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The Curl of Magnetic Potential

  1. Feb 3, 2013 #1
    Hi guys! I recently saw on Wiki that given a magnetic potential A(r)=(u/4π)(mXR/r^3) ,( where u is the permeability of free space, m is the magnetic dipole moment of a magnetic field, R is the position vector, and r is the distance from the magnetic field ) upon taking the curl of the magnetic potential A(r), one obtains (u/4π)([3(m•r)R/r^5]-[m/r^3]). Can someone show me how they ended up with this result? I would like a step by step process please.(: the wiki article is called Magnetic Moment under the section called Effects of an External Magnetic Field on a Magnetic Moment.
     
  2. jcsd
  3. Feb 3, 2013 #2

    SammyS

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    That's not how we do things in the Homework section of PF. We're not even allowed to give such help. But we can try to help you work your way through such a problem.

    One way to get the result, is to write A(r) in terms of it's Cartesian components, then take the curl. I suppose you could also take the curl directly in spherical coordinates.
     
  4. Feb 3, 2013 #3
    I'm sorry! Well I thought about doing it in spherical coordinates but I was not sure if I should use the whole three partials in the del operator.
     
  5. Feb 3, 2013 #4

    SammyS

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    Yes, use all three partials.
     
  6. Feb 3, 2013 #5

    Dick

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    I don't think there is any advantage is using spherical coordinates. Try using the identity:

    [tex]\nabla \times (a \times b)=(b \cdot \nabla) a-b(\nabla \cdot a)
    -(a \cdot \nabla) b+a(\nabla \cdot b)
    [/tex]

    If you set a to be m and b to be ##\frac{\vec r}{r^3}##, the first two terms vanish because m is a constant. The last one vanishes because ##\frac{\vec r}{r^3}## is divergence free. So you only have to puzzle out what the third term means.
     
  7. Feb 3, 2013 #6
    Thank you very much! (:
     
  8. Feb 3, 2013 #7
    Is there any other identity I could use for (a(div))b? I did the calculation but I think I did it wrong. I'm missing the second term in the curl of the magnetic potential. The (m/r^3) term
     
  9. Feb 3, 2013 #8

    Dick

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    Show what you did. You'll need to use the quotient rule on the dervivatives ##\frac{\vec r}{r^3}##.
     
  10. Feb 3, 2013 #9
    Well, at first I used the identity (a(div))b= (div(a))b-b(div(a)). Was that a correct move?
     
  11. Feb 3, 2013 #10
    Oh! I get it! Nevermind! I didn't think to carry out the quotient rule. That makes a lot more sense! Thank you!
     
  12. Feb 3, 2013 #11
    So I should calculate using the radial term of the spherical version of the del operator?
     
  13. Feb 3, 2013 #12
    That's the only way I see to take the derivative of R/r^3.
     
  14. Feb 3, 2013 #13

    Dick

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    R/r^3 is just (x,y,z)/(x^2+y^2+z^2)^(3/2). It worked out pretty ok for me in cartesian coordinates.
     
  15. Feb 3, 2013 #14
    Would it be a little easier with spherical coordinates?
     
  16. Feb 3, 2013 #15

    Dick

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    Maybe. I didn't try it that way. I just know it's not hard in cartesian.
     
  17. Feb 3, 2013 #16
    I will try it! Thank you!
     
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