91影视

Write the expression for the Lorenz gauge condition.

$\mathbf{鈭�}\mathbf{路}\mathit{A}\mathbf{=}\mathbf{-}{\mathit{渭}}_{\mathbf{0}}{\mathit{蔚}}_{\mathbf{0}}\left(\frac{鈭�V}{鈭�t}\right)$ 鈥︹�� (1)

Here,${渭}_0$ is the magnetic permeability${蔚}_0$ is the magnetic permittivity, A is the vector potential, and V is the scalar potential.

Write the expression for the vector potential (using equation 11.17 ).

$$\begin{gathered} A=-\frac{{渭}_0{p}_0蝇}{4蟺}\frac{1}{r}\sin \left(蝇\left(t-\frac{r}{c}\right)\right)\hat{z} \\ A=-\frac{{渭}_0{p}_0蝇}{4蟺}\frac{1}{r}\sin \left(蝇\left(t-\frac{r}{c}\right)\right)\left(\cos 胃\hat{r}-\sin 胃\widehat{胃}\right) \end{gathered}$$

Calculate the value of $鈭�路A$.

$$\begin{gathered} 鈭�路A=\frac{1}{r^2}\frac{鈭�}{鈭�r}\left(r^2{A}_r\right)+\frac{1}{r\sin 胃}\frac{鈭�}{鈭�胃}\left(\sin 胃A_{胃}\right)+\frac{1}{r\sin 胃}\frac{鈭�蠒}{鈭�蠒} \\ 鈭�路A=\frac{1}{r^2}\frac{鈭�}{鈭�r}\left[\begin{array}{c}{r}^2\left[\left(-\frac{{渭}_0{p}_0蝇}{4蟺}\right)\frac{1}{r}\sin \left(蝇\left(t-\frac{r}{c}\right)\right)\cos 胃\right]\\ \frac{1}{r\sin 胃}\frac{鈭�}{鈭�胃}-\left[\left(\frac{{渭}_0{p}_0蝇}{4蟺}\right)\frac{1}{r}\sin \left(蝇\left(t-\frac{r}{c}\right)\right)\left(-{\sin }^2胃\right)\right]\end{array}+\right] \\ 鈭�路A=-\frac{{渭}_0{p}_0蝇}{4蟺}\left[\begin{array}{c}\frac{1}{r^2}\frac{鈭�}{鈭�r}\frac{1}{r}{r}^2\sin \left(蝇\left(t-\frac{r}{c}\right)\right)\cos 胃-\left[\frac{蝇}{rc}\cos \left(蝇t-\frac{蝇r}{c}\right)\cos 胃\right]-\\ \frac{2\sin 胃\cos 胃}{r^3\sin 胃}\sin \left(蝇t-\frac{蝇r}{c}\right)\end{array}\right] \\ 鈭�路A=-\frac{{渭}_0{p}_0蝇}{4蟺}\left[\frac{1}{r^2}\sin \left(蝇t-\frac{蝇r}{c}\right)\frac{蝇}{rc}\cos \left(蝇t-\frac{蝇r}{c}\right)-\frac{2}{r^2}\sin \left(蝇t-\frac{蝇r}{c}\right)\right]\cos 胃 \end{gathered}$$

On further solving, the above equation becomes,

localid="1653907297258" $$\begin{gathered} 鈭�路A=-\frac{{渭}_0{p}_0蝇}{4蟺}\left[\frac{\left(2-1\right)}{r^2}\sin \left(蝇t-\frac{蝇r}{c}\right)+\frac{蝇}{rc}\cos \left(蝇t-\frac{蝇r}{c}\right)\right]\cos 胃 \\ 鈭�路A=-{渭}_0蝇\left[\frac{p_0蝇}{4蟺{蔚}_0}\frac{1}{r^2}\sin 蝇\left(t-\frac{r}{c}\right)+\frac{蝇}{rc}\cos 蝇\left(蝇-\frac{r}{c}\right)\right]\cos 胃....\left(1\right) \end{gathered}$$

Write the expression for the scalar potential for an oscillating dipole potential (using equation 11.12 ).

$V=\frac{p_0\cos 胃}{4蟺{蔚}_{0}r}\left[-\frac{蝇}{c}\sin \left[蝇\left(t-\frac{r}{c}\right)\right]+\frac{1}{r}\cos 蝇\left(t-\frac{r}{c}\right)\right]$

Calculate the value of $\frac{鈭�V}{鈭�t}$.

$$\begin{gathered} \frac{鈭�V}{鈭�t}=\frac{p_0\cos 胃}{4蟺{蔚}_{0}r}\left[-\frac{{蝇}^2}{c}\cos 蝇\left(t-\frac{r}{c}\right)-\frac{蝇}{r}\sin 蝇\left(t-\frac{r}{c}\right)\right] \\ \frac{鈭�V}{鈭�t}=\frac{p_0蝇}{4蟺{蔚}_0}\left[\frac{1}{r^2}\sin 蝇\left(t-\frac{r}{c}\right)+\frac{蝇}{rc}\cos 蝇\left(t-\frac{r}{c}\right)\right]\cos 胃.......\left(2\right) \end{gathered}$$

From equations (1) and (2),

$鈭�路A=-{渭}_0蝇\frac{鈭�V}{鈭�t}$

Therefore, the retarded potentials of an oscillating dipole satisfy the Lorenz gauge condition.