Jωρ 0 these are all examples of the inhomogeneous

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jωρ = 0 . These are all examples of the inhomogeneous Helmholtz equation, which has a well-known solution. Notice that the equation for the vector potential separates into three scalar equations for the three vector components of A and J in Cartesian coordinates. In the case of the elemental dipole where the current flows in one Cartesian direction, it is sufficient therefore to solve one scalar equation. 2 A z + k 2 A z = μ J z (2.1) We view the right side of (2.1) as a source and the left side as a response. The general method of solution involves finding the response to a point source excitation, called a Greens function, and then constructing the solution to a particular problem from an appropriate superposition of point sources. In this case, the excitation is already a point source, and the Greens function is itself the solution we seek. We first solve the homogeneous (source free) problem. The overall symmetry of the problem suggests that the solution should depend only on the variable r . In spherical coordinates, this implies the following ordinary differential equation: 1 r 2 d dr r 2 dA z dr + k 2 A z = 0 The solution is facilitated with the change of variables given by A z = ψ/r , yielding d 2 ψ dr 2 + k 2 ψ = 0 28
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with the solutions ψ ( r ) = c 1 e jkr + c 2 e jkr The two solutions represent spherically expanding and contracting wavefronts, and while both are mathematically valid, only the former is physically meaningful. The solution to the homogeneous problem, which holds everywhere except the origin where the elemental electric dipole is located, is then A z = c 1 e jkr r All that remains is to determine the constant c 1 . This is done by reconsidering the inhomogeneous equation (with the source restored). Integrating the equation over a small spherical volume surrounding the origin with a radius r yields contintegraldisplay s A z · ˆ rds + integraldisplay v k 2 A z dv = μ Idl where the divergence theorem has been used to convert one volume integral to a surface integral over the sphere and where the volume integral over the totally enclosed current element is Idl . By considering the limit where the radius r is made arbitrarily small, it is possible to make the term immediately to the left of the equal sign vanish, since the product A z dV r 2 . The surface integral term meanwhile evaluates to 4 πc 1 in the limit that r λ . The constant is thereby set, and the solution for the vector potential due to an elemental dipole at the origin is found to be A z = μ Idl e jkr 4 πr (2.2) 2.1.2 Electromagnetic fields The electric and magnetic fields can both be found from the vector potential, since H = ∇ × A and E = A + ∇∇ · A /jωμ ǫ (invoking the potential definitions and the Lorentz condition in phasor form — see the appendix for details). A little more computation gives H φ = Idl sin θ 1 4 π bracketleftbigg jk r + 1 r 2 bracketrightbigg e j ( ωt kr ) E r = Idl cos θ Z 2 πk bracketleftbigg k r 2 + 1 jr 3 bracketrightbigg
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