91Ó°ÊÓ

Consider this, incidentally, the simplest possible spherical wave. For notational convenience, let $(\text{kr}−\mathrm{Ó\neg }\text{t})≡\text{u}$in your calculations.

Write the formula of electric field is,

$∇â‹\dots \mathbf{E}=\frac{1}{{\mathbf{r}}^2}\frac{∂}{∂\mathbf{r}}\left({\mathbf{r}}^2{\mathbf{E}}_r\right)+\frac{1}{\mathbf{rsin}\mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\mathbf{sin}\mathrm{θ}{\mathbf{Ε}}_{\mathrm{θ}}\right)+\frac{1}{\mathbf{rsin}\mathrm{θ}}\frac{∂{\mathbf{Ε}}_{\mathrm{Ï•}}}{∂\mathrm{Ï•}}$ …… (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, ${\mathbf{E}}_{\mathbf{θ}}$electric field component of a spherical wave and ${\mathbf{E}}_{\mathbf{ϕ}}$is electric field component of a spherical wave.

Write the formula of curl of electric field.

localid="1658817884404" $∇×\mathbf{E}=\frac{−∂\mathbf{B}}{∂\mathbf{r}}$ …… (2)

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

Write the formula of magnetic field.

$\begin{array}{l}\mathbf{B}\mathbf{=}\frac{1}{\mathbf{r²õ¾\pm ²Ôθ}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{θ}}\left[\frac{{\mathrm{Asin}}^2\mathrm{θ}}{}\left(∫\mathrm{cosu}−\frac{}{}∫\mathrm{sinu}\right)\right]\widehat{\mathbf{r}}\\ \mathbf{−}\frac{1}{r}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left[\mathrm{A²õ¾\pm ²Ôθ}\left(∫\mathrm{cosu}−\frac{}{}∫\mathrm{sinu}\right)\right]\widehat{\mathbf{θ}}\end{array}$ …… (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" $∇â‹\dots \mathbf{B}=\mathbf{0}$ …… (4)

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

Write the formula of Ampere’s law.

$\mathbf{∇}\mathbf{×}\mathbf{B}\mathbf{=}\mathbf{μσ·¡}\mathbf{+}\frac{1}{{\mathbf{c}}^2}\frac{\mathbf{∂}\mathbf{E}}{\mathbf{∂}\mathbf{t}}$ …… (5)

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

Write the formula of intensity vector.

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

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

Write the formula of total power radiated.

$\mathbf{P}\mathbf{=}∫Iâ‹\dots \mathrm{da}$ …… (7)

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

From Gauss’s law,

$∇â‹\dots \text{E}=\frac{{\mathrm{ÒÏ}}_f}{{\mathrm{Î\mu }}_0}$

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

Determine the divergence of electric field is,

Substitute $0$ for $E_r,E_{\mathrm{θ}}$and $\frac{A\sin \mathrm{θ}}{r}\left[\cos (kr−\mathrm{Ó\neg }t)−\frac{1}{kr}\sin (kr−\mathrm{Ó\neg }t)\right]$ for $E_{\mathrm{Ï•}}$.

$\begin{array}{c}∇â‹\dots \text{E}=\frac{1}{{\mathrm{r}}^2}\frac{∂}{∂\mathrm{r}}\left(\begin{array}{l}{\mathrm{r}}^2\\ \left(0\right)\end{array}\right)+\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\frac{∂}{∂\mathrm{θ}}\left(\begin{array}{l}\mathrm{²õ¾\pm ²Ôθ}\\ \left(0\right)\end{array}\right)+\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\frac{∂\left(\begin{array}{l}\frac{\mathrm{A²õ¾\pm ²Ôθ}}{\mathrm{r}}\\ \left[\cos (\mathrm{kr}−\mathrm{Ó\neg ³Ù})−\frac{1}{\mathrm{kr}}\sin (\mathrm{kr}−\mathrm{Ó\neg ³Ù})\right]\end{array}\right)}{∂\mathrm{Ï•}}\\ =\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\frac{∂{\mathrm{E}}_{\mathrm{Ï•}}}{∂\mathrm{Ï•}}\\ =0\end{array}$

As there is no free charge density here, therefore$∇â‹\dots \text{E}=0$.

Hence, Gauss’s law is obeyed.

According to Faraday’s Law.

Determine the curl of electric field is,

$∇×\text{E}=\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\left[\frac{∂}{∂\mathrm{θ}}\left({\mathrm{²õ¾\pm ²ÔθE}}_{\mathrm{Ï•}}\right)−\frac{∂{\mathrm{E}}_{\mathrm{θ}}}{∂\mathrm{Ï•}}\right]\widehat{\mathrm{r}}+\frac{1}{\mathrm{r}}\left[\frac{1}{\mathrm{²õ¾\pm ²Ôθ}}\frac{∂{\mathrm{E}}_{\mathrm{r}}}{∂\mathrm{r}}\left({\mathrm{rE}}_{\mathrm{Ï•}}\right)\right]\widehat{\mathrm{θ}}+\frac{1}{\mathrm{r}}\left[\frac{∂}{∂\mathrm{r}}\left({\mathrm{rE}}_{\mathrm{θ}}\right)−\frac{∂{\mathrm{E}}_{\mathrm{r}}}{∂\mathrm{θ}}\right]\widehat{\mathrm{Ï•}}$

Therefore, the value of curl of electric field is

$∇×\text{E}=\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}{E}_{\mathrm{ϕ}}\right)\widehat{r}−\frac{1}{r}\frac{∂}{∂r}\left(r{E}_{\mathrm{ϕ}}\right)\widehat{\mathrm{θ}}$

Substitute $\frac{−∂B}{∂t}$ for $∇×\text{E}$.

$\begin{array}{l}\frac{−∂\mathrm{B}}{∂\mathrm{t}}=\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\frac{∂}{∂\mathrm{θ}}\left[\frac{{\mathrm{Asin}}^2\mathrm{θ}}{\mathrm{r}}\left(\mathrm{cosu}−\frac{1}{\mathrm{kr}}\mathrm{sinu}\right)\right]\widehat{\mathrm{r}}\\ −\frac{1}{\mathrm{r}}\frac{∂}{∂\mathrm{r}}\left[\mathrm{A²õ¾\pm ²Ôθ}\left(\mathrm{cosu}−\frac{1}{\mathrm{kr}}\mathrm{sinu}\right)\right]\widehat{\mathrm{θ}}\end{array}$ …… (8)

Here, $\mathrm{u}=(\mathrm{kr}−\mathrm{Ó\neg ³Ù})$.

Integrate equation (8),

$\begin{array}{l}\text{B}=\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left[\frac{A{\sin }^2\mathrm{θ}}{r}\left(∫\cos u−\frac{1}{kr}∫\sin u\right)\right]\widehat{r}\\ −\frac{1}{r}\frac{∂}{∂r}\left[A\sin \mathrm{θ}\left(∫\cos u−\frac{1}{kr}∫\sin u\right)\right]\widehat{\mathrm{θ}}\end{array}$ …… (9)

Substitute $−\frac{1}{\mathrm{Ó\neg }}\sin u$ for $∫\cos u\text{ }dt$and $\frac{1}{\mathrm{Ó\neg }}\cos u$ for $∫\sin u\text{ }dt$into equation (9).

$\text{B}=\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(\sin u+\frac{1}{kr}\cos u\right)\widehat{r}+\frac{A\sin \mathrm{θ}}{\mathrm{Ó\neg }r}\left(−k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\widehat{\mathrm{θ}}$

Therefore, the value of magnetic field is .

$\text{B}=\frac{2\mathrm{´¡³¦´Ç²õθ}}{{\mathrm{Ó\neg °ù}}^2}\left(\mathrm{sinu}+\frac{1}{\mathrm{kr}}\mathrm{cosu}\right)\widehat{\mathrm{r}}+\frac{\mathrm{A²õ¾\pm ²Ôθ}}{\mathrm{Ó\neg °ù}}\left(−\mathrm{kcosu}\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}+\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\widehat{\mathrm{θ}}$

Determine the Gauss’s law of magnetism,

Substitute $\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(\sin u+\frac{1}{kr}\cos u\right)\widehat{r}+\frac{A\sin \mathrm{θ}}{\mathrm{Ó\neg }r}\left(−k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\widehat{\mathrm{θ}}$ for $B$ into equation (4).

$\begin{array}{c}∇â‹\dots \text{B}=\frac{1}{{\mathrm{r}}^2}\frac{∂}{∂\mathrm{r}}\left({\mathrm{r}}^2\mathrm{B}\right)+\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\frac{∂}{∂\mathrm{θ}}\left({\mathrm{²õ¾\pm ²ÔθB}}_{\mathrm{θ}}\right)\\ =\frac{1}{{\mathrm{r}}^2}\frac{∂}{∂\mathrm{r}}\left[\frac{2\mathrm{´¡³¦´Ç²õθ}}{\mathrm{Ó\neg }}\left(\mathrm{sinu}+\frac{1}{\mathrm{kr}}\mathrm{cosu}\right)\right]\\ +\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\frac{∂}{∂\mathrm{θ}}\left[\frac{{\mathrm{Asin}}^2\mathrm{θ}}{\mathrm{Ó\neg °ù}}\left(−\mathrm{kcosu}+\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}+\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\right]\\ =\frac{1}{{\mathrm{r}}^2}\frac{2\mathrm{´¡³¦´Ç²õθ}}{\mathrm{Ó\neg }}\left(\mathrm{kcosu}−\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}−\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\\ +\frac{1}{\mathrm{r²õ¾\pm ²Ôθ}}\frac{2\mathrm{A²õ¾\pm ²Ô賦´Ç²õθ}}{\mathrm{Ó\neg °ù}}\left(−\mathrm{kcosu}+\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}+\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\end{array}$

Solve further as

$\begin{array}{c}∇â‹\dots \text{B}=\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(k\cos u−\frac{1}{k{r}^2}\cos u−\frac{1}{r}\sin u−k\cos u+\frac{1}{r}\sin u\right)\\ =0\end{array}$

Hence, the Gauss law of magnetism is obeyed.

Determine the Ampere’s law,

As $\mathrm{σ}=0$, therefore,

$\begin{array}{c}∇×\text{B}=\frac{1}{c^2}\frac{∂\text{E}}{∂t}\\ =\frac{1}{r}\left[\frac{∂}{∂r}\left(r{B}_{\mathrm{θ}}\right)−\frac{∂B}{∂\mathrm{θ}}\right]\mathrm{ϕ}\end{array}$

Substitute $\frac{2\mathrm{´¡³¦´Ç²õθ}}{{\mathrm{Ó\neg °ù}}^2}\left(\mathrm{sinu}+\frac{1}{\mathrm{kr}}\mathrm{cosu}\right)\widehat{\mathrm{r}}+\frac{\mathrm{A²õ¾\pm ²Ôθ}}{\mathrm{Ó\neg °ù}}\left(−\mathrm{kcosu}\frac{1}{{\mathrm{kr}}^2}\mathrm{cosu}+\frac{1}{\mathrm{r}}\mathrm{sinu}\right)\widehat{\mathrm{θ}}$ for $B$.

$\begin{array}{c}∇×\text{B}=\frac{1}{r}\left\{\begin{array}{c}\frac{∂}{∂r}\left[\frac{A\sin \mathrm{θ}}{\mathrm{Ó\neg }}\left(−k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\right]\\ −\frac{∂}{∂\mathrm{θ}}\left[\left(\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(\sin u+\frac{1}{kr}\cos u\right)\right)\right]\widehat{\mathrm{Ï•}}\end{array}\right\}\\ =\frac{k}{\mathrm{Ó\neg }}\frac{A\sin \mathrm{θ}}{r}\left(k\sin u+\frac{1}{r}\cos u\right)\widehat{\mathrm{Ï•}}\\ =\frac{A\sin \mathrm{θ}}{cr}\left(k\sin u+\frac{1}{r}\cos u\right)\widehat{\mathrm{Ï•}}\end{array}$

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

$\begin{array}{c}\frac{1}{c^2}\frac{∂\text{E}}{∂t}=\frac{1}{c^2}\frac{A\sin \mathrm{θ}}{r}\left(\mathrm{Ó\neg }\sin u+\frac{\mathrm{Ó\neg }}{kr}\cos u\right)\widehat{\mathrm{Ï•}}\\ =\frac{1}{c^2}\frac{\mathrm{Ó\neg }}{k}\frac{A\sin \mathrm{θ}}{r}\left(k\sin u+\frac{1}{r}\cos u\right)\mathrm{Ï•}\\ =\frac{1}{c}\frac{A\sin \mathrm{θ}}{r}\left(k\sin u+\frac{1}{r}\cos u\right)\mathrm{Ï•}\\ =∇×\text{B}\end{array}$

Hence, Ampere’s law is obeyed.

Determine the Poynting vector is given by the following equation.

$S=\frac{1}{{\mathrm{μ}}_0}(\text{E}×\text{B})$ …… (10)

Substitute $\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(\sin u+\frac{1}{kr}\cos u\right)\widehat{r}+\frac{A\sin \mathrm{θ}}{\mathrm{Ó\neg }r}\left(−k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\widehat{\mathrm{θ}}$ for $B$ and $\frac{A\sin \mathrm{θ}}{r}\left[\cos u−\frac{1}{kr}\sin u\right]\widehat{\mathrm{Ï•}}$ for $E$ into above equation (10).

$\begin{array}{c}\text{S}=\frac{1}{{\mathrm{μ}}_0}\left[\frac{A\sin \mathrm{θ}}{r}\left[\cos u−\frac{1}{kr}\sin u\right]\widehat{\mathrm{Ï•}}\right]×\\ \left[\begin{array}{l}\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(\sin u+\frac{1}{kr}\cos u\right)\widehat{r}\\ +\frac{A\sin \mathrm{θ}}{\mathrm{Ó\neg }r}\left(−k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\widehat{\mathrm{θ}}\end{array}\right]\\ =\frac{A^2\sin \mathrm{θ}}{{\mathrm{μ}}_0\mathrm{Ó\neg }{r}^2}\left\{\begin{array}{l}\frac{2\cos \mathrm{θ}}{r}\left[\sin u\cos u+\frac{1}{kr}({\cos }^{2}u−{\sin }^{2}u)−\frac{1}{k^2{r}^2}\sin u\cos u\right]\widehat{\mathrm{θ}}\\ −\sin \mathrm{θ}\left(\begin{array}{l}−k{\cos }^{2}u+\frac{1}{k{r}^2}{\cos }^{2}u+\frac{1}{r}\sin u\cos u+\frac{1}{r}\sin u\cos u\\ −\frac{1}{k^2{r}^3}\sin u\cos u−\frac{1}{k{r}^2}{\sin }^{2}u\end{array}\right)\widehat{r}\end{array}\right\}\end{array}$

Average over a full cycle is,

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

Determine theIntensity vector.

Substitute $\frac{A^2\sin }{{\mathrm{μ}}_0\mathrm{Ó\neg }{r}^2}\left(\frac{k}{2}\sin \mathrm{θ}\right)\widehat{r}$ for $⟨\text{S}âŸ\copyright$into equation (6).

$\begin{array}{c}\text{I}=\frac{A^2\sin }{{\mathrm{μ}}_0\mathrm{Ó\neg }{r}^2}\left(\frac{k}{2}\sin \mathrm{θ}\right)\widehat{r}\\ =\frac{A^2{\sin }^2\mathrm{θ}}{2{\mathrm{μ}}_{0}c{r}^2}\widehat{r}\end{array}$

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 $\frac{A^2{\sin }^2\mathrm{θ}}{2{\mathrm{μ}}_{0}c{r}^2}\widehat{r}$.

Determine the total power radiated is,

Substitute $\frac{A^2{\sin }^2\mathrm{θ}}{2{\mathrm{μ}}_{0}c{r}^2}\widehat{r}$ for $I$.

$\begin{array}{c}P=\frac{A^2}{2{\mathrm{μ}}_{0}c}∫\frac{{\sin }^2\mathrm{θ}}{r^2}{r}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ}\\ =\frac{A^2}{2{\mathrm{μ}}_{0}c}2\mathrm{π}{∫}_0^{\mathrm{π}}{\sin }^2\mathrm{θ}d\mathrm{θ}\\ =\frac{4\mathrm{π}}{3}\frac{A^2}{{\mathrm{μ}}_{0}c}\end{array}$

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