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Definition of magnetic moments and torques

  1. Oct 6, 2011 #1
    When one finds the torque on a rectangular current loop one finds that it's equal to

    [tex]\vec m \times \vec B[/tex]

    where m is the magnetic moment of the loop. I want to generalize this to arbirary current loops with constant current and I found that the torque would be equal to

    [tex] \left(I\oint \vec r \times \vec dl\right) \times \vec B[/tex]

    so I would think it would be natural to define

    [tex] \vec m =I\oint \vec r \times \vec dl [/tex]

    for these current loops. However when it up at http://en.wikipedia.org/wiki/Magnetic_moment" [Broken] I find that the magnetic moment is defined as

    [tex]\vec m =\frac{1}{2} I\oint \vec r \times \vec dl. [/tex]

    Where does this factor of a half come into the picture? Is'nt 'moments' supposed to be that what generates torques and not just be proportional to it?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 6, 2011 #2
    Look closely at your formula, does it give the right area when you take for example a circle? (The factor 1/2 is needed to calculate the area correctly)
  4. Oct 6, 2011 #3
    That's true. But I do not want the right area, just the right torque. I must admit I am a bit confused. I just saw a derivation of a statement that

    [tex] \vec r \times (d\vec r \times \vec B) = \frac{1}{2} \left( d\left[ \vec r \times ( \vec r \times \vec B\right] - \vec B \times ( \vec r \times d \vec r ) \right)[/tex]

    Which would indeed solve the problem. But does this then imply that

    [tex] \oint \vec r \times (d\vec r \times \vec B) = \frac{1}{2} \oint \left( d\left[ \vec r \times ( \vec r \times \vec B\right] - \vec B \times ( \vec r \times d \vec r ) \right) = \frac{1}{2} (\oint \vec r \times d \vec r) \times \vec B?[/tex]

    if so that would imply that

    [tex]\oint \vec r \times ( d \vec r \times \vec B ) - \frac{1}2 (\vec r \times d \vec r) \times \vec B = 0 [/tex]

    and can we not then conclude that

    [tex]\vec r \times ( d \vec r \times \vec B ) = \frac{1}2 (\vec r \times d \vec r) \times \vec B?[/tex]

    which seems to be a contradiction when I would think that

    [tex]\vec A \times ( \vec B \times \vec C) = (\vec A \times \vec B) \times \vec C.[/tex]

    I would really appreciate if someone could enligthen me.
  5. Oct 6, 2011 #4
    When two integrals are the same, that does not necessarily mean the integrated functions are the same. Also, in general Ax(BxC) is not equal to (AxB)xC, always watch out with that. (these are just general remarks)

    How did you arrive at your first expression for the torque? (in post #1)

    edit: I was a little too fast with this reply, I should add that in this case of course you need the factor 1/2 to get to the right answer, and therefore you will need some identity ;). I suspect you (mis)used that Ax(BxC) = (AxB)xC for your first result?
    Last edited: Oct 6, 2011
  6. Oct 6, 2011 #5
    You're right! That exactly what I did. And that solves my problem. Thank you! :)
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