91Ó°ÊÓ

Write the expression for the total radiated power.

$\mathit{P}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}{\stackrel{\mathbf{¨}}{\mathbf{p}}}^{\mathbf{2}}}{\mathbf{6}\mathbf{Π}\mathbf{c}}$ …… (1)

Here,${\mathrm{μ}}_0$ is the magnetic permeability, p is the dipole moment, and c is the speed of light.

Write the expression for the rotating dipole moment.

$p\left(t\right)=p_0\left[\cos \left(\mathrm{Ó\neg }t\right)\hat{y}-\sin \left(\mathrm{Ó\neg }t\right)\hat{x}\right]$

Here, $p_0$is the dipole moment in the oscillating electric dipole.

Take the double differentiation of the above equation.

$$\begin{gathered} \stackrel{Ë™}{p}\left(t\right)=p_0\left[-\mathrm{Ó\neg }\sin \left(\mathrm{Ó\neg }t\right)\hat{y}-\mathrm{Ó\neg }\cos \left(\mathrm{Ó\neg }t\right)\hat{x}\right] \\ \stackrel{¨}{p}\left(t\right)=p_0\left[-{\mathrm{Ó\neg }}^2\cos \left(\mathrm{Ó\neg }t\right)\hat{y}+{\mathrm{Ó\neg }}^2\sin \left(\mathrm{Ó\neg }t\right)\hat{x}\right] \\ \stackrel{¨}{p}\left(t\right)=-{\mathrm{Ó\neg }}^2{p}_0\left[\cos \left(\mathrm{Ó\neg }t\right)\hat{y}-\sin \left(\mathrm{Ó\neg }t\right)\hat{x}\right] \\ \stackrel{¨}{p}\left(t\right)=-{\mathrm{Ó\neg }}^{2}p\left(t\right) \end{gathered}$$

Squaring on both sides,

$$\begin{gathered} {\left[\stackrel{¨}{p}\left(t\right)\right]}^2={\left[-{\mathrm{Ó\neg }}^{2}p\left(t\right)\right]}^2 \\ {\left[\stackrel{¨}{p}\left(t\right)\right]}^2={\mathrm{Ó\neg }}^4{p}_{0........\left(2\right)}^2 \end{gathered}$$

Write the expression for the dipole moment of a circular ring that lies in the xy plane.

$p_0=∫\mathrm{λ}rdI$

Here,$\mathrm{λ}$ is the linear charge density.

Substitute $\mathrm{λ}={\mathrm{λ}}_0\sin \mathrm{ϕ},r=b\sin \mathrm{ϕ}\hat{y}+b\cos \mathrm{ϕ}\hat{x}$and$dI=bd\mathrm{ϕ}$ in the above expression.

$$\begin{gathered} p_0=∫\left({\mathrm{λ}}_0\sin \mathrm{ϕ}\right)\left(b\sin \mathrm{ϕ}\hat{y}+b\cos \mathrm{ϕ}\hat{x}\right)bd\mathrm{ϕ} \\ {p}_0=∫\left(b{\mathrm{λ}}_0{\sin }^2\mathrm{ϕ}\hat{y}+b{\mathrm{λ}}_0\sin \mathrm{ϕ}\cos \mathrm{ϕ}\hat{x}\right)bd\mathrm{ϕ} \\ {p}_0=b^2{\mathrm{λ}}_0{∫}_0^{2\mathrm{π}}{\sin }^2\mathrm{ϕ}d\mathrm{ϕ}\hat{y}+b^2{\mathrm{λ}}_0{∫}_0^{2\mathrm{π}}\sin \mathrm{ϕ}\cos \mathrm{ϕ}d\mathrm{ϕ}\hat{x} \\ {p}_0=b^2{\mathrm{λ}}_0\left[\hat{y}{∫}_0^{2\mathrm{π}}{\sin }^2\mathrm{ϕ}d\mathrm{ϕ}+\hat{x}{∫}_0^{2\mathrm{π}}\sin \mathrm{ϕ}\cos \mathrm{ϕ}d\mathrm{ϕ}\right] \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} p_0=b^2{\mathrm{λ}}_0\left[\hat{y}{∫}_0^{2\mathrm{π}}\left(\frac{1-\cos 2\mathrm{ϕ}}{2}\right)d\mathrm{ϕ}+\hat{x}{∫}_0^{2\mathrm{π}}\frac{1}{2}2\sin \mathrm{ϕ}\cos \mathrm{ϕ}d\mathrm{ϕ}\right] \\ {p}_0=b^2{\mathrm{λ}}_0\left[\hat{y}{∫}_0^{2\mathrm{π}}\left(\frac{1}{2}-\frac{\cos 2\mathrm{ϕ}}{2}\right)d\mathrm{ϕ}+\hat{x}{∫}_0^{2\mathrm{π}}\frac{1}{2}\left(\sin 2\mathrm{ϕ}\right)d\mathrm{ϕ}\right] \\ {p}_0=b^2{\mathrm{λ}}_0\left[\hat{y}{\left(\frac{\mathrm{ϕ}}{2}\right)}_0^{2\mathrm{π}}-{\left(\frac{\sin 2\mathrm{ϕ}}{4}\right)}_0^{2\mathrm{π}}+\hat{x}{\left[-\frac{\cos 2\mathrm{ϕ}}{4}\right]}_0^{2\mathrm{π}}\right] \\ {p}_0=b^2{\mathrm{λ}}_0\left[\hat{y}\left(\frac{2\mathrm{π}}{2}-0\right)-\left(\frac{\sin 4\mathrm{π}-\sin 0}{4}\right)+\hat{x}\left[-\frac{\cos 4\mathrm{π}-\cos 0}{4}\right]\right] \end{gathered}$$

Again on further solving,

$$\begin{gathered} p_0=b^2{\mathrm{λ}}_0\left[\hat{y}\left(\frac{2\mathrm{π}}{2}-0\right)-\left(0\right)+\hat{x}\left[-\frac{1-1}{4}\right]\right] \\ {p}_0=b^2{\mathrm{λ}}_0\left[\mathrm{π}\hat{y}+0\hat{x}\right] \\ {p}_0=\mathrm{π}{b}^2{\mathrm{λ}}_0\hat{y} \end{gathered}$$

Substitute$p_0=\mathrm{π}{b}^2{\mathrm{λ}}_0\hat{y}$ in equation (2).

$\begin{array}{rcl}{\left[\stackrel{¨}{p}\left(t\right)\right]}^2& =& {\mathrm{Ó\neg }}^4{\left(\mathrm{Ï€}{b}^2{\mathrm{λ}}_0\hat{y}\right)}^2\\ {\stackrel{¨}{p}}^2& =& {\mathrm{Ó\neg }}^4{\mathrm{Ï€}}^2{b}^4{\mathrm{λ}}_0^2{\hat{y}}^2\end{array}$

Substitute${\stackrel{¨}{p}}^2={\mathrm{Ó\neg }}^4{\mathrm{Ï€}}^2{\mathrm{b}}^4{\mathrm{λ}}_0^2{\hat{\mathrm{y}}}^2$ in equation (1).

$$\begin{gathered} P=\frac{{\mathrm{μ}}_0\left({\mathrm{Ó\neg }}^4{\mathrm{Ï€}}^2{b}^4{\mathrm{λ}}_0^2\right)}{6\mathrm{Ï€}c} \\ P=\frac{\mathrm{Ï€}{\mathrm{μ}}_0{\mathrm{Ó\neg }}^4{b}^4{\mathrm{λ}}_0^2}{6c} \end{gathered}$$

Therefore, the power radiated is $P=\frac{\mathrm{Ï€}{\mathrm{μ}}_0{\mathrm{Ó\neg }}^4{b}^4{\mathrm{λ}}_0^2}{6c}$.