# Force on a superconducting cube

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1. Aug 9, 2015

### IanBerkman

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. Aug 15, 2015