91Ó°ÊÓ

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

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

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

$S=\frac{{\mathrm{μ}}_0{\left|\stackrel{¨}{p}\left(t\right)\right|}^2}{16{\mathrm{π}}^{2}c}\frac{{\sin }^2\mathrm{θ}}{r^2}\hat{r}$ …… (2)

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

$P_{rad}\left(t_0\right)â‰\dots \frac{{\mathrm{μ}}_0}{6\mathrm{Ï€}c}{\left[\stackrel{¨}{p}\left(t_0\right)\right]}^2$ …… (3)

Take the double differentiation of equation (1).

$$\begin{gathered} \stackrel{Ë™}{p}\left(t\right)=-p_0\mathrm{Ó\neg }\sin \left(\mathrm{Ó\neg }t\right)\hat{x}+p_0\mathrm{Ó\neg }\cos \left(\mathrm{Ó\neg }t\right)\hat{y} \\ \stackrel{¨}{p}\left(t\right)=-p_0{\mathrm{Ó\neg }}^2\cos \left(\mathrm{Ó\neg }t\right)\hat{x}-p_0{\mathrm{Ó\neg }}^2\sin \left(\mathrm{Ó\neg }t\right)\hat{y} \\ \stackrel{¨}{p}\left(t\right)=-p_0{\mathrm{Ó\neg }}^2\left[\cos \left(\mathrm{Ó\neg }t\right)\hat{x}+\sin \left(\mathrm{Ó\neg }t\right)\hat{y}\right] \end{gathered}$$

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

$$\begin{gathered} S=\frac{{\mathrm{μ}}_0{\left|-p_0{\mathrm{Ó\neg }}^2\left[\cos \left(\mathrm{Ó\neg }t\right)\hat{x}+\sin \left(\mathrm{Ó\neg }t\right)\hat{y}\right]\right|}^2}{16{\mathrm{Ï€}}^{2}c}\frac{{\sin }^2\mathrm{θ}}{r^2}\hat{r} \\ S=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{16{\mathrm{Ï€}}^{2}c}\left({\cos }^2\mathrm{Ó\neg }t+{\sin }^2\mathrm{Ó\neg }t\right)\frac{{\sin }^2\mathrm{θ}}{r^2}\hat{r} \\ S=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{16{\mathrm{Ï€}}^{2}c}\frac{{\sin }^2\mathrm{θ}}{r^2}\hat{r} \end{gathered}$$

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 $\stackrel{¨}{p}\left(t_0\right)$.

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

$$\begin{gathered} P_{rad}\left(t_0\right)â‰\dots \frac{{\mathrm{μ}}_0}{6\mathrm{Ï€}c}{\left[-p_0{\mathrm{Ó\neg }}^2\left[\cos \left(\mathrm{Ó\neg }t\right)\hat{x}+\sin \left(\mathrm{Ó\neg }t\right)\hat{y}\right]\right]}^2 \\ {P}_{rad}\left(t_0\right)=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}c} \end{gathered}$$

For the case of total radiated power, the value of$P_{rad}\left(t_0\right)=\frac{{\mathrm{μ}}_0{p}_0^2{\mathrm{Ó\neg }}^4}{6\mathrm{Ï€}c}$ 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.