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I Volume magnetization currents - derivation confusion

  1. Aug 15, 2016 #1
    Hello there,

    I've just been learning about surface magnetization currents circulating around hypothetical square loops. Since the magnetization is uniform the circulation currents cancel where the square loops are adjacent to one another and it can therefore be said that the current circulates around the edge of the magnet where the current is a bound current. I understand this and the derivation.

    I am struggling a bit with the derivation of the same situation but now without uniform magnetization. I'll explain where I can get up to.

    First looking at two loops, A and B. Loop A lies perpendicular to the z-direction and is centred on the point [itex](x,y,z)[/itex]. It represents part of a slice of thickness [itex]\delta z[/itex], and has sides of length [itex]\delta x[/itex] and [itex]\delta y[/itex]. The z-component of the magnetic moment of this volume element is [itex]M_x (x,y,z) \delta x \delta y \delta z[/itex], so the current that flows around the loop to generate this moment is
    [tex]I(x,y,z)=(magnetic moment)/area=M_z (x,y,z) \delta z[/tex].
    Loop B has the same dimensions but is centred at [itex](x+ \delta x, y, z)[/itex]. We can write the current in this loop as
    [tex]I(x+ \delta x, y, z)=M_z (x + \delta x , y, z) \delta z [/tex].
    The net current in the y-direction at the boundary between the two loops is
    [tex]I(x,y,z)-I(x+ \delta x, y, z)=[M_z (x,y,z)-M_z(x + \delta x,y,z)] \delta z=-(\partial M_z/ \partial x) \delta x \delta z[/tex]
    This is the bit that I'm getting stuck on now. Can someone please explain how we get the minus partial derivative and where the [itex]\delta x[/itex] comes from please? If you could break it down into smaller steps it would be greatly appreciated. Thanks in advance for any help.

    Regards

    Brian
     
  2. jcsd
  3. Aug 15, 2016 #2

    Charles Link

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    The equation that gets you from magnetization vector to magnetic current density is ## \nabla \times M=\mu_o J_m ##. Going from a solid with uniform magnetization ## M ## to a vacuum, you get, (using Stoke's theorem), magnetic surface current per unit length ## K_m=M \times \hat{n}/\mu_o ##.( ## \hat{n} ## is an outward pointing unit vector normal to the surface.) The actual proof of these two statements is somewhat difficult, but the second one is a little easier to prove than the first.## \\ ## Griffiths E&M textbook actually proves them both together by starting with a distribution of magnetic moments and computing the vector potential ## A ## for that distribution and using the result ## A(x)= \int \frac{ \mu_o J(x')}{|x-x'|} \, d^3x' ## . ## \\ ## The vector potential ## A ## satisfies ## \ ## ##B= \nabla \times A ## ## \ ## and is not to be confused with the letter ## A ## for area below. ## \\ ## Hopefully this is at least somewhat helpful. editing..the magnetization ## M ## consists of microscopic current loops of magnetic moment ## \mu=IA ##. (## A ## is the area of the microscopic loop and points perpendicular to the loop.) Magnetization ## M=n \mu ## where ## n ## is the number of microscopic magnetic moments ## \mu ## per unit volume. I would recommend using the above results without necessarily working through completely detailed proofs. ..and notice also for uniform magnetization ## M ## that there is no net bulk current (only surface currents) because ## \nabla \times M=0 ##.
     
    Last edited: Aug 15, 2016
  4. Aug 15, 2016 #3

    Twigg

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    It's not obvious until you draw the two loops next to each other and right-hand-rule it to the end. A positive value of ##M_{z}## means the current flows counterclockwise around the loop. That means the current on the left hand side of loop B is going to be flowing against the current of the right hand side of loop A. That's why the current in the y-direction for the side shared by both loops is ##I(x,y,z) - I(x + \delta x,y,z)##, which is where the negative derivative comes from. The math used there is just ##I(x+\delta x, y, z) - I(x,y,z) = \frac{\partial I}{\partial x} \delta x = \frac{\partial}{\partial x}(M_{z}\delta z) \delta x##, but with the minus sign I talked about due to the orientation.
     
  5. Aug 15, 2016 #4

    Charles Link

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    It's a good start to showing the ## J_y ## current, but you also get a term ## dM_x/dz ##. (Of course we're just working with ## M_z ##). No doubt there are a couple of E&M textbooks that present a very rigorous proof for the equation ## \nabla \times M=\mu_o J_m ##. Griffiths E&M textbook does a roundabout method using vector potentials along with deriving the vector potential for a single microscopic magnetic moment, which is an equally effective proof. editing... and a couple minor corrections to my post #2 above: I think the integral expression for vector potential ## A ## needs a ## 4 \pi ## in the denominator, and also the microscopic magnetic moment using ## B=\mu_o H +M ## will have ## \mu=\mu_o I A ##. When ## M ## is defined so that ## B=\mu_o (H +M) ##, then ## \mu=I A ##. For this alternative definition of ## M ##, ## \nabla \times M=J_m ## without any ## \mu_o ##.
     
    Last edited: Aug 15, 2016
  6. Aug 16, 2016 #5

    vanhees71

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    Let's concentrate on magnetostatics. I use Heaviside-Lorentz units. Then you have Ampere's Law
    $$\vec{\nabla} \times \vec{H}=\frac{1}{c} \vec{j}.$$
    With the magnetization this can be written as
    $$\vec{H}=\vec{B}-\vec{M},$$
    which leads to the microscopic law
    $$\vec{\nabla} \times \vec{B} = \frac{1}{c} \vec{j}+\vec{\nabla} \times \vec{M},$$
    i.e.,
    $$\vec{\nabla} \times \vec{B} = \frac{1}{c}(\vec{j}+\vec{j}_{\text{m}}), \quad \vec{j}_{\text{m}}=c \vec{\nabla} \times \vec{M}.$$
    As usual, using the differential form of Maxwell's equations, leads to tremendous simplification, because it's the natural form of a relativistic field theory (i.e., laws local in space and time).
     
  7. Aug 20, 2016 #6
    Regarding surface current, I believe Rutherford demonstrated that if a permanent magnet is placed in Nitric Acid and a very thin outer layer is etched away, the magnetism is lost.
    This does not seem to stack up with the idea of multiple loops inside the magnet, which would presumably re-assert their influence on the outer layer.
     
  8. Aug 20, 2016 #7

    Charles Link

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    I am very skeptical of this. I would like to see more detail of any actual experiment. In general, in the magnetic surface currents, there is no actual charge transport. Oxidizing their conducting surface wouldn't affect them in the least. Nor does the insulated laminations in the layers of a transformer block them, while blocking the unwated eddy currents from the Faraday EMF inside the iron.
     
  9. Aug 20, 2016 #8
    I am still searching for my original source, but there is an article in New Scientist Magazine about it below. Sorry it is a long URL, don't know how to avoid this.
    It also says that he discovered, by the method of etching with acid, that when a magnet is created using high frequency AC, there are layers of opposite polarity as we go deeper in the material, gradually becoming weaker as we go deeper. It sounds as if the wave propagates into the material leaving a signature behind it.
    https://books.google.co.uk/books?id...epage&q=rutherford magnetised needles&f=false
     
  10. Aug 20, 2016 #9
    I am thinking now, in the absence of the original paper of 1895/6, that the "skin magnetism" I originally mentioned might have been for the case of a magnet created using high frequency AC. This was something Rutherford experimented with, finding that magnets could be created as well as destroyed using high frequencies.
     
  11. Aug 20, 2016 #10

    Charles Link

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    Yes. Thank you. This article mentions Rutherford created "surface magnetization" with high frequency oscillations whose magnetization he was able to remove with nitric acid. He was not using a "bulk" permanent magnet with magnetization throughout the material along with its "magnetic surface currents".
     
    Last edited: Aug 20, 2016
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