91Ó°ÊÓ

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.

$\mathbf{E}\mathbf{=}\frac{\mathbf{∂}\mathbf{A}}{\mathbf{∂}\mathbf{t}}$ …… (1)

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

Write the formula ofmagnetic fields of an oscillating magnetic dipole.

$\mathbf{B}\mathbf{=}\mathbf{∇}\mathbf{×}\mathbf{A}$ …… (2)

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

Write the formula ofaverage value of Poynting vector intensity.

. $\mathbf{S}\mathbf{=}\frac{1}{C}\mathbf{(}\mathbf{E}\mathbf{×}\mathbf{B}\mathbf{)}$ …… (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:

$A(r,\mathrm{θ},t)=\frac{{\mathrm{μ}}_0{m}_0}{4\mathrm{Ï€}}\frac{\sin \mathrm{θ}}{r}\left\{\frac{1}{r}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)−\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\right]\right\}\widehat{\mathrm{φ}}$

Here, ${\mathrm{μ}}_0$is permeability of free space, $m_0$is the mass, $K$ is the wave number, $C$is the speed of sound, $\mathrm{Ó\neg }$is the angular frequency, and $t$ is the time.

Determine theelectric fields of an oscillating magnetic dipole.

Substitute $\frac{{\mathrm{μ}}_0{m}_0}{4\mathrm{Ï€}}\frac{\sin \mathrm{θ}}{r}\left\{\frac{1}{r}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)−\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\right]\right\}\widehat{\mathrm{φ}}$ for $A$into equation (1).

$\begin{array}{c}E=\frac{{\mathrm{μ}}_0{m}_0}{4\mathrm{Ï€}}\frac{\sin \mathrm{θ}}{r}\left\{\frac{1}{r}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)−\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\right]\right\}\widehat{\mathrm{φ}}\\ =\frac{−{\mathrm{μ}}_0{m}_0}{4\mathrm{Ï€}}\frac{\sin \mathrm{θ}}{r}\left\{\left[\frac{1}{r}\mathrm{Ó\neg }\left(\sin \mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right)\right]−\frac{{\mathrm{Ó\neg }}^2}{c}\cos \left(\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right)\right\}\widehat{\mathrm{Ï•}}\\ =\frac{{\mathrm{μ}}_0{m}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\frac{\sin \mathrm{θ}}{r}\left[\frac{1}{r}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]+\frac{\mathrm{Ó\neg }}{c}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\right]\widehat{\mathrm{Ï•}}\end{array}$

Therefore, the electric fields of an oscillating magnetic dipole is

Determine the magnetic fields of an oscillating magnetic dipole.

Substitute $\frac{{\mathrm{μ}}_0{m}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\frac{\sin \mathrm{θ}}{r}\left[\frac{1}{r}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]+\frac{\mathrm{Ó\neg }}{c}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\right]\mathrm{Ï•}$ for $A$into equation (2).

$\begin{array}{l}∇×A=\frac{1}{r\sin \mathrm{θ}}\left[\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}A\mathrm{ϕ}\right)−\frac{∂A\mathrm{θ}}{r\mathrm{θ}}\right]\widehat{r}+\left[\frac{1}{\sin \mathrm{θ}}\frac{∂A_r}{∂\mathrm{ϕ}}−\frac{∂}{∂r}\left(rA\mathrm{ϕ}\right)\right]\widehat{\mathrm{ϕ}}\\ +\frac{1}{r}\left[\frac{∂}{∂r}\left(rA\mathrm{θ}1−\frac{d{A}_r}{∂\mathrm{θ}}\right)\right]\widehat{\mathrm{ϕ}}\end{array}$

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

role="math" localid="1658931445531" $\begin{array}{c}∇×A=\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}{A}_{\mathrm{Ï•}}\right)\widehat{r}\frac{1}{r}\frac{∂}{∂r}\left(r{A}_{\mathrm{Ï•}}\right)\widehat{\mathrm{θ}}\\ ∇×A=\frac{{\mathrm{μ}}_0{m}_0}{4\mathrm{Ï€}}\left[\begin{array}{l}\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\sin \mathrm{θ}\left(\frac{\sin \mathrm{θ}}{r}\right)\left\{\frac{1}{r}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]−\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\right\}\widehat{r}\\ −\frac{1}{r}\frac{∂}{∂r}\left\{râ‹\dots \left(\frac{\sin \mathrm{θ}}{r}\right)\left[\frac{1}{r}\cos [\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)−\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]]\right]\right\}\widehat{\mathrm{θ}}\end{array}\right]\\ B=\frac{{\mathrm{μ}}_0{m}_0}{4\mathrm{Ï€}}\left\{\begin{array}{l}\frac{2\cos \mathrm{θ}}{r^2}\left[\frac{1}{r}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]−\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\widehat{r}\right]\\ −\frac{\sin \mathrm{θ}}{r}\left[\begin{array}{l}−\frac{1}{r^2}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]+\frac{\mathrm{Ó\neg }}{rc}\sin \mathrm{Ó\neg }\left[\left(t−\frac{r}{c}\right)\right]\\ +{\left(\frac{\mathrm{Ó\neg }}{c}\right)}^2\cos \mathrm{Ó\neg }\left[\left(t−\frac{r}{c}\right)\right]\end{array}\right]\widehat{\mathrm{θ}}\end{array}\right\}\end{array}$

In problem 9.33 the result is $A=\frac{{\mathrm{μ}}_0{m}_0{\mathrm{Ó\neg }}^2}{4\mathrm{Ï€}c}$

Determine the average value of Poynting vector intensity.

Substitute $\frac{{\mathrm{μ}}_0{m}_0\mathrm{Ó\neg }}{4\mathrm{Ï€}}\frac{\sin \mathrm{θ}}{r}\left[\frac{1}{r}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]+\frac{\mathrm{Ó\neg }}{c}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\right]]\widehat{\mathrm{Ï•}}$ for $E$and

$B=\frac{{\mathrm{μ}}_0{m}_0}{4\mathrm{Ï€}}\left\{\begin{array}{l}\frac{2\cos \mathrm{θ}}{r^2}\left[\frac{1}{r}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]−\frac{\mathrm{Ó\neg }}{c}\sin \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]\widehat{r}\right]\\ −\frac{\sin \mathrm{θ}}{r}\left[\begin{array}{l}−\frac{1}{r^2}\cos \left[\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)\right]+\frac{\mathrm{Ó\neg }}{rc}\sin \mathrm{Ó\neg }\left[\left(t−\frac{r}{c}\right)\right]\\ +{\left(\frac{\mathrm{Ó\neg }}{c}\right)}^2\cos \mathrm{Ó\neg }\left[\left(t−\frac{r}{c}\right)\right]\end{array}\right]\end{array}\right\}$ for $B$into equation (3).

$\begin{array}{l}S=\frac{1}{c}(\mathbf{E}×\mathbf{B})\\ S=\frac{{\mathrm{μ}}_0{m}_0^2{\mathrm{Ó\neg }}^3}{16{\mathrm{Ï€}}^2{c}^2}\left(\frac{\sin \mathrm{θ}}{r^2}\right)\left\{\begin{array}{c}\frac{2\cos \mathrm{θ}}{r}\left[\left(1−\frac{c^2}{{\mathrm{Ó\neg }}^2{r}^2}\right)\sin u\cos u+\frac{c}{\mathrm{Ó\neg }r}({\cos }^{2}u−{\sin }^{2}u)\right]\widehat{\mathrm{θ}}\\ +\sin \mathrm{θ}\left[\begin{array}{l}\left(−\frac{2}{r}+\frac{c^2}{{\mathrm{Ó\neg }}^2{r}^3}\right)\sin u\cos u+\frac{\mathrm{Ó\neg }}{c}{\cos }^{2}u\\ +\frac{c}{\mathrm{Ó\neg }{r}^2}({\sin }^{2}u−{\cos }^{2}u)\end{array}\right]\widehat{r}\end{array}\right\}\end{array}$

Here, $u=−\mathrm{Ó\neg }\left(t−\frac{r}{c}\right)$

Then average value of Poynting vector is intensity

$⟨\text{S}âŸ\copyright =\frac{{\mathrm{μ}}_0{m}_0^2{\mathrm{Ó\neg }}^4}{32{\mathrm{Ï€}}^2{c}^3}\frac{{\sin }^2\mathrm{θ}}{r^2}\widehat{r}$

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

Therefore, the average value of Poynting vector intensity is $<\text{S}>=\frac{{\mathrm{μ}}_0{m}_0^2{\mathrm{Ó\neg }}^4}{32{\mathrm{Ï€}}^2{c}^3}\frac{{\sin }^2\mathrm{θ}}{r^2}\widehat{r}$ .