of free currents. Something like [tex]\hat n \times (\vec M_2 - \vec M_1 ) = \vec K.[/tex]
     
  13. Feb 16, 2013 #12

    fluidistic

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    Hmm fine but I still have the same problem. While I know ##\vec M_1## is constant inside the sphere and it's equal to ##\vec M_2## at both poles, I don't know how ##\vec M_2## varies along the surface of the sphere.

    Hmm now that I think about it, the direction of ##\vec H_2## is always perpendicular to the surface of the sphere. It means that ##\vec K## is worth 0 at both poles and it's maximum along the equator. Does this imply that ##H_2=H_1 \cos \theta## where I'm considering magnitudes here?
    If so, then ##K=H_1(1-\cos \theta )##, for ##\theta=n \pi## this gives no current (which makes sense because it corresponds to both poles, n is any integer). And the current is maximum at ##\theta =\pi ( n+1/2)## which corresponds to the equator and it's worth ##H_1##. Hmm the units are not matching at all, something is definitely wrong I guess.
     
  14. Feb 16, 2013 #13

    marcusl

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    A) M1 is not equal to M2 at the poles. M2 exists in the air outside of the sphere and has the same value everywhere. What is it?

    B) Spoiler alert: The answer follows.

    No, it should go as sin(theta)
     
  15. Feb 16, 2013 #14

    fluidistic

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    Thanks for the help... I'm quite confused. :frown:
    Here's the picture I have in mind of the problem: http://www.aanda.org/index.php?opti...rl=/articles/aa/full/2002/12/aa1471/img28.gif.
    From it, I don't understand how can M2 have the same value everywhere.
    Also I don't understand why M1 is not equal to M2 right on the poles.
     
  16. Feb 17, 2013 #15

    marcusl

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    M_1 (inside the magnetized sphere) is constant, and M_2 (outside the sphere) must be zero because air is non-magnetic and cannot support a magnetization M. To put that mathematically, its relative permeability is one, mu_r = 1.

    EDIT: I suggested this approach because I thought it might be a quick and intuitive way to establish the sinθ dependence of surface current density. It's just one line [tex]\vec K = \hat n \times (\vec M_2-\vec M_1) = -\hat n \times \vec M_1 = - M_1\sin(\theta)\hat\phi.[/tex] I see that, instead, I confused you--very sorry. Perhaps you should return to your original approach? Here is a link to a solution using that approach
    http://web.mit.edu/6.013_book/www/chapter8/8.5.html
     
    Last edited: Feb 17, 2013
  17. Feb 17, 2013 #16

    fluidistic

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    Ok thank you very much, what you wrote makes sense. I'm a newbie in magnetostatics (just started to self study it). I previously thought that M was proportional to H which in turn was proportional to B and since there's an external magnetic field, I thought that M could be different from 0 even in air/vacuum.
    I'll try to follow carefully the link you gave me.

    I'm left with calculating the external magnetic field. I believe I must use the formula posted in the first post, ##\nabla \times \vec{B} = \mu_0 \vec{J}## though I should replace J by K.
     
  18. Feb 17, 2013 #17

    marcusl

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    Fluidustic, you are doing yourself a disservice by attempting to learn E&M for the first time from a difficult graduate-school text like Jackson. I strongly suggest starting with a standard 1st course text like Halliday & Resnick. Used copies of the old blue and orange edition are very inexpensive. Then move to Griffiths, which is the gold standard 2nd course (jr.year). At that point you will know E&M well and will have the background to move to Jackson if you wish.
     
  19. Feb 17, 2013 #18

    fluidistic

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    I already took the freshman course using Halliday and Resnick. We jumped from that level to Jackson in only 1 semester, we never passed by Griffith's level (they gave us problems in electrostatics that had no azimuthal symmetry and sadly Griffith's book was useless there). I had to withdraw from the upper level undergrad course since at that time we had a very inadequate mathematical background (solved only some basics ODE's but without having taken a course on them, nothing about the Dirac's delta distribution, nothing about PDE's) and I wasn't understanding anything. But since then I've taken a mathematical methods in physics course (covering PDE's, ODE's, Green functions although not the Dirac's delta) and I've just retaken the study of the upper level EM. The math should not be that much of a problem, though knowing Jackson's problems, nothing is easy either.
    We're given several possible days per year for the final exam which is worth 100% of our grade. I'm aiming to take the final exam in early March, I'm currently in holidays so I really hope I have time to catch up by then. It's the only course I'm currently studying.
    I'll likely post a lot of problems during the next days. :smile:
     
  20. Feb 17, 2013 #19

    marcusl

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    Requiring Jackson for an undergrad course is criminal, IMO. It sounds like you have a copy of Griffiths. I might suggest that you check a copy of Reitz and Milford out of the library. It's a fairly compact and straightforward treatment at an intermediate level that could help you with concepts like B, M and H.
     
  21. Feb 17, 2013 #20

    fluidistic

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    I don't have Griffith's, I had borrowed it from the library. But I'm going to borrow it again tomorrow as well as your suggestion by Reitz and Milford. I just checked online and my university seems to have 2 copies of it available.
    Just to be sure, the book is called "Foundations of Electromagnetic Theory", right?
    Thanks a lot!
     
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