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I Does a Magnetic Moment μ exist for any closed current loop?

  1. Aug 30, 2016 #1
    I'm asking the above question as I imagined a current loop made out of a curving wire that cannot fit into a flat plane.

    Where does the direction of μ point then?

    Also, is there only one μ value for any current loop, independent of the chosen axis of tilting?
    Or can a current loop have different values of μ depending on the angular position of the axis of tilting?

    The angular positions lie in a plane perpendicular to the external uniform magnetic field.
    The axis is parallel to that plane and also perpendicular to the B field. A current loop will experience a torque about the axis.
     
  2. jcsd
  3. Aug 30, 2016 #2
    An arbitrary shaped and oriented current loop can be thought of as a sum of small flat current loops. Each small current loop has a magnetic dipole moment with a magnitude equal to the curremt times the area and a direction given by the right hand rule. The total magnetic dipole moment is then the vector sum of the individual moments
     
  4. Aug 31, 2016 #3

    vanhees71

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    Well, let's do the calculation. I guess from the context you mean magnetostatic fields. So let ##\vec{j}## be the current density, fulfilling the static continuity equation ##\vec{\nabla} \cdot \vec{j}=0##. Then the Maxwell equations for the magnetic field simplify to Ampere's and Gauss's Laws (in Heaviside-Lorentz units)
    $$\vec \nabla \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0.$$
    The latter equation means that you can introduce a vector potential
    $$\vec{B}=\vec{\nabla} \times \vec{A},$$
    which is (for given ##\vec{B}##) determined only up to a gradient field (gauge invariance), and you can impose a simplifying constraint by fixing the gauge. In magnetostatics Coulomb gauge is most convenient, i.e.,
    $$\vec{\nabla} \cdot \vec{A}=0.$$
    Then using Cartesian coordinates you have
    $$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=-\Delta \vec{A}=\frac{1}{c} \vec{j}.$$
    Now from electrostatics you know the Green's function of the Laplace operator, which now can be applied to each Cartesian component separately, so that you get
    $$\vec{A}(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{4 \pi c |\vec{x}-\vec{x}'|}.$$
    Now suppose that the entire current distribution is within a sphere of radius ##R## around the origin, and you look at the potential for locations far away, i.e., for ##|\vec{x}|=r \gg R##. Then you can Taylor-expand the Green's function around ##\vec{x}'=0##. Here we take only the first two terms
    $$\frac{1}{|\vec{x}-\vec{x}'|}=\frac{1}{r} +\frac{\vec{x}' \cdot \vec{x}}{r^3}+\ldots.$$
    What you then get is a potential in powers of ##1/r##. The first terms gives the monopole contribution to the vector potential
    $$\vec{A}_0(\vec{x})=\frac{1}{4 \pi r} \int_{K_R} \mathrm{d}^3 \vec{x}' \vec{j}(\vec{x}').$$
    Here ##K_R## is the sphere of radius ##R##. Now we show that this vanishes. To this end we note that (Carstesian Einstein summation convention used)
    $$\partial_b (x_a j_b)=j_a + x_a \partial_b j_b=j_a,$$
    because ##\vec{\nabla} \cdot \vec{j} = \partial_b j_b=0##.
    From this we get using Gauss's integral theorem
    $$A_a=\frac{1}{4 \pi c r} \int_{K_R} \mathrm{d}^3 \vec{x}' \partial_b' (x_a' j_b)= \frac{1}{4 \pi c r} \int_{\partial K_R} \mathrm{d}^3 \vec{F}_b' x_a' j_b=0,$$
    because along the spherical shell ##|\vec{x}'|=R## by assumption ##\vec{j}=0##. Thus, there is no monopole term in the magnetic field, which is quite clear already from Gauss's Law ##\vec{\nabla} \cdot \vec{B}=0## (no magnetic monopoles!).

    Thus the leading order in the ##1/r## exansion is from the 2nd term in the expansion of the Green's function, i.e.,
    $$\vec{A}_{1a}(\vec{x})=\frac{\vec{x}_b}{4 \pi c r^3} \int_{K_R} \mathrm{d}^3 \vec{x}' j_a(\vec{x}') x_b'. \qquad(*)$$
    To make this a bit simpler we can use a similar trick as above and evaluate
    $$\partial_{c}'( x_a' x_b' j_c)=x_b' j_a + x_a' j_b.$$
    On the other hand we have
    $$\epsilon_{abc} \epsilon_{cde} x_d' j_e=(\delta_{ad} \delta_{be}-\delta_{ae} \delta_{bd}) x_d' j_e=x_a' j_b-x_b' j_a.$$
    subtracting the last two equations thus gives
    $$\partial_{c'}(x_a' x_b' j_c)-\epsilon_{abc} \epsilon_{cde} x_d' j_e=2 x_b' j_a,$$
    and we can write (*) as
    $$\vec{A}_{1a}(\vec{x})=-\frac{1}{2} \frac{x_b}{4 \pi c r^3} \int_{K_R} \mathrm{d}^3 \vec{x}' \epsilon_{abc} \epsilon_{cde} x_d' j_e$$
    or
    $$\vec{A}_{1a}(\vec{x})=-\frac{1}{2} \frac{x_b}{4 \pi c r^3} \int_{K_R} \mathrm{d}^3 \vec{x}' \epsilon_{abc} (\vec{x}' \times \vec{j})_c$$
    or finally
    $$\vec{A}_1(\vec{x})=-\frac{\vec{x}}{8 \pi c r^3} \times \int_{K_R} \mathrm{d}^3 \vec{x}' \vec{x}' \times \vec{j}(\vec{x}')=\frac{\vec{m} \times \vec{x}}{4 \pi r^3}.$$
    This leads to
    $$\vec{m}=+\frac{1}{2 c} \int_{K_R} \mathrm{d}^3 \vec{x}' \vec{x}' \times \vec{j}(\vec{x}'). \qquad (**)$$
    Evaluating the curl of ##\vec{A}_1## indeed leads to a dipole field:
    $$\vec{B}_1=\vec{\nabla} \times \vec{A}_1=\frac{3 (\vec{m} \cdot \vec{x}) \vec{x}-r^2 \vec{m}}{4 \pi r^5}.$$
    Thus (**) is the magnetic dipole moment for a general current distribution.

    For a filament-like current you have to substitute
    $$\mathrm{d}^3 \vec{x}' \vec{j} \rightarrow I \mathrm{d} \vec{r}'.$$
    Then the dipole moment gets
    $$\vec{m}=-\frac{I}{2 c} \int_{\ell} \mathrm{d} \vec{x}' \times \vec{x}' .$$
    Now we use Stokes's integral theorem. You can write for an arbitrary field ##\phi## and a constant vector ##\vec{n}##
    $$\int_F \mathrm{d} \vec{F} \cdot \vec{\nabla} \times (\vec{n} \phi) = \int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{n} \phi.$$
    Now for ##\vec{n}=\vec{e}_a## (##a \in \{1,2,3\}##) we get
    $$\int_F \mathrm{d} \vec{F} \cdot \vec{\nabla} \times (\vec{e}_a \phi) = \int_F \mathrm{d} \vec{F} \cdot (\vec{\nabla} \phi \times \vec{e}_a) = \int_F (\mathrm{d} \vec{F} \times \vec{\nabla} \phi)_a=\int_{\partial F} \mathrm{d} x_a \phi.$$
    This formula can be used to write (after some manipulations with the Levi-Civita symbols
    $$m_a=\frac{I}{c} \int_{F} \mathrm{d} F_a,$$
    where ##F## is an arbitrary surface with the boundary given by the current loop.
     
  5. Sep 2, 2016 #4
    Am I right to say that from here we can see that it is impossible for one current loop to have two different values of mu?
     
  6. Sep 2, 2016 #5
    I am not sure what you mean by two different values of μ. Any loop has one magnetic dipole moment. Different parts of the loop can be considered to effectively have a separate dipole moment, but the entire loop has one dipole moment. I have attached a Word doc. of a simple case of a non-planar loop showing how the total dipole moment can be calculated.
     

    Attached Files:

  7. Sep 3, 2016 #6
    think of a single loop, but you tilt it at different axes. Is the mu the same for all axes? For axes at all angles possible.
     
  8. Sep 3, 2016 #7
    The definition of the magnetic dipole moment of a simple loop is: μ = I A, where A is the vector area perpendicular to the plane of the loop. So if you tilt the loop, the vector changes direction, and the dipole moment is no longer the same.
     
  9. Sep 3, 2016 #8
    Okay.
    How do we prove that the mu of two loops combined is just the vector sum of their individual mu?
     
  10. Sep 3, 2016 #9

    vanhees71

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    It doesn't make sense to consider different parts of a current loop. You must always have a closed loop, because otherwise you don't get a magnetic field obeying the magnetostatic equations.
     
  11. Sep 3, 2016 #10

    jtbell

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    I suspect that Chandra is thinking of something like this:
    loops.gif
    where the loops on the right are infinitesimally close to each other.
     
  12. Sep 3, 2016 #11
    Absolutely.
     
  13. Sep 3, 2016 #12

    vanhees71

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    I don't get what you are after. Of course you can always use such a construction, but it doesn't change anything. The inner paths cancel each other. It's used in the standard proof of Stokes's Law and for the definition of the vector-derivative operator "curl=##\vec{\nabla} \times ##".
     
  14. Sep 4, 2016 #13
    So,...
     
  15. Oct 2, 2016 #14
    So Stoke's Law is essential to the proof that the mu of two loops combined is just the vector sum of their individual mu?
     
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