91Ó°ÊÓ

Write the expression for the Lorenz gauge condition.

$\mathbf{∇}\mathbf{·}\mathit{A}\mathbf{=}\mathbf{-}{\mathit{μ}}_{\mathbf{0}}{\mathit{Î\mu }}_{\mathbf{0}}\left(\frac{∂V}{∂t}\right)$ …… (1)

Here,${\mathrm{μ}}_0$ is the magnetic permeability${\mathrm{Î\mu }}_0$ 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 ).

$$\begin{gathered} A=-\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\frac{1}{r}\sin \left(\mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right)\hat{z} \\ A=-\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\frac{1}{r}\sin \left(\mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right)\left(\cos \mathrm{θ}\hat{r}-\sin \mathrm{θ}\widehat{\mathrm{θ}}\right) \end{gathered}$$

Calculate the value of $∇·A$.

$$\begin{gathered} ∇·A=\frac{1}{r^2}\frac{∂}{∂r}\left(r^2{A}_r\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}{A}_{\mathrm{θ}}\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂\mathrm{Ï•}}{∂\mathrm{Ï•}} \\ ∇·A=\frac{1}{r^2}\frac{∂}{∂r}\left[\begin{array}{c}{r}^2\left[\left(-\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\right)\frac{1}{r}\sin \left(\mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right)\cos \mathrm{θ}\right]\\ \frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}-\left[\left(\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\right)\frac{1}{r}\sin \left(\mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right)\left(-{\sin }^2\mathrm{θ}\right)\right]\end{array}+\right] \\ ∇·A=-\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\left[\begin{array}{c}\frac{1}{r^2}\frac{∂}{∂r}\frac{1}{r}{r}^2\sin \left(\mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right)\cos \mathrm{θ}-\left[\frac{\mathrm{Ó\neg }}{rc}\cos \left(\mathrm{Ó\neg }t-\frac{\mathrm{Ó\neg }r}{c}\right)\cos \mathrm{θ}\right]-\\ \frac{2\sin \mathrm{θ}\cos \mathrm{θ}}{r^3\sin \mathrm{θ}}\sin \left(\mathrm{Ó\neg }t-\frac{\mathrm{Ó\neg }r}{c}\right)\end{array}\right] \\ ∇·A=-\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\left[\frac{1}{r^2}\sin \left(\mathrm{Ó\neg }t-\frac{\mathrm{Ó\neg }r}{c}\right)\frac{\mathrm{Ó\neg }}{rc}\cos \left(\mathrm{Ó\neg }t-\frac{\mathrm{Ó\neg }r}{c}\right)-\frac{2}{r^2}\sin \left(\mathrm{Ó\neg }t-\frac{\mathrm{Ó\neg }r}{c}\right)\right]\cos \mathrm{θ} \end{gathered}$$

On further solving, the above equation becomes,

localid="1653907297258" $$\begin{gathered} ∇·A=-\frac{{\mathrm{μ}}_0{p}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\left[\frac{\left(2-1\right)}{r^2}\sin \left(\mathrm{Ó\neg }t-\frac{\mathrm{Ó\neg }r}{c}\right)+\frac{\mathrm{Ó\neg }}{rc}\cos \left(\mathrm{Ó\neg }t-\frac{\mathrm{Ó\neg }r}{c}\right)\right]\cos \mathrm{θ} \\ ∇·A=-{\mathrm{μ}}_0\mathrm{Ó\neg }\left[\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\sin \mathrm{Ó\neg }\left(t-\frac{r}{c}\right)+\frac{\mathrm{Ó\neg }}{rc}\cos \mathrm{Ó\neg }\left(\mathrm{Ó\neg }-\frac{r}{c}\right)\right]\cos \mathrm{θ}....\left(1\right) \end{gathered}$$

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

$V=\frac{p_0\cos \mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}\left[-\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right]+\frac{1}{r}\cos \mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right]$

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

$$\begin{gathered} \frac{∂V}{∂t}=\frac{p_0\cos \mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}\left[-\frac{{\mathrm{Ó\neg }}^2}{c}\cos \mathrm{Ó\neg }\left(t-\frac{r}{c}\right)-\frac{\mathrm{Ó\neg }}{r}\sin \mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right] \\ \frac{∂V}{∂t}=\frac{p_0\mathrm{Ó\neg }}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[\frac{1}{r^2}\sin \mathrm{Ó\neg }\left(t-\frac{r}{c}\right)+\frac{\mathrm{Ó\neg }}{rc}\cos \mathrm{Ó\neg }\left(t-\frac{r}{c}\right)\right]\cos \mathrm{θ}.......\left(2\right) \end{gathered}$$

From equations (1) and (2),

$∇·A=-{\mathrm{μ}}_0\mathrm{Ó\neg }\frac{∂V}{∂t}$

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