An insulating circular ring (radius b) lies in the xy plane, centered at the origin. It carries a linear charge density $\mathit{\lambda }\mathbf{=}{\mathit{\lambda }}_{\mathbf{0}}\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi }$, where${\mathit{\lambda }}_{\mathbf{0}}$ is constant and$\mathit{\varphi }$ is the usual azimuthal angle. The ring is now set spinning at a constant angular velocity ω about the z axis. Calculate the power radiated

Write the expression for the total radiated power.

$\mathit{P}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\ddot{\mathbf{p}}}^{\mathbf{2}}}{\mathbf{6}\mathbf{\Pi }\mathbf{c}}$ …… (1)

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

Write the expression for the rotating dipole moment.

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

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

Take the double differentiation of the above equation.

$$\begin{gathered} \dot{p}\left(t\right)=p_0\left[-\omega \sin \left(\omega t\right)\hat{y}-\omega \cos \left(\omega t\right)\hat{x}\right] \\ \ddot{p}\left(t\right)=p_0\left[-{\omega }^2\cos \left(\omega t\right)\hat{y}+{\omega }^2\sin \left(\omega t\right)\hat{x}\right] \\ \ddot{p}\left(t\right)=-{\omega }^2{p}_0\left[\cos \left(\omega t\right)\hat{y}-\sin \left(\omega t\right)\hat{x}\right] \\ \ddot{p}\left(t\right)=-{\omega }^{2}p\left(t\right) \end{gathered}$$

Squaring on both sides,

$$\begin{gathered} {\left[\ddot{p}\left(t\right)\right]}^2={\left[-{\omega }^{2}p\left(t\right)\right]}^2 \\ {\left[\ddot{p}\left(t\right)\right]}^2={\omega }^4{p}_{0........\left(2\right)}^2 \end{gathered}$$

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

$p_0=\int \lambda rdI$

Here,$\lambda$ is the linear charge density.

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

$$\begin{gathered} p_0=\int \left({\lambda }_0\sin \varphi \right)\left(b\sin \varphi \hat{y}+b\cos \varphi \hat{x}\right)bd\varphi \\ {p}_0=\int \left(b{\lambda }_0{\sin }^2\varphi \hat{y}+b{\lambda }_0\sin \varphi \cos \varphi \hat{x}\right)bd\varphi \\ {p}_0=b^2{\lambda }_0{\int }_0^{2\pi }{\sin }^2\varphi d\varphi \hat{y}+b^2{\lambda }_0{\int }_0^{2\pi }\sin \varphi \cos \varphi d\varphi \hat{x} \\ {p}_0=b^2{\lambda }_0\left[\hat{y}{\int }_0^{2\pi }{\sin }^2\varphi d\varphi +\hat{x}{\int }_0^{2\pi }\sin \varphi \cos \varphi d\varphi \right] \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} p_0=b^2{\lambda }_0\left[\hat{y}{\int }_0^{2\pi }\left(\frac{1-\cos 2\varphi }{2}\right)d\varphi +\hat{x}{\int }_0^{2\pi }\frac{1}{2}2\sin \varphi \cos \varphi d\varphi \right] \\ {p}_0=b^2{\lambda }_0\left[\hat{y}{\int }_0^{2\pi }\left(\frac{1}{2}-\frac{\cos 2\varphi }{2}\right)d\varphi +\hat{x}{\int }_0^{2\pi }\frac{1}{2}\left(\sin 2\varphi \right)d\varphi \right] \\ {p}_0=b^2{\lambda }_0\left[\hat{y}{\left(\frac{\varphi }{2}\right)}_0^{2\pi }-{\left(\frac{\sin 2\varphi }{4}\right)}_0^{2\pi }+\hat{x}{\left[-\frac{\cos 2\varphi }{4}\right]}_0^{2\pi }\right] \\ {p}_0=b^2{\lambda }_0\left[\hat{y}\left(\frac{2\pi }{2}-0\right)-\left(\frac{\sin 4\pi -\sin 0}{4}\right)+\hat{x}\left[-\frac{\cos 4\pi -\cos 0}{4}\right]\right] \end{gathered}$$

Again on further solving,

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

Substitute$p_0=\pi b^2{\lambda }_0\hat{y}$ in equation (2).

$\begin{array}{rcl}{\left[\ddot{p}\left(t\right)\right]}^2& =& {\omega }^4{\left(\pi b^2{\lambda }_0\hat{y}\right)}^2\\ {\ddot{p}}^2& =& {\omega }^4{\pi }^2{b}^4{\lambda }_0^2{\hat{y}}^2\end{array}$

Substitute${\ddot{p}}^2={\omega }^4{\mathrm{\pi }}^2{\mathrm{b}}^4{\mathrm{\lambda }}_0^2{\hat{\mathrm{y}}}^2$ in equation (1).

$$\begin{gathered} P=\frac{{\mu }_0\left({\omega }^4{\pi }^2{b}^4{\lambda }_0^2\right)}{6\pi c} \\ P=\frac{\pi {\mu }_0{\omega }^4{b}^4{\lambda }_0^2}{6c} \end{gathered}$$

Therefore, the power radiated is $P=\frac{\pi {\mu }_0{\omega }^4{b}^4{\lambda }_0^2}{6c}$.