Analytic Solution

Purpose

Here, Maxwell’s equations are solved for a harmonic magnetic dipole source. This is accomplished by using the method of Schelkunoff potentials, as shown in Ward and Hohmann ([WH88]). Analytic expressions for the electric field, the magnetic field and the corresponding vector potential are provided.

For a magnetic source (\(\mathbf{J_m^s}\)), Maxwell’s equations in the frequency domain are given by:

(260)\[\nabla \times \mathbf{E_m} + i\omega \mu \mathbf{H_m} = - \mathbf{J_m^s}\]

(261)\[\nabla \times \mathbf{H_m} - (\sigma + i\omega \varepsilon) \mathbf{E_m} = 0\]

where the subscripts \(_m\) remind us that we are using a magnetic source. For a magnetic source (\(\mathbf{J_m^s}\)), the electric and magnetic fields can be expressed in terms of the Schelkunoff vector potential (\(\mathbf{F}\)), where:

(262)\[\mathbf{H}_m = -(\sigma + i \omega \varepsilon) \mathbf{F} + \frac{1}{(i \omega \mu)} \nabla (\nabla \cdot \mathbf{F})\]

and

(263)\[\mathbf{E}_m = - \nabla \times \mathbf{F}\]

Eq. (263) can be obtained simply by taking the divergence of Eq. (261). Eq. (262) is obtained by manipulating Eqs. (260), (261) and (263), and choosing an appropriate Gauge. We can see from Eqs. (262) and (263) that \(\mathbf{F}\) contains all the information corresponding to the electric and magnetic fields. Therefore, Maxwell’s equations will be manipulated to solve for \(\mathbf{F}\); which can then be used to obtain \(\mathbf{E_m}\) and \(\mathbf{H_m}\).

By manipulating Eqs. (260), (261) and (263) and choosing an appropriate Gauge, we find that \(\mathbf{F}\) can be expressed using the Helmholtz equation:

(264)\[\nabla^2 \mathbf{F} + k^2 \mathbf{F} = - \mathbf{J}_m^s, \ \ \ \ \text{where} \ \ k^2 = \omega^2\mu\varepsilon -i\omega\mu\sigma\]

The Helmholtz equation with boundary conditions can be solved to generate \(\mathbf{F}\). For infinite media, the boundary condition is such that \(\mathbf{F} \rightarrow 0\) as \(r \rightarrow \infty\). From the Helmholtz equation, we can see that \(\mathbf{F}\) will only have a component along the direction of \(\mathbf{J_m^s}\). The scalar Green’s function for the Helmholtz equation is:

(265)\[G(r) = \frac{e^{-ikr}}{4\pi r}.\]

and hence the vector potential for an arbitrary magnetic source is:

(266)\[\mathbf{F}(\mathbf{r}) = \int_{V^\prime} \frac{e^{-ik|\mathbf{r}-\mathbf{r}'|}}{4\pi |\mathbf{r}-\mathbf{r}'|} \mathbf{J}_m^s (\mathbf{r}') dV^\prime\]

where \(\mathbf{r}\) is the observation location, \(\mathbf{r^\prime}\) refers to locations within the source region and \(V^\prime\) is the volume of the source region. For a magnetic dipole oriented in the \(\mathbf{\hat x}\) direction, the source term is given by:

(267)\[\mathbf{J}_m^s = i \omega \mu I S \delta(x) \delta(y) \delta(z) \mathbf{\hat x}\]

and the solution to Eq. (266) is:

(268)\[\mathbf{F} = \frac{i \omega \mu m}{4\pi r} e^{-ikr} \mathbf{\hat x}\]

Recall the \(\mathbf{F}\) can be used to obtain the electric and magnetic field according to Eqs. (262) and (263). Thus the electric field for a magnetic dipole in the \(\hat x\) direction is:

(269)\[\mathbf{E_m} = \frac{i \omega \mu m}{4 \pi r^2} \left( ikr + 1 \right) e^{-ikr} \left( -\frac{z}{r} \mathbf{\hat y} + \frac{y}{r} \mathbf{\hat z} \right).\]

While the magnetic field is given by:

(270)\[\mathbf{H_m} = \frac{m}{4 \pi r^3} e^{-ikr} \left[ \left(\frac{x^2}{r^2} \mathbf{\hat x} + \frac{xy}{r^2} \mathbf{\hat y} + \frac{xz}{r^2} \mathbf{\hat z} \right) \left(-k^2 r^2 + 3ikr +3 \right) + \left(k^2 r^2 - ikr -1 \right) \mathbf{\hat x} \right].\]

On the following page, we show how Eqs. (269) and (270) can be simplified for various cases.