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# 9.8: Radiation from Surface and Volume Distributions of Current

In Section 9.3, a solution was developed for radiation from current located at a single point $${\bf r}'$$. This current distribution was expressed mathematically as a current moment, as follows:

$\widetilde{\bf J}({\bf r}) = \hat{\bf l}~\widetilde{I}~\Delta l~\delta({\bf r}-{\bf r}') \label{m0221_eCM}$

where $$\hat{\bf l}$$ is the direction of current flow, $$\widetilde{I}$$ has units of current (SI base units of A), $$\Delta l$$ has units of length (SI base units of m), and $$\delta({\bf r})$$ is the volumetric sampling function (Dirac “delta” function; SI base units of m$$^{-3}$$). In this formulation, $$\widetilde{\bf J}$$ has SI base units of A/m$$^2$$. The magnetic vector potential radiated by this current, observed at the field point $${\bf r}$$, was found to be

$\widetilde{\bf A}({\bf r}) = \hat{\bf l}~\mu~\widetilde{I}~\Delta l~\frac{e^{-\gamma \left|{\bf r}-{\bf r}'\right|}}{4\pi \left|{\bf r}-{\bf r}'\right|} \label{m0221_eA}$

This solution was subsequently generalized to obtain the radiation from any distribution of line current; i.e., any distribution of current that is constrained to flow along a single path through space, as along an infinitesimally-thin wire.

In this section, we derive an expression for the radiation from current that is constrained to flow along a surface and from current which flows through a volume. The solution in both cases can be obtained by “recycling” the solution for line current as follows. Letting $$\Delta l$$ shrink to the differential length $$dl$$, Equation \ref{m0221_eCM} becomes:

$d\widetilde{\bf J}({\bf r}) = \hat{\bf l}~\widetilde{I}~dl~\delta({\bf r}-{\bf r}') \label{m0221_eCMd}$

Subsequently, Equation \ref{m0221_eA} becomes:

$d\widetilde{\bf A}({\bf r};{\bf r}') = \hat{\bf l}~\mu~\widetilde{I}~dl~\frac{e^{-\gamma \left|{\bf r}-{\bf r}'\right|}}{4\pi \left|{\bf r}-{\bf r}'\right|} \label{m0221_eAd}$

where the notation $$d\widetilde{\bf A}({\bf r};{\bf r}')$$ is used to denote the magnetic vector potential at the field point $${\bf r}$$ due to the differential-length current moment at the source point $${\bf r}'$$. Next, consider that any distribution of current can be described as a distribution of current moments. By the principle of superposition, the radiation from this distribution of current moments can be calculated as the sum of the radiation from the individual current moments. This is expressed mathematically as follows:

$\widetilde{\bf A}({\bf r}) = \int{ d\widetilde{\bf A}({\bf r};{\bf r}') } \label{m0221_eAi}$

where the integral is over $${\bf r}'$$; i.e., summing over the source current.

If we substitute Equation \ref{m0221_eAd} into Equation \ref{m0221_eAi}, we obtain the solution derived in Section 9.3, which is specific to line distributions of current. To obtain the solution for surface and volume distributions of current, we reinterpret the definition of the differential current moment in Equation \ref{m0221_eCMd}. Note that this current is completely specified by its direction ($$\hat{\bf l}$$) and the quantity $$\widetilde{I}~dl$$, which has SI base units of A$$\cdot$$m. We may describe the same current distribution alternatively as follows:

$d\widetilde{\bf J}({\bf r}) = \hat{\bf l}~\widetilde{J}_s~ds~\delta({\bf r}-{\bf r}') \label{m0221_eCM2}$

where $$\widetilde{J}_s$$ has units of surface current density (SI base units of A/m) and $$ds$$ has units of area (SI base units of m$$^2$$). To emphasize that this is precisely the same current moment, note that $$\widetilde{J}_s~ds$$, like $$\widetilde{I}~dl$$, has units of A$$\cdot$$m. Similarly,

$d\widetilde{\bf J}({\bf r}) = \hat{\bf l}~\widetilde{J}~dv~\delta({\bf r}-{\bf r}') \label{m0221_eCM3}$

where $$\widetilde{J}$$ has units of volume current density (SI base units of A/m$$^2$$) and $$dv$$ has units of volume (SI base units of m$$^3$$). Again, $$\widetilde{J}~dv$$ has units of A$$\cdot$$m. Summarizing, we have found that

$\widetilde{I}~dl = \widetilde{J}_s~ds = \widetilde{J}~dv \label{m0221_eCDE}$

all describe the same differential current moment.

Thus, we may obtain solutions for surface and volume distributions of current simply by replacing $$\widetilde{I}~dl$$ in Equation \ref{m0221_eAd}, and subsequently in Equation \ref{m0221_eAi}, with the appropriate quantity from Equation \ref{m0221_eCDE}. For a surface current distribution, we obtain:

$\boxed{ \widetilde{\bf A}({\bf r}) = \frac{\mu}{4\pi}~\int_{\mathcal{S}}~\widetilde{\bf J}_s({\bf r}') ~\frac{e^{-\gamma \left|{\bf r}-{\bf r}'\right|}}{\left|{\bf r}-{\bf r}'\right|} ~ds } \label{m0221_eAS}$

where $$\widetilde{\bf J}_s({\bf r}') \triangleq \hat{\bf l}({\bf r}')\widetilde{J}_s({\bf r}')$$ and where $$\mathcal{S}$$ is the surface over which the current flows. Similarly for a volume current distribution, we obtain:

$\boxed{ \widetilde{\bf A}({\bf r}) = \frac{\mu}{4\pi}~\int_{\mathcal{V}} ~\widetilde{\bf J}({\bf r}') ~\frac{e^{-\gamma \left|{\bf r}-{\bf r}'\right|}}{\left|{\bf r}-{\bf r}'\right|} ~dv } \label{m0221_eAV}$

where $$\widetilde{\bf J}({\bf r}') \triangleq \hat{\bf l}({\bf r}')\widetilde{J}({\bf r}')$$ and where $$\mathcal{V}$$ is the volume in which the current flows.

The magnetic vector potential corresponding to radiation from a surface and volume distribution of current is given by Equations \ref{m0221_eAS} and \ref{m0221_eAV}, respectively.

Given $$\widetilde{\bf A}({\bf r})$$, the magnetic and electric fields may be determined using the procedure developed in Section 9.2.