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After we integrate over the whole wire, (assume it's infinite), we get the well known expression [itex]B=\frac{\mu_0 I}{2\pi r}[/itex].

**My question is:**What happens to the magnetic field, in case we "rotate" the current carrying wire, so it's no longer perpendicular to the xy (or [itex]r\varphi[/itex]) plane, but it has an angle theta ?

My first idea was that the only change is at the [itex]dl[/itex] element of the law, since it will no longer be along the z axis, but "analyzed" in two perpendicular vectors that will be added... vector-like. The new dlz will be the only element that will be giving a dBz field, whereas the new "dlx" (or "dlr") will not give a field, since the cross product dlr x r will be zero (the two vectors are along the same axis). So, the field B' in comparison with the initial field B will be B' = Bsin(theta). (Only dlz contributes to the magnetic field)

My second idea (which was chronologically first, though :P) was pretty simplistic: I rotated the whole well known image of the current wire and the "rings" of magnetic field lines (the contours), well given by the "right hand rule", so every tiny "part" of the field, which was only Br first, will be now analyzed in a Br' and a Bz' components. So, this means that Br' will be Bcos(theta) and Bz' = Bsin(theta). If theta = 0 (ie our wire is straight and perp again, as it was initially) we have Br' = B and Bz' = 0 (pretty convenient...)

Is any of these two ideas correct? If not, does anybody have a really good solution for this? :)