(a) Use the Neumann formula (Eq. 7.23) to calculate the mutual inductance of the configuration in Fig. 7.37, assuming a is very small (a ≪ b, a ≪ z). Compare your answer to Prob. 7.22.
M_{21}=\frac{\mu_{0}}{4 \pi} \oint \oint \frac{d l _{1} \cdot d l _{2}}{ᴫ} (7.23)
(b) For the general case (not assuming a is small), show that
M=\frac{\mu_{0} \pi \beta}{2} \sqrt{a b \beta}\left(1+\frac{15}{8} \beta^{2}+\ldots\right) ,
where
\beta \equiv \frac{a b}{z^{2}+a^{2}+b^{2}} .
A point on the upper loop: r _{2}=\left(a \cos \phi_{2}, a \sin \phi_{2}, z\right) \text {; a point on the lower loop: } r _{1}=\left(b \cos \phi_{1}, b \sin \phi_{1}, 0\right) .
ᴫ^{2}=\left( r _{2}- r _{1}\right)^{2}=\left(a \cos \phi_{2}-b \cos \phi_{1}\right)^{2}+\left(a \sin \phi_{2}-b \sin \phi_{1}\right)^{2}+z^{2} =a^{2} \cos ^{2} \phi_{2}-2 a b \cos \phi_{2} \cos \phi_{1}+b^{2} \cos ^{2} \phi_{1}+a^{2} \sin ^{2} \phi_{2}-2 a b \sin \phi_{1} \sin \phi_{2}+b^{2} \sin ^{2} \phi_{1}+z^{2} =a^{2}+b^{2}+z^{2}-2 a b\left(\cos \phi_{2} \cos \phi_{1}+\sin \phi_{2} \sin \phi_{1}\right)=a^{2}+b^{2}+z^{2}-2 a b \cos \left(\phi_{2}-\phi_{1}\right)=\left(a^{2}+b^{2}+z^{2}\right)\left[1-2 \beta \cos \left(\phi_{2}-\phi_{1}\right)\right]=\frac{a b}{\beta}\left[1-2 \beta \cos \left(\phi_{2}-\phi_{1}\right)\right] .
d l_{1}=b d \phi_{1} \hat{ \phi }_{1}=b d \phi_{1}\left[-\sin \phi_{1} \hat{ x }+\cos \phi_{1} \hat{ y }\right] ; d l_{2}=a d \phi_{2} \hat{ \phi }_{2}=a d \phi_{2}\left[-\sin \phi_{2} \hat{ x }+\cos \phi_{2} \hat{ y }\right] , so
d l_{1} \cdot d l_{2}=a b d \phi_{1} d \phi_{2}\left[\sin \phi_{1} \sin \phi_{2}+\cos \phi_{1} \cos \phi_{2}\right]=a b \cos \left(\phi_{2}-\phi_{1}\right) d \phi_{1} d \phi_{2} .
M=\frac{\mu_{0}}{4 \pi} \oint \oint \frac{d l_{1} \cdot d l_{2}}{ᴫ}=\frac{\mu_{0}}{4 \pi} \frac{a b}{\sqrt{a b / \beta}} \iint \frac{\cos \left(\phi_{2}-\phi_{1}\right)}{\sqrt{1-2 \beta \cos \left(\phi_{2}-\phi_{1}\right)}} d \phi_{2} d \phi_{1} .
Both integrals run from 0 to 2 \pi \text {. Do the } \phi_{2} \text { integral first, letting } u \equiv \phi_{2}-\phi_{1} :
\int_{-\phi_{1}}^{2 \pi-\phi_{1}} \frac{\cos u}{\sqrt{1-2 \beta \cos u}} d u=\int_{0}^{2 \pi} \frac{\cos u}{\sqrt{1-2 \beta \cos u}} d u(since the integral runs over a complete cycle of cos u, we may as well change the limits to 0 → 2π).
Then the \phi_{1} integral is just 2π, and
M=\frac{\mu_{0}}{4 \pi} \sqrt{a b \beta} 2 \pi \int_{0}^{2 \pi} \frac{\cos u}{\sqrt{1-2 \beta \cos u}} d u=\frac{\mu_{0}}{2} \sqrt{a b \beta} \int_{0}^{2 \pi} \frac{\cos u}{\sqrt{1-2 \beta \cos u}} d u.
(a) If a is small, then \beta \ll 1, so (using the binomial theorem)
\frac{1}{\sqrt{1-2 \beta \cos u}} \cong 1+\beta \cos u, \text { and } \int_{0}^{2 \pi} \frac{\cos u}{\sqrt{1-2 \beta \cos u}} d u \cong \int_{0}^{2 \pi} \cos u d u+\beta \int_{0}^{2 \pi} \cos ^{2} u d u=0+\beta \pi ,
and hence M=\left(\mu_{0} \pi / 2\right) \sqrt{a b \beta^{3}} . \text { Moreover, } \beta \cong a b /\left(b^{2}+z^{2}\right), \text { so } M \cong \frac{\mu_{0} \pi a^{2} b^{2}}{2\left(b^{2}+z^{2}\right)^{3 / 2}} (same as in Prob. 7.22).
(b) More generally,
(1+\epsilon)^{-1 / 2}=1-\frac{1}{2} \epsilon+\frac{3}{8} \epsilon^{2}-\frac{5}{16} \epsilon^{3}+\cdots \Longrightarrow \frac{1}{\sqrt{1-2 \beta \cos u}}=1+\beta \cos u+\frac{3}{2} \beta^{2} \cos ^{2} u+\frac{5}{2} \beta^{3} \cos ^{3} u+\cdots ,
so
M=\frac{\mu_{0}}{2} \sqrt{a b \beta}\left\{\int_{0}^{2 \pi} \cos u d u+\beta \int_{0}^{2 \pi} \cos ^{2} u d u+\frac{3}{2} \beta^{2} \int_{0}^{2 \pi} \cos ^{3} u d u+\frac{5}{2} \beta^{3} \int_{0}^{2 \pi} \cos ^{4} u d u+\cdots\right\}=\frac{\mu_{0}}{2} \sqrt{a b \beta}\left[0+\beta(\pi)+\frac{3}{2} \beta^{2}(0)+\frac{5}{2} \beta^{3}\left(\frac{3}{4} \pi\right)+\cdots\right]=\frac{\mu_{0} \pi}{2} \sqrt{a b \beta^{3}}\left(1+\frac{15}{8} \beta^{2}+( ) \beta^{4}+\cdots\right) . qed