91Ó°ÊÓ

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

Write the formula of electric field is,

$\mathbf{∇}\mathbf{.}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left(r^2{E}_r\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{θ}}\left(sin\mathrm{θ}{E}_{\mathrm{θ}}\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}}\frac{\mathbf{∂}\mathbf{E}}{\mathbf{∂}\mathbf{ϕ}}$ …… (1)

Here,role="math" localid="1658485458510" $\mathbf{E}$is the electric field component of a spherical wave, role="math" localid="1658485452985" ${\mathit{E}}_{\mathbf{r}}$is electric field component of a spherical wave, $\mathit{r}$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.

$\mathbf{∇}\mathbf{×}\mathbf{E}\mathbf{=}\frac{\mathbf{-}\mathbf{∂}\mathbf{B}}{\mathbf{∂}\mathbf{r}}$ …… (2)

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

Write the formula of magnetic field.

$$\begin{gathered} \mathit{B}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{θ}}\left[\frac{Asi{n}^2\mathrm{θ}}{r}∫cosu-\frac{1}{kr}∫sinu\right]\stackrel{\mathbf{Áåœ}}{\mathbf{r}} \\ \mathbf{-}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left[Asin\mathrm{θ}\left(∫cosu-\frac{1}{kr}∫sinu\right)\right]\mathbf{}\stackrel{\mathbf{Áåœ}}{\mathbf{θ}}\mathbf{}\mathbf{} \end{gathered}$$ …… (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.

$\mathbf{∇}\mathbf{.}\mathbf{B}\mathbf{=}\mathbf{0}$ …… (4)

Here, isthe magnetic field strength.

Write the formula of Ampere’s law.

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

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

Write the formula of intensity vector.

$\mathbf{I}\mathbf{=}\left(S\right)$ …… (6)

Here, S is Poynting vector.

Write the formula of total power radiated.

$\mathbf{P}\mathbf{=}\mathbf{∫}\mathbf{I}\mathbf{.}\mathbf{da}$ …… (7)

Here,I is intensity vector.

From Gauss’s law,

$∇.E=\frac{p_f}{{\mathrm{Î\mu }}_0}$

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

Determine the divergence of electric field is,

Substitute$$\begin{gathered} 0\mathrm{for}{\mathrm{E}}_{\mathrm{r}},{\mathrm{E}}_{\mathrm{θ}}\mathrm{and}\frac{\mathrm{´¡²õ¾\pm ²Ôθ}}{\mathrm{r}}\left[\cos \left(\mathrm{kr}-\mathrm{Ó\neg t}\right)-\frac{1}{\mathrm{kr}}\sin \left(\mathrm{kr}-\mathrm{Ó\neg t}\right)\right]\mathrm{for}{\mathrm{E}}_{\mathrm{Ï•}} \\ ∇.\mathrm{E}=\frac{1}{{\mathrm{r}}^2}\frac{∂}{∂\mathrm{r}}\left(\begin{array}{c}{\mathrm{r}}^2\\ \left(0\right)\end{array}\right)+\frac{1}{\mathrm{rsin}\mathrm{θ}}\frac{∂}{∂\mathrm{r}}\left(\begin{array}{c}\mathrm{²õ¾\pm ²Ôθ}\\ \left(0\right)\end{array}\right)+\frac{1}{\mathrm{rsin}\mathrm{θ}}\frac{∂\left(\begin{array}{c}{\displaystyle \frac{\mathrm{´¡²õ¾\pm ²Ôθ}}{\mathrm{r}}}\\ \left[\cos \left(\mathrm{kr}-\mathrm{Ó\neg t}\right)-{\displaystyle \frac{1}{\mathrm{kr}}}\sin \left(\mathrm{kr}-\mathrm{Ó\neg t}\right)\right]\end{array}\right)}{∂\mathrm{Ï•}} \\ =\frac{1}{\mathrm{r}\sin \mathrm{θ}}\frac{∂\mathrm{E}}{∂\mathrm{Ï•}} \\ =0 \end{gathered}$$.

As there is no free charge density here, therefore $∇.E=0$.

Hence, Gauss’s law is obeyed.

According to Faraday’s Law.

Determine the curl of electric field is,

role="math" localid="1658487525637" $∇×E=\frac{1}{r\sin \mathrm{θ}}\left[\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}{E}_{\mathrm{Ï•}}\right)-\frac{∂E_{\mathrm{θ}}}{∂\mathrm{Ï•}}\right]\stackrel{Áåœ}{r}+\frac{1}{r}\left[\frac{∂}{∂r}\right]\left(r{E}_{\mathrm{Ï•}}\right)\stackrel{Áåœ}{\mathrm{Ï•}}+\frac{1}{r}\left[\frac{∂}{∂r}\left(r{E}_{\mathrm{θ}}\right)-\frac{∂E_r}{∂\mathrm{θ}}\right]\stackrel{Áåœ}{\mathrm{Ï•}}$

Therefore, the value of curl of electric field is

$∇×E=\frac{1}{r\sin \mathrm{θ}}\left[\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}{E}_{\mathrm{Ï•}}\right)-\frac{∂E_{\mathrm{θ}}}{∂\mathrm{Ï•}}\right]\stackrel{Áåœ}{r}+\frac{1}{r}\frac{∂}{∂r}\left(r{E}_{\mathrm{Ï•}}\right)\stackrel{Áåœ}{\mathrm{Ï•}}$

Substitute $\frac{-∂B}{∂t}$for $∇×E$.

$$\begin{gathered} \frac{-∂B}{∂t}=\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left[\frac{A{\sin }^2\mathrm{θ}}{r}\left(\cos u-\frac{1}{kr}\sin u\right)\right]\stackrel{Áåœ}{r} \\ -\frac{1}{r}\frac{∂}{∂t}\left[A\sin \mathrm{θ}\left(\cos u-\frac{1}{kr}-\frac{1}{kr}\sin u\right)\right]\stackrel{Áåœ}{\mathrm{θ}} \end{gathered}$$ …… (8)

Here,$u=(kr-\mathrm{Ó\neg }t)$.

Integrate equation (8),

$$\begin{gathered} B=\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left[\frac{A{\sin }^2\mathrm{θ}}{r}\left(∫\cos u-\frac{1}{kr}∫\sin u\right)\right]\stackrel{Áåœ}{r} \\ -\frac{1}{r}\frac{∂}{∂\mathrm{θ}}\left[A\sin \mathrm{θ}\left(∫\cos u-\frac{1}{kr}∫\sin u\right)\right]\stackrel{Áåœ}{\mathrm{θ}} \end{gathered}$$ …… (9)

Substitute role="math" localid="1658488205000" $-\frac{1}{\mathrm{Ó\neg }}\sin u∫\cos udt$for $∫\cos udt$and $\frac{1}{\mathrm{Ó\neg }}\cos u$for $∫\sin udt$into equation (9).

$B=\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(\sin u+\frac{1}{kr}\cos u\right)\stackrel{Áåœ}{r}+\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}-kcosu\frac{1}{k{r}^2}cosu+\frac{1}{r}sinu)\stackrel{Áåœ}{\mathrm{θ}}$

Therefore, the value of magnetic field is

$B=\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}sinu+\frac{1}{kr}cosu)\stackrel{Áåœ}{r}+\frac{A\sin \mathrm{θ}}{\mathrm{Ó\neg }r}-kcosu\frac{1}{k{r}^2}cosu+\frac{1}{r}sinu)\stackrel{Áåœ}{\mathrm{θ}}$

Determine the Gauss’s law of magnetism,

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

$$\begin{gathered} ∇.B=\frac{1}{r^2}\frac{∂}{∂r}\left(r^{2}B\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}{B}_{\mathrm{θ}}\right) \\ =\frac{1}{r^2}\frac{∂}{∂r}\left[\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }r}\left(\sin u+\frac{1}{kr}\cos u\right)\right] \\ +\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂r}\left[\frac{A{\sin }^2\mathrm{θ}}{\mathrm{Ó\neg }r}\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\right] \\ =\frac{1}{r^2}\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }}\left(k\cos u-\frac{1}{k{r}^2}\cos u-\frac{1}{r}\sin u\right) \\ +\frac{1}{r\sin \mathrm{θ}}\frac{2A\sin \mathrm{θ}\cos \mathrm{θ}}{\mathrm{Ó\neg }r}\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right) \end{gathered}$$

Solve further as

$$\begin{gathered} ∇.B=\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(k\cos u-\frac{1}{k{r}^2}\cos u-\frac{1}{r}\sin u-k\cos u+\frac{1}{r}\sin u\right) \\ =0 \end{gathered}$$

Hence, the Gauss law of magnetism is obeyed.

Determine the Ampere’s law,

As$\mathrm{σ}=0$, therefore,

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

Substitute $\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(\sin u+\frac{1}{kr}\cos u\right)\stackrel{Áåœ}{r}+\frac{A\sin \mathrm{θ}}{\mathrm{Ó\neg }r}\left(-k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\stackrel{Áåœ}{B}$for B .

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

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

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

Hence, Ampere’s law is obeyed.

Determine the Poynting vector is given by the following equation.

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

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

$$\begin{gathered} S=\frac{1}{{\mathrm{μ}}_0}\left[\frac{A\sin \mathrm{θ}}{r}\left[\cos u-\frac{1}{kr}\sin u\right]\stackrel{Áåœ}{\mathrm{Ï•}}\right]x \\ \left[\begin{array}{c}\frac{2A\cos \mathrm{θ}}{\mathrm{Ó\neg }{r}^2}\left(\sin u+\frac{1}{kr}\cos u\right)\stackrel{Áåœ}{r}\\ +\frac{A\sin \mathrm{θ}}{\mathrm{Ó\neg }r}\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\stackrel{Áåœ}{\mathrm{θ}}\end{array}\right] \\ =\frac{A^2\sin \mathrm{θ}}{{\mathrm{μ}}_0\mathrm{Ó\neg }{r}^2}\left\{\begin{array}{c}\frac{2\cos \mathrm{θ}}{r}\left[\sin u\cos u+\frac{1}{r}\left({\cos }^{2}u-{\sin }^{2}u\right)-\frac{1}{k^2{r}^2}\sin u\cos u\right]\stackrel{Áåœ}{\mathrm{θ}}\\ -\sin \left(\begin{array}{c}-k{\cos }^{2}u+\frac{1}{r}\sin u\cos u+\frac{1}{r}\sin u\cos u\\ -\frac{1}{k^2{r}^2}\sin u\cos u-\frac{1}{k{r}^2}{\sin }^{2}u\end{array}\right)\stackrel{Áåœ}{r}\end{array}\right\} \end{gathered}$$

Average over a full cycle is,

$$\begin{gathered} \left(\sin u\cos u\right)=0 \\ \left({\sin }^{2}u\right)=\left({\cos }^{2}u\right)=\frac{1}{2} \end{gathered}$$

Determine theIntensity vector.

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

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

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

Therefore, the intensity vector is$\frac{A^2{\sin }^2\mathrm{θ}}{2{\mathrm{μ}}_{0}c{r}^2}\stackrel{Áåœ}{r}$.

Determine the total power radiated is,

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

role="math" localid="1658492229808" $$\begin{gathered} P=\frac{A^2}{2{\mathrm{μ}}_{0}c}∫\frac{{\sin }^2\mathrm{θ}}{r^2}{r}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•} \\ =\frac{A^2}{2{\mathrm{μ}}_{0}c}2\mathrm{Ï€}{∫}_0^{\mathrm{Ï€}}{\sin }^2\mathrm{θ»åθ} \\ =\frac{4\mathrm{Ï€}}{3}\frac{{\mathrm{A}}^2}{{\mathrm{μ}}_0\mathrm{c}} \end{gathered}$$

Therefore, the value of total power radiated is $\frac{4\mathrm{π}}{3}\frac{{\mathrm{A}}^2}{{\mathrm{μ}}_0\mathrm{c}}$ .