91Ó°ÊÓ

Consider the limitations of this site $l$ will be using letter $l$ as the distance from the charge to the point of interest.

Write the formula of partial time derivative of the vector potential.

$\frac{∂\stackrel{\mathbf{→}}{\mathbf{A}}}{∂\mathbf{t}}=\frac{{\mathrm{μ}}_0\mathbf{qc}}{\mathbf{4}\mathrm{Ï€}}\mathbf{[}\frac{}{\stackrel{\mathbf{→}}{\mathbf{ι}}\mathbf{â‹\dots }\stackrel{\mathbf{→}}{\mathbf{u}}}\frac{\mathbf{∂}\stackrel{\mathbf{→}}{\mathbf{Ï\dots }}}{\mathbf{∂}\mathbf{t}}\mathbf{−}\frac{\stackrel{\mathbf{→}}{\mathbf{Ï\dots }}}{{\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{ι}}\mathbf{â‹\dots }\stackrel{\mathbf{→}}{\mathbf{u}}\mathbf{)}}^{\mathbf{2}}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{t}}\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{ι}}\mathbf{â‹\dots }\stackrel{\mathbf{→}}{\mathbf{u}}\mathbf{)}\mathbf{]}$ …… (1)

Here, ${\mathrm{μ}}_0$ is permeability, $\mathbf{q}$ is test charge,$\stackrel{\mathbf{→}}{\mathbf{ι}}$ is retarded time and $\stackrel{\mathbf{→}}{\mathbf{Ï\dots }}$ is retarded time evaluated.

The Lienard-Wiechert vector potential of the charge in motion is:

$\begin{array}{c}\stackrel{→}{\mathrm{A}}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\mathrm{qc}\stackrel{→}{\mathrm{Ï\dots }}}{\mathrm{鳦}−\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{Ï\dots }}}\\ =\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{\mathrm{qc}\stackrel{→}{\mathrm{Ï\dots }}}{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}}\end{array}$

All quantities evaluated at retarded time.

Consider the case:

$\begin{array}{c}\mathrm{ι}=\mathrm{c}(\mathrm{t}−{\mathrm{t}}_{\mathrm{r}})\\ {\mathrm{ι}}^2={\mathrm{c}}^2{(\mathrm{ι}−{\mathrm{ι}}_{\mathrm{r}})}^2\end{array}$

Solve for the partial time derivative as:

$\begin{array}{c}2\stackrel{→}{\mathrm{ι}}â‹\dots \frac{∂\stackrel{→}{\mathrm{ι}}}{∂\mathrm{t}}=2{\mathrm{c}}^2(\mathrm{ι}−{\mathrm{ι}}_{\mathrm{r}})\left(1−\frac{∂{\mathrm{ι}}_{\mathrm{r}}}{∂\mathrm{t}}\right)\\ \frac{∂{\mathrm{ι}}_{\mathrm{r}}}{∂\mathrm{t}}=1−\frac{\stackrel{→}{\mathrm{ι}}}{\mathrm{c}}â‹\dots \frac{∂\stackrel{→}{\mathrm{ι}}}{∂\mathrm{t}}\\ \frac{∂{\mathrm{ι}}_{\mathrm{r}}}{∂\mathrm{t}}=\frac{\mathrm{³¦Î¹}}{\stackrel{→}{\mathrm{ι}}â‹\dots (\mathrm{c}\widehat{\mathrm{ι}}−\stackrel{→}{\mathrm{Ï\dots }})}\\ \frac{∂{\mathrm{ι}}_{\mathrm{r}}}{∂\mathrm{t}}=\frac{\mathrm{³¦Î¹}}{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}}\end{array}$

Since, $\stackrel{→}{\mathrm{ι}}=\stackrel{→}{\mathrm{r}}−\stackrel{→}{\mathrm{Ó\neg }}\left({\mathrm{t}}_{\mathrm{r}}\right)$solve as:

$\begin{array}{c}\frac{∂\stackrel{→}{\mathrm{ι}}}{∂\mathrm{t}}=−\frac{∂\stackrel{→}{\mathrm{Ó\neg }}}{∂{\mathrm{t}}_{\mathrm{r}}}\frac{∂{\mathrm{t}}_{\mathrm{r}}}{∂\mathrm{t}}=−\stackrel{→}{\mathrm{Ï\dots }}\left({\mathrm{t}}_{\mathrm{r}}\right)\frac{∂{\mathrm{t}}_{\mathrm{r}}}{∂\mathrm{t}}\\ \frac{∂{\mathrm{t}}_{\mathrm{r}}}{∂\mathrm{t}}=1+\frac{\widehat{\mathrm{ι}}}{\mathrm{c}}â‹\dots \stackrel{→}{\mathrm{Ï\dots }}\left({\mathrm{t}}_{\mathrm{r}}\right)\frac{∂{\mathrm{t}}_{\mathrm{r}}}{∂\mathrm{t}}\end{array}$

With this we can get the partial time derivative of the vector potential:

Determine the $\frac{∂\stackrel{→}{\mathrm{Ï\dots }}}{∂t}$.

$\begin{array}{c}\frac{∂\stackrel{→}{\mathrm{Ï\dots }}}{∂\mathrm{t}}=\frac{∂\stackrel{→}{\mathrm{Ï\dots }}}{∂{\mathrm{t}}_{\mathrm{r}}}\frac{∂{\mathrm{t}}_{\mathrm{r}}}{∂\mathrm{t}}\\ =\stackrel{→}{\mathrm{a}}\frac{\mathrm{³¦Î¹}}{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}}\end{array}$

Determine$\frac{∂}{∂\mathrm{t}}(\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}})\frac{∂\stackrel{→}{\mathrm{Ï\dots }}}{∂\mathrm{t}}$.

$\begin{array}{c}\frac{∂}{∂\mathrm{t}}(\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}})=\mathrm{c}\frac{∂\mathrm{ι}}{∂\mathrm{t}}−\frac{∂\stackrel{→}{\mathrm{ι}}}{∂\mathrm{t}}â‹\dots \stackrel{→}{\mathrm{Ï\dots }}−\frac{∂\stackrel{→}{\mathrm{Ï\dots }}}{∂\mathrm{t}}â‹\dots \stackrel{→}{\mathrm{ι}}\\ =\mathrm{c}\frac{∂\mathrm{ι}}{∂\mathrm{t}}+{\mathrm{Ï\dots }}^2\frac{\mathrm{³¦Î¹}}{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}}−(\stackrel{→}{\mathrm{a}}â‹\dots \stackrel{→}{\mathrm{ι}})\frac{\mathrm{³¦Î¹}}{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}}\end{array}$

Determine the$\frac{∂\mathrm{ι}}{∂\mathrm{t}}$.

$\begin{array}{c}\frac{∂\mathrm{ι}}{∂\mathrm{t}}=\frac{∂}{∂\mathrm{t}}\left(\mathrm{c}(\mathrm{t}−{\mathrm{t}}_{\mathrm{r}})\right)\\ =\mathrm{c}\frac{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}−\mathrm{鳦}}{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}}\end{array}$

Solve further as:

$\begin{array}{c}\frac{∂}{∂\mathrm{t}}(\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}})={\mathrm{c}}^2\frac{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}−\mathrm{鳦}}{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}}+{\mathrm{Ï\dots }}^2\frac{\mathrm{³¦Î¹}}{\stackrel{→}{\mathrm{I}}â‹\dots \stackrel{→}{\mathrm{u}}}−(\stackrel{→}{\mathrm{a}}â‹\dots \stackrel{→}{\mathrm{ι}})\frac{\mathrm{³¦Î¹}}{\stackrel{→}{\mathrm{I}}â‹\dots \stackrel{→}{\mathrm{u}}}\\ =−\frac{1}{\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}}(\mathrm{³¦Î¹}({\mathrm{c}}^2−{\mathrm{Ï\dots }}^2+\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{a}})−{\mathrm{c}}^2(\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}))\end{array}$

Hence:

Determine the partial time derivative of the vector potential.

Substitute$\frac{∂\stackrel{→}{\mathrm{Ï\dots }}}{∂\mathrm{t}}$, $\frac{∂\stackrel{→}{\mathrm{Ï\dots }}}{∂\mathrm{t}}$, $\frac{∂\mathrm{ι}}{∂\mathrm{t}}$ into equation (1).

$\begin{array}{c}\frac{∂\stackrel{→}{\mathrm{A}}}{∂\mathrm{t}}=\frac{{\mathrm{μ}}_0\mathrm{qc}}{4\mathrm{Ï€}}\frac{1}{{(\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}})}^3}[\mathrm{鳦}\stackrel{→}{\mathrm{a}}(\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}})+\stackrel{→}{\mathrm{Ï\dots }}(\mathrm{cl}({\mathrm{c}}^2−{\mathrm{Ï\dots }}^2+\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{a}})+{\mathrm{c}}^2(\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}}))]\\ =\frac{\mathrm{qc}}{4{\mathrm{Ï€Î\mu }}_0}\left[(\stackrel{→}{\mathrm{ι}}â‹\dots \stackrel{→}{\mathrm{u}})\left(\frac{1}{\mathrm{c}}\stackrel{→}{\mathrm{a}}−\stackrel{→}{\mathrm{Ï\dots }}\right)+\frac{\mathrm{ι}}{\mathrm{c}}\stackrel{→}{\mathrm{Ï\dots }}({\mathrm{c}}^2−{\mathrm{Ï\dots }}^2+\stackrel{→}{\mathrm{r}}â‹\dots \stackrel{→}{\mathrm{a}})\right]\end{array}$

Therefore, the value of partial time derivative of the vector potential is $\frac{∂\stackrel{→}{\mathrm{A}}}{∂\mathrm{t}}=\frac{\mathrm{qc}}{4{\mathrm{Ï€Î\mu }}_0}\left[(\stackrel{→}{\mathrm{l}}â‹\dots \stackrel{→}{\mathrm{u}})\left(\left(\frac{1}{\mathrm{c}}\stackrel{→}{\mathrm{a}}−\stackrel{→}{\mathrm{Ï\dots }}\right)\right)+\frac{\mathrm{l}}{\mathrm{c}}\stackrel{→}{\mathrm{Ï\dots }}({\mathrm{c}}^2−{\mathrm{Ï\dots }}^2+\stackrel{→}{\mathrm{r}}â‹\dots \stackrel{→}{\mathrm{a}})\right]$.