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Assume that the uniformly distributed bound surface and bound volume current, $\vec{K}_{b}$,$\vec{J}_{b}$ respectively, are in the azimuthal direction $\phi$ within an infinitely long solenoid with a current passing through the wounded coils centered on the z-axis.

Clearly, the magnetic field field has no z and $\phi$ dependence due to the symmetry of current about the z-axis and the infinitely long solenoid. Hence,

$\vec{B}=\left \langle B_{s}\left ( s,\phi,\theta \right ),B_{\phi}\left ( s,\phi,\theta \right ),B_{z}\left ( s,\phi,\theta \right ) \right \rangle\rightarrow \vec{B}=\left ( B_{s}\left ( s \right ),B_{\phi}\left ( s \right ) ,B_{\theta}\left ( s \right )\right )$

Evidently, $B_{\phi}$ cannot exists for if it did it would violate the fact that a magnetic field cannot in a parallel to the current that generates it.

I know that using the Right hand rule in context of a current carrying solenoid, if the current runs in a direction 'towards' me, the thumb points in the positive z-direction. If the current runs in a direction 'away' from me, the thumb points in the negative z-direction. That thumb tells me the direction of the magnetic field. In both cases, it points outward/ away from the solenoid.

Here is where a bit of explanation would help:

How does this ties in with the fact that it violates the Biot-Savard law?

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First, the last $B_\theta$ should be a $B_z$.

Your assumption of $B_\phi=0$ is wrong since the current is going upward. The fact that the coil is wound is insignificant since on the large scale of things the solenoid is infinitesimally thin. This then gives $B_\phi\neq0$ and $B_s=B_r=0$.

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