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Generalization of the magnetic moment

  1. Nov 14, 2006 #1
    I have searched various websites... but all they have about magnetic moments is:

    but all the the sources only showed that the equation is true for rectangular loop of current.

    but I need a more satisfactory answer. So I try to prove this in a general case:

    I want to prove that the torque is indeed [tex]\vec\mu\times\vec{B}[/tex], when magnetic moment is defined as:
    [tex]\vec\mu=I\int d\vec{A}[/tex]
    (after proving that, i will try to prove the energy relation: [tex]E=-\vec\mu\cdot{\vec{B}}[/tex])

    So, I forge ahead, let a closed current loop be the curve r, and a constant magnetic field B exists (I is current in the loop, tau is torque):
    [tex]d\vec\tau=\vec{r}\times d\vec{F}=I \vec{r}\times (d\vec{r}\times\vec{B})[/tex]
    where I is current in the wire, assume that it is constant.


    I used the triple product formula and some manipulations, then I got:

    the first term looks like [tex]\vec\mu\times\vec{B}[/tex] but then I can't tell if the second term is zero...well, i checked the manipulation numerous times... i basically used:


    when integrating, B(r * dr) goes to zero since it is a closed loop.

    I just changed the integrand.

    now I'm stuck...supposedly, torque=mu cross magnetic field. but what is the second term of the integral doing there? perhaps the formula for the torque on any closed current loop is not simply mu cross B?

    Questions: what frame should I use to calculate the torque? does it matter? should I choose the center of mass frame? Is the torque always the same regardless of what reference frame one is in?

    edited: clarified my intentions.
    Last edited: Nov 14, 2006
  2. jcsd
  3. Nov 14, 2006 #2


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    Hi, I don't know what you're trying to do here but if as the title of the thread suggest, you're looking for the general definition of the magnetic moment of a current loop, it is the following thing:

    Given a current loop in which circulates a current I, the magnetic moment is the vector


    where the integral is taken over any surface whose boundaries are the current loop itself (the result is independant of the choice of the surface). [itex]\hat{n}[/itex] denotes the vector normal to the surface.
  4. Nov 14, 2006 #3
    I'm trying to prove that the torque is indeed mu cross A. I should've specified that. sorry about that.
  5. Nov 15, 2006 #4

    Meir Achuz

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    [itex]\tau =I\oint{\bf r\times(dr\times B)}\\
    =I\oint{\bf dr(r\cdot B)}
    -I\oint{\bf B(r\cdot dr)}.
    The second integral vanishes for constant B.
  6. Nov 15, 2006 #5

    Meir Achuz

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    For the first integral:
    \oint{\bf (r\cdot B)dr}=\int_S{\bf dS\times\nabla(r\cdot B)}
    ={\bf S}\times{\bf B}.
  7. Nov 15, 2006 #6
    Thx, Meir, it looks rather simple... Can you tell me what theorem you used? I am currently learning vector calculus and I'm not that familiar with all those different theorems (stokes, greens...).
  8. Nov 15, 2006 #7


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    What Meir used in his last post is a corollary of Stokes theorem. It's a nice exercise proving it.
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