ElectroQuasiStatic (EQS) regime - struggling with understanding the 1st order correction - computation of $\vec{H}$

Question

I’ve the following problem:

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Given a dipole with the moment $\vec{p} = p\hat{z}\cos(\omega t)$ I was tasked first with finding the field $\vec{E}$, justifying ElectroQuasiStatic (EQS) regime and then finding the first order correction: $\vec{H}$. I’ve managed to find the dipole with the dot product formula, and generally am able to compute curls - however their solution with ampers law baffles me a little.

first of all if $\nabla \times \vec{H} = j \omega \epsilon_0 \vec{E}$ won’t that mean that $\vec{E} = \frac{-j}{\omega \epsilon_0} \cdot \nabla \times \vec{H}$ as opposed to $+j$ like we see in the answers? Also -why does $\vec{H}$ only have a component in the $\hat{\phi}$ direction?

Answer 1

Physically, you have and oscillating electric dipole, so you would expect EM radiation. However, by the method you are employing, you are interested in the "near" field, namely, $$ r\gg d \quad r\ll \frac{c}{\omega} $$ In this case, you can indeed set the electric field given by the quasi-static approximation, namely calculate it as in the static case with the instantaneous distribution of charge.

For the magnetic field, in this regime, it is zero since instantaneously, there is no current. The next leading term is therefore induced by the electric field. The Maxwell displacement current has at every point the plane symmetry with respect to the plane perpendicular to $\hat\phi$, so $H$ is parallel to $\hat \phi$ being a pseudo vector respecting the same symmetries. There is no sign issue, the fourth and fifth line are just: $$ E_r = \frac{-j}{\epsilon_0\omega}(\nabla\times H)_r \\ E_\theta = \frac{-j}{\epsilon_0\omega}(\nabla\times H)_\theta $$ where the expressions for $E$ and $\nabla H$ are substitutes.

Note that to get the full solution, you'll need to look at the induced electric field with Faraday's law giving a correction to the electric field and repeat the process indefinitely. What you are effectively doing is writing the solution as a power series in $c$, or to make it dimensionalness, in powers of $\frac{c}{r\omega}$. It turns out that you can exactly solve this problem, it is given in the section Dipole radiation, and you can check that your result match the leading order terms in powers of $c$.

Hope this helps.