91影视

Consider a $鈭�$ denotes derivatives with respect to , and ${鈭�}^{\text{'}}$ denotes derivatives with respect to $r^{\text{'}}$. Next, noting that $\text{J}(鈭�\text{r}/\text{c})$depends on $r^{\text{'}}$both explicitly and through , whereas it depends on r only through$r$ .

Write the formula of divergence of.

$\mathbf{鈭�}\mathbf{鈰�}\mathbf{A}\mathbf{=}\frac{{\mathbf{渭}}_0}{\mathbf{4}\mathbf{蟺}}\mathbf{鈭�}\mathbf{鈭�}\mathbf{鈰�}\mathbf{\left(}\frac{J}{r}\mathbf{\right)}$鈥︹�� (1)

Here, ${\mathit{渭}}_0$is the permeability of free space $\mathbf{鈭�}$denotes derivatives,$\mathrm{J}$ is surface density and $\mathrm{r}$ is resistance.

We know that from the product rule.

$鈭�鈰�\left(\frac{\text{J}}{\text{R}}\right)=\frac{1}{\text{r}}(鈭�鈰�\text{J})+\text{J}鈰�(鈭�\frac{1}{\text{r}})$ 鈥︹�� (2)

Then

localid="1658814372004" ${鈭�}^{\text{'}}鈰�\left(\frac{\text{J}}{\text{R}}\right)=\frac{1}{\text{r}}({鈭�}^{\text{'}}鈰�\text{J})+\text{J}鈰�\left({鈭�}^{\text{'}}\frac{1}{\text{r}}\right)$ 鈥︹�� (3)

As it knows that

$\text{r}=\text{r}鈭�$$r^{\text{'}}$

Then,

$鈭�\left(\frac{1}{\text{r}}\right)=鈭�{鈭�}^{\text{'}}\left(\frac{1}{\text{r}}\right)$

Substitute $鈭�{鈭�}^{\text{'}}\left(\frac{1}{\text{r}}\right)$for$鈭�\left(\frac{1}{\text{r}}\right)$ in the equation (2).

$鈭�鈰�\left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}(鈭�鈰�\text{J})鈭�\text{J}鈰�\left({鈭�}^{\text{'}}\frac{1}{\text{r}}\right)$

From the equation (2)

$\text{J}鈰�\left({鈭�}^{\text{'}}\frac{1}{\text{r}}\right)={鈭�}^{\text{'}}鈰�\left(\frac{\text{J}}{\text{r}}\right)鈭�\frac{1}{\text{r}}({鈭�}^{\text{'}}鈰�\text{J})$

Then, equation (1) reduced as follows:

$\begin{array}{c}鈭�鈰�\left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}(鈭�鈰�\text{J})鈭�({鈭�}^{\text{'}}鈰�\left(\frac{\text{J}}{\text{r}}\right)鈭�\frac{1}{\text{r}}({鈭�}^{\text{'}}鈰�\text{J}))\\ =\frac{1}{\text{r}}(鈭�鈰�\text{J})鈭�{鈭�}^{\text{'}}\left(\frac{\text{J}}{\text{r}}\right)+\frac{1}{\text{r}}({鈭�}^{\text{'}}鈰�\text{J})\end{array}$

We know that

$\begin{array}{c}鈭�鈰�\text{J}=\frac{鈭�J_x}{鈭�x}+\frac{鈭�J_y}{鈭�y}+\frac{鈭�J_z}{鈭�z}\\ =\frac{鈭�J_x}{鈭�t_r}\frac{鈭�t_r}{鈭�x}+\frac{鈭�J_y}{鈭�t_r}\frac{鈭�t_r}{鈭�y}+\frac{鈭�J_z}{鈭�t_r}\frac{鈭�t_r}{鈭�z}\end{array}$ 鈥︹�� (4)

But it is known that

$\begin{array}{l}\frac{鈭�t_r}{鈭�x}=\frac{鈭�1}{c}\frac{鈭�r}{鈭�x}\\ \frac{鈭�t_r}{鈭�y}=\frac{鈭�1}{c}\frac{鈭�r}{鈭�y}\\ \frac{鈭�t_r}{鈭�z}=\frac{鈭�1}{c}\frac{鈭�r}{鈭�z}\end{array}$

Substitute $\frac{鈭�1}{c}\frac{鈭�r}{鈭�x}$for$\frac{鈭�t_r}{鈭�x}$, $\frac{鈭�1}{c}\frac{鈭�r}{鈭�y}$for$\frac{鈭�t_r}{鈭�y}$, and$\frac{鈭�1}{c}\frac{鈭�r}{鈭�z}$ for $\frac{鈭�t_r}{鈭�z}$ in the equation (4).

$鈭�鈰�J=鈭�\frac{1}{c}[\frac{鈭�J_x}{鈭�t_r}鈰�\frac{鈭�r}{鈭�x}+\frac{鈭�J_y}{鈭�t_r}鈰�\frac{鈭�r}{鈭�y}+\frac{鈭�J_z}{鈭�t_r}鈰�\frac{鈭�r}{鈭�z}]$

We know that

$鈭�鈰�r=\frac{未r}{未x}+\frac{未r}{未y}+\frac{未r}{未z}$

Then

$鈭�鈰�\text{J}=鈭�\frac{1}{c}[\left(\frac{鈭�J}{鈭�t_r}\right)鈰�\left({鈭�}^{\text{'}}r\right)]$

Similarly

${鈭�}^{\text{'}}鈰�\text{J}=鈭�\frac{鈭�蚁}{鈭�t}鈭�\frac{1}{c}\frac{鈭�\text{J}}{鈭�t_r}鈰�\left({鈭�}^{\text{'}}r\right)$

Now substitute $\frac{鈭�1}{c}\frac{鈭�\text{J}}{鈭�t_r}鈰�(鈭�r)$ for$(鈭�鈰�\text{J})$ and $鈭�\frac{鈭�蚁}{鈭�t}鈭�\frac{1}{c}\frac{鈭�\text{J}}{鈭�t_r}鈰�\left({鈭�}^{\text{'}}r\right)$for$({鈭�}^{\text{'}}鈰�\text{J})$ into the above equation.

$\begin{array}{c}鈭�鈰�\left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}(鈭�鈰�\text{J})鈭�{鈭�}^{\text{'}}\left(\frac{\text{J}}{\text{r}}\right)+\frac{1}{\text{r}}({鈭�}^{\text{'}}鈰�\text{J})\\ 鈭�鈰�\left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}(鈭�鈰�\text{J})+\frac{1}{\text{r}}\left({鈭�}^{\text{'}}\text{J}\right)鈭�{鈭�}^{\text{'}}\frac{\text{J}}{\text{r}}\\ =\frac{1}{\text{r}}[\frac{鈭�1}{c}\frac{鈭�\text{J}}{鈭�t_r}鈰�(鈭�\text{r})]+\frac{1}{\text{r}}[鈭�\frac{鈭�蚁}{鈭�t}鈭�\frac{1}{c}\frac{鈭�\text{J}}{鈭�t_r}鈰�\left({鈭�}^{\text{'}}\text{r}\right)]鈭�{鈭�}^{\text{'}}\left(\frac{J}{\text{r}}\right)\end{array}$

Here, $鈭�\text{r}=鈭�{鈭�}^{\text{'}}\text{r}$, then

$鈭�鈰�\left(\frac{\text{J}}{\text{r}}\right)=\frac{1}{\text{r}}\frac{鈭�蚁}{鈭�t}鈭�{鈭�}^{\text{'}}鈰�\left(\frac{\text{J}}{\text{r}}\right)$

We know that vector potential is calculated by using the formulae

$\text{A}=\frac{{渭}_0}{4蟺}鈭�\frac{I\left(t_r\right)}{r}dI$

Here, $I$is the current through the loop, $dI$is the length of the elementary part and $k$is the surface current.

$\text{A}(r,t)=\frac{{渭}_0}{4蟺}鈭�\text{J}\left(\frac{r,t}{r}\right)d蟿$

Then

$鈭�鈰�\text{A}=\frac{{渭}_0}{4蟺}鈭�鈭�鈰�\left(\frac{\text{J}}{\text{r}}\right)d蟿$

Substitute$\frac{\text{1}}{\text{r}}\frac{鈭�蚁}{鈭�t}鈭�{鈭�}^{\text{'}}鈰�\left(\frac{\text{J}}{\text{r}}\right)$ for $鈭�鈰�\left(\frac{\text{J}}{\text{r}}\right)$ into the above equation.

$\begin{array}{c}鈭�鈰�\text{A}=\frac{{渭}_0}{4蟺}[\frac{鈭�鈭�}{鈭�t}鈭�\frac{蚁}{\text{r}}d蟿鈭�鈭�{鈭�}^{\text{'}}鈰�\left(\frac{J}{r}\right)d蟿]\\ =鈭�{渭}_0{蔚}_0\frac{鈭�}{鈭�t}[\frac{1}{4蟺{蔚}_0}鈭�\frac{蚁}{\text{r}}d蟿]鈭�\frac{{渭}_0}{4蟺}鈭�螡\frac{1}{\text{r}}鈰�da\end{array}$

The$\text{J}=0$ across the surface at infinity, as far as we are aware. The last term then becomes zero, and the $鈭�鈰�\text{A}$equation follows the following reduction:

$鈭�鈰�\text{A}=鈭�{渭}_0{蔚}_0\frac{鈭�}{鈭�t}[\frac{1}{4蟺{蔚}_0}鈭�\frac{蚁}{r}d蟿]$

Here, $V=\frac{1}{4蟺{蔚}_0}鈭�\frac{蚁}{\text{r}}d蟿$

Then, substitute $\frac{1}{4蟺{蔚}_0}鈭�\frac{蚁}{\text{r}}d蟿$ for $V$into the above equation.

$\begin{array}{l}鈭�鈰�\text{A}=鈭�{渭}_0{蔚}_0\frac{鈭�}{鈭�t}[\frac{1}{4蟺{蔚}_0}鈭�\frac{蚁}{\text{r}}d蟿]\\ 鈭�鈰�\text{A}=鈭�{渭}_0{蔚}_0\frac{鈭�}{鈭�t}V\end{array}$

Therefore, the value of divergence of$\text{A}$ can be expressed as$鈭�{渭}_0{蔚}_0\frac{鈭�}{鈭�t}V$.