Question

Question: (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 $\left(a<<b,a<<z\right)$. Compare your answer to Pro b. 7 .22.

(b) For the general case (not assuming is small), show that

$\mathit{M}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{\pi }\mathbf{\beta }}{\mathbf{2}}\sqrt{\mathbf{a}\mathbf{b}\mathbf{\beta }}\left(1+\frac{15}{8}{\beta }^2+.....\right)$

where

$\mathit{\beta }\mathbf{=}\frac{\mathbf{a}\mathbf{b}}{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

Short answer

Answer

(a) The expression for the mutual inductance is $\frac{{\mu }_0\pi a^2{b}^2}{2{\left(b^2+z^2\right)}^{\frac{3}{2}}}$.

(b) The mutual inductance for the general case is$M=\frac{{\mu }_0\pi \beta }{2}\sqrt{ab\beta }\left(1+\frac{15}{8}{\beta }^2\pi +.......\right)$

Step-by-step solution

Write the given data from the question

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 .

Determine the formula to calculate the mutual inductance of configuration.

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

$M=\frac{{\mu }_0}{4\pi }\oint \oint \frac{d{l}_1·d{l}_2}{r}$ …… (1)

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

 Step 3: Determine the formulas to calculate the mutual inductance of configuration.

(a)

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

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

Consider a point on the lower loop and the coordinates.

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

The distance between the two points is given by,

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

Solve further as,

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

The small segment of the loop is given by,

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

The small segment of the loop is given by,

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

Calculate the product segment and .

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

Calculate the mutual inductance.

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

$$\begin{gathered} M=\frac{{\mu }_0}{4\pi }\oint \oint \frac{ab\cos \left({\varphi }_2-{\varphi }_1\right)}{\sqrt{\frac{ab}{\beta }}\sqrt{1-2\beta \cos \left({\varphi }_2-{\varphi }_1\right)}}d{\varphi }_{1}d{\varphi }_2 \\ M=\frac{{\mu }_{0}ab}{4\pi \sqrt{\frac{ab}{\beta }}}\oint \oint \frac{\cos \left({\varphi }_2-{\varphi }_1\right)}{\sqrt{1-2\beta \cos \left({\varphi }_2-{\varphi }_1\right)}}d{\varphi }_{1}d{\varphi }_2 \end{gathered}$$

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

${\int }_{-{\varphi }_1}^{2\pi -{\varphi }_1}\frac{\cos \left(u\right)}{\sqrt{1-2\beta \cos \left(u\right)}}du={\int }_0^{2\pi }\frac{\cos \left(u\right)}{\sqrt{1-2\beta \cos \left(u\right)}}du$

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

The integration over is just .

$$\begin{gathered} M=\frac{{\mu }_0}{4\pi }\sqrt{ab\beta }2\pi {\int }_0^{2\pi }\frac{\cos \left(u\right)}{\sqrt{1-2\beta \cos \left(u\right)}}du \\ M=\frac{{\mu }_0}{2}\sqrt{ab\beta }{\int }_0^{2\pi }\frac{\cos \left(u\right)}{\sqrt{1-2\beta \cos \left(u\right)}}du \end{gathered}$$ ……. (2)

If is small then,,

$\frac{1}{\sqrt{1-2\beta \cos \left(u\right)}}\cong 1+\beta \cos \left(u\right)$

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

By integrating the above equation,

$$\begin{gathered} M=\frac{{\mu }_0}{2}\sqrt{ab\beta }{\int }_0^{2\pi }\cos \left(u\right)\left[1+\beta \cos \left(u\right)\right]du \\ M=\frac{{\mu }_0}{2}\sqrt{ab\beta }{\int }_0^{2\pi }\cos \left(u\right)du+\beta {\cos }^2\left(u\right)du \end{gathered}$$

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

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

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

Show the equation for mutual inductance for the general case.

(b)

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

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

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

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

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

$$\begin{gathered} M=\frac{{\mu }_0}{2}\sqrt{ab\beta }{\int }_0^{2\pi }\cos \left(u\right)\left(1+\frac{2\beta }{2}\cos u+\frac{3}{8}\times 2{\beta }^2{\cos }^{2}u-\frac{5}{16}{\beta }^3{\cos }^{3}u+.....\right)du \\ M=\frac{{\mu }_0}{2}\sqrt{ab\beta }{\int }_0^{2\pi }\left(\cos u+\frac{2\beta }{2}{\cos }^{2}u+\frac{3}{8}\times 2{\beta }^2{\cos }^{3}u-\frac{5}{16}{\beta }^3{\cos }^{4}u+.....\right)du \\ M=\frac{{\mu }_0}{2}\sqrt{ab\beta }\left[0+\beta \left(\pi \right)+\frac{3}{2}{\beta }^2\left(0\right)+\frac{5}{2}{\beta }^3\left(\frac{3}{4}\pi \right)+.......\right] \\ M=\frac{{\mu }_0}{2}\sqrt{ab\beta }\left(\beta \pi +\frac{15}{8}{\beta }^3\pi \right) \end{gathered}$$

Solve further as,

$M=\frac{{\mu }_0\pi \beta }{2}\sqrt{ab\beta }\left(1+\frac{15}{8}{\beta }^2\pi +.......\right)$

Hence the mutual inductance for the general case is .

$M=\frac{{\mu }_0\pi \beta }{2}\sqrt{ab\beta }\left(1+\frac{15}{8}{\beta }^2\pi +.......\right)$