91Ó°ÊÓ

The radius of the small loop is a.

The radius of the large loop is b.

The distance between the large and small loop is z.

The current is the large loop isI .

The expression to calculate the self-inductance is given as follows.

$M=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}âˆ\circledR âˆ\circledR \frac{d{l}_1·d{l}_2}{r}$ …… (1)

Here, $d{l}_1$,$d{l}_2$ is the small segment of the loops.

(a)

Let us consider a point on the upper loop and the coordinates.

$r_2=\left(a\cos {\mathrm{Ï•}}_2,a\sin {\mathrm{Ï•}}_2,z\right)$

Consider a point on the lower loop and the coordinates.

$r_1=\left(b\cos {\mathrm{Ï•}}_1,b\sin {\mathrm{Ï•}}_1,0\right)$

The distance between the two points is given by,

$$\begin{gathered} r^2={\left(r_2-r_1\right)}^2 \\ {r}^2={\left(a\cos {\mathrm{Ï•}}_2-b\cos {\mathrm{Ï•}}_1\right)}^2+{\left(a\sin {\mathrm{Ï•}}_2-b\sin {\mathrm{Ï•}}_1\right)}^2+z^2 \\ {r}^2=a^2{\cos }^2{\mathrm{Ï•}}_2-2b\cos {\mathrm{Ï•}}_2\cos {\mathrm{Ï•}}_1+b^2{\cos }^2{\mathrm{Ï•}}_1+a^2{\sin }^2{\mathrm{Ï•}}_2-2ab\sin {\mathrm{Ï•}}_1\sin {\mathrm{Ï•}}_2+b^2{\sin }^2{\mathrm{Ï•}}_1+z^2 \\ {r}^2=a^2+b^2+z^2-2ab\left(\cos {\mathrm{Ï•}}_2\cos {\mathrm{Ï•}}_1+\sin {\mathrm{Ï•}}_2\sin {\mathrm{Ï•}}_1\right) \end{gathered}$$

Solve further as,

$$\begin{gathered} r^2=a^2+b^2+z^2-2ab\cos \left({\mathrm{ϕ}}_2-{\mathrm{ϕ}}_1\right) \\ {r}^2=\left(a^2+b^2+z^2\right)\left[1-2\mathrm{β}\cos \left({\mathrm{ϕ}}_2-{\mathrm{ϕ}}_1\right)\right] \\ {r}^2=\frac{ab}{\mathrm{β}}\left[1-2\mathrm{β}\cos \left({\mathrm{ϕ}}_2-{\mathrm{ϕ}}_1\right)\right] \\ r=\sqrt{\frac{ab}{\mathrm{β}}\left[1-2\mathrm{β}\cos \left({\mathrm{ϕ}}_2-{\mathrm{ϕ}}_1\right)\right]} \end{gathered}$$

The small segment of the loop is given by,

$$\begin{gathered} \backslash \mathrm{begingathered}d{l}_1=bd\widehat{{\mathrm{Ï•}}_1} \\ d{l}_1=bd{\mathrm{Ï•}}_1\left[-\sin {\mathrm{Ï•}}_1\hat{x}+\cos {\mathrm{Ï•}}_1\hat{y}\right] \end{gathered}$$

The small segment of the loop is given by,

$$\begin{gathered} d{l}_2=ad\mathrm{Ï•}\widehat{{\mathrm{Ï•}}_2} \\ d{l}_2=ad{\mathrm{Ï•}}_2\left[-\sin {\mathrm{Ï•}}_2\hat{x}+\cos {\mathrm{Ï•}}_2\hat{y}\right] \end{gathered}$$

Calculate the product segment and .

localid="1658401850706" $$\begin{gathered} d{l}_1·d{l}_2=bd{\mathrm{ϕ}}_1\left[-\sin {\mathrm{ϕ}}_1\hat{x}+\cos {\mathrm{ϕ}}_1\hat{y}\right]·ad{\mathrm{ϕ}}_2\left[-\sin {\mathrm{ϕ}}_2\hat{x}+\cos {\mathrm{ϕ}}_2\hat{y}\right] \\ d{l}_1·d{l}_2=abd{\mathrm{ϕ}}_{1}d{\mathrm{ϕ}}_2\left[\sin {\mathrm{ϕ}}_1\sin {\mathrm{ϕ}}_2+\cos {\mathrm{ϕ}}_1\cos {\mathrm{ϕ}}_2\right] \\ d{l}_1·d{l}_2=ab\cos \left({\mathrm{ϕ}}_2-{\mathrm{ϕ}}_1\right)d{\mathrm{ϕ}}_{1}d{\mathrm{ϕ}}_2 \end{gathered}$$

Calculate the mutual inductance.

Substitute $ab\cos \left({\mathrm{ϕ}}_2-{\mathrm{ϕ}}_1\right)d{\mathrm{ϕ}}_{1}d{\mathrm{ϕ}}_2$for , and $\sqrt{\frac{ab}{\mathrm{β}}\left[1-2\mathrm{β}\cos \left({\mathrm{ϕ}}_2-{\mathrm{ϕ}}_1\right)\right]}$for into equation (1).

$$\begin{gathered} M=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}âˆ\circledR âˆ\circledR \frac{ab\cos \left({\mathrm{Ï•}}_2-{\mathrm{Ï•}}_1\right)}{\sqrt{\frac{ab}{\mathrm{β}}}\sqrt{1-2\mathrm{β}\cos \left({\mathrm{Ï•}}_2-{\mathrm{Ï•}}_1\right)}}d{\mathrm{Ï•}}_{1}d{\mathrm{Ï•}}_2 \\ M=\frac{{\mathrm{μ}}_{0}ab}{4\mathrm{Ï€}\sqrt{\frac{ab}{\mathrm{β}}}}âˆ\circledR âˆ\circledR \frac{\cos \left({\mathrm{Ï•}}_2-{\mathrm{Ï•}}_1\right)}{\sqrt{1-2\mathrm{β}\cos \left({\mathrm{Ï•}}_2-{\mathrm{Ï•}}_1\right)}}d{\mathrm{Ï•}}_{1}d{\mathrm{Ï•}}_2 \end{gathered}$$

Integrate the above integration from to for the both integrations and let assume $u={\mathrm{Ï•}}_2-{\mathrm{Ï•}}_1$.

${∫}_{-{\mathrm{ϕ}}_1}^{2\mathrm{π}-{\mathrm{ϕ}}_1}\frac{\cos \left(u\right)}{\sqrt{1-2\mathrm{β}\cos \left(u\right)}}du={∫}_0^{2\mathrm{π}}\frac{\cos \left(u\right)}{\sqrt{1-2\mathrm{β}\cos \left(u\right)}}du$

Since the integration is runs over the complete cycle of , the limits can be changed from .

The integration over is just .

$$\begin{gathered} M=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\sqrt{ab\mathrm{β}}2\mathrm{π}{∫}_0^{2\mathrm{π}}\frac{\cos \left(u\right)}{\sqrt{1-2\mathrm{β}\cos \left(u\right)}}du \\ M=\frac{{\mathrm{μ}}_0}{2}\sqrt{ab\mathrm{β}}{∫}_0^{2\mathrm{π}}\frac{\cos \left(u\right)}{\sqrt{1-2\mathrm{β}\cos \left(u\right)}}du \end{gathered}$$ ……. (2)

If is small then,,

$\frac{1}{\sqrt{1-2\mathrm{β}\cos \left(u\right)}}â‰\dots 1+\mathrm{β}\cos \left(u\right)$

Substitute$1+\mathrm{β}\cos \left(u\right)for\frac{1}{\sqrt{1-2\mathrm{β}\cos \left(u\right)}}$into equation (2).

By integrating the above equation,

$$\begin{gathered} M=\frac{{\mathrm{μ}}_0}{2}\sqrt{ab\mathrm{β}}{∫}_0^{2\mathrm{π}}\cos \left(u\right)\left[1+\mathrm{β}\cos \left(u\right)\right]du \\ M=\frac{{\mathrm{μ}}_0}{2}\sqrt{ab\mathrm{β}}{∫}_0^{2\mathrm{π}}\cos \left(u\right)du+\mathrm{β}{\cos }^2\left(u\right)du \end{gathered}$$

Substitute$\frac{ab}{b^2+z^2}$for$\mathrm{β}$into above equation.

localid="1658311173529" $$\begin{gathered} M=\left(\frac{{\mathrm{μ}}_0\mathrm{π}}{2}\right)\sqrt{ab{\left(\frac{ab}{b^2+z^2}\right)}^3} \\ M=\frac{{\mathrm{μ}}_0\mathrm{π}{a}^2{b}^2}{2{\left(b^2+z^2\right)}^{\frac{3}{2}}} \end{gathered}$$

Hence the expression for the mutual inductance is $\frac{{\mathrm{μ}}_0\mathrm{π}{a}^2{b}^2}{2{\left(b^2+z^2\right)}^{\frac{3}{2}}}$.

(b)

The term${\left(1+\mathrm{Î\mu }\right)}^{-1}$ can be expand as,

${\left(1+\mathrm{Î\mu }\right)}^{-1}=1-\frac{1}{2}\mathrm{Î\mu }+\frac{3}{8}{\mathrm{Î\mu }}^2-\frac{5}{16}{\mathrm{Î\mu }}^3+......$

The expand the term $\frac{1}{\sqrt{1-2\mathrm{β}\cos \left(u\right)}}$by using the above series.

${\left(1-2\mathrm{β}\cos \left(u\right)\right)}^{-1}=1+\frac{2\mathrm{β}}{2}\cos u+\frac{3}{8}×2\mathrm{β}{\cos }^{2}u-\frac{5}{16}{\mathrm{β}}^3{\cos }^{3}u+.....$

Substitute $1+\frac{2\mathrm{β}}{2}\cos u+\frac{3}{8}×2\mathrm{β}{\cos }^{2}u-\frac{5}{16}{\mathrm{β}}^3{\cos }^{3}u+.....$for $\frac{1}{\sqrt{1-2\mathrm{β}\cos \left(u\right)}}$ into equation (2).

$$\begin{gathered} M=\frac{{\mathrm{μ}}_0}{2}\sqrt{ab\mathrm{β}}{∫}_0^{2\mathrm{π}}\cos \left(u\right)\left(1+\frac{2\mathrm{β}}{2}\cos u+\frac{3}{8}×2{\mathrm{β}}^2{\cos }^{2}u-\frac{5}{16}{\mathrm{β}}^3{\cos }^{3}u+.....\right)du \\ M=\frac{{\mathrm{μ}}_0}{2}\sqrt{ab\mathrm{β}}{∫}_0^{2\mathrm{π}}\left(\cos u+\frac{2\mathrm{β}}{2}{\cos }^{2}u+\frac{3}{8}×2{\mathrm{β}}^2{\cos }^{3}u-\frac{5}{16}{\mathrm{β}}^3{\cos }^{4}u+.....\right)du \\ M=\frac{{\mathrm{μ}}_0}{2}\sqrt{ab\mathrm{β}}\left[0+\mathrm{β}\left(\mathrm{π}\right)+\frac{3}{2}{\mathrm{β}}^2\left(0\right)+\frac{5}{2}{\mathrm{β}}^3\left(\frac{3}{4}\mathrm{π}\right)+.......\right] \\ M=\frac{{\mathrm{μ}}_0}{2}\sqrt{ab\mathrm{β}}\left(\mathrm{β}\mathrm{π}+\frac{15}{8}{\mathrm{β}}^3\mathrm{π}\right) \end{gathered}$$

Solve further as,

$M=\frac{{\mathrm{μ}}_0\mathrm{π}\mathrm{β}}{2}\sqrt{ab\mathrm{β}}\left(1+\frac{15}{8}{\mathrm{β}}^2\mathrm{π}+.......\right)$

Hence the mutual inductance for the general case is .

$M=\frac{{\mathrm{μ}}_0\mathrm{π}\mathrm{β}}{2}\sqrt{ab\mathrm{β}}\left(1+\frac{15}{8}{\mathrm{β}}^2\mathrm{π}+.......\right)$