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I Magnetic Potential from rod

  1. Nov 13, 2016 #1
    I am trying to calculate the vector magnetic potential for a rod of length L extending along the Z-axis. I was asked to find the magnetic potential at a point P which is a distance r from the center of the rod in the XY plane.

    I know the formula I have to use is the vector Poisson's equation A = u / 4pi * integral ( J / R') dv', where the bounds are in respect to v'. The dv' refers to the volume of the object (the rod).

    I am not sure how I might start this problem. I have thought about using J = I * (pi * a^2), where I is the current of the rod and a is the radius. As well, R' can be calculated using r = (z^2 + r^2)^1/2. Besides that, I am confused on how I might take into account of the volume and current of the rod. I would greatly appreciate any help that can be offered.
     
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  3. Nov 13, 2016 #2

    Charles Link

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    Presumably the rod has some magnetization ## \vec{M} ## per unit volume. The rod can be modeled as a magnetic dipole with dipole strength ## \vec{m}=\vec{M}V ##. If you google "magnetic dipole", Wikipedia has a good write-up that even gives the formula for the magnetic vector potential ## \vec{A} ## for a magnetic dipole.
     
  4. Nov 13, 2016 #3
    Unfortunately, this does not help me. My book does have the formulae for the example, but they find it using the Biot-Savart method and they want me to compare it by using a different method it seems. My professor as well indicated that I should use the formula I listed above, but I am not sure how I can convert the quantities into something I can use.
     
  5. Nov 14, 2016 #4

    Charles Link

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    Griffith's E&M textbook computes the vector potential ## A ## for an arbitrary distribution of magnetization of microscopic magnetic dipoles and shows by a couple of vector identities for integrals that it is equivalent to magnetic surface currents per unit length ## K_m=M \times \hat{n}/\mu_o ## plus bulk magnetic currents ## J_m=\nabla \times M/\mu_o ##. ## \\ ## Griffiths simply presents the derivation along with the result, but it is really quite an important one , and he might do well to give it additional emphasis. I highly recommend you read carefully through the derivation in Griffiths textbook. ## \\ ## ## \\ ## The vector potential ## A ## can then be calculated from the surface currents : ## A(x)=\frac{\mu_o}{4 \pi} \int \frac{K_m(x')}{|x-x'|} \, dS' ##. (To get from this integral to the Wikipedia result for ## A ## might take a little work, but if you keep just the lowest order terms, the results should agree.) By taking ## B=\nabla \times A ## you get the same answer for ## B ## that you would by doing a Biot-Savart integral with the surface currents. ## \\ ## (Please check the ## \mu_o/(4 \pi ) ## in the above formula, but I think I got it right. I often do these E&M calculations in c.g.s. units where the constants are different.) ## \\ ## Computing the vector potential ## A ## and then taking ##B= \nabla \times A ## is often easier than using Biot-Savart integrals to compute ## B ##.
     
    Last edited: Nov 14, 2016
  6. Nov 15, 2016 #5

    Charles Link

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    @Vaclav Just an additional input: Oftentimes, the magnetic surface currents are shown by using ## J_m=\nabla \times M/\mu_o ## along with Stokes theorem at a surface boundary to give surface current per unit length ## K_m=M \times \hat{n}/\mu_o ##. One can expect bulk currents ## J_m ## to depend upon some vector gradient of the magnetization ## M ##. In general, a proof of the equation ## J_m=\nabla \times M/\mu_o ## would be quite detailed and it can be expected that the student accepts this formula without a detailed proof. .. Griffiths alternatively takes a rather unique approach and computes the vector potential ## A ## for an arbitrary distribution of magnetic dipoles. The vector potential from a single microscopic dipole is already known, and he uses superposition along with a couple vector identities to generate the result that the bulk currents and surface currents must be what is given above by using the result that ## A(x)=\frac{\mu_o}{4 \pi} \int \frac{J_m(x')}{|x-x'|} \, d^3x' ##. The result of this derivation is of much significance. Alternatively, you can accept the equations without proof that ## J_m=\nabla \times M/\mu_o ## which from Stokes theorem results in ## K_m=M \times \hat{n}/\mu_o ## at the surface. Griffiths proves these in a very unique way and his is quite an interesting derivation.
     
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