Calculate the electric and magnetic fields of an oscillating magnetic dipole without using approximation . [Do they look familiar? Compare Prob. 9.35.] Find the Poynting vector, and show that the intensity of the radiation is exactly the same as we got using approximation .

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.

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

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

Write the formula ofmagnetic fields of an oscillating magnetic dipole.

$\mathbf{B}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\mathbf{A}$ …… (2)

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

Write the formula ofaverage value of Poynting vector intensity.

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

$A(r,\theta ,t)=\frac{{\mu }_0{m}_0}{4\pi }\frac{\sin \theta }{r}\left\{\frac{1}{r}\cos \left[\omega \left(t-\frac{r}{c}\right)-\frac{\omega }{c}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\right]\right\}\widehat{\phi }$

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

Determine theelectric fields of an oscillating magnetic dipole.

Substitute $\frac{{\mu }_0{m}_0}{4\pi }\frac{\sin \theta }{r}\left\{\frac{1}{r}\cos \left[\omega \left(t-\frac{r}{c}\right)-\frac{\omega }{c}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\right]\right\}\widehat{\phi }$ for $A$into equation (1).

$\begin{array}{c}E=\frac{{\mu }_0{m}_0}{4\pi }\frac{\sin \theta }{r}\left\{\frac{1}{r}\cos \left[\omega \left(t-\frac{r}{c}\right)-\frac{\omega }{c}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\right]\right\}\widehat{\phi }\\ =\frac{-{\mu }_0{m}_0}{4\pi }\frac{\sin \theta }{r}\left\{\left[\frac{1}{r}\omega \left(\sin \omega \left(t-\frac{r}{c}\right)\right)\right]-\frac{{\omega }^2}{c}\cos \left(\omega \left(t-\frac{r}{c}\right)\right)\right\}\widehat{\varphi }\\ =\frac{{\mu }_0{m}_0\omega }{4\pi }\frac{\sin \theta }{r}\left[\frac{1}{r}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]+\frac{\omega }{c}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\right]\widehat{\varphi }\end{array}$

Therefore, the electric fields of an oscillating magnetic dipole is

Determine the magnetic fields of an oscillating magnetic dipole.

Substitute $\frac{{\mu }_0{m}_0\omega }{4\pi }\frac{\sin \theta }{r}\left[\frac{1}{r}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]+\frac{\omega }{c}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\right]\varphi$ for $A$into equation (2).

$\begin{array}{l}\nabla \times A=\frac{1}{r\sin \theta }\left[\frac{\partial }{\partial \theta }\left(\sin \theta A\varphi \right)-\frac{\partial A\theta }{r\theta }\right]\widehat{r}+\left[\frac{1}{\sin \theta }\frac{\partial A_r}{\partial \varphi }-\frac{\partial }{\partial r}\left(rA\varphi \right)\right]\widehat{\varphi }\\ +\frac{1}{r}\left[\frac{\partial }{\partial r}\left(rA\theta 1-\frac{d{A}_r}{\partial \theta }\right)\right]\widehat{\varphi }\end{array}$

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

role="math" localid="1658931445531" $\begin{array}{c}\nabla \times A=\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta A_{\varphi }\right)\widehat{r}\frac{1}{r}\frac{\partial }{\partial r}\left(r{A}_{\varphi }\right)\widehat{\theta }\\ \nabla \times A=\frac{{\mu }_0{m}_0}{4\pi }\left[\begin{array}{l}\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\sin \theta \left(\frac{\sin \theta }{r}\right)\left\{\frac{1}{r}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]-\frac{\omega }{c}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\right\}\widehat{r}\\ -\frac{1}{r}\frac{\partial }{\partial r}\left\{r\cdot \left(\frac{\sin \theta }{r}\right)\left[\frac{1}{r}\cos [\omega \left(t-\frac{r}{c}\right)-\frac{\omega }{c}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]]\right]\right\}\widehat{\theta }\end{array}\right]\\ B=\frac{{\mu }_0{m}_0}{4\pi }\left\{\begin{array}{l}\frac{2\cos \theta }{r^2}\left[\frac{1}{r}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]-\frac{\omega }{c}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\widehat{r}\right]\\ -\frac{\sin \theta }{r}\left[\begin{array}{l}-\frac{1}{r^2}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]+\frac{\omega }{rc}\sin \omega \left[\left(t-\frac{r}{c}\right)\right]\\ +{\left(\frac{\omega }{c}\right)}^2\cos \omega \left[\left(t-\frac{r}{c}\right)\right]\end{array}\right]\widehat{\theta }\end{array}\right\}\end{array}$

In problem 9.33 the result is $A=\frac{{\mu }_0{m}_0{\omega }^2}{4\pi c}$

Determine the average value of Poynting vector intensity.

Substitute $\frac{{\mu }_0{m}_0\omega }{4\pi }\frac{\sin \theta }{r}\left[\frac{1}{r}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]+\frac{\omega }{c}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]\right]]\widehat{\varphi }$ for $E$and

$B=\frac{{\mu }_0{m}_0}{4\pi }\left\{\begin{array}{l}\frac{2\cos \theta }{r^2}\left[\frac{1}{r}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]-\frac{\omega }{c}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]\widehat{r}\right]\\ -\frac{\sin \theta }{r}\left[\begin{array}{l}-\frac{1}{r^2}\cos \left[\omega \left(t-\frac{r}{c}\right)\right]+\frac{\omega }{rc}\sin \omega \left[\left(t-\frac{r}{c}\right)\right]\\ +{\left(\frac{\omega }{c}\right)}^2\cos \omega \left[\left(t-\frac{r}{c}\right)\right]\end{array}\right]\end{array}\right\}$ for $B$into equation (3).

$\begin{array}{l}S=\frac{1}{c}(\mathbf{E}\times \mathbf{B})\\ S=\frac{{\mu }_0{m}_0^2{\omega }^3}{16{\pi }^2{c}^2}\left(\frac{\sin \theta }{r^2}\right)\left\{\begin{array}{c}\frac{2\cos \theta }{r}\left[\left(1-\frac{c^2}{{\omega }^2{r}^2}\right)\sin u\cos u+\frac{c}{\omega r}({\cos }^{2}u-{\sin }^{2}u)\right]\widehat{\theta }\\ +\sin \theta \left[\begin{array}{l}\left(-\frac{2}{r}+\frac{c^2}{{\omega }^2{r}^3}\right)\sin u\cos u+\frac{\omega }{c}{\cos }^{2}u\\ +\frac{c}{\omega r^2}({\sin }^{2}u-{\cos }^{2}u)\end{array}\right]\widehat{r}\end{array}\right\}\end{array}$

Here, $u=-\omega \left(t-\frac{r}{c}\right)$

Then average value of Poynting vector is intensity

$⟨\text{S}⟩=\frac{{\mu }_0{m}_0^2{\omega }^4}{32{\pi }^2{c}^3}\frac{{\sin }^2\theta }{r^2}\widehat{r}$

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

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