Find the force of attraction between two magnetic dipoles, ${\mathit{m}}_{\mathbf{1}}$and ${\mathbf{m}}_{\mathbf{2}}$, oriented as shown in Fig. 6.7, a distance r apart, (a) using Eq. 6.2, and (b) using Eq.6.3.

The two dipoles are${\mathrm{m}}_1$ and ${\mathrm{m}}_2$.

The distance between the two dipoles is r .

From equation 6.2 of textbook.

The net force on the circular loop due to another loop,

$\mathrm{F}=2\mathrm{\pi IRB}\mathrm{cos\theta }$

From the equation 6.3 of textbook.

$\mathit{F}\mathbf{=}\mathbf{\nabla }\mathbf{(}\mathit{m}\mathbf{\times }\mathit{B}\mathbf{)}\mathbf{}$

(a)

The expression of magnetic field due to dipole is,

$B=\frac{{\mu }_0}{4\pi r^3}\left[3\right(m\hat{r})\cdot \hat{r}-m]$ …… (1)

Here m is the magnetic dipole.

Consider the diagram that shows the magnetic dipole moment at distance r from the axis.

Consider the dipole moment $m_1$.

The magnetic field due to dipole $m_1$at the distance r .

$B_1=\frac{{\mu }_0}{4\pi }\frac{1}{r^3}\left[3\right(m_1\hat{r})\cdot \hat{r}-m_1]$

The y component of the$B_1$is expressed as,

$$\begin{gathered} B.\hat{y}=\left|B\right|\left|\hat{y}\right|\cos \theta \\ B.\hat{y}=B\cos \theta \\ B\cos \theta =B.\hat{y} \end{gathered}$$

Substitute $\frac{{\mu }_0}{4\pi }\frac{1}{r^3}\left[3\right(m_1\hat{r})\cdot \hat{r}-m_1]$for B into above equation.

localid="1657693067703" $$\begin{gathered} Bcos\theta =\frac{{\mu }_0}{4\pi }\frac{1}{r^3}\left[3\right(m_1\hat{r})\cdot \hat{r}-m_1]\hat{y} \\ Bcos\theta =\frac{{\mu }_0}{4\pi }\frac{1}{r^3}\left[3\right(m_1\hat{r})\hat{r}\cdot \hat{y}-m_1\cdot \hat{y}]\dots \dots \left(2\right) \end{gathered}$$

The dot products of the equation (2) are solved as,

$$\begin{gathered} {\mathrm{m}}_1\cdot \hat{\mathrm{y}}=0 \\ \hat{\mathrm{r}}\cdot \hat{\mathrm{y}}=\sin \mathrm{\varphi } \\ {\mathrm{m}}_1\cdot \hat{\mathrm{r}}={\mathrm{m}}_1\cos \mathrm{\varphi } \end{gathered}$$

Substitute 0 for ${\mathrm{m}}_1\cdot \hat{\mathrm{y}},\mathrm{sin\varphi }\mathrm{for}\hat{\mathrm{r}}\cdot \hat{\mathrm{y}}\mathrm{and}{\mathrm{m}}_1\cos \mathrm{\varphi }\mathrm{for}{\mathrm{m}}_1\cdot \hat{\mathrm{r}}$into equation (2).

$$\begin{gathered} \mathrm{Bcos\theta }=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{r}}^3}\left[3\right({\mathrm{m}}_1\mathrm{cos\varphi })\mathrm{sin\varphi }-0] \\ \mathrm{Bcos\theta }=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{r}}^3}\left[3\right({\mathrm{m}}_1\mathrm{cos\varphi }\left)\mathrm{sin\varphi }\right]\dots \dots \left(3\right) \end{gathered}$$

From the figure,

$$\begin{gathered} \sin \mathrm{\varphi }=\frac{\mathrm{R}}{\mathrm{r}} \\ \mathrm{cos\varphi }=\frac{\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{\mathrm{r}} \end{gathered}$$

Substitute $\frac{R}{r}$for$\sin \varphi$and$\frac{\sqrt{r^2-R^2}}{r}$for $\cos \varphi$into equation (3).

$$\begin{gathered} \mathrm{Bcos\theta }=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{r}}^3}\left[3\left({\mathrm{m}}_1\frac{\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{\mathrm{r}}\right)\frac{\mathrm{R}}{\mathrm{r}}\right] \\ \mathrm{Bcos\theta }=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left[\frac{3{\mathrm{m}}_1\mathrm{R}\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{{\mathrm{r}}^5}\right] \end{gathered}$$

The net force on the circular loop due to another loop,

$\mathrm{F}=2\mathrm{\pi IRB}\mathrm{cos\theta }$

Substitute$\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left[\frac{3{\mathrm{m}}_1\mathrm{R}\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{{\mathrm{r}}^5}\right]$for $\mathrm{B}\mathrm{cos\theta }$into above equation.

localid="1657690027964" $$\begin{gathered} \mathrm{F}=2\mathrm{\pi IR}\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left[\frac{3{\mathrm{m}}_1\mathrm{R}\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{{\mathrm{r}}^5}\right] \\ \mathrm{F}=2{\mathrm{\pi IR}}^2\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left[\frac{3{\mathrm{m}}_1\mathrm{R}\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{{\mathrm{r}}^5}\right] \\ \mathrm{F}=\left(\frac{3{\mathrm{\mu }}_0}{2\mathrm{\pi }}\right)\left({\mathrm{\pi IR}}^2\right)\frac{{\mathrm{m}}_1\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{{\mathrm{r}}^5}\dots \dots \left(4\right) \end{gathered}$$

The magnetic moment for another dipole,

${\mathrm{m}}_2={\mathrm{\pi IR}}^2$

Substitute$\pi I{R}^2$for $m_2$into equation (4).

$$\begin{gathered} \mathrm{F}=\frac{3{\mathrm{\mu }}_0}{2\mathrm{\pi }}{\mathrm{m}}_2\frac{{\mathrm{m}}_1\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{{\mathrm{r}}^5} \\ \mathrm{F}=\frac{3{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{{\mathrm{m}}_2{\mathrm{m}}_1\sqrt{{\mathrm{r}}^2-{\mathrm{R}}^2}}{{\mathrm{r}}^5} \end{gathered}$$

Since r is very large as compare to $\mathrm{R},\mathrm{r}>>\mathrm{R}.$

$$\begin{gathered} \mathrm{F}=\frac{3{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{{\mathrm{m}}_2{\mathrm{m}}_1\sqrt{{\mathrm{r}}^2}}{{\mathrm{r}}^5} \\ \mathrm{F}=\frac{3{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{{\mathrm{m}}_2{\mathrm{m}}_1\mathrm{r}}{{\mathrm{r}}^5} \\ \mathrm{F}=\frac{3{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{{\mathrm{m}}_2{\mathrm{m}}_1}{\mathrm{r}4} \end{gathered}$$

Hence the force of attraction between the two dipoles $m_1$and $m_2$is $\frac{3{\mu }_0}{2\pi }\frac{m_2{m}_1}{r^4}$

(b)

The expression of magnetic field whose radial part is along the z direction.

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{z}}^3}\left[3\right({\mathrm{m}}_1\hat{\mathrm{z}})\cdot \hat{\mathrm{z}}-{\mathrm{m}}_1] \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{z}}^3}[3{\mathrm{m}}_1-{\mathrm{m}}_1] \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{z}}^3}2{\mathrm{m}}_1 \end{gathered}$$

The expression of force acting on the dipole of magnetic moment placed in magnetic field,

$$\begin{gathered} \mathrm{F}=\nabla ({\mathrm{m}}_2\cdot \mathrm{B}) \\ \mathrm{F}=({\mathrm{m}}_2\cdot \nabla )\mathrm{B}+(\mathrm{B}\cdot \nabla ){\mathrm{m}}_2+{\mathrm{m}}_2\times (\nabla \times \mathrm{B})+\mathrm{B}\times (\nabla \times {\mathrm{m}}_2) \\ \mathrm{F}=({\mathrm{m}}_2\cdot \nabla )\mathrm{B}+0+0+0 \\ \mathrm{F}=({\mathrm{m}}_2\cdot \nabla )\mathrm{B} \end{gathered}$$

Substitute$\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{z}}^3}2{\mathrm{m}}_1$for B and$\hat{\mathrm{x}}\frac{\partial }{\partial \mathrm{x}}+\hat{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}+\hat{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}$ for$\nabla$ into above equation.

role="math" localid="1657694752574" $$\begin{gathered} \mathrm{F}=\left[{\mathrm{m}}_2\cdot \left(\hat{\mathrm{x}}\frac{\partial }{\partial \mathrm{x}}+\hat{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}+\hat{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}\right)\right]\left(\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{1}{{\mathrm{z}}^3}2{\mathrm{m}}_1\right) \\ \mathrm{F}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\left(2{\mathrm{m}}_1{\mathrm{m}}_2\right)\left[\hat{\mathrm{x}}\frac{\partial }{\partial \mathrm{x}}\left(\frac{1}{{\mathrm{z}}^3}\right)+\hat{\mathrm{y}}\frac{\partial }{\partial \mathrm{y}}\left(\frac{1}{{\mathrm{z}}^3}\right)+\hat{\mathrm{z}}\frac{\partial }{\partial \mathrm{z}}\frac{1}{{\mathrm{z}}^3}\right] \\ \mathrm{F}=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\left({\mathrm{m}}_1{\mathrm{m}}_2\right)\left[0+0+\hat{\mathrm{z}}\left(\frac{-3}{{\mathrm{z}}^4}\right)\right] \\ \mathrm{F}=-\frac{3{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{z}}^4}\hat{\mathrm{z}} \end{gathered}$$

Hence the attraction force between the dipoles ${\mathrm{m}}_1$and ${\mathrm{m}}_2$is$-\frac{3{\mu }_0}{2\pi }\frac{m_2{m}_1}{z^4}\hat{z}$ .