Addendum: Babic, S., et al. Self-Inductance of the Circular Coils of the Rectangular Cross-Section with the Radial and Azimuthal Current Densities. Physics 2020, 2, 352–367

In [1] some ambiguities and imperfections were subsequently identified and are correspondingly modified and simplified below so that they correspond to those given in our software package [2]. Thus, people that have an interest in this domain could easily calculate the self-inductances presented here using the Mathematica code for them.

On page 354, Equation (5) reads:

$$L_R=-\frac{2{\mu }_0{N}^2{R}_1}{b^2{\ln }^2(\alpha )}{\displaystyle \sum }_{n=1}^{11}\underset{0}{\overset{\frac{\pi }{2}}{{\displaystyle \int }}}\cos (2\beta ) T_{n}d\beta ,$$

(5)

where

$$T_1=\frac{b^3}{3\sin (2\beta )}[2\arctan (q)-\arctan (q_1)-\arctan (q_2)],$$

$$T_2=\frac{8}{3}({\alpha }^3+1){\cos }^3(\beta )-\frac{4}{3}{\cos }^2(\beta )({\alpha }^2{r}_2+r_1)+\frac{2}{3}[({\alpha }^2+1)\cos (2\beta )2\alpha ](r-r_0),$$

$$\begin{gathered} T_3=4b{\alpha }^2 {\cos }^2(\beta )\mathrm{arsinh}[\frac{b}{2\alpha \cos (\beta )}]+4b{\cos }^2(\beta )\mathrm{arsinh}\left[\frac{b}{2\cos (\beta )}\right] \\ -2b[({\alpha }^2+1)\cos (2\beta )+2\alpha ]\mathrm{arsinh}(\frac{b}{r_0}), \end{gathered}$$

$$T_4=2b^2[\alpha \mathrm{arsinh}(v_{22})+\mathrm{arsinh}(v_{11})],$$

$$T_5=-2b^2[\alpha \mathrm{arsinh}(v_2)+\mathrm{arsinh}(v_1)],$$

$$T_6=-2b\sin (2\beta )[{\alpha }^2\arctan (p_{22})+\arctan (p_{11})],$$

$$T_7=2b\sin (2\beta )[{\alpha }^2\arctan (p_2)+\arctan (p_1)],$$

$$T_8=\frac{{\alpha }^3}{3}{\sin }^2(2\beta )\ln (\frac{m_2+1}{m_2-1})+\frac{1}{3}{\sin }^2(2\beta )\ln (\frac{m_1+1}{m_1-1}),$$

$$T_9=-\frac{{\alpha }^3}{3}{\sin }^2(2\beta )\ln (\frac{m_{20}+1}{m_{20}-1})-\frac{1}{3}{\sin }^2(2\beta )\ln (\frac{m_{10}+1}{m_{10}-1}),$$

$$T_{10}=-\frac{{\alpha }^3}{3}{\sin }^2(2\beta )\ln (\frac{m_{22}+1}{m_{22}-1})-\frac{1}{3}{\sin }^2(2\beta )\ln (\frac{m_{11}+1}{m_{11}-1}),$$

$$T_{11}=\frac{2({\alpha }^3+1)}{3}{\sin }^2(2\beta )\ln \left[\frac{\cos (\beta /2)}{\cos (\beta /2)}\right],$$

where

$$r=\sqrt{b^2+{\alpha }^2+1+2\alpha \cos (2\beta )}, r_0=\sqrt{{\alpha }^2+1+2\alpha \cos (2\beta )}, r_1=\sqrt{b^2+4{\cos }^2(\beta )},$$

$$r_2=\sqrt{b^2+4{\alpha }^2{\cos }^2(\beta )} , r_{01}=\sqrt{4{\cos }^2(\beta )}=2\cos (\beta ), r_{02}=\sqrt{4{\alpha }^2{\cos }^2(\beta )}=2\alpha \cos (\beta ),$$

$$q=\frac{\alpha {\sin }^2(2\beta )-b^2\cos (2\beta )}{b\sin (2\beta )r}, q_1=\frac{{\alpha }^2{\sin }^2(2\beta )-b^2\cos (2\beta )}{b\sin (2\beta )r_2},$$

$$q_2=\frac{{\sin }^2(2\beta )-b^2\cos (2\beta )}{b\sin (2\beta )r_1}, v_1=\frac{\alpha +\cos (2\beta )}{\sqrt{b^2+{\sin }^2(2\beta )}}, v_2=\frac{1+\alpha \cos (2\beta )}{\sqrt{b^2+{\alpha }^2{\sin }^2(2\beta )}},$$

$$\begin{gathered} v_{11}=\frac{1+\cos (2\beta )}{\sqrt{b^2+{\sin }^2(2\beta )}}, v_{22}=\frac{\alpha +\alpha \cos (2\beta )}{\sqrt{b^2+{\alpha }^2{\sin }^2(2\beta )}}, p_1=\frac{b\left[\alpha +\cos \left(2\beta \right)\right]}{\sin \left(2\beta \right)r}, \\ {p}_2=\frac{b\left[1+\alpha \cos \left(2\beta \right)\right]}{\alpha \sin \left(2\beta \right)r}, p_{11}=\frac{b\left[1+\cos \left(2\beta \right)\right]}{\sin \left(2\beta \right)r_1}, p_{22}=\frac{b\left[\alpha +\alpha \cos \left(2\beta \right)\right]}{\alpha \sin \left(2\beta \right)r_2} , \\ m_1=\frac{r}{\alpha +\cos (2\beta )}, m_2=\frac{r}{1+\alpha \cos (2\beta )}, m_{10}=\frac{r_0}{\alpha +\cos (2\beta )}, m_{20}=\frac{ r_0}{1+\alpha \cos (2\beta )} , \\ m_{11}=\frac{r_1}{1+\cos (2\beta )} , m_{22}=\frac{r_2}{\alpha +\alpha \cos (2\beta )}. \end{gathered}$$

Expressions for $T_n, n=1,2,\dots ,11,$ are given as in [2] and Appendix A where we calculate the self-inductance for the coil of the rectangular cross-section with radial current in Example 2 [1].

On page 355, Equation (6) reads:

$$L_{R-\mathrm{disk}}=\frac{4{\mu }_0{N}^2{R}_1(\alpha +1)}{{\ln }^2\alpha }[E(k_0)-1],$$

(6)

where

$$k_0^2=\frac{4\alpha }{{(\alpha +1)}^2}.$$

On page 355, Equation (7) reads:

$$L_A=-\frac{{\mu }_0{N}^2{R}_1}{15b^2{(\alpha -1)}^2}{\displaystyle \sum }_{n=1}^6\underset{0}{\overset{\frac{\pi }{2}}{{\displaystyle \int }}}\cos (2\beta ) S_{n}d\beta ,$$

(7)

where

$$\begin{gathered} S_1=\frac{b^4}{{\sin }^2\left(2\beta \right)}\left[r_2-\frac{b\cos \left(2\beta \right)}{\sin \left(2\beta \right)}\arctan \left(q_2\right)\right]+\frac{b^4}{{\sin }^2\left(2\beta \right)}\left[r_1-\frac{b\cos \left(2\beta \right)}{\sin \left(2\beta \right)}\arctan \left(q_1\right)\right] \\ -\frac{2b^4}{{\sin }^2(2\beta )}[r-\frac{b\cos (2\beta )}{\sin (2\beta )}\arctan (q)], \end{gathered}$$

$$\begin{gathered} S_2=9a^2{b}^2{r}_2+9b^2{r}_1-9(a^2+1)b^{2}r+2\left[6a^4{\cos }^2\left(2\beta \right)-2a^4\cos \left(2\beta \right)-8a^4\right]\left(r_2-r_{02}\right) \\ +2[6{\cos }^2(2\beta )-2\cos (2\beta )-8](r_1-r_{01})-4[3(a^4+1){\cos }^2(2\beta )-\alpha (a^2+1)\cos (2\beta ) \\ -2{({\alpha }^2+1)}^2](r-r_0) , \end{gathered}$$

$$S_3=30b\sin (2\beta )\cos (2\beta )[a^4\arctan (p_{22})+\arctan (p_{11})-a^4\arctan (p_2)-\arctan (p_1)],$$

$$\begin{gathered} S_4=30b\{{\alpha }^4{\sin }^2(2\beta )\ln (\frac{r_2+b}{r_2-b})+{\sin }^2(2\beta )\ln (\frac{r_1+b}{r_1-b}) \\ -\frac{1}{2}[{\left({\alpha }^2+1\right)}^2-2({\alpha }^4+1){\cos }^2(2\beta )]\ln \left(\frac{r+b}{r-b}\right)\}, \end{gathered}$$

$$\begin{gathered} S_5=12\cos (2\beta ){\sin }^2(2\beta )\{{\alpha }^5\ln [r_0+1+\alpha \cos (2\beta )]+\ln [r_0+\alpha +\cos (2\beta )] \\ -({\alpha }^5+1)\ln [4\cos (\beta ){\cos }^2(\beta /2)]-{\alpha }^5\ln (\alpha )\}, \end{gathered}$$

$$\begin{gathered} S_6=4\cos (2\beta )[5b^2-3{\sin }^2(2\beta )]\ln \left[\frac{r+\alpha +\cos (2\beta )}{r_1+1+\cos (2\beta )}\right] \\ +4{\alpha }^3\cos (2\beta )[5b^2-3{\alpha }^2{\sin }^2(2\beta )]\ln \left[\frac{r+1+\alpha \cos (2\beta )}{r_2+\alpha +\alpha \cos (2\beta )}\right]. \end{gathered}$$

Expressions for $S_{n },n=1,2,\dots ,6,$ are given as in [2] and Appendix B where we calculate the self-inductance for the coil of the rectangular cross-section with azimuthal current in Example 4 (case $\alpha =3, b=2)$ [1].

On page 359, the self-inductance of the thin Bitter disk (pancake) is to be replaced by:

$$L_{R-\mathrm{disk}}=3.56991288673 \mathrm{H}.$$

At the request of many interested people, who have contacted us, we give the Mathematica codes for calculating the self-inductances with modified expressions for $T_n$ and $S_n$ [2], which are more friendly for the calculations than those given in [1].