91影视

Consider that the intensity of the radiation is exactly the same as we got using approximation $3$.

Write the formula ofelectric fields of an oscillating magnetic dipole.

$\mathbf{E}\mathbf{=}\frac{\mathbf{鈭�}\mathbf{A}}{\mathbf{鈭�}\mathbf{t}}$ 鈥︹�� (1)

Here,$\mathbf{A}$ is vector potential of an Oscillating magnetic dipole.

Write the formula ofmagnetic fields of an oscillating magnetic dipole.

$\mathbf{B}\mathbf{=}\mathbf{鈭�}\mathbf{脳}\mathbf{A}$ 鈥︹�� (2)

Here, $\mathbf{A}$ is vector potential of an Oscillating magnetic dipole.

Write the formula ofaverage value of Poynting vector intensity.

. $\mathbf{S}\mathbf{=}\frac{1}{C}\mathbf{(}\mathbf{E}\mathbf{脳}\mathbf{B}\mathbf{)}$ 鈥︹�� (3)

Here, $\mathbf{E}$ is electric field of an oscillating magnetic dipole, $\mathbf{C}$ is speed of sound and $\mathbf{B}$ is magnetic field of an oscillating magnetic dipole.

The vector potential of an Oscillating magnetic dipole is:

$A(r,胃,t)=\frac{{渭}_0{m}_0}{4蟺}\frac{\sin 胃}{r}\left\{\frac{1}{r}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)鈭�\frac{蝇}{c}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\right]\right\}\widehat{蠁}$

Here, ${渭}_0$is permeability of free space, $m_0$is the mass, $K$ is the wave number, $C$is the speed of sound, $蝇$is the angular frequency, and $t$ is the time.

Determine theelectric fields of an oscillating magnetic dipole.

Substitute $\frac{{渭}_0{m}_0}{4蟺}\frac{\sin 胃}{r}\left\{\frac{1}{r}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)鈭�\frac{蝇}{c}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\right]\right\}\widehat{蠁}$ for $A$into equation (1).

$\begin{array}{c}E=\frac{{渭}_0{m}_0}{4蟺}\frac{\sin 胃}{r}\left\{\frac{1}{r}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)鈭�\frac{蝇}{c}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\right]\right\}\widehat{蠁}\\ =\frac{鈭�{渭}_0{m}_0}{4蟺}\frac{\sin 胃}{r}\left\{\left[\frac{1}{r}蝇\left(\sin 蝇\left(t鈭�\frac{r}{c}\right)\right)\right]鈭�\frac{{蝇}^2}{c}\cos \left(蝇\left(t鈭�\frac{r}{c}\right)\right)\right\}\widehat{蠒}\\ =\frac{{渭}_0{m}_0蝇}{4蟺}\frac{\sin 胃}{r}\left[\frac{1}{r}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]+\frac{蝇}{c}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\right]\widehat{蠒}\end{array}$

Therefore, the electric fields of an oscillating magnetic dipole is

Determine the magnetic fields of an oscillating magnetic dipole.

Substitute $\frac{{渭}_0{m}_0蝇}{4蟺}\frac{\sin 胃}{r}\left[\frac{1}{r}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]+\frac{蝇}{c}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\right]蠒$ for $A$into equation (2).

$\begin{array}{l}鈭�脳A=\frac{1}{r\sin 胃}\left[\frac{鈭�}{鈭�胃}\left(\sin 胃A蠒\right)鈭�\frac{鈭�A胃}{r胃}\right]\widehat{r}+\left[\frac{1}{\sin 胃}\frac{鈭�A_r}{鈭�蠒}鈭�\frac{鈭�}{鈭�r}\left(rA蠒\right)\right]\widehat{蠒}\\ +\frac{1}{r}\left[\frac{鈭�}{鈭�r}\left(rA胃1鈭�\frac{d{A}_r}{鈭�胃}\right)\right]\widehat{蠒}\end{array}$

Here, $A_r$and $A_0$are not taken then:

role="math" localid="1658931445531" $\begin{array}{c}鈭�脳A=\frac{1}{r\sin 胃}\frac{鈭�}{鈭�胃}\left(\sin 胃A_{蠒}\right)\widehat{r}\frac{1}{r}\frac{鈭�}{鈭�r}\left(r{A}_{蠒}\right)\widehat{胃}\\ 鈭�脳A=\frac{{渭}_0{m}_0}{4蟺}\left[\begin{array}{l}\frac{1}{r\sin 胃}\frac{鈭�}{鈭�胃}\sin 胃\left(\frac{\sin 胃}{r}\right)\left\{\frac{1}{r}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)\right]鈭�\frac{蝇}{c}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\right\}\widehat{r}\\ 鈭�\frac{1}{r}\frac{鈭�}{鈭�r}\left\{r鈰�\left(\frac{\sin 胃}{r}\right)\left[\frac{1}{r}\cos [蝇\left(t鈭�\frac{r}{c}\right)鈭�\frac{蝇}{c}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]]\right]\right\}\widehat{胃}\end{array}\right]\\ B=\frac{{渭}_0{m}_0}{4蟺}\left\{\begin{array}{l}\frac{2\cos 胃}{r^2}\left[\frac{1}{r}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)\right]鈭�\frac{蝇}{c}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\widehat{r}\right]\\ 鈭�\frac{\sin 胃}{r}\left[\begin{array}{l}鈭�\frac{1}{r^2}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)\right]+\frac{蝇}{rc}\sin 蝇\left[\left(t鈭�\frac{r}{c}\right)\right]\\ +{\left(\frac{蝇}{c}\right)}^2\cos 蝇\left[\left(t鈭�\frac{r}{c}\right)\right]\end{array}\right]\widehat{胃}\end{array}\right\}\end{array}$

In problem 9.33 the result is $A=\frac{{渭}_0{m}_0{蝇}^2}{4蟺c}$

Determine the average value of Poynting vector intensity.

Substitute $\frac{{渭}_0{m}_0蝇}{4蟺}\frac{\sin 胃}{r}\left[\frac{1}{r}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]+\frac{蝇}{c}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\right]]\widehat{蠒}$ for $E$and

$B=\frac{{渭}_0{m}_0}{4蟺}\left\{\begin{array}{l}\frac{2\cos 胃}{r^2}\left[\frac{1}{r}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)\right]鈭�\frac{蝇}{c}\sin \left[蝇\left(t鈭�\frac{r}{c}\right)\right]\widehat{r}\right]\\ 鈭�\frac{\sin 胃}{r}\left[\begin{array}{l}鈭�\frac{1}{r^2}\cos \left[蝇\left(t鈭�\frac{r}{c}\right)\right]+\frac{蝇}{rc}\sin 蝇\left[\left(t鈭�\frac{r}{c}\right)\right]\\ +{\left(\frac{蝇}{c}\right)}^2\cos 蝇\left[\left(t鈭�\frac{r}{c}\right)\right]\end{array}\right]\end{array}\right\}$ for $B$into equation (3).

$\begin{array}{l}S=\frac{1}{c}(\mathbf{E}脳\mathbf{B})\\ S=\frac{{渭}_0{m}_0^2{蝇}^3}{16{蟺}^2{c}^2}\left(\frac{\sin 胃}{r^2}\right)\left\{\begin{array}{c}\frac{2\cos 胃}{r}\left[\left(1鈭�\frac{c^2}{{蝇}^2{r}^2}\right)\sin u\cos u+\frac{c}{蝇r}({\cos }^{2}u鈭�{\sin }^{2}u)\right]\widehat{胃}\\ +\sin 胃\left[\begin{array}{l}\left(鈭�\frac{2}{r}+\frac{c^2}{{蝇}^2{r}^3}\right)\sin u\cos u+\frac{蝇}{c}{\cos }^{2}u\\ +\frac{c}{蝇r^2}({\sin }^{2}u鈭�{\cos }^{2}u)\end{array}\right]\widehat{r}\end{array}\right\}\end{array}$

Here, $u=鈭�蝇\left(t鈭�\frac{r}{c}\right)$

Then average value of Poynting vector is intensity

$鉄�\text{S}鉄�=\frac{{渭}_0{m}_0^2{蝇}^4}{32{蟺}^2{c}^3}\frac{{\sin }^2胃}{r^2}\widehat{r}$

This is the same radiation intensity that we estimated using approximations.

Therefore, the average value of Poynting vector intensity is $<\text{S}>=\frac{{渭}_0{m}_0^2{蝇}^4}{32{蟺}^2{c}^3}\frac{{\sin }^2胃}{r^2}\widehat{r}$ .