In Sect, 6.2.1, we began with the potential of a perfect dipole (Eq. 6.10), whereas in fact we are dealing with physical dipoles. Show, by the method of Sect. 4.2.3, that we nonetheless get the correct macroscopic field.

We began with the potential of a perfect dipole (Eq. 6.10), whereas in fact we are dealing with physical dipoles.

Write the formula ofmacroscopic vector potential due to physical dipole.

$\overrightarrow{\mathbf{A}}\left(\overrightarrow{\mathbf{r}}\right)={\overrightarrow{\mathbf{A}}}_{\mathrm{in}}+{\overrightarrow{\mathbf{A}}}_{\mathrm{out}}$ …… (1)

Here,${\overrightarrow{\mathbf{A}}}_{\mathrm{in}}$ is small sphere magnetization and${\overrightarrow{\mathbf{A}}}_{\mathrm{out}}$ is ideal dipole.

We choose a random location inside the content. We can locate the origin at that position without losing any generality. Therefore, we must demonstrate that the macroscopic vector potential resulting from physical dipoles is given by

$\overrightarrow{A}\left(\overrightarrow{r}\right)={\overrightarrow{A}}_{in}+{\overrightarrow{A}}_{out}$

If we average the tiny field $\overrightarrow{B}=\frac{{\mu }_0}{4\pi }{\int }_{\nu }d{r}^{\text{'}}\overrightarrow{J}\left({\overrightarrow{r}}^{\text{'}}\right)\times \frac{\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}}}{{|\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}}|}^3}$over a sphere of radius R, we obtain the following, which makes working with the macroscopic $B$field easier:

$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }{\int }_{\nu }d{r}^{\text{'}}\overrightarrow{J}\left({\overrightarrow{r}}^{\text{'}}\right)\times \frac{1}{{\nu }_{in}}{\int }_{{\nu }_{in}}dr\frac{\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}}}{{|\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}}|}^3}$

${\nu }_{in}$is the volume of a sphere with a radius of $R$, and $\nu$ is the volume of the substance. It is clear that integral ${\int }_{{\nu }_{in}}dr\frac{\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}}}{{|\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}}|}^3}$and the sphere's uniform charge have many similarities.

${\int }_{{\nu }_{in}}dr\frac{\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}}}{{|\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}}|}^3}=\left\{\begin{array}{l}-{\nu }_{in}\frac{{\overrightarrow{r}}^{\text{'}}}{{r^{\text{'}}}^3}{r}^{\text{'}}>R\\ -\frac{4\pi }{3}{\overrightarrow{r}}^{\text{'}}{r}^{\text{'}}<R\end{array}\right\}$

So, we split the $\overrightarrow{B}={\overrightarrow{B}}_{in}+{\overrightarrow{B}}_{out}$field into two contributions for different values of $r^{\text{'}}$. With ${\nu }_{in}=\frac{4\pi R^3}{3}$.

$\overrightarrow{B}\left(0\right)=\frac{{\mu }_0}{4\pi }{\int }_{{\nu }_{out}}d{r}^{\text{'}}\overrightarrow{J}\left({\overrightarrow{r}}^{\text{'}}\right)\times \frac{-{\overrightarrow{r}}^{\text{'}}}{{r^{\text{'}}}^3}+\frac{{\mu }_0}{4\pi R^3}{\int }_{{\nu }_{in}}d{r}^{\text{'}}\overrightarrow{J}\left({\overrightarrow{r}}^{\text{'}}\right)\times (-{\overrightarrow{r}}^{\text{'}})$

The microscopic $B_{out}$field calculated at the origin is the first integral. And lastly, we know that $\overrightarrow{m}=\frac{1}{2}{\int }_{{\nu }_{in}}{\overrightarrow{r}}^{\text{'}}\times \overrightarrow{J}d{r}^{\text{'}}$for the second integral.

For ${\overrightarrow{B}}_{out}^{\left(micro\right)}$all physical dipoles are far from the sphere’s center, ergo we can treat them as ideal dipoles.

${\overrightarrow{A}}_{out}=\frac{{\mu }_0}{4\pi }{\int }_{{\nu }_{out}}\frac{\overrightarrow{M}\left({\overrightarrow{r}}^{\text{'}}\right)\times (\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}})}{{r^{\text{'}}}^3}d{r}^{\text{'}}$

We determine that average field${\overrightarrow{B}}_{in}$is equal to that of a uniformly magnetized sphere i.e.

$\begin{array}{c}{\overrightarrow{B}}_{in}=\frac{{\mu }_{0}2\overrightarrow{m}}{4\pi R^3}\\ =\frac{2}{3}{\mu }_0\overrightarrow{M}\\ {\overrightarrow{A}}_{in}=\frac{{\mu }_0}{3}\overrightarrow{M}\times \overrightarrow{r}\end{array}$

Determine a small sphere magnetized can be considered constant, it follows:

$\begin{array}{c}{\overrightarrow{A}}_{in}=\frac{{\mu }_0}{3}\overrightarrow{M}\times \overrightarrow{r}\\ =\frac{{\mu }_0}{4\pi }{\int }_{{\nu }_{in}}\frac{\overrightarrow{M}\left({\overrightarrow{r}}^{\text{'}}\right)\times ({\overrightarrow{r}}^{\text{'}}-{\overrightarrow{r}}^{\text{'}})}{{r^{\text{'}}}^3}d{r}^{\text{'}}\end{array}$

Putting it all together into equation (1).

$\overrightarrow{A}\left(\overrightarrow{r}\right)=\frac{{\mu }_0}{4\pi }{\int }_{}\frac{\overrightarrow{M}\left({\overrightarrow{r}}^{\text{'}}\right)\times (\overrightarrow{r^{\text{'}}}-\overrightarrow{r^{\text{'}}})}{{r^{\text{'}}}^3}d{r}^{\text{'}}$

Therefore, the value of macroscopic vector potential due to physical dipole is given by$\overrightarrow{A}\left(\overrightarrow{r}\right)=\frac{{\mu }_0}{4\pi }{\int }_{\nu }\frac{\overrightarrow{M}\left({\overrightarrow{r}}^{\text{'}}\right)\times (\overrightarrow{r}-{\overrightarrow{r}}^{\text{'}})}{{r^{\text{'}}}^3}d{r}^{\text{'}}$