Q22P - Introduction to Electrodynamics

Question

A small loop of wire (radius a) is held a distance z above the center of a large loop (radius b ), as shown in Fig. 7.37. The planes of the two loops are parallel, and perpendicular to the common axis. 

(a) Suppose current I flows in the big loop. Find the flux through the little loop. (The little loop is so small that you may consider the field of the big loop to be essentially constant.) 

(b) Suppose current I flows in the little loop. Find the flux through the big loop. (The little loop is so small that you may treat it as a magnetic dipole.) 

(c) Find the mutual inductances, and confirm that ${\mathbf{M}}_{\mathbf{12}}\mathbf{=}{\mathbf{M}}_{\mathbf{21}}$ ·

Step-by-Step Solution

Answer

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{The} \mathrm{flux} \mathrm{passing} \mathrm{through} \mathrm{the} \mathrm{small} \mathrm{loop} \mathrm{is}\frac{{\mathrm{\mu }}_0\mathrm{I}}{2}\frac{{\mathrm{\pi }}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}. \\ \left(\mathrm{b}\right) \mathrm{The} \mathrm{flux} \mathrm{passing} \mathrm{through} \mathrm{big} \mathrm{loop} \mathrm{is}\frac{{\mathrm{\mu }}_0\mathrm{I}}{2}\frac{{\mathrm{\pi }}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} . \\ \left(\mathrm{c}\right) \mathrm{The} \mathrm{mutual} \mathrm{inductance} \mathrm{is} \frac{{\mathrm{\mu }}_0\mathrm{I}}{2}\frac{{\mathrm{\pi }}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \mathrm{and} \mathrm{it} \mathrm{is} \mathrm{proved} \mathrm{that} {\mathrm{M}}_{12}={\mathrm{M}}_{21.} \end{gathered}$$

Step 1: 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 is I .

Step 2: Calculate the flus in the small loop.

(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 $\sin {\theta }_0$ can be calculated as,

 $\sin {\theta }_0=\frac{b}{r}$

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

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

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

${\mathrm{dB}}_{\mathrm{z}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\frac{{\mathrm{sin\theta }}_0\mathrm{dI}}{{\mathrm{r}}^2}$ 

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

$$\begin{gathered} \oint {\mathrm{dB}}_{\mathrm{Z}}=\oint \frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\frac{{\mathrm{sin\theta }}_0\mathrm{dI}}{{\mathrm{r}}^2} \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\oint \frac{{\mathrm{sin\theta }}_0\mathrm{dI}}{{\mathrm{r}}^2} \\ \mathrm{Substitute} \frac{\mathrm{b}}{\left(\sqrt{{\mathrm{b}}^2+{\mathrm{z}}^2}\right)} \mathrm{for} {\mathrm{sin\theta }}_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{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\oint \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{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\oint \frac{\mathrm{dI}}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{\mu }}_0\mathrm{Ib}}{4\mathrm{\pi }{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}\oint \mathrm{dI} \\ \mathrm{Substitute} 2\mathrm{\pi b} \mathrm{for} \oint \mathrm{dI} \mathrm{into} \mathrm{above} \mathrm{equation}. \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{\mu }}_0\mathrm{Ib}}{4\mathrm{\pi }{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}\left(2\mathrm{\pi b}\right) \\ {\mathrm{B}}_{\mathrm{z}}=\frac{{\mathrm{\mu }}_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{\varphi }=\int \mathrm{B}.\mathrm{A} \\ \mathrm{\varphi }=\mathrm{B}.\mathrm{A} \\ \mathrm{Substitute} \frac{{\mathrm{\mu }}_0{\mathrm{Ib}}^2}{2{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \mathrm{for} \mathrm{B} \mathrm{and} {\mathrm{\pi a}}^2 \mathrm{for} \mathrm{A} \mathrm{into} \mathrm{above} \mathrm{equation}. \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0{\mathrm{Ib}}^2}{2{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}\times {\mathrm{\pi a}}^2 \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2}\frac{{\mathrm{\pi a}}^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{\mu }}_0\mathrm{I}}{2}\frac{{\mathrm{\pi a}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} . \end{gathered}$$

Step 3: Calculate the flux in the big loop.

(b)

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

$m=I\pi a^2$ 

The magn

etic field due the small loop is given by,

 $B=\frac{{\mu }_0}{4\pi }\frac{m}{r^3}\left(2\cos \theta \hat{r}+\sin \theta \hat{\theta }\right)$

Substitute $I\pi a^2$ for m into above equation.

$B=\frac{{\mu }_0}{4\pi }\frac{I\pi a^2}{r^3}\left(2\cos \theta \hat{r}+\sin \theta \hat{\theta }\right)$

 

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

$\varphi =\int B.da$

$$\begin{gathered} \mathrm{Substitute} \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{{\mathrm{I\pi a}}^2}{{\mathrm{r}}^3}\left(2\mathrm{cos\theta }\hat{\mathrm{r}}+\mathrm{sin\theta }\hat{\mathrm{\theta }}\right) \mathrm{for} \mathrm{B} \mathrm{into} \mathrm{above} \mathrm{equation}. \\ \mathrm{\varphi }=\int \left(\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{{\mathrm{I\pi a}}^2}{{\mathrm{r}}^3}\left(2\mathrm{cos\theta }\hat{\mathrm{r}}+\mathrm{sin\theta }\hat{\mathrm{\theta }}\right)\right).\mathrm{da} \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2}{4}{\int }_0^{2\mathrm{\pi }}\mathrm{d\varphi }{\int }_0^{{\mathrm{\theta }}_0}\left(\frac{1}{{\mathrm{r}}^3}\right)\left(2 \mathrm{cos\theta }\times {\mathrm{r}}^2\mathrm{sin\theta d\theta }\right) \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2}{2\mathrm{r}}\times 2\mathrm{\pi }{\int }_0^{{\mathrm{\theta }}_0}\mathrm{cos\theta sin\theta d\theta } \\ =\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2}{\mathrm{r}}{\int }_0^{{\mathrm{\theta }}_0}\mathrm{cos\theta sin\theta d\theta } \\ \mathrm{Solve} \mathrm{further} \mathrm{as}, \end{gathered}$$

 $$\begin{gathered} \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2\mathrm{\pi }}{\mathrm{r}}{\left(\frac{{\sin }^2\mathrm{\theta }}{2}\right)}^{\mathrm{\theta }0} \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2\mathrm{\pi }}{\mathrm{r}}\left(\frac{{\sin }^2{\mathrm{\theta }}_0}{2}-0\right) \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2\mathrm{\pi }}{2\mathrm{r}}\left({\sin }^2{\mathrm{\theta }}_0\right) \\ \mathrm{Substitute} \sqrt{{\mathrm{b}}^2+{\mathrm{z}}^2} \mathrm{for} \mathrm{r} \mathrm{and} \frac{\mathrm{b}}{\sqrt{{\mathrm{b}}^2+{\mathrm{z}}^2}} \mathrm{for} {\mathrm{sin\theta }}_0 \mathrm{into} \mathrm{above} \mathrm{equation}. \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2\mathrm{\pi }}{2\sqrt{{\mathrm{b}}^2+{\mathrm{z}}^2}}\times {\left(\frac{\mathrm{b}}{\sqrt{{\mathrm{b}}^2+{\mathrm{z}}^2}}\right)}^2 \\ \mathrm{\varphi }=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2}\frac{{\mathrm{\pi a}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ \mathrm{Hence} \mathrm{the} \mathrm{flux} \mathrm{passing} \mathrm{through} \mathrm{big} \mathrm{loop} \mathrm{is} \frac{{\mathrm{\mu }}_0\mathrm{I}}{2}\frac{{\mathrm{\pi a}}^2{\mathrm{b}}^2}{{\left({\mathrm{b}}^2+{\mathrm{z}}^2\right)}^{{\displaystyle \frac{3}{2}}}}. \end{gathered}$$

Step 4: Find the mutual inductances, and prove M 12 = M 21

(c)

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

${\varphi }_{small}=M_{12}I$

Substitute $\frac{{\mu }_{0}I}{2}\frac{\pi a^2{b}^2}{{\left(b^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}$for ${\varphi }_{small}$into above equation.

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

${\varphi }_{big}=M_{21}I$

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

From the equation (1) and (2),

$M_{12}=M_{21}$

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