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For this question use the normalized magnetic scalar potential $\mathcal{U}$ defined by $\mathbf{B}=({\mu }_0/4\pi )\nabla \mathcal{U}$. The previous question instead includes ${\mu }_0/4\pi$ in its scalar potential; its $U$ equals $({\mu }_0/4\pi )\mathcal{U}$. This normalization change in the paper must be kept separate from the kernel calculation.

Take the radial component of the Geselowitz formula. On the spherical boundary, ${\mathbf{n}}^{\prime }$ is parallel to ${\mathbf{r}}^{\prime }$, andThus the surface-current term contributes no radial magnetic field. The primary Biot-Savart law term gives, along a fixed direction $\hat{\mathbf{r}}$,Choose $\mathcal{U}\to 0$ at infinity. Integrating this equation fixes the full potential along every ray, givingTo express it as a source-position derivative, defineWith $\mathbf{W}=\mathbf{Q}\times {\mathbf{r}}_0$, differentiating the integrand givesso $\mathcal{U}=\mathbf{W}\cdot {\nabla }_{{\mathbf{r}}_0}\varphi$ as required.

Let $b=|{\mathbf{r}}_0|$ and $P=|\mathbf{r}-{\mathbf{r}}_0|$. An elementary antiderivative gives the spherical current-dipole logarithmic kernelFor $r>a>b$, both logarithm arguments are positive. Its radial derivative is $-1/(rP)$ and its limit at infinity is zero, which verifies the preceding integral. At $b=0$ use the continuous limit $\varphi =1/r$.

For a direct check, differentiating along $\mathbf{W}$ leaves $b$ unchanged and gives $\mathbf{W}\cdot {\nabla }_{{\mathbf{r}}_0}P=-\mathbf{W}\cdot \mathbf{r}/P$. Sinceone obtainsIts gradient is the Sarvas formula. The sphere radius and electrical conductivity disappear from this exterior magnetic result because spherical symmetry removed the radial contribution of the return currents. The original derivation and spherical-conductor assumptions are discussed in Sarvas's paper.