91影视

Consider this, incidentally, the simplest possible spherical wave. For notational convenience, let $\left(kr-蝇t\right)鈮�u$in your calculations.

Write the formula of electric field is,

$\mathbf{鈭�}\mathbf{.}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}}\left(r^2{E}_r\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{胃}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{胃}}\left(sin胃E_{胃}\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{胃}}\frac{\mathbf{鈭�}\mathbf{E}}{\mathbf{鈭�}\mathbf{蠒}}$ 鈥︹�� (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, ${\mathbf{E}}_{\mathbf{胃}}$electric field component of a spherical wave and${\mathbf{E}}_{\mathbf{蠒}}$is electric field component of a spherical wave.

Write the formula of curl of electric field.

$\mathbf{鈭�}\mathbf{脳}\mathbf{E}\mathbf{=}\frac{\mathbf{-}\mathbf{鈭�}\mathbf{B}}{\mathbf{鈭�}\mathbf{r}}$ 鈥︹�� (2)

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

Write the formula of magnetic field.

$$\begin{gathered} \mathit{B}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{胃}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{胃}}\left[\frac{Asi{n}^2胃}{r}鈭�cosu-\frac{1}{kr}鈭�sinu\right]\stackrel{\mathbf{铃�}}{\mathbf{r}} \\ \mathbf{-}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}}\left[Asin胃\left(鈭�cosu-\frac{1}{kr}鈭�sinu\right)\right]\mathbf{}\stackrel{\mathbf{铃�}}{\mathbf{胃}}\mathbf{}\mathbf{} \end{gathered}$$ 鈥︹�� (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.

$\mathbf{鈭�}\mathbf{.}\mathbf{B}\mathbf{=}\mathbf{0}$ 鈥︹�� (4)

Here, isthe magnetic field strength.

Write the formula of Ampere鈥檚 law.

$\mathbf{鈭�}\mathbf{脳}\mathbf{B}\mathbf{=}\mathbf{渭蟽贰}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\frac{\mathbf{鈭�}\mathbf{E}}{\mathbf{鈭�}\mathbf{t}}$ 鈥︹�� (5)

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

Write the formula of intensity vector.

$\mathbf{I}\mathbf{=}\left(S\right)$ 鈥︹�� (6)

Here, S is Poynting vector.

Write the formula of total power radiated.

$\mathbf{P}\mathbf{=}\mathbf{鈭�}\mathbf{I}\mathbf{.}\mathbf{da}$ 鈥︹�� (7)

Here,I is intensity vector.

From Gauss鈥檚 law,

$鈭�.E=\frac{p_f}{{蔚}_0}$

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

Determine the divergence of electric field is,

Substitute$$\begin{gathered} 0\mathrm{for}{\mathrm{E}}_{\mathrm{r}},{\mathrm{E}}_{\mathrm{胃}}\mathrm{and}\frac{\mathrm{础蝉颈苍胃}}{\mathrm{r}}\left[\cos \left(\mathrm{kr}-\mathrm{蝇t}\right)-\frac{1}{\mathrm{kr}}\sin \left(\mathrm{kr}-\mathrm{蝇t}\right)\right]\mathrm{for}{\mathrm{E}}_{\mathrm{蠒}} \\ 鈭�.\mathrm{E}=\frac{1}{{\mathrm{r}}^2}\frac{鈭�}{鈭�\mathrm{r}}\left(\begin{array}{c}{\mathrm{r}}^2\\ \left(0\right)\end{array}\right)+\frac{1}{\mathrm{rsin}\mathrm{胃}}\frac{鈭�}{鈭�\mathrm{r}}\left(\begin{array}{c}\mathrm{蝉颈苍胃}\\ \left(0\right)\end{array}\right)+\frac{1}{\mathrm{rsin}\mathrm{胃}}\frac{鈭�\left(\begin{array}{c}{\displaystyle \frac{\mathrm{础蝉颈苍胃}}{\mathrm{r}}}\\ \left[\cos \left(\mathrm{kr}-\mathrm{蝇t}\right)-{\displaystyle \frac{1}{\mathrm{kr}}}\sin \left(\mathrm{kr}-\mathrm{蝇t}\right)\right]\end{array}\right)}{鈭�\mathrm{蠒}} \\ =\frac{1}{\mathrm{r}\sin \mathrm{胃}}\frac{鈭�\mathrm{E}}{鈭�\mathrm{蠒}} \\ =0 \end{gathered}$$.

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

Hence, Gauss鈥檚 law is obeyed.

According to Faraday鈥檚 Law.

Determine the curl of electric field is,

role="math" localid="1658487525637" $鈭�脳E=\frac{1}{r\sin 胃}\left[\frac{鈭�}{鈭�胃}\left(\sin 胃E_{蠒}\right)-\frac{鈭�E_{胃}}{鈭�蠒}\right]\stackrel{铃�}{r}+\frac{1}{r}\left[\frac{鈭�}{鈭�r}\right]\left(r{E}_{蠒}\right)\stackrel{铃�}{蠒}+\frac{1}{r}\left[\frac{鈭�}{鈭�r}\left(r{E}_{胃}\right)-\frac{鈭�E_r}{鈭�胃}\right]\stackrel{铃�}{蠒}$

Therefore, the value of curl of electric field is

$鈭�脳E=\frac{1}{r\sin 胃}\left[\frac{鈭�}{鈭�胃}\left(\sin 胃E_{蠒}\right)-\frac{鈭�E_{胃}}{鈭�蠒}\right]\stackrel{铃�}{r}+\frac{1}{r}\frac{鈭�}{鈭�r}\left(r{E}_{蠒}\right)\stackrel{铃�}{蠒}$

Substitute $\frac{-鈭�B}{鈭�t}$for $鈭�脳E$.

$$\begin{gathered} \frac{-鈭�B}{鈭�t}=\frac{1}{r\sin 胃}\frac{鈭�}{鈭�胃}\left[\frac{A{\sin }^2胃}{r}\left(\cos u-\frac{1}{kr}\sin u\right)\right]\stackrel{铃�}{r} \\ -\frac{1}{r}\frac{鈭�}{鈭�t}\left[A\sin 胃\left(\cos u-\frac{1}{kr}-\frac{1}{kr}\sin u\right)\right]\stackrel{铃�}{胃} \end{gathered}$$ 鈥︹�� (8)

Here,$u=(kr-蝇t)$.

Integrate equation (8),

$$\begin{gathered} B=\frac{1}{r\sin 胃}\frac{鈭�}{鈭�胃}\left[\frac{A{\sin }^2胃}{r}\left(鈭�\cos u-\frac{1}{kr}鈭�\sin u\right)\right]\stackrel{铃�}{r} \\ -\frac{1}{r}\frac{鈭�}{鈭�胃}\left[A\sin 胃\left(鈭�\cos u-\frac{1}{kr}鈭�\sin u\right)\right]\stackrel{铃�}{胃} \end{gathered}$$ 鈥︹�� (9)

Substitute role="math" localid="1658488205000" $-\frac{1}{蝇}\sin u鈭�\cos udt$for $鈭�\cos udt$and $\frac{1}{蝇}\cos u$for $鈭�\sin udt$into equation (9).

$B=\frac{2A\cos 胃}{蝇r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\stackrel{铃�}{r}+\frac{2A\cos 胃}{蝇r^2}-kcosu\frac{1}{k{r}^2}cosu+\frac{1}{r}sinu)\stackrel{铃�}{胃}$

Therefore, the value of magnetic field is

$B=\frac{2A\cos 胃}{蝇r^2}sinu+\frac{1}{kr}cosu)\stackrel{铃�}{r}+\frac{A\sin 胃}{蝇r}-kcosu\frac{1}{k{r}^2}cosu+\frac{1}{r}sinu)\stackrel{铃�}{胃}$

Determine the Gauss鈥檚 law of magnetism,

Substitute $\frac{2A\cos 胃}{蝇r^2}\left(\sin u+\frac{1}{kr}\cos u)\stackrel{铃�}{r}+\frac{A\sin 胃}{蝇r}-k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r\sin u}\right)\stackrel{铃�}{胃}$for B into equation (4).

$$\begin{gathered} 鈭�.B=\frac{1}{r^2}\frac{鈭�}{鈭�r}\left(r^{2}B\right)+\frac{1}{r\sin 胃}\frac{鈭�}{鈭�胃}\left(\sin 胃B_{胃}\right) \\ =\frac{1}{r^2}\frac{鈭�}{鈭�r}\left[\frac{2A\cos 胃}{蝇r}\left(\sin u+\frac{1}{kr}\cos u\right)\right] \\ +\frac{1}{r\sin 胃}\frac{鈭�}{鈭�r}\left[\frac{A{\sin }^2胃}{蝇r}\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\right] \\ =\frac{1}{r^2}\frac{2A\cos 胃}{蝇}\left(k\cos u-\frac{1}{k{r}^2}\cos u-\frac{1}{r}\sin u\right) \\ +\frac{1}{r\sin 胃}\frac{2A\sin 胃\cos 胃}{蝇r}\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right) \end{gathered}$$

Solve further as

$$\begin{gathered} 鈭�.B=\frac{2A\cos 胃}{蝇r^2}\left(k\cos u-\frac{1}{k{r}^2}\cos u-\frac{1}{r}\sin u-k\cos u+\frac{1}{r}\sin u\right) \\ =0 \end{gathered}$$

Hence, the Gauss law of magnetism is obeyed.

Determine the Ampere鈥檚 law,

As$蟽=0$, therefore,

$$\begin{gathered} 鈭�脳B=\frac{1}{c^2}\frac{鈭�E}{鈭�t} \\ =\frac{1}{r}\left[\frac{鈭�}{鈭�r}\left(r{B}_{胃}\right)-\frac{鈭�B}{鈭�胃}\right]蠒 \end{gathered}$$

Substitute $\frac{2A\cos 胃}{蝇r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\stackrel{铃�}{r}+\frac{A\sin 胃}{蝇r}\left(-k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\stackrel{铃�}{B}$for B .

$$\begin{gathered} 鈭�脳B=\frac{1}{r}\left\{\begin{array}{c}\frac{鈭�}{鈭�r}\left[\frac{A\sin 胃}{蝇}\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\right]\\ -\frac{鈭�}{鈭�胃}\left[\frac{2A\cos 胃}{蝇r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\right]\stackrel{铃�}{蠒}\end{array}\right\} \\ =\frac{k}{蝇}\frac{A\sin 胃}{蝇}\left(k\sin u+\frac{1}{r}\cos u\right)\stackrel{铃�}{蠒} \\ =\frac{A\sin 胃}{蝇}\left(k\sin u+\frac{1}{r}\cos u\right)\stackrel{铃�}{蠒} \end{gathered}$$

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

$$\begin{gathered} \frac{1}{c^2}\frac{鈭�E}{鈭�t}=\frac{1}{c^2}\frac{A\sin 胃}{r}\left(蝇\sin u+\frac{蝇}{kr}\cos u\right)\stackrel{铃�}{蠒} \\ =\frac{1}{c^2}\frac{蝇}{k}\frac{A\sin 胃}{r}\left(k\sin u+\frac{1}{r}\cos u\right)\stackrel{铃�}{蠒} \\ =\frac{1}{c}\frac{A\sin 胃}{r}\left(k\sin u+\frac{1}{r}\cos u\right)\stackrel{铃�}{蠒} \\ =鈭�脳B \end{gathered}$$

Hence, Ampere鈥檚 law is obeyed.

Determine the Poynting vector is given by the following equation.

$S=\frac{1}{{渭}_0}\left(E脳B\right)$ 鈥︹�� (10)

Substitute $\frac{2A\cos 胃}{蝇r^2}\left(\sin u+\frac{1}{KR}\cos u\right)\stackrel{铃�}{r}+\frac{A\sin 胃}{蝇r}\left(-k\cos u\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\stackrel{铃�}{胃}$ for B and $\frac{A\sin 胃}{r}\left[\cos u-\frac{1}{kr}\sin u\right]蠒$for E into above equation (10).

$$\begin{gathered} S=\frac{1}{{渭}_0}\left[\frac{A\sin 胃}{r}\left[\cos u-\frac{1}{kr}\sin u\right]\stackrel{铃�}{蠒}\right]x \\ \left[\begin{array}{c}\frac{2A\cos 胃}{蝇r^2}\left(\sin u+\frac{1}{kr}\cos u\right)\stackrel{铃�}{r}\\ +\frac{A\sin 胃}{蝇r}\left(-k\cos u+\frac{1}{k{r}^2}\cos u+\frac{1}{r}\sin u\right)\stackrel{铃�}{胃}\end{array}\right] \\ =\frac{A^2\sin 胃}{{渭}_0蝇r^2}\left\{\begin{array}{c}\frac{2\cos 胃}{r}\left[\sin u\cos u+\frac{1}{r}\left({\cos }^{2}u-{\sin }^{2}u\right)-\frac{1}{k^2{r}^2}\sin u\cos u\right]\stackrel{铃�}{胃}\\ -\sin \left(\begin{array}{c}-k{\cos }^{2}u+\frac{1}{r}\sin u\cos u+\frac{1}{r}\sin u\cos u\\ -\frac{1}{k^2{r}^2}\sin u\cos u-\frac{1}{k{r}^2}{\sin }^{2}u\end{array}\right)\stackrel{铃�}{r}\end{array}\right\} \end{gathered}$$

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 $\frac{A^2\sin }{{渭}_0蝇r^2}\left(\frac{k}{2}\sin 胃\right)\stackrel{铃�}{r}$for $\left(S\right)$into equation (6).

$$\begin{gathered} I=\frac{A^2\sin }{{渭}_0蝇r^2}\left(\frac{k}{2}\sin 胃\right)\stackrel{铃�}{r} \\ =\frac{A^2{\sin }^2胃}{2{渭}_{0}c{r}^2}\stackrel{铃�}{r} \end{gathered}$$

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

Therefore, the intensity vector is$\frac{A^2{\sin }^2胃}{2{渭}_{0}c{r}^2}\stackrel{铃�}{r}$.

Determine the total power radiated is,

Substitute $\frac{A^2{\sin }^2胃}{2{渭}_{0}c{r}^2}\stackrel{铃�}{r}$for I.

role="math" localid="1658492229808" $$\begin{gathered} P=\frac{A^2}{2{渭}_{0}c}鈭�\frac{{\sin }^2胃}{r^2}{r}^2\sin 胃d胃d蠒 \\ =\frac{A^2}{2{渭}_{0}c}2\mathrm{蟺}{鈭�}_0^{\mathrm{蟺}}{\sin }^2\mathrm{胃诲胃} \\ =\frac{4\mathrm{蟺}}{3}\frac{{\mathrm{A}}^2}{{\mathrm{渭}}_0\mathrm{c}} \end{gathered}$$

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