Magnetic Dipole Radiation

#Physics
400

$\displaystyle V=0$

There's no charge, so no potential

$\displaystyle \vec{A}(r,\theta,t)=- \frac{{\mu}{0}m{0}\omega}{4\pi c}\left( \frac{\sin \theta}{r} \right)\sin[\omega(t-r /c)]\hat{\phi}$

Vector potential around a loop of wire with radius $\displaystyle b\ll r$, where $\displaystyle r$ is the distance from the magnetic dipole
$\displaystyle \omega$ is the angular frequency of the current running through wire loop

$\displaystyle \vec{E}=\frac{{\mu}{0}m{0}\omega ^{2}}{4\pi c}\left( \frac{\sin \theta}{r} \right)\cos[\omega(t-r /c)]\hat{\phi}$

$\displaystyle \vec{B}=-\frac{{\mu}{0}m{0}\omega ^{2}}{4\pi c^{2}}\left( \frac{\sin \theta}{r} \right)\cos[\omega(t-r /c)]\hat{\theta}$

$\displaystyle {\left\langle{S}\right\rangle}=\left( \frac{{\mu}{0}m{0}^{2}\omega^{4}}{32\pi ^{2}c^{3}} \right) \frac{\sin ^{2}\theta}{r^{2}}\hat{r}$

Same formula as in the electric dipole radiation but we replace $\displaystyle p_{0}$ with $\displaystyle m_{0}$

$\displaystyle {\left\langle{P}\right\rangle}=\frac{{\mu}{0}m{0}^{2}\omega^{4}}{12\pi c^{3}}$

Same formula as in the electric dipole radiation but we replace $\displaystyle p_{0}$ with $\displaystyle m_{0}$ and divide by $\displaystyle c^{2}$