# §19.34 Mutual Inductance of Coaxial Circles

The mutual inductance $M$ of two coaxial circles of radius $a$ and $b$ with centers at a distance $h$ apart is given in cgs units by

 19.34.1 $\frac{{c^{2}}M}{2\pi}=ab\int_{0}^{2\pi}(h^{2}+a^{2}+b^{2}-2ab\mathop{\cos\/}% \nolimits\theta)^{-1/2}\mathop{\cos\/}\nolimits\theta\mathrm{d}\theta=2ab\int_% {-1}^{1}\frac{t\mathrm{d}t}{\sqrt{(1+t)(1-t)(a_{3}-2abt)}}=2abI(\mathbf{e}_{5}),$

where $c$ is the speed of light, and in (19.29.11),

 19.34.2 $\displaystyle a_{3}$ $\displaystyle=h^{2}+a^{2}+b^{2},$ $\displaystyle a_{5}$ $\displaystyle=0,$ $\displaystyle b_{5}$ $\displaystyle=1.$ Symbols: $h$: distance Permalink: http://dlmf.nist.gov/19.34.E2 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 19.34

The method of §19.29(ii) uses (19.29.18), (19.29.16), and (19.29.15) to produce

 19.34.3 $2abI(\mathbf{e}_{5})=a_{3}I(\boldsymbol{{0}})-I(\mathbf{e}_{3})=a_{3}I(% \boldsymbol{{0}})-r_{+}^{2}r_{-}^{2}I(-\mathbf{e}_{3})=2ab(I(\boldsymbol{{0}})% -r_{-}^{2}I(\mathbf{e}_{1}-\mathbf{e}_{3})),$ Symbols: $I(\mathbf{m})$: integral and $r_{\pm}$ Permalink: http://dlmf.nist.gov/19.34.E3 Encodings: TeX, pMML, png See also: Annotations for 19.34

where $a_{1}+b_{1}t=1+t$ and

 19.34.4 $r_{\pm}^{2}=a_{3}\pm 2ab=h^{2}+(a\pm b)^{2}$ Symbols: $h$: distance and $r_{\pm}$ Permalink: http://dlmf.nist.gov/19.34.E4 Encodings: TeX, pMML, png See also: Annotations for 19.34

is the square of the maximum (upper signs) or minimum (lower signs) distance between the circles. Application of (19.29.4) and (19.29.7) with $\alpha=1$, $a_{\beta}+b_{\beta}t=1-t$, $\delta=3$, and $a_{\gamma}+b_{\gamma}t=1$ yields

 19.34.5 $\frac{3{c^{2}}}{8\pi ab}M=3\!\mathop{R_{F}\/}\nolimits\!\left(0,r_{+}^{2},r_{-% }^{2}\right)-2r_{-}^{2}\mathop{R_{D}\/}\nolimits\!\left(0,r_{+}^{2},r_{-}^{2}% \right),$

or, by (19.21.3),

 19.34.6 $\frac{{c^{2}}}{2\pi}M=(r_{+}^{2}+r_{-}^{2})\mathop{R_{F}\/}\nolimits\!\left(0,% r_{+}^{2},r_{-}^{2}\right)-4\!\mathop{R_{G}\/}\nolimits\!\left(0,r_{+}^{2},r_{% -}^{2}\right).$

A simpler form of the result is

 19.34.7 $M=(2/{c^{2}})(\pi a^{2})(\pi b^{2})\mathop{R_{-\frac{3}{2}}\/}\nolimits\!\left% (\tfrac{3}{2},\tfrac{3}{2};r_{+}^{2},r_{-}^{2}\right).$

References for other inductance problems solvable in terms of elliptic integrals are given in Grover (1946, pp. 8 and 283).