Use the result of Prob. 10.34 to determine the power radiated by an ideal electric dipole, , at the origin. Check that your answer is consistent with Eq. 11.22, in the case of sinusoidal time dependence, and with Prob. 11.26, in the case of quadratic time dependence.

Write the expression for the power radiated by an ideal electric dipole.

$\mathit{P}\mathbf{=}\mathbf{\oint }{\mathit{S}}_{\mathbf{r}}\mathbf{‐}\mathit{d}\mathit{a}$ …… (1)

Here, $S_r$is the Poynting vector of an ideal dipole which is given as:

$S_r=\frac{1}{{\mu }_0}\left(E_r\times B_r\right)$ …… (2)

Here,$S_r$ is the electric field of the ideal dipole and is the magnetic field of the ideal dipole.

Write the expression for a non-static ideal dipole in an electric field.

$E\left(r,t\right)=-\frac{{\mu }_0}{4\pi }\left\{\frac{{\displaystyle \stackrel{,,}{p}}{\displaystyle \stackrel{}{-\hat{r}}}\left(\hat{r}-{\displaystyle \stackrel{,,}{p}}\right)}{r}+C^2\frac{\left[p+\left({\displaystyle \frac{r}{c}}\right){\displaystyle \stackrel{,}{p}}\right]\left(\hat{r}‐\left[p+{\displaystyle \frac{r}{c}}\stackrel{,}{p}\right]\right)}{r}\right\}$

Here, ${\mu }_0$is the permeability of free space, p is the dipole moment, r is the distance of the source, and c is the speed of light.

Write the expression for a non-static ideal dipole in the magnetic field.

$B\left(r,t\right)=-\frac{{\mu }_0}{4\pi }\left\{\frac{\hat{r}\times \left[{\displaystyle \stackrel{,}{p}}+\left({\displaystyle \frac{r}{c}}\right){\displaystyle \stackrel{,,}{p}}\right]}{r^2}\right\}$

Determine the total power radiated over the sphere with the radius r is,

Write the radiation for the fixed time at the origin $t=t-\frac{r}{c}$from the electric and magnetic dipole moment.

$$\begin{gathered} E_r=-\frac{{\mu }_0}{4\pi }\left\{\frac{{\displaystyle \stackrel{,,}{p-\hat{r}}}\left(\hat{r}‐{\displaystyle \stackrel{,,}{p}}\right)}{r^2}\right\} \\ {B}_r=-\frac{{\mu }_0}{4\pi }\left\{\frac{\hat{r}\times {\displaystyle \stackrel{,,}{p}}}{rc}\right\} \end{gathered}$$

Substitute the value of $E_r$and $B_r$in equation (2).

$$\begin{gathered} S_r=\frac{1}{{\mu }_0}\left(-\frac{{\mu }_0}{4\pi }\left\{\frac{{\displaystyle \stackrel{,,}{p}}-\hat{r}\left(\hat{r}‐{\displaystyle \stackrel{,,}{p}}\right)}{r}\right\}\times -\frac{{\mu }_0}{4\pi }\left\{\frac{\hat{r}‐\stackrel{,,}{p}}{rc}\right\}\right) \\ {S}_r=\frac{{\mu }_0}{16{\pi }^{2}c{r}^2}\left[\left(\stackrel{,,}{p}-\hat{r}\left(\hat{r}‐\stackrel{,,}{p}\right)\right)\times \left(\hat{r}‐\stackrel{,,}{p}\right)\right] \\ {S}_r=\frac{{\mu }_0}{16{\pi }^{2}c{r}^2}\left[\left(\stackrel{,,}{p}-\hat{r}\left(\hat{r}‐\stackrel{,,}{p}\right)\right)-\left(\hat{r}‐\stackrel{,,}{p}\right)\left[\hat{r}\times \left(\hat{r}‐\stackrel{,,}{p}\right)\right]\right] \\ {S}_r=\frac{{\mu }_0}{16{\pi }^{2}c{r}^2}\left[{\stackrel{,,}{p}}^2-\left(\hat{r}‐\stackrel{,,}{p}\right)\right]\hat{r} \end{gathered}$$

Substitute the value of$S_r$in equation (1).

$$\begin{gathered} P=\oint \frac{{\mu }_0}{16{\pi }^{2}c{r}^2}\left[{\stackrel{,,}{p^2-\left(\hat{r}‐\stackrel{,,}{p}\right)}}^2\right]\hat{r}-da \\ P=\oint \frac{{\mu }_0}{16{\pi }^{2}c{r}^2}\left[{\stackrel{,,}{p^2-\left(\hat{r}‐\stackrel{,,}{p}\right)}}^2\right]\hat{r}-\left(r^2\sin \theta d\theta d\varphi \right) \end{gathered}$$

Here,$\left[{\stackrel{,,}{p^2-\left(\hat{r}‐\stackrel{,,}{p}\right)}}^2\right]=\stackrel{,,}{p^2{\sin }^2\theta }$

Hence, the equation becomes,

$$\begin{gathered} P=\frac{{\mu }_0}{16{\pi }^{2}c{r}^2}\stackrel{..}{p}\oint \frac{1}{r^2}{\sin }^2\theta r^2\sin \theta d\theta d\varphi \\ P=\frac{{\mu }_0}{16{\pi }^{2}c{r}^2}\stackrel{..}{p^2}\left(4\pi \right){\int }_0^{\pi }\left({\sin }^3\theta d\theta \right) \\ P=\frac{{\mu }_{0}p{\displaystyle \stackrel{..}{p^2}}}{6\pi c} \end{gathered}$$ …… (3)

Write the expression for the electric dipole moment in the case of sinusoidal time dependence.

$p=p_0\cos \omega t$

Take the double differentiation of the above equation.

$$\begin{gathered} \stackrel{}{\stackrel{.}{p}=-}\omega p_0\sin \omega t \\ \stackrel{}{\stackrel{..}{p}=-{\omega }^2{p}_0}\cos \omega t \end{gathered}$$

Squaring on both sides,

$\stackrel{}{\stackrel{..}{p^2}=-{\omega }^4{p^2}_0}{\cos }^2\omega t$

Take the time average of .

Substitute the value of $\stackrel{.{.}^2}{p}$in equation (3).

$$\begin{gathered} P=\frac{{\mu }_0\left({\displaystyle \frac{1}{2}}{\omega }^4{p}_0^2\right)}{6\pi c} \\ P=\frac{{\mu }_0{p}_0^2{\omega }^4}{12\pi c} \end{gathered}$$

Hence, the result is consistent with equation 11.22.

Therefore, the power radiated by an ideal electric dipole at the origin is, $P=\frac{{\mu }_0{p}_0^2{\omega }^4}{12\pi c}$0and the obtained result is consistent with equation 11.22.