Past exam of the electromagnetism course of the University of Oxford / 2023 / 3 / d

Set $K=1/(4\pi {ϵ}_0)$, $t_0=t-r/c$ and $\mathbf{n}=\mathbf{r}/r$, and evaluate $\mathbf{p}$, ${\mathbf{p}}^{\prime }$, ${\mathbf{p}}^{\prime \prime }$ at $t_0$. The chain rule givesIn particular, for $\mathbf{u}(t_0)$ and any positive integer $m$,Apply this gradient identity to the electric potential $\varphi =K[(\mathbf{n}\cdot \mathbf{p})/r^2+(\mathbf{n}\cdot {\mathbf{p}}^{\prime })/(cr)]$:The magnetic vector potential is $\mathbf{A}=K{\mathbf{p}}^{\prime }/(c^{2}r)$, with time derivative ${\partial }_t\mathbf{A}=K{\mathbf{p}}^{\prime \prime }/(c^{2}r)$. Thus the electric field $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ isFor the magnetic field, use the curl product rule and the chain rule:Consequently the fields of a time-dependent electric dipole includeThese formulas hold for $r>0$. The $r^{-3}$ electric field is the retarded electric dipole field, the $r^{-2}$ terms are induction contributions, and the $r^{-1}$ terms describe electric dipole radiation. Retaining the dependence of $t_0$ on position in every spatial derivative is what produces both the induction and radiation contributions.

The leading electric dipole radiation fields areThe radiation electric field is transverse, $\mathbf{n}\cdot {\mathbf{E}}_{\mathrm{rad}}=0$. The vector triple product therefore gives the leading Poynting vectorIf $\theta$ is the angle between ${\mathbf{p}}^{\prime \prime }(t-r/c)$ and $\mathbf{n}$, thenUsing ${\mu }_0{ϵ}_0{c}^2=1$, the final result isThe leading Poynting flux is outward, vanishes along the instantaneous retarded ${\mathbf{p}}^{\prime \prime }$ axis, and is maximal in its perpendicular plane. If ${\mathbf{p}}^{\prime \prime }=0$, the radiation term is zero and no angle needs to be assigned. The large-$r$ expansion assumes bounded retarded electric dipole moment and time derivatives; it may also be read at fixed retarded time.