\dfrac{\int_0^{2\pi}
   \int_0^\pi \,
   \int_0^{r_c} \,  
r^2 \sin \theta |\Vec{H}(r,\theta)| dr d\theta d\varphi}
   {V},
   \label{Eq:H_c}\end{aligned}$$ where we choose $r_c = 0.01 r_e$ and $V$ is the volume of the central region with $r \le r_c$, defined as $$\begin{aligned}
   V = \int_0^{2\pi} \, 
\int_0^\pi 
\int_0^{r_c} r^2\, 
\sin \theta dr 
\, d\theta
d\varphi \ .\end{aligned}$$ This central region seems to be very small, but we can resolve it sufficiently because we use non-uniform and centrally concentrated meshes (see. Fig. \[fig:mesh\] and Eq. \[Eq:r\_mesh\] in Appendix). We have $77$ meshes to resolve the region in actual numerical computations.

In order to know the contributions of the [*poloidal*]{} magnetic field and the [*toroidal*]{} magnetic field separately, we define the [*poloidal*]{} magnetic energy ${\cal H}_p$ and the [*toroidal*]{} magnetic energy ${\cal H}_t$ as $$\begin{aligned}
   {\cal H}_p = \frac{1}{8\pi}
    \int_0^{2 \pi} \int_0^{\pi} \int_0^{\infty}
r^2 \, \sin \theta \,  
 |H_r(r,\theta)^2 + H_\theta(r,\theta)^2| dr d\theta d\varphi \ ,\end{aligned}$$ $$\begin{aligned}
   {\cal H}_t = \frac{1}{8\pi}
    \int_0^{2 \pi} \int_0^{\pi} \int_0^{\infty}
r^2 \, \sin \theta \,  
 |H_\varphi(r,\theta)^2| dr d\theta d\varphi \ ,\end{aligned}$$ As for the magnetic multipole moment seen outside a star, we compute each multipole component by solving the following equation in a vacuum: $$\begin{aligned}
   \Delta \left(A_\varphi  \sin \varphi\right) = 0 \ 