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In the case of TE polarization the magnetic field near the edge is, by(4.375),

                                                      ∞
                                                     #
                                          ˜
                                         H z (ρ, φ, ω) =  B n cos ν n φJ ν n (k 0 ρ),  ρ < ρ 0 .
                                                     n=0
                        The current at φ = 0 is
                                                                      ˜
                                                            ˜
                                                       ˆ
                                              ˜
                                              J s (ρ, ω) = φ × ˆ zH z | φ=0 = ˆρH z (ρ, 0,ω)
                        or
                                                              ∞
                                                  ˜
                                                             #
                                                  J s (ρ, ω) = ˆρ  B n J ν n (k 0 ρ).
                                                             n=0
                        For ρ → 0 we use (E.51) to write
                                                        ∞        1       k 0    ν n
                                             ˜
                                                        #
                                             J s (ρ, ω) = ˆρ  B n           ρ .
                                                                              ν n
                                                              (ν n + 1)  2
                                                        n=0
                        The n = 0 term gives a constant contribution, so we keep the first two terms to see how
                        the current behaves near ρ = 0:
                                                                   π
                                                        ˜
                                                       J s ∼ b 0 + b 1 ρ .
                                                                   ψ
                        Here b 0 and b 1 depend on the form of the impressed field. For a thin plate where ψ = 2π
                        this becomes
                                                                  √
                                                       ˜ J s ∼ b 0 + b 1 ρ.
                        This is the companion square-root behavior to (4.377). When perpendicular to a sharp
                        edge, the current grows awayfrom the edge as ρ 1/2 . In most cases b 0 = 0 since there is
                        no mechanism to store charge along a sharp edge.

                        4.11.8   Propagation of spherical waves in a conducting medium

                          We cannot obtain uniform spherical wave solutions to Maxwell’s equations. Anyfield
                        dependent onlyon r produces the null field external to the source region, as shown in
                        § 4.11.9. Nonuniform spherical waves are in general complicated and most easilyhandled
                        using potentials. We consider here onlythe simple problem of fields dependent on r and
                        θ. These waves displaythe fundamental properties of all spherical waves: theydiverge
                        from a localized source and expand with finite velocity.
                          Consider a homogeneous, source-free region characterized by ˜ (ω), ˜µ(ω), and ˜σ(ω).
                                                                                    ˜
                                                                                         ˜
                        We seek wave solutions that are TEM r in spherical coordinates (H r = E r = 0) and
                        φ-independent. Thus we write
                                              ˜
                                                                    ˆ ˜
                                                       ˆ ˜
                                              E(r,ω) = θE θ (r,θ,ω) + φE φ (r,θ,ω),
                                                                    ˆ ˜
                                              ˜
                                                       ˆ ˜
                                              H(r,ω) = θH θ (r,θ,ω) + φH φ (r,θ,ω).
                        To determine the behavior of these fields we first examine Faraday’s law
                                          1   ∂                   1 ∂                1 ∂
                                                                                           ˜
                                                                                    ˆ
                                                                        ˜
                             ˜
                                                     ˜
                        ∇× E(r,θ,ω) = ˆ r       [sin θ E φ (r,θ,ω)] − θ ˆ  [r E φ (r,θ,ω)] + φ  [r E θ (r,θ,ω)]
                                        r sin θ ∂θ                r ∂r               r ∂r
                                             ˜
                                     =− jω ˜µH(r,θ,ω).                                        (4.378)
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