91Ó°ÊÓ

There are two dipoles having moments ${\stackrel{→}{p}}_1$and${\stackrel{→}{p}}_2$perpendicular to each other.

The force on a dipole having moment $\stackrel{→}{p}$in the presence of an electric field $\stackrel{→}{E}$is

$\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\left(\stackrel{→}{p}.\stackrel{→}{∇}\right)\stackrel{\mathbf{→}}{\mathbf{E}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The cross product of dipole moment of the second dipole and the electric field due to the first dipole is

${\stackrel{\mathbf{→}}{\mathbf{P}}}_{\mathbf{2}}\mathbf{×}{\stackrel{\mathbf{→}}{\mathbf{E}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{-}{\mathbf{p}}_{\mathbf{1}}{\mathbf{p}}_{\mathbf{2}}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here,${\mathrm{Î\mu }}_0$ is the permittivity of free space.

Let the second dipole moment be

$\stackrel{→}{p}=p_2\hat{y}$

The electric field due to the first dipole at the position of the second dipole is

$\stackrel{→}{E}=-\frac{p_1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{y}^3}\hat{z}$

Substitute this in equation (1) to get the force on the second dipole due to the first dipole

localid="1657775080935" $$\begin{gathered} \stackrel{→}{F}=p_2\frac{∂}{∂y}\left(-\frac{p_1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{y}^3}\hat{z}\right) \\ =\frac{3p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{y}^4}\hat{z} \\ =\frac{3p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^4}\hat{z} \end{gathered}$$

To find the electric field due to the second dipole, it is kept at the origin and pointed towards z axis. The expression for the electric field due to the first dipole at the position of the second dipole is

$\stackrel{→}{E}=\frac{p_2}{4\mathrm{Ï€}\mathrm{Î\mu }}\left[\frac{3xz\hat{x}+3yz\hat{y}-\left(x^2+y^2-2z^2\right)\hat{z}}{{\left(x^2+y^2+z^2\right)}^{5/2}}\right]$

Thus from equation (1), he force on the first dipole due to the second dipole is

${\stackrel{→}{F}}_2={\overline{)-p_1\frac{∂{\stackrel{→}{E}}_2}{∂y}}}_{x=0,y=0.z=r}$

Substitute the values in the above equation006E

localid="1657775867177" $$\begin{gathered} {\stackrel{→}{F}}_2=-p_1\frac{∂}{∂y}\frac{p_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\overline{)\left[\frac{3xz\hat{x}+3yz\hat{y}-\left(x^2+y^2-2z^2\right)\hat{z}}{{\left(x^2+y^2+z^2\right)}^{5/2}}\right]}}_{x=0,y=0.z=-r} \\ =\frac{-p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left[\frac{-5}{2}2y\frac{3xz\hat{x}+3yz\hat{y}-\left(x^2+y^2-2z^2\right)\hat{z}}{{\left(x^2+y^2+z^2\right)}^{7/2}}+\frac{3yz\hat{y}-2y\hat{z}}{{\left(x^2+y^2+z^2\right)}^{5/2}}\right]}_{x=0,y=0.z=r} \\ =\frac{3p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^4}\hat{y} \end{gathered}$$

$\hat{y}$in the turned coordinate is equal to $\hat{z}$in the original coordinate.

Thus, the forces are$\frac{3p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^4}\hat{z}$ and $-\frac{3p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^4}\hat{z}$. They are equal and opposite and hence are consistent with Newton's third law of motion.

The expression for the torque on the second dipole is

$\widehat{\mathrm{τ}}={\stackrel{→}{p}}_2×{\stackrel{→}{E}}_1+\stackrel{→}{r}×{\stackrel{→}{F}}_2$

Here, $\stackrel{→}{r}=-r\hat{y}$

Use equation (2) and substitute values in the above equation

$$\begin{gathered} \stackrel{→}{\mathrm{Ï„}}=\frac{-p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\hat{x}+r\hat{y}×\frac{3p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^4}\hat{z} \\ =\left(\frac{-p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}+\frac{3p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\right)\hat{x} \\ =\frac{2p_1{p}_2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^3}\hat{x} \end{gathered}$$

This is equal and opposite to the torque on the first dipole.