I want to compute the magnetic field on any point $\mathbf{r}$ inside a spherical shell whose surface holds some uniform current that goes circularly around an axis. If the rotation axis corresponded to $z$, then the surface current would be $$ \mathbf{K} = \sigma v(-\sin{\phi}\mathbf{\hat{x}}+\cos{\phi}\mathbf{\hat{y}}) $$

outside of the axis poles, and $\mathbf{0}$ on the axis poles.

This is sort of similar to Griffiths Example 5.11 (3rd edition), where there is a rotating charged sphere. He chooses to put $\mathbf{r}$ on the $z$ axis, and the rotation axis lays anywhere on the $xz$ plane.

I am having trouble trying to write the current density on the case where the axis lays on the $xz$ plane. I suppose I could use the Griffiths result for the velocity of the charges and change the modulus from $R\omega\sin{\angle(\mathbf{w},\mathbf{\mathbf{r}'})}$ to constant $v$. But that seems to be pretty complicated.

To illustrate what I mean, here is the Griffiths result for the velocity $$ \mathbf{v}_G=\mathbf{\omega}\times\mathbf{r}'= R\omega\left[-\cos{\psi}\sin{\theta'}\sin{\phi'}\mathbf{\hat{x}} + (\cos{\psi}\sin{\theta'}\cos{\phi'} - \sin{\psi}\cos{\theta'})\mathbf{\hat{y}} + \sin{\psi}\sin{\theta'}\sin{\phi'}\mathbf{\hat{z}}\right] $$

where $R$ is the sphere radius, $\omega$ is the angular velocity and $\psi$ is the angle between the rotation axis and the $z$ axis. From then on I could compute the vector potential rather straightforwardly.

  • $\begingroup$ I’m not sure it’s the best method for your problem. The reason why Griffiths could perform the integral was because the end result was simple. In your case, it isn’t. I think it’s best for you to stay in spherical coordinates and solve the problem by separation of variables. Your original expression $K=ke_\phi$ suffices. $\endgroup$
    – LPZ
    Jan 3 at 18:25


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