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Force on a superconducting cube

  1. Aug 9, 2015 #1
    Hi everyone,

    I need some help to look if I did these calculations right.


    Let us assume a three dimensional magnetic field:

    ##\vec{B}(x,y,z) = B_x(x,y,z)\hat{x} + B_y(x,y,z)\hat{y} + B_z(x,y,z)\hat{z}##

    The equation for the force on a superconducting particle in a magnetic field is given by:

    ##\vec{F} = -\frac{2}{\mu_0}\oint[\vec{n}||\vec{B}||^2-\vec{B}(\vec{n}\cdot\vec{B})]\,\mathrm{d}S##

    Where the integration takes place over the surface of the superconducting particle.

    Let us assume a cube with edges of a length ##2l## and its center at ##(x',y',z')##
    For the front face of the cube (at ##x = x'+l## and parallel to the yz-plane) the normal vector is just ##\hat{x}##

    The force on this face becomes:
    ##F_{x1} = -\frac{2}{\mu_0}\oint[\hat{x}(B_x^2+B_y^2+B_z^2)-B_x(\hat{x}\cdot B_x\hat{x})]\,\mathrm{d}S =\\
    -\frac{2}{\mu_0}\hat{x}\oint B_x^2+B_y^2+B_z^2- B_x^2\,\mathrm{d}S =\\
    -\frac{2}{\mu_0}\hat{x}\oint B_y^2+B_z^2\,\mathrm{d}S##

    Switching to a double integral:
    ##F_{x1} =-\frac{2}{\mu_0}\hat{x}\int_{y'-l}^{y'+l}\int_{z'-l}^{z'+l}B_y(x'+l,y,z)^2+B_z(x'+l,y,z)^2\,\mathrm{d}z\mathrm{d}y##

    For the back face (at x = x'-l) I just replaced the normal vector with ##-\hat{x}## and end up with:
    ##F_{x2} = \frac{2}{\mu_0}\hat{x}\int_{y'-l}^{y'+l}\int_{z'-l}^{z'+l}B_y(x'-l,y,z)^2+B_z(x'-l,y,z)^2\,\mathrm{d}z\mathrm{d}y##

    And the resulting force in the x direction is then ##F_{x1}+F_{x2}##
    Same story with the forces on the y and z faces.

    Did I do this right? Because I want to run these equations through Mathematica but I keep ending up getting wrong plots.

    Thanks in advance, Ian
     
  2. jcsd
  3. Aug 15, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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