A current $\mathbf{I}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$flows around the circular ring in Fig. 11.8. Derive the general formula for the power radiated (analogous to Eq. 11.60), expressing your answer in terms of the magnetic dipole moment, $\mathbf{m}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ , of the loop.

The given equation 11.60

${\mathrm{P}}_{\mathrm{rad}}\left({\mathrm{t}}_0\right)\cong \frac{{\mathrm{\mu }}_0}{6\mathrm{\pi c}}{\left[\ddot{\mathrm{p}}\left({\mathrm{t}}_0\right)\right]}^2$

The expression for the pointing vector is

$\mathbf{S}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\mu }}_{\mathbf{0}}}\mathbf{(}\mathbf{E}\mathbf{\times }\mathbf{B}\mathbf{)}$ ………. (1)

Here, ${\mathbf{\mu }}_{\mathbf{0}}$is the permeability of free space, $\mathbf{E}$is the electric field, $\mathbf{B}$is the magnetic field.

The retarded vector potential is

role="math" localid="1658736946261" $\mathbf{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\mathbf{\int }\frac{\mathbf{J}\left(t-\frac{r}{c}\right)}{\mathbf{r}}\mathbf{d\tau }\mathbf{\text{'}}$ ………. (2)

Here,$\mathbf{J}$is the current density, $\mathbf{C}$is the speed of the light, $\mathbf{t}$ is the time, and role="math" localid="1658737028413" $\mathbf{r}$is the position vector.

The change in current is

$\mathbf{dI}\mathbf{\text{'}}\mathbf{=}\mathbf{bd\varphi }\mathbf{\text{'}}\widehat{\mathbf{\varphi }}$ ………. (3)

Here, $\mathbf{b}$is the radius of the wire loop, and $\widehat{\mathbf{\varphi }}$ is the direction vector.

The maxwell equation for the electric field is

$\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{V}\mathbf{-}\frac{\mathbf{\partial }\mathbf{A}}{\mathbf{\partial }\mathbf{t}}$ ………. (4)

Here, $\mathbf{E}$ is the electric field, role="math" localid="1658737189047" $\mathbf{V}$is the voltage.

The maxwell equation for the magnetic field is

$\mathbf{B}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\mathbf{A}$ ………. (5)

Here, $\mathbf{\nabla }\mathbf{\times }\mathbf{A}$ is the curl of role="math" localid="1658737263083" $\mathbf{A}$.

The total radiated power is

role="math" localid="1658737288650" $\mathbf{P}\mathbf{=}\mathbf{\int }\mathbf{S}\mathbf{.}\mathbf{da}$ ………. (6)

Here,role="math" localid="1658737336377" $\mathbf{S}$is the pointing vector.

The magnetic dipole moment is

$\mathbf{\mu }\mathbf{=}\mathbf{IA}$ ………. (7)

Here, $\mathbf{I}$ is the current and$\mathbf{A}$ is the area of the ring.

From the equation (2)

$\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\int \frac{\mathrm{J}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}\mathrm{d\tau }\text{'}$

Rearrange the equation,

$\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\int \frac{\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}(\mathrm{Jd\tau }\text{'})$

As we know,

$\mathrm{Jd\tau }\text{'}=\mathrm{IdI}\text{'}$

Substitute $\mathrm{IdI}\text{'}$ for $\mathrm{Jd\tau }\text{'}$in the above equation.

$A(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\int \frac{\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)}{\mathrm{r}}(\mathrm{IdI}\text{'})$ ………. (8)

Here,

$\frac{1}{\mathrm{r}}=\frac{1}{\mathrm{r}}\left(1+\frac{\mathrm{b}}{\mathrm{r}}\mathrm{sin\theta cos\varphi }\text{'}\right)$

Now substitute $(-\mathrm{sin\varphi }\text{'}\widehat{\mathrm{x}}+\mathrm{cos\varphi }\text{'}\widehat{\mathrm{y}})$ for $\mathrm{d\varphi }\text{'}$ in the equation (3)

$\mathrm{dI}\text{'}=\mathrm{b}(-\mathrm{sin\varphi }\text{'}\widehat{\mathrm{x}}+\mathrm{cos\varphi }\text{'}\widehat{\mathrm{y}})\widehat{\mathrm{f}}$

Also,

$\mathrm{I}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)=\mathrm{I}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}+\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{sin\theta cos\varphi }\text{'}\right)$

Substitute ${\mathrm{t}}_0$for $\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}$ in the above equation.

$\begin{array}{c}\mathrm{I}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)=\mathrm{I}\left({\mathrm{t}}_0+\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{sin\theta cos\varphi }\text{'}\right)\\ =\mathrm{I}\left({\mathrm{t}}_0\right)+\dot{\mathrm{I}}\left({\mathrm{t}}_0\right)\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{sin\theta cos\varphi }\text{'}\\ =\mathrm{I}+\dot{\mathrm{I}}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{sin\theta cos\varphi }\text{'}\end{array}$

Substitute $\mathrm{I}+\dot{\mathrm{I}}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{sin\theta cos\varphi }\text{'}$for $\mathrm{I}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)$ in the equation (8)

$\begin{array}{c}\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\oint \frac{1}{\mathrm{r}}\left(1+\left(\frac{\mathrm{b}}{\mathrm{r}}\right)\mathrm{sin\theta cosf}\text{'}\right)\left((\mathrm{I}+\mathrm{I}\text{'}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{sin\theta cos\varphi }\text{'})\right)\mathrm{b}(-\mathrm{sin\varphi }\text{'}\widehat{\mathrm{x}}+\mathrm{cos\varphi }\text{'}\widehat{\mathrm{y}})\mathrm{d\varphi }\text{'}\\ =\frac{{\mathrm{\mu }}_0\mathrm{b}}{4\mathrm{\pi r}}\underset{0}{\overset{2\mathrm{\pi }}{\int }}\left(\left(\mathrm{I}+\mathrm{I}\text{'}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{sin\theta cosf}\text{'}\right)+\mathrm{I}\left(\frac{\mathrm{b}}{\mathrm{r}}\right)\mathrm{sin\theta cos\varphi }\text{'}\right)(-\mathrm{sin\varphi }\text{'}\widehat{\mathrm{x}}+\mathrm{cos\varphi }\text{'}\widehat{\mathrm{y}})\mathrm{d\varphi }\text{'}\end{array}$

From the trigonometry,

$\begin{array}{c}\underset{0}{\overset{2\mathrm{\pi }}{\int }}\mathrm{sin\varphi }\text{'}\mathrm{d\varphi }\text{'}=\underset{0}{\overset{2\mathrm{\pi }}{\int }}\mathrm{cos\varphi }\text{'}\mathrm{d\varphi }\text{'}=0\\ \underset{0}{\overset{2\mathrm{\pi }}{\int }}\mathrm{sin\varphi cos\varphi }\text{'}\mathrm{d\varphi }\text{'}=0\\ \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\cos }^2\mathrm{\varphi }\text{'}\mathrm{d\varphi }\text{'}=\mathrm{\pi }\end{array}$

Therefore, the equation (9) becomes

$\begin{array}{c}\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0\mathrm{b}}{4\mathrm{\pi r}}\mathrm{\pi }\widehat{\mathrm{y}}\left[\dot{\mathrm{I}}\left(\frac{\mathrm{b}}{\mathrm{c}}\right)\mathrm{sin\theta }+\mathrm{I}\left(\frac{\mathrm{b}}{\mathrm{r}}\right)\mathrm{sin\theta }\right]\\ \mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0{\mathrm{b}}^2}{4{\mathrm{r}}^2}\left[\dot{\mathrm{I}}\left(\frac{\mathrm{r}}{\mathrm{c}}\right)\mathrm{sin\theta }+\mathrm{Isin\theta }\right]\widehat{\mathrm{y}}\end{array}$

For points not lying (z-x) plane

$\widehat{y}\to \widehat{f}$

$\mathrm{A}(\mathrm{r},\mathrm{t})=\frac{{\mathrm{\mu }}_0{\mathrm{b}}^2}{4\mathrm{c}}\left[\dot{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\right]\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{f}}$ ………. (10)

From the equation (4)

$\mathrm{E}(\mathrm{r},\mathrm{t})=-\nabla \mathrm{V}-\frac{\partial \mathrm{A}}{\partial \mathrm{t}}$

As the voltage, $\mathrm{V}=0$, the electric field is

$\mathrm{E}(\mathrm{r},\mathrm{t})=-\frac{\partial \mathrm{A}}{\partial \mathrm{t}}$

Substitute equation (10) in the above expression.

$\begin{array}{c}\mathrm{E}(\mathrm{r},\mathrm{t})=\frac{\partial }{\partial \mathrm{t}}\left(\frac{{\mathrm{\mu }}_0{\mathrm{b}}^2}{4\mathrm{c}}\left[\dot{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\right]\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{\varphi }}\right)\\ \mathrm{E}(\mathrm{r},\mathrm{t})=\frac{-{\mathrm{\mu }}_0{\mathrm{b}}^2}{4\mathrm{c}}\frac{\partial }{\partial \mathrm{t}}\dot{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{\varphi }}\\ \mathrm{E}(\mathrm{r},\mathrm{t})=\frac{-{\mathrm{\mu }}_0{\mathrm{b}}^2}{4\mathrm{c}}\ddot{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{\varphi }}\end{array}$

From the equation (5)

$\mathrm{B}=\nabla \times \mathrm{A}$

The curl of the A is

$\nabla \times \mathrm{A}=\frac{1}{\mathrm{rsin\theta }}\frac{\partial }{\partial \mathrm{\theta }}\left({\mathrm{A}}_{\mathrm{\varphi }}\mathrm{sin\theta }\right)\widehat{\mathrm{r}}-\frac{1}{\mathrm{r}}\frac{\partial }{\partial \mathrm{r}}\left({\mathrm{rA}}_{\mathrm{\varphi }}\right)\widehat{\mathrm{\theta }}$

Substitute equation (10) in the above expression

Therefore, the magnetic field is$\frac{{\mathrm{\mu }}_0{\mathrm{b}}^2}{4{\mathrm{c}}^2}\ddot{\mathrm{I}}\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{\theta }}$

From the equation (1)

$\mathrm{S}=\frac{1}{{\mathrm{\mu }}_0}(\mathrm{E}\times \mathrm{B})$

Substitute$\frac{{\mathrm{\mu }}_0{\mathrm{b}}^2}{4{\mathrm{c}}^2}\ddot{\mathrm{I}}\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{\theta }}$ for $\mathrm{B}$ and $\frac{-{\mathrm{\mu }}_0{\mathrm{b}}^2}{4\mathrm{c}}\ddot{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{\varphi }}$ for $\mathrm{E}$

$\begin{array}{c}\mathrm{S}=\frac{1}{{\mathrm{\mu }}_0}\left(\frac{-{\mathrm{\mu }}_0{\mathrm{b}}^2}{4\mathrm{c}}\ddot{\mathrm{I}}\left(\mathrm{t}-\frac{\mathrm{r}}{\mathrm{c}}\right)\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{f}}2\times \frac{{\mathrm{\mu }}_0{\mathrm{b}}^2}{4{\mathrm{c}}^2}\ddot{\mathrm{I}}\frac{\mathrm{sin\theta }}{\mathrm{r}}\widehat{\mathrm{\theta }}\right)\\ \mathrm{S}=\frac{1}{{\mathrm{\mu }}_0\mathrm{c}}{\left(\frac{{\mathrm{\mu }}_0{\mathrm{b}}^2}{4\mathrm{c}}\ddot{\mathrm{I}}\frac{\mathrm{sin\theta }}{\mathrm{r}}\right)}^2(-\widehat{\mathrm{f}}\times \widehat{\mathrm{\theta }})\\ \mathrm{S}=\frac{{\mathrm{\mu }}_0\mathrm{b}}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\ddot{\mathrm{I}}\right)}^2\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{r}}^2}\widehat{\mathrm{r}}\end{array}$

The total radiated power is

$\mathrm{P}=\int \mathrm{S}.\mathrm{da}$

Substitute$\frac{{\mathrm{\mu }}_0\mathrm{b}}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\ddot{\mathrm{I}}\right)}^2\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{r}}^2}\widehat{\mathrm{r}}$ for$\mathrm{S}$.

$\begin{array}{c}\mathrm{P}=\int \mathrm{S}.\mathrm{da}\\ \mathrm{P}=\underset{\mathrm{f}=0}{\overset{2\mathrm{\pi }}{\int }}\underset{\mathrm{\theta }=0}{\overset{\mathrm{\pi }}{\int }}\frac{{\mathrm{\mu }}_0\mathrm{b}}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\ddot{\mathrm{I}}\right)}^2\frac{{\sin }^2\mathrm{\theta }}{{\mathrm{r}}^2}{\mathrm{r}}^2\mathrm{sin\theta d\theta df}\\ \mathrm{P}=2\mathrm{\pi }\frac{{\mathrm{\mu }}_0\mathrm{b}}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\ddot{\mathrm{I}}\right)}^2\underset{0}{\overset{\mathrm{\pi }}{\int }}{\sin }^2\mathrm{\theta d\theta }\\ \mathrm{P}=\frac{{\mathrm{\mu }}_0}{16{\mathrm{c}}^3}{\left({\mathrm{b}}^2\ddot{\mathrm{I}}\right)}^2\times 2\mathrm{\pi }\times \frac{4}{3}\end{array}$

Hence,

$\mathrm{P}=\frac{{\mathrm{\mu }}_0\mathrm{\pi }}{6{\mathrm{c}}^3}{\left({\mathrm{b}}^2\ddot{\mathrm{I}}\right)}^2$ ………… (11)

From the equation (7)

$\mathrm{\mu }=\mathrm{IA}$

Substitute ${\mathrm{\pi b}}^2$ for $\mathrm{A}$

$\mathrm{\mu }=\mathrm{I}\times \left({\mathrm{\pi b}}^2\right)$

Here, $I$is the current and $b$is the radius of the ring.

Differentiate the above equation twice.

$\begin{array}{c}\ddot{\mathrm{m}}=\ddot{\mathrm{I}}{\mathrm{\pi b}}^2\\ \frac{\ddot{\mathrm{m}}}{\mathrm{\pi }}=\ddot{\mathrm{I}}{\mathrm{b}}^2\end{array}$

Substitute $\frac{\ddot{\mathrm{m}}}{\mathrm{\pi }}$ for $\ddot{\mathrm{I}}{\mathrm{b}}^2$in the equation (11)

$\begin{array}{c}\mathrm{P}=\frac{{\mathrm{\mu }}_0\mathrm{\pi }}{6{\mathrm{c}}^3}{\left(\frac{\ddot{\mathrm{m}}}{\mathrm{\pi }}\right)}^2\\ \mathrm{P}=\frac{{\mathrm{\mu }}_0\mathrm{\pi }{\ddot{\mathrm{m}}}^2}{6{\mathrm{c}}^3\times {\mathrm{\pi }}^2}\\ \mathrm{P}=\frac{{\mathrm{\mu }}_0{\ddot{\mathrm{m}}}^2}{6{\mathrm{\pi c}}^3}\end{array}$

Thus, the general formula for the power radiated is $\mathrm{P}=\frac{{\mathrm{\mu }}_0{\ddot{\mathrm{m}}}^2}{6{\mathrm{\pi c}}^3}$.