- #36
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A much simpler way to see, what's described is to calculate the current density, using
$$\vec{\nabla} \times \vec{B}=\mu \vec{j}.$$
The only subtlety is that you have to be careful how to deal with the singularity along the ##z##-axis!
$$\vec{\nabla} \times \vec{B}=\mu \vec{j}.$$
The only subtlety is that you have to be careful how to deal with the singularity along the ##z##-axis!