Hertzian Dipole

Editors：何偉碩、黃旭成、林雋哲

Advisor：江簡富  教授

   Consider a Hertzian dipole which consists of two point charges oscillating around their center of mass with extreme separation of $\ell$  . The dipole is located at the origin , with the current $I(t)=I_0\cos\left(\omega t\right)$ The associated vector potential is   $\bar{A}\left(\bar{r},t\right)=\hat{z}\displaystyle \frac{\mu I\ell}{4\pi r}=\hat{z}\displaystyle \frac{\mu I_0 \ell}{4\pi r}\cos\left(\omega t-kr\right)$ . Appling the relations $\bar{H}=\displaystyle \frac{1}{\mu}\triangledown\times \bar{A}$  and  $\hat{z}=\hat{r}\cos\theta-\hat{\theta}\sin\theta$  , the magnetic field can be derived as $\bar{H}\left(\bar{r},t\right)=\hat{\phi}\displaystyle \frac{I_0\ell\sin\theta}{4\pi}\left[\displaystyle \frac{\cos\left(\omega t-kr\right)}{r^2}-\displaystyle \frac{k\sin\left(\omega t-kr\right)}{r}\right]$ . The electric field can then be derived by Ampere’s law $\triangledown\times\bar{H}=\epsilon\displaystyle \frac{\partial\bar{E}}{\partial t}$  as $\bar{E}\left(\bar{r},t\right)=\hat{r}\displaystyle \frac{2I_0\ell\cos\theta}{4\pi\epsilon\omega}\left[\displaystyle \frac{\sin\left(\omega t-kr\right)}{r^3}+\displaystyle \frac{k\cos\left(\omega t-kr\right)}{r^2}\right]+\hat{\theta}\displaystyle \frac{I_0\ell\sin\theta}{4\pi\epsilon\omega}\left[\displaystyle \frac{\sin\left(\omega t-kr\right)}{r^3}+\displaystyle \frac{k\cos\left(\omega t-kr\right)}{r^2}-\displaystyle \frac{k^2\sin\left(\omega t-kr\right)}{r}\right]$ .

In the far field ( $kr\gg1$ ), the magnetic field and the electric field will be dominated by the terms $\bar{H}=-\hat{\phi}\left(\displaystyle \frac{I_0\ell k\sin\theta}{4\pi r}\right)\sin\left(\omega t-kr\right)$ and $\hat{E}=-\hat{\theta}\displaystyle \frac{I_0\ell k^2\sin\theta}{4\pi r\epsilon\omega}\sin\left(\omega t-kr\right)=-\hat{\theta}\eta\displaystyle \frac{I_0\ell k\sin\theta}{4\pi r}\sin\left(\omega t-kr\right)$ . The Poynting vector $\bar{S}\left(\bar{r},t\right)=\bar{E}\left(\bar{r},t\right)\times\bar{H}\left(\bar{r},t\right)=\hat{r}S_r+\hat{\theta}S_\theta$  in the far-field region can thus be derived as $\bar{S}\left(\bar{r},t\right)=\hat{r}\eta\left(\displaystyle \frac{I_0\ell k\sin\theta}{4\pi r}\right)^2\sin^2\left(\omega t-kr\right)$ , which radiates in the $\hat{r}$  direction.