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 is I .

(a)

Consider the diagram shown below with the small loop and large loop separated by the distance z .

From the above diagram the value of the r can be calculated as,

$$\begin{gathered} r^2=b^2+z^2 \\ r=\sqrt{b^2+z^2} \end{gathered}$$

The value of the$\sin {\mathrm{θ}}_0$can be calculated as,

$\sin {\mathrm{θ}}_0=\frac{b}{r}$

Substitute $\sqrt{b^2+z^2}$for r into above equation.

$\sin {\mathrm{θ}}_0=\frac{b}{{\left(\sqrt{b^2+z^2}\right)}^2}$

The expression for the magnetic field for the big loop is given by,

${\mathrm{dB}}_{\mathrm{z}}=\frac{{\mathrm{μ}}_0\mathrm{I}}{4\mathrm{Ï€}}\frac{{\mathrm{²õ¾\pm ²Ôθ}}_0\mathrm{dI}}{{\mathrm{r}}^2}$

The magnetic field due to entire loop can be calculated by integrating the above equation.

$$\begin{gathered} âˆ\circledR {\mathrm{dB}}_{\mathrm{Z}}=âˆ\circledR \frac{{\mathrm{μ}}_0\mathrm{I}}{4\mathrm{Ï€}}\frac{{\mathrm{²õ¾\pm ²Ôθ}}_0\mathrm{dI}}{{\mathrm{r}}^2} \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{μ}}_0\mathrm{I}}{4\mathrm{Ï€}}âˆ\circledR \frac{{\mathrm{²õ¾\pm ²Ôθ}}_0\mathrm{dI}}{{\mathrm{r}}^2} \\ \mathrm{Substitute}\frac{\mathrm{b}}{\left(\sqrt{{\mathrm{b}}^2+{\mathrm{z}}^2}\right)}\mathrm{for}{\mathrm{²õ¾\pm ²Ôθ}}_0\mathrm{and}{\mathrm{b}}^2+{\mathrm{z}}^2\mathrm{for}{\mathrm{r}}^2\mathrm{into}\mathrm{above}\mathrm{equation}. \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{μ}}_0\mathrm{I}}{4\mathrm{Ï€}}âˆ\circledR \frac{\mathrm{bdI}}{\left(\sqrt{{\mathrm{b}}^2+{\mathrm{z}}^2}\right)\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)} \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{μ}}_0\mathrm{I}}{4\mathrm{Ï€}}âˆ\circledR \frac{\mathrm{dI}}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{μ}}_0\mathrm{Ib}}{4\mathrm{Ï€}{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}âˆ\circledR \mathrm{dI} \\ \mathrm{Substitute}2\mathrm{Ï€²ú}\mathrm{for}âˆ\circledR \mathrm{dI}\mathrm{into}\mathrm{above}\mathrm{equation}. \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{μ}}_0\mathrm{Ib}}{4\mathrm{Ï€}{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}\left(2\mathrm{Ï€²ú}\right) \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{μ}}_0{\mathrm{Ib}}^2}{2{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \mathrm{The}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{flux}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{Ï•}=∫\mathrm{B}.\mathrm{A} \\ \mathrm{Ï•}=\mathrm{B}.\mathrm{A} \\ \mathrm{Substitute}\frac{{\mathrm{μ}}_0{\mathrm{Ib}}^2}{2{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}\mathrm{for}\mathrm{B}\mathrm{and}{\mathrm{Ï€²¹}}^2\mathrm{for}\mathrm{A}\mathrm{into}\mathrm{above}\mathrm{equation}. \\ \mathrm{Ï•}=\frac{{\mathrm{μ}}_0{\mathrm{Ib}}^2}{2{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}×{\mathrm{Ï€²¹}}^2 \\ \mathrm{Ï•}=\frac{{\mathrm{μ}}_0\mathrm{I}}{2}\frac{{\mathrm{Ï€²¹}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \mathrm{Hence}\mathrm{the}\mathrm{flux}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{small}\mathrm{loop}\mathrm{is}\frac{{\mathrm{μ}}_0\mathrm{I}}{2}\frac{{\mathrm{Ï€²¹}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}. \end{gathered}$$

(b)

The magnetic dipole moment due to small loop is given by,

$m=I\mathrm{Ï€}{a}^2$

The magn

etic field due the small loop is given by,

$B=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{m}{r^3}\left(2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)$

Substitute $I\mathrm{Ï€}{a}^2$for m into above equation.

$B=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{I\mathrm{π}{a}^2}{r^3}\left(2\cos \mathrm{θ}\hat{r}+\sin \mathrm{θ}\widehat{\mathrm{θ}}\right)$

The magnetic flux passing through area of the loop is given by,

$\mathrm{ϕ}=∫B.da$

$$\begin{gathered} \mathrm{Substitute}\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{{\mathrm{IÏ€²¹}}^2}{{\mathrm{r}}^3}\left(2\mathrm{³¦´Ç²õθ}\hat{\mathrm{r}}+\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{θ}}\right)\mathrm{for}\mathrm{B}\mathrm{into}\mathrm{above}\mathrm{equation}. \\ \mathrm{Ï•}=∫\left(\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{{\mathrm{IÏ€²¹}}^2}{{\mathrm{r}}^3}\left(2\mathrm{³¦´Ç²õθ}\hat{\mathrm{r}}+\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{θ}}\right)\right).\mathrm{da} \\ \mathrm{Ï•}=\frac{{\mathrm{μ}}_0{\mathrm{Ia}}^2}{4}{∫}_0^{2\mathrm{Ï€}}\mathrm{»åÏ•}{∫}_0^{{\mathrm{θ}}_0}\left(\frac{1}{{\mathrm{r}}^3}\right)\left(2\mathrm{³¦´Ç²õθ}×{\mathrm{r}}^2\mathrm{²õ¾\pm ²Ôθdθ}\right) \\ \mathrm{Ï•}=\frac{{\mathrm{μ}}_0{\mathrm{Ia}}^2}{2\mathrm{r}}×2\mathrm{Ï€}{∫}_0^{{\mathrm{θ}}_0}\mathrm{³¦´Ç²õθ²õ¾\pm ²Ôθdθ} \\ =\frac{{\mathrm{μ}}_0{\mathrm{Ia}}^2}{\mathrm{r}}{∫}_0^{{\mathrm{θ}}_0}\mathrm{³¦´Ç²õθ²õ¾\pm ²Ôθdθ} \\ \mathrm{Solve}\mathrm{further}\mathrm{as}, \end{gathered}$$

(c)

The relationship between the flux of small loop and mutual inductance is given by,

${\mathrm{Ï•}}_{small}=M_{12}I$

Substitute $\frac{{\mathrm{μ}}_{0}I}{2}\frac{\mathrm{π}{a}^2{b}^2}{{\left(b^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}$for ${\mathrm{ϕ}}_{small}$into above equation.

localid="1658146761019" $$\begin{gathered} \frac{{\mathrm{μ}}_{0}I}{2}\frac{\mathrm{π}{a}^2{b}^2}{{\left(b^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}=M_{12}I \\ \frac{{\mathrm{μ}}_0}{2}\frac{\mathrm{π}{a}^2{b}^2}{{\left(b^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}=M_{12} \\ {M}_{12}=\frac{{\mathrm{μ}}_0}{2}\frac{\mathrm{π}{a}^2{b}^2}{{\left(b^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}...\left(1\right) \end{gathered}$$
The relationship between the flux of big loop and mutual inductance is given by,

${\mathrm{Ï•}}_{big}=M_{21}I$

$$\begin{gathered} \mathrm{Substitute}\frac{{\mathrm{μ}}_0\mathrm{I}}{2}\frac{{\mathrm{Ï€²¹}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}\mathrm{for}{\mathrm{Ï•}}_{\mathrm{big}}\mathrm{into}\mathrm{above}\mathrm{equation}. \\ \frac{{\mathrm{μ}}_0\mathrm{I}}{2}\frac{{\mathrm{Ï€²¹}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}={\mathrm{M}}_{21}\mathrm{I} \\ \frac{{\mathrm{μ}}_0}{2}\frac{{\mathrm{Ï€²¹}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}={\mathrm{M}}_{21} \\ {\mathrm{M}}_{21}=\frac{{\mathrm{μ}}_0}{2}\frac{{\mathrm{Ï€²¹}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}....\left(2\right) \end{gathered}$$

From the equation (1) and (2),

$M_{12}=M_{21}$

$\mathrm{Hence}\mathrm{the}\mathrm{mutual}\mathrm{inductance}\mathrm{is}\frac{{\mathrm{μ}}_0}{2}\frac{{\mathrm{Ï€²¹}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}\mathrm{and}\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{that}{\mathrm{M}}_{12}={\mathrm{M}}_{21}.$