# Griffiths Electrodynamics rotating sphere magnetic potential

I've been working on this problem for some time now and can't manage to understand why it is wrong, although I know it evidently is. I've also found questions on the same exercise in the Stack Exchange but none of them seem to be asking how to calculate the magnetic potential through direct integration. Here is the problem statement:

A spherical shell of radius R, carrying a uniform surface charge $$\sigma$$, is set spinning at an angular velocity $$\omega$$. Find the vector potential it produces at point $$\vec{r}$$.

My reasoning is as follows: I begin by stating the definition of the vector potential, acknowledging the fact that when the sphere rotates with a uniform charge density, the system is very much like a static sphere with a stationary surface current $$\vec{K}$$.

$$\vec{A} = \frac{\mu_0}{4\pi}\int \frac{\vec{K(\vec{r}')}}{|\vec{r}-\vec{r}'|}da'$$

I firstly find a suitable expression for $$\vec{K}$$. Taking into account its definition:

$$\vec{K} = \sigma \vec{v} = \sigma \vec{\omega}\times \vec{r'} = \sigma R\omega sin(\theta)\vec{e}_\varphi$$

Where $$\theta \equiv \Psi$$ for common notation and $$\vec{e}_\varphi$$ is the unit vector of the "circumferencial" component.

As to the denominator, since it is positions that are being subtracted, I can write:

$$\vec{r'} = R\vec{e}_r; \ \ \ \ \vec{r} = r\vec{e}_r$$

Where $$\vec{e}_r$$ is the unit vector of the radial component. Therefore, I can write the numerator as:

$$|\vec{r}-\vec{r'}| = |(r-R)\cdot \vec{e}_r| = r-R$$

Also, my surface differential should be given by: $$da' = R^2sin(\theta)d\varphi d\theta$$, so I know have all the elements I need for my integral:

$$\vec{A} = \frac{\mu_0}{4\pi}\int_0^{\pi}\int_{0}^{2\pi}\frac{\sigma R\omega sin(\theta)\vec{e}_\varphi}{r-R}R^2sin(\theta)d\varphi d\theta$$

Notice that unit vector $$\vec{e}_\varphi$$ on the numerator. Since I'm integrating $$\varphi$$ from $$0$$ to $$2\pi$$ and the only element of my integral depending on $$\varphi$$ is precisely said unit vector, then I get:

$$\int_0^{2\pi} \vec{e}_\varphi d\varphi = \int_0^{2\pi} (-sin(\varphi), cos(\varphi), 0)d\varphi = \vec{0}$$

Which cancels all my integral. Where did I go wrong?

The problem is that $$(\vec e_r,\vec e_\theta,\vec e_\varphi)$$ is a local basis, i.e. it depends on the point where it is constructed. In your calculation, you need to compute $$|\vec r-\vec r'|$$. $$\vec r$$ is the point, say $$M$$ where the vector potential $$\vec A$$ is measured. $$\vec r'$$ is the point, say $$P$$, where is located the surface element $$da'$$ with a current density $$K(\vec r')$$. At the two points M and P, we have two different local basis that one should denote $$(\vec e_r(M),\vec e_\theta(M),\vec e_\varphi(M))$$ and $$(\vec e_r(P),\vec e_\theta(P),\vec e_\varphi(P))$$. Therefore $$\vec r-\vec r'=\overrightarrow{\rm PM}=r\vec e_r(M)-r\vec e_r(P)$$ Your calculation is wrong because you assumed that $$\vec e_r(M)=\vec e_r(P)$$ which is wrong in the general case.
• It is possible to compute the magnetic field on the $z$ axis ($A$ vanishes) Away from the $z$ axis, as far as I know, one can compute B and A only as a Taylor expansion in powers of $r/R$. Commented Nov 18, 2023 at 19:12