Suppose

$\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{,}\mathbf{\varphi }\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{A}\frac{\mathbf{sin\theta }}{r}\mathbf{[}\mathbf{cos}\mathbf{(}\mathbf{kr}\mathbf{-}\mathbf{\omega t}\mathbf{)}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{kr}\mathbf{)}\mathbf{sin}\mathbf{(}\mathbf{kr}\mathbf{-}\mathbf{\omega t}\mathbf{)}\mathbf{]}\widehat{\mathbf{\varphi }}$

(This is, incidentally, the simplest possible spherical wave. For notational convenience, let role="math" localid="1658817164296" $\mathbf{(}\mathbf{kr}\mathbf{-}\mathbf{\omega t}\mathbf{)}\mathbf{\equiv }\mathbf{u}$ in your calculations.)

(a) Show that $\mathbf{E}$ obeys all four of Maxwell's equations, in vacuum, and find the associated magnetic field.

(b) Calculate the Poynting vector. Average $\mathbf{S}$ over a full cycle to get the intensity vector $\mathbf{I}$. (Does it point in the expected direction? Does it fall off like${\mathbf{r}}^{\mathbf{-}\mathbf{2}}$, as it should?)

(c) Integrate role="math" localid="1658817283737" $\mathbf{I}\mathbf{\cdot }\mathbf{da}$ over a spherical surface to determine the total power radiated. [Answer: $\mathbf{4}{\mathbf{\pi A}}^2\mathbf{/}\mathbf{3}{\mathbf{\mu }}_0\mathbf{c}$ ]

Consider this, incidentally, the simplest possible spherical wave. For notational convenience, let $(\text{kr}-\omega \text{t})\equiv \text{u}$in your calculations.

Write the formula of electric field is,

$\nabla \cdot \mathbf{E}=\frac{1}{{\mathbf{r}}^2}\frac{\partial }{\partial \mathbf{r}}\left({\mathbf{r}}^2{\mathbf{E}}_r\right)+\frac{1}{\mathbf{rsin}\mathrm{\theta }}\frac{\partial }{\partial \mathrm{\theta }}\left(\mathbf{sin}\mathrm{\theta }{\mathbf{{\rm E}}}_{\mathrm{\theta }}\right)+\frac{1}{\mathbf{rsin}\mathrm{\theta }}\frac{\partial {\mathbf{{\rm E}}}_{\mathrm{\varphi }}}{\partial \mathrm{\varphi }}$ …… (1)

Here, $\mathbf{E}$ is the electric field component of a spherical wave, ${\mathbf{E}}_{\mathbf{r}}$ is electric field component of a spherical wave, $\mathbf{r}$ is radius, ${\mathbf{E}}_{\mathbf{\theta }}$electric field component of a spherical wave and ${\mathbf{E}}_{\mathbf{\varphi }}$is electric field component of a spherical wave.

Write the formula of curl of electric field.

localid="1658817884404" $\nabla \times \mathbf{E}=\frac{-\partial \mathbf{B}}{\partial \mathbf{r}}$ …… (2)

Here, localid="1658817920962" $\mathbf{B}$ is the magnetic field strength and $\mathbf{r}$is radius.

Write the formula of magnetic field.

$\begin{array}{l}\mathbf{B}\mathbf{=}\frac{1}{\mathbf{rsin\theta }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\theta }}\left[\frac{{\mathrm{Asin}}^2\theta }{}\left(\int \mathrm{cosu}-\frac{}{}\int \mathrm{sinu}\right)\right]\widehat{\mathbf{r}}\\ \mathbf{-}\frac{1}{r}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left[\mathrm{Asin\theta }\left(\int \mathrm{cosu}-\frac{}{}\int \mathrm{sinu}\right)\right]\widehat{\mathbf{\theta }}\end{array}$ …… (3)

Here, $\mathbf{r}$ is radius, $\mathbf{k}$represent the wave number and $\mathbf{A}$ is constant.

Write the formula of Gauss law of magnetism.

localid="1658818098642" $\nabla \cdot \mathbf{B}=\mathbf{0}$ …… (4)

Here, $\mathbf{B}$isthe magnetic field strength.

Write the formula of Ampere’s law.

$\mathbf{\nabla }\mathbf{\times }\mathbf{B}\mathbf{=}\mathbf{\mu \sigma E}\mathbf{+}\frac{1}{{\mathbf{c}}^2}\frac{\mathbf{\partial }\mathbf{E}}{\mathbf{\partial }\mathbf{t}}$ …… (5)

Here, $\mathbf{E}$ is the electric field component of a spherical wave, $\mathbf{\mu }$ is permeability, $\mathbf{c}$denotes the speed of light.

Write the formula of intensity vector.

$\mathbf{I}\mathbf{=}\mathbf{⟨}\mathbf{S}\mathbf{⟩}$ …… (6)

Here, localid="1658818241979" $\mathbf{S}$ is Poynting vector.

Write the formula of total power radiated.

$\mathbf{P}\mathbf{=}\int I\cdot \mathrm{da}$ …… (7)

Here, $\mathbf{I}$ is intensity vector.

From Gauss’s law,

$\nabla \cdot \text{E}=\frac{{\rho }_f}{{\epsilon }_0}$

Here, ${\rho }_f$is the free charge density.

Determine the divergence of electric field is,

Substitute $0$ for $E_r,E_{\theta }$and $\frac{A\sin \theta }{r}\left[\cos (kr-\omega t)-\frac{1}{kr}\sin (kr-\omega t)\right]$ for $E_{\varphi }$.

$\begin{array}{c}\nabla \cdot \text{E}=\frac{1}{{\mathrm{r}}^2}\frac{\partial }{\partial \mathrm{r}}\left(\begin{array}{l}{\mathrm{r}}^2\\ \left(0\right)\end{array}\right)+\frac{1}{\mathrm{rsin\theta }}\frac{\partial }{\partial \mathrm{\theta }}\left(\begin{array}{l}\mathrm{sin\theta }\\ \left(0\right)\end{array}\right)+\frac{1}{\mathrm{rsin\theta }}\frac{\partial \left(\begin{array}{l}\frac{\mathrm{Asin\theta }}{\mathrm{r}}\\ \left[\cos (\mathrm{kr}-\mathrm{\omega t})-\frac{1}{\mathrm{kr}}\sin (\mathrm{kr}-\mathrm{\omega t})\right]\end{array}\right)}{\partial \mathrm{\varphi }}\\ =\frac{1}{\mathrm{rsin\theta }}\frac{\partial {\mathrm{E}}_{\mathrm{\varphi }}}{\partial \mathrm{\varphi }}\\ =0\end{array}$

As there is no free charge density here, therefore$\nabla \cdot \text{E}=0$.

Hence, Gauss’s law is obeyed.

According to Faraday’s Law.

Determine the curl of electric field is,

$\nabla \times \text{E}=\frac{1}{\mathrm{rsin\theta }}\left[\frac{\partial }{\partial \mathrm{\theta }}\left({\mathrm{sin\theta E}}_{\mathrm{\varphi }}\right)-\frac{\partial {\mathrm{E}}_{\mathrm{\theta }}}{\partial \mathrm{\varphi }}\right]\widehat{\mathrm{r}}+\frac{1}{\mathrm{r}}\left[\frac{1}{\mathrm{sin\theta }}\frac{\partial {\mathrm{E}}_{\mathrm{r}}}{\partial \mathrm{r}}\left({\mathrm{rE}}_{\mathrm{\varphi }}\right)\right]\widehat{\mathrm{\theta }}+\frac{1}{\mathrm{r}}\left[\frac{\partial }{\partial \mathrm{r}}\left({\mathrm{rE}}_{\mathrm{\theta }}\right)-\frac{\partial {\mathrm{E}}_{\mathrm{r}}}{\partial \mathrm{\theta }}\right]\widehat{\mathrm{\varphi }}$

Therefore, the value of curl of electric field is

$\nabla \times \text{E}=\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta E_{\varphi }\right)\widehat{r}-\frac{1}{r}\frac{\partial }{\partial r}\left(r{E}_{\varphi }\right)\widehat{\theta }$

Substitute $\frac{-\partial B}{\partial t}$ for $\nabla \times \text{E}$.

$\begin{array}{l}\frac{-\partial \mathrm{B}}{\partial \mathrm{t}}=\frac{1}{\mathrm{rsin\theta }}\frac{\partial }{\partial \mathrm{\theta }}\left[\frac{{\mathrm{Asin}}^2\mathrm{\theta }}{\mathrm{r}}\left(\mathrm{cosu}-\frac{1}{\mathrm{kr}}\mathrm{sinu}\right)\right]\widehat{\mathrm{r}}\\ -\frac{1}{\mathrm{r}}\frac{\partial }{\partial \mathrm{r}}\left[\mathrm{Asin\theta }\left(\mathrm{cosu}-\frac{1}{\mathrm{kr}}\mathrm{sinu}\right)\right]\widehat{\mathrm{\theta }}\end{array}$ …… (8)

Here, $\mathrm{u}=(\mathrm{kr}-\mathrm{\omega t})$.

Integrate equation (8),

$\begin{array}{l}\text{B}=\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left[\frac{A{\sin }^2\theta }{r}\left(\int \cos u-\frac{1}{kr}\int \sin u\right)\right]\widehat{r}\\ -\frac{1}{r}\frac{\partial }{\partial r}\left[A\sin \theta \left(\int \cos u-\frac{1}{kr}\int \sin u\right)\right]\widehat{\theta }\end{array}$ …… (9)

Substitute $-\frac{1}{\omega }\sin u$ for $\int \cos udt$and $\frac{1}{\omega }\cos u$ for $\int \sin udt$into equation (9).

$\text{B}=\frac{2A\cos \theta }{\omega r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\widehat{r}+\frac{A\sin \theta }{\omega r}\left(-k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\widehat{\theta }$

Therefore, the value of magnetic field is .

$\text{B}=\frac{2\mathrm{Acos\theta }}{{\mathrm{\omega r}}^2}\left(\mathrm{sinu}+\frac{1}{\mathrm{kr}}\mathrm{cosu}\right)\widehat{\mathrm{r}}+\frac{\mathrm{Asin\theta }}{\mathrm{\omega r}}\left(-\mathrm{kcosu}\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}+\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\widehat{\mathrm{\theta }}$

Determine the Gauss’s law of magnetism,

Substitute $\frac{2A\cos \theta }{\omega r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\widehat{r}+\frac{A\sin \theta }{\omega r}\left(-k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\widehat{\theta }$ for $B$ into equation (4).

$\begin{array}{c}\nabla \cdot \text{B}=\frac{1}{{\mathrm{r}}^2}\frac{\partial }{\partial \mathrm{r}}\left({\mathrm{r}}^2\mathrm{B}\right)+\frac{1}{\mathrm{rsin\theta }}\frac{\partial }{\partial \mathrm{\theta }}\left({\mathrm{sin\theta B}}_{\mathrm{\theta }}\right)\\ =\frac{1}{{\mathrm{r}}^2}\frac{\partial }{\partial \mathrm{r}}\left[\frac{2\mathrm{Acos\theta }}{\mathrm{\omega }}\left(\mathrm{sinu}+\frac{1}{\mathrm{kr}}\mathrm{cosu}\right)\right]\\ +\frac{1}{\mathrm{rsin\theta }}\frac{\partial }{\partial \mathrm{\theta }}\left[\frac{{\mathrm{Asin}}^2\mathrm{\theta }}{\mathrm{\omega r}}\left(-\mathrm{kcosu}+\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}+\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\right]\\ =\frac{1}{{\mathrm{r}}^2}\frac{2\mathrm{Acos\theta }}{\mathrm{\omega }}\left(\mathrm{kcosu}-\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}-\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\\ +\frac{1}{\mathrm{rsin\theta }}\frac{2\mathrm{Asin\theta cos\theta }}{\mathrm{\omega r}}\left(-\mathrm{kcosu}+\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}+\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\end{array}$

Solve further as

$\begin{array}{c}\nabla \cdot \text{B}=\frac{2A\cos \theta }{\omega r^2}\left(k\cos u-\frac{1}{k{r}^2}\cos u-\frac{1}{r}\sin u-k\cos u+\frac{1}{r}\sin u\right)\\ =0\end{array}$

Hence, the Gauss law of magnetism is obeyed.

Determine the Ampere’s law,

As $\mathrm{\sigma }=0$, therefore,

$\begin{array}{c}\nabla \times \text{B}=\frac{1}{c^2}\frac{\partial \text{E}}{\partial t}\\ =\frac{1}{r}\left[\frac{\partial }{\partial r}\left(r{B}_{\theta }\right)-\frac{\partial B}{\partial \theta }\right]\varphi \end{array}$

Substitute $\frac{2\mathrm{Acos\theta }}{{\mathrm{\omega r}}^2}\left(\mathrm{sinu}+\frac{1}{\mathrm{kr}}\mathrm{cosu}\right)\widehat{\mathrm{r}}+\frac{\mathrm{Asin\theta }}{\mathrm{\omega r}}\left(-\mathrm{kcosu}\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}+\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\widehat{\mathrm{\theta }}$ for $B$.

$\begin{array}{c}\nabla \times \text{B}=\frac{1}{r}\left\{\begin{array}{c}\frac{\partial }{\partial r}\left[\frac{A\sin \theta }{\omega }\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\right]\\ -\frac{\partial }{\partial \theta }\left[\left(\frac{2A\cos \theta }{\omega r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\right)\right]\widehat{\varphi }\end{array}\right\}\\ =\frac{k}{\omega }\frac{A\sin \theta }{r}\left(k\sin u+\frac{1}{r}\cos u\right)\widehat{\varphi }\\ =\frac{A\sin \theta }{cr}\left(k\sin u+\frac{1}{r}\cos u\right)\widehat{\varphi }\end{array}$

Solve the term $\frac{1}{c^2}\frac{\partial \text{E}}{\partial t}$,

$\begin{array}{c}\frac{1}{c^2}\frac{\partial \text{E}}{\partial t}=\frac{1}{c^2}\frac{A\sin \theta }{r}\left(\omega \sin u+\frac{\omega }{kr}\cos u\right)\widehat{\varphi }\\ =\frac{1}{c^2}\frac{\omega }{k}\frac{A\sin \theta }{r}\left(k\sin u+\frac{1}{r}\cos u\right)\varphi \\ =\frac{1}{c}\frac{A\sin \theta }{r}\left(k\sin u+\frac{1}{r}\cos u\right)\varphi \\ =\nabla \times \text{B}\end{array}$

Hence, Ampere’s law is obeyed.

Determine the Poynting vector is given by the following equation.

$S=\frac{1}{{\mu }_0}(\text{E}\times \text{B})$ …… (10)

Substitute $\frac{2A\cos \theta }{\omega r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\widehat{r}+\frac{A\sin \theta }{\omega r}\left(-k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\widehat{\theta }$ for $B$ and $\frac{A\sin \theta }{r}\left[\cos u-\frac{1}{kr}\sin u\right]\widehat{\varphi }$ for $E$ into above equation (10).

$\begin{array}{c}\text{S}=\frac{1}{{\mu }_0}\left[\frac{A\sin \theta }{r}\left[\cos u-\frac{1}{kr}\sin u\right]\widehat{\varphi }\right]\times \\ \left[\begin{array}{l}\frac{2A\cos \theta }{\omega r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\widehat{r}\\ +\frac{A\sin \theta }{\omega r}\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\widehat{\theta }\end{array}\right]\\ =\frac{A^2\sin \theta }{{\mu }_0\omega r^2}\left\{\begin{array}{l}\frac{2\cos \theta }{r}\left[\sin u\cos u+\frac{1}{kr}({\cos }^{2}u-{\sin }^{2}u)-\frac{1}{k^2{r}^2}\sin u\cos u\right]\widehat{\theta }\\ -\sin \theta \left(\begin{array}{l}-k{\cos }^{2}u+\frac{1}{k{r}^2}{\cos }^{2}u+\frac{1}{r}\sin u\cos u+\frac{1}{r}\sin u\cos u\\ -\frac{1}{k^2{r}^3}\sin u\cos u-\frac{1}{k{r}^2}{\sin }^{2}u\end{array}\right)\widehat{r}\end{array}\right\}\end{array}$

Average over a full cycle is,

$\begin{array}{c}⟨\sin u\cos u⟩=0\\ ⟨{\sin }^{2}u⟩=⟨{\cos }^{2}u⟩=\frac{1}{2}\end{array}$

Determine theIntensity vector.

Substitute $\frac{A^2\sin }{{\mu }_0\omega r^2}\left(\frac{k}{2}\sin \theta \right)\widehat{r}$ for $⟨\text{S}⟩$into equation (6).

$\begin{array}{c}\text{I}=\frac{A^2\sin }{{\mu }_0\omega r^2}\left(\frac{k}{2}\sin \theta \right)\widehat{r}\\ =\frac{A^2{\sin }^2\theta }{2{\mu }_{0}c{r}^2}\widehat{r}\end{array}$

The intensity fluctuates as $\frac{1}{r^2}$ and faces in the direction of $\widehat{r}$. A spherical wave is predicted to behave in this way.

Therefore, the intensity vector is $\frac{A^2{\sin }^2\theta }{2{\mu }_{0}c{r}^2}\widehat{r}$.

Determine the total power radiated is,

Substitute $\frac{A^2{\sin }^2\theta }{2{\mu }_{0}c{r}^2}\widehat{r}$ for $I$.

$\begin{array}{c}P=\frac{A^2}{2{\mu }_{0}c}\int \frac{{\sin }^2\theta }{r^2}{r}^2\sin \theta d\theta d\varphi \\ =\frac{A^2}{2{\mu }_{0}c}2\pi {\int }_0^{\pi }{\sin }^2\theta d\theta \\ =\frac{4\pi }{3}\frac{A^2}{{\mu }_{0}c}\end{array}$

Therefore, the value of total power radiated is $\frac{4\pi }{3}\frac{A^2}{{\mu }_{0}c}$.