# Magnetic field of a Herzian dipole antenna

If I am given the dipole moment of very short dipole antenna as $P = P_0 sin (\omega t)$, what will be the magnetic field and polarization of far field radiation?

Do I need to consider the time variation when calculating near and far fields, as in how many half or full cycles will exists along the dipole ?

• Dipole moment is a vector, so unless you have been given its vector direction, then you can't answer the question. Commented Jan 16, 2017 at 14:37

You can use phasors instead of dealing with time domain.

For a Hertzian dipole oriented along the $z$ axis, the magnetic field will have only one ($\phi$) component, both in near-field and in far-field, which will be equal to

$$H_{\phi}=\frac{i\omega P}{4\pi}\left[\frac{ik}{r}+\frac{1}{r^2}\right]\sin\theta e^{-i k r}$$

where $(r,\theta,\phi)$ are the spherical coordinates, $i$ is the imaginary unit, $\omega$ is the angular frequency, $k$ is the wavenumber, $P$ is the dipole moment and the $\exp(i\omega t)$ time convention has been adopted. The far magnetic field can be easily obtained by dropping the $1/r^2$ term, i.e.

$$H_{\phi}\simeq\frac{i\omega P}{4\pi}\frac{ik}{r}\sin\theta e^{-i k r}$$

Opposite to that, the electric field will have two components (i.e., $r$ and $\theta$) in the near field equal to

$$E_{\theta}=\zeta\frac{i\omega P}{4\pi}\left[\frac{ik}{r}+\frac{1}{r^2}+\frac{1}{i k r^3}\right]\sin\theta e^{-i k r}$$

and

$$E_{r}=\zeta\frac{i\omega P}{4\pi}\left[\frac{1}{r^2}+\frac{1}{i k r^3}\right]\cos\theta e^{-i k r}$$

The far field can be obtained by dropping the $1/r^2$ and $1/r^3$ terms. Accordingly,

$$E_{\theta}\simeq\zeta\frac{i\omega P}{4\pi}\frac{ik}{r}\sin\theta e^{-i k r}$$

and

$$E_{r}\simeq0$$

The electric far field is then linearly polarized along $\hat{i}_\theta$, while the magnetic far field is linearly polarized along $\hat{i}_\phi$.

• Hello Jack, I thought in my calculations I missed a very trivial part related to time variant current in the dipole, which created havoc in my calculations, however I would like to thank you very much for your precious time and efforts :). Commented May 22, 2014 at 18:51