# Magnetic field inside a spherical shell with "uniform" surface current

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.

• 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.
– LPZ
Jan 3 at 18:25