I wrote a C++ program that can calculate the magnetic field $\bar{B}$ generated by a circular coil that is placed in the origin, for a given point $\bar{P}$ in 3D space.
The coil is in the $x$,$y$-surface, with $z=0$.
I want to describe multipoles now so I would have to be able to transform the coordinates of $\bar{P}$ to the coordinate system of a coil that is shifted and rotated $\bar{P}^\prime$.
The coils that are shifted and rotated would have their coordinate systems like so:
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the origin of the coil system is at $O^\prime=(R\cos\phi,R\sin\phi,0)$
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The $z^\prime$-axis points to the origin of the normal system, is also in the $x$,$y$-surface and makes an angle $\phi$ with the $x$-axis of the normal system.
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The $y^\prime$-axis is parallel to the $z$-axis
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the $x^\prime$-axis follows (all systems being orthonormal).
Thus all coils (the number of coils is arbitrary) lie on a circle with radius $R$ and their central axis $z$.
My approach has been this so far:
- transform $\bar{P}$ to $\bar{P}^\prime$ in the coil's system
- calculate $\bar{B}^\prime$ using my program
- transform $\bar{B}^\prime$ back to the normal system (only rotation has to be undone, since $\bar{B}$ is a free vector and I only need it's components).
I've been using a transformation matrix $M=R_y(\phi)R_x(\pi/2)$ that looks like this:
$$M=\begin{bmatrix}\cos\phi&0&\sin\phi\\0&1&0\\-\sin\phi&0&\cos\phi\end{bmatrix}\begin{bmatrix}1&0&0\\0&0&-1\\0&1&0\end{bmatrix}=\begin{bmatrix}
\cos\phi&\sin\phi&0\\0&0&-1\\-\sin\phi&\cos\phi&0\end{bmatrix}$$
so $\bar{P}^\prime=M\bar{P}$, after first translating $\bar{P}$ to $\bar{P}_t$ by:
$$\bar{P}_t=\bar{P}-O^\prime$$
And $\bar{B}=M^\prime \bar{B}^\prime$ with $M^\prime=M^T$.
My results are fautly, so what am I doing wrong?
Also, let me know if this question is too concrete to post on a math forum.
Added image for clarity:
The gray circle is where the coil always is in the system that's being used for my C++ program, the dark circle is the coil where it should be.
Best Answer
I don't understand how you arrived at your rotation matrix, so I'll just go by your description of the transformed system and the image, which I think I understand.
If we choose $\phi=0$, $O'$ lies on the $x$-axis, so the $z'$-axis should be the negative $x$-axis, the $y'$-axis should be the $z$-axis, and the $x'$-axis should correspondingly be the negative $y$-axis. That corresponds to the transformation matrix
$$\left(\begin{array}{ccc}0&-1&0\\0&0&1\\-1&0&0\end{array}\right)\;.$$
This is not what your expression for the rotation matrix yields for $\phi=0$. Since the $y'$-axis always points along the $z$-axis and only the $x'$-axis and the $z'$-axis are rotated into each other if we change $\phi$, we can immediately infer the general form of the rotation matrix as
$$\left(\begin{array}{ccc}\sin\phi&-\cos\phi&0\\0&0&1\\-\cos\phi&-\sin\phi&0\end{array}\right)\;,$$
where only the sign of the sine was still to be determined and can be fixed by considering $\phi=\pi/2$.