Flux of magnetic moment into a coil

[flying_fred]

Hi,

I've an experiment in which I've some magnetic material place into a coil that's used as sensor. Say it's a cylindrical one for the sake of simplicity.
I'm trying to evaluate the flux induced by the material into the coil so that I can optimize the coil parameters (size, position and later its shape).
The material magnetization is given as a function of position.
To start as simple as possible I'm trying to evaluate the flux induced by a small magnetic moment placed at the center of a circular loop, the moment being perpendicular to the loop plane. Then I'll integrate it over the whole coil, then over the whole material volume.

The moment axial component is given at:
http://en.wikipedia.org/wiki/Magnetic_moment#Magnetic_field_produced_by_a_magnetic_moment"

The problem I'm facing is that the integral over the loop surface doesn't converge.
Setting z=0, y=0 and x=r, then integrating 2 pi r Bz[r,0,0] over the [0,R] interval we get
\Phi =\int_0^R \frac{\mu \mu_{0}}{2 r^2} \, dr

which obviously does not converge.

What am I missing?

Thanks for any hint,
Fred.

 

[flying_fred]

OK, replying to myself...

then the
\Phi =\int_0^R \frac{\mu \mu_{0}}{2 r^2} \, dr

integral does not converge but the full function Bz[x,y,z] (with x=r and y=0) is integrable on the disk and gives:

\Phi =\int_0^R \frac{{3\mu \mu_{0}}}{4\pi } \cdot \frac{z^2-\frac{1}{3}\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^{\frac{5}{2}}} \cdot 2\pi r \, dr =\int_0^R \frac{{3\mu \mu_{0}}}{4\pi } \cdot \frac{z^2-\frac{1}{3}\left(r^2+z^2\right)}{\left(r^2+z^2\right)^{\frac{5}{2}}} \cdot 2\pi r \, dr

which simplifies to :

\Phi =\int_0^R \frac{{\mu \mu_{0}}}{2} \cdot \frac{\left2 z^2-r^2\right}{\left(r^2+z^2\right)^{\frac{5}{2}}} \cdot r \, dr

and nicely integrates to:

\Phi = \frac{ {\mu \mu_0}}{2} \frac{R^2}{\left(R^2+z^2\right)^{3/2}}

yielding the simple

\Phi = \frac{ {\mu \mu_0}}{2R} at z=0.
If now the magnetic moment is set at a d offset from the loop center, say on the x axis, the expression is much more complex and, still integrating over a disk centered at x=y=0, the Bz[x,y,z] field component becomes:

{Bz}[r {Cos}[\theta ]-d,r {Sin}[\theta ],z]

and the surface integral is now:

\Phi = \int_0^R \int_0^{2\,\pi\right} \frac{{3\mu \mu_0}}{4\pi } \cdot \frac{z^2-\frac{1}{3}\left((r{Cos}[\theta ]-d)^2+r^2 {Sin^2}[\theta ]+z^2\right)}{\left((r{Cos}[\theta ]-d)^2+r^2 {Sin^2}[\theta ]+z^2\right)^{\frac{5}{2}}} \cdot 2\pi r \, dr

which simplifies to:

\Phi = \int_0^R \int_0^{2\,\pi\right} \frac{\mu\,\mu_0}{2} \cdot \frac{\left d^2+r^2-2 z^2+2\, d\, r\, \text{Cos}[\theta ]\right}{\left(d^2+r^2+z^2+2\, d\, r\, \text{Cos}[\theta ]\right)^{5/2}}\cdot r \, d\theta\, dr

At this point I'm stuck and I can't find any closed form for that double integral.

Anyone knowing if it's already been solved (and where) or if numeric integration is the only resort there?

Thanks in advance,
Fred.