Given a surface current distribution, $\textbf{J}(r)$, we can calculate the magnetic potential in the Lorenz gauge using $$ \textbf{A}(\textbf{r})=-μ_{0} ∬g(\textbf{r},\textbf{r}')\textbf{J}(\textbf{r}')dA' $$ where $$ g(\textbf{r},\textbf{r}')=-\frac{e^{ik_{0}|r-r'|}}{4π|r-r'|} $$ is the Green’s function (I’m using an $e^{-iωt}$ temporal dependence throughout). The magnetic field can then be calculated using $$ \textbf{H}(\textbf{r})=\frac{1}{\mu_{0}}∇×\textbf{A}(\textbf{r}) $$ Doing this for a perfectly conducting infinite planar sheet located at $x=0$ supporting a uniform surface current $J$ in the $z$ direction leads to a magnetic field $$ \textbf{H}(\textbf{r})=\frac{sign(x)J}{2}e^{ik_{0}|x|}y $$ If we impose that the whole region x<0 is occupied by a conductor, then, by observation, the magnetic field is $$ H(r)=Je^{ik_{0}x} $$ in $x>0$ and vanishing in $x<0$. My question is whether there is a way of deriving this latter magnetic field using an equation of the type $$ \textbf{A}(\textbf{r})=-μ_{0} ∬g(\textbf{r},\textbf{r}')\textbf{J}(\textbf{r}')dA' $$ i.e. using the Green’s function. Essentially I want to impose that the radiation due to the current locally propagates only in one direction (in reality I don’t have an infinite sheet so I can’t work out the answer just by observation).
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$\begingroup$ > "then, by observation, the magnetic field is...vanishing in $x<0$." -- Can you elaborate? Do you mean observation in real world? There is no infinite conductive half-spaces in real world... $\endgroup$– Ján LalinskýCommented May 24, 2022 at 17:00
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$\begingroup$ I fail to see how a harmonic current density tangential to and oriented along an infinite conducting surface would create any field at all... Am I missing something? $\endgroup$– Cyrus TirbandCommented May 24, 2022 at 19:30
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1 Answer
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In the case of an infinite perfectly conducting plane, you can use the image theorem.
That is, since there is a perfectly conductive half space for $x<0$, the current density $\mathbf{J}(x',y',z')$ in the other half space $x\geq 0$ can be modified to include an image current. The x-component becomes of this modified current density becomes $J_x(x',y',z') + J_x(-x', y',z') $ and the tranverse component becomes $\mathbf{J}_T(x',y',z') - \mathbf{J}_T(-x',y',z')$. If you're using a dyadic Green's function you can instead adjust the respective components of the Green's function.
You can then set the fields to zero for $x<0$.
