Apply Eqs. 11.59 and 11.60 to the rotating dipole of Prob. 11.4. Explain any apparent discrepancies with your previous answer

Write the expression for the dipole moment of the rotating dipole.

$\mathit{p}\left(t\right)\mathbf{=}{\mathit{p}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\left(\omega t\right)\hat{\mathbf{x}}\mathbf{+}{\mathit{p}}_{\mathbf{0}}\mathit{s}\mathit{i}\mathit{n}\left(\omega t\right)\hat{\mathbf{y}}$ …… (1)

Write the expression for the Poynting vector (using equation 11.59 ).

$S=\frac{{\mu }_0{\left|\ddot{p}\left(t\right)\right|}^2}{16{\pi }^{2}c}\frac{{\sin }^2\theta }{r^2}\hat{r}$ …… (2)

Write the expression for the total radiated power (using equation ).

$P_{rad}\left(t_0\right)\cong \frac{{\mu }_0}{6\pi c}{\left[\ddot{p}\left(t_0\right)\right]}^2$ …… (3)

Take the double differentiation of equation (1).

$$\begin{gathered} \dot{p}\left(t\right)=-p_0\omega \sin \left(\omega t\right)\hat{x}+p_0\omega \cos \left(\omega t\right)\hat{y} \\ \ddot{p}\left(t\right)=-p_0{\omega }^2\cos \left(\omega t\right)\hat{x}-p_0{\omega }^2\sin \left(\omega t\right)\hat{y} \\ \ddot{p}\left(t\right)=-p_0{\omega }^2\left[\cos \left(\omega t\right)\hat{x}+\sin \left(\omega t\right)\hat{y}\right] \end{gathered}$$

Substitute$\ddot{p}\left(t\right)=-p_0{\omega }^2\left[\cos \left(\omega t\right)\hat{x}+\sin \left(\omega t\right)\hat{y}\right]$ in the above value in equation (2).

$$\begin{gathered} S=\frac{{\mu }_0{\left|-p_0{\omega }^2\left[\cos \left(\omega t\right)\hat{x}+\sin \left(\omega t\right)\hat{y}\right]\right|}^2}{16{\pi }^{2}c}\frac{{\sin }^2\theta }{r^2}\hat{r} \\ S=\frac{{\mu }_0{p}_0^2{\omega }^4}{16{\pi }^{2}c}\left({\cos }^2\omega t+{\sin }^2\omega t\right)\frac{{\sin }^2\theta }{r^2}\hat{r} \\ S=\frac{{\mu }_0{p}_0^2{\omega }^4}{16{\pi }^{2}c}\frac{{\sin }^2\theta }{r^2}\hat{r} \end{gathered}$$

Hence, the above equation disagrees with the value of Poynting vector S in problem 11.4 because, in this problem, the polar axis is along the direction of $\ddot{p}\left(t_0\right)$.

Substitute$\ddot{p}\left(t\right)=-p_0{\omega }^2\left[\cos \left(\omega t\right)\hat{x}+\sin \left(\omega t\right)\hat{y}\right]$ in equation (3).

$$\begin{gathered} P_{rad}\left(t_0\right)\cong \frac{{\mu }_0}{6\pi c}{\left[-p_0{\omega }^2\left[\cos \left(\omega t\right)\hat{x}+\sin \left(\omega t\right)\hat{y}\right]\right]}^2 \\ {P}_{rad}\left(t_0\right)=\frac{{\mu }_0{p}_0^2{\omega }^4}{6\pi c} \end{gathered}$$

For the case of total radiated power, the value of$P_{rad}\left(t_0\right)=\frac{{\mu }_0{p}_0^2{\omega }^4}{6\pi c}$ agrees with the value of P in problem 11.4 because, in this problem, the integration of all angles is calculated, and the orientation of the polar axis is irrelevant.