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Scattering of a Plane Electromagnetic Wave by a Small Conducting Cylinder Kirk T

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Scattering of a Plane Electromagnetic Wave by a Small Conducting Cylinder Kirk T Scattering of a Plane Electromagnetic Wave by a Small Conducting Cylinder Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (October 4, 2011) 1Problem Discuss the scattering of a plane electromagnetic wave of angular frequency ω that is normally incident on a perfectly conducting cylinder of radius a (in vacuum) when the wavelength obeys λ a (ka 1). Comment also on the limit of large cylinders, ka 1, via the optical theorem. Extend the discussion to the case of small conducting elliptical cylinders, which have (small) flat strips as a limit. Relate the case of a small strip to that of a screen with a small slit using the electromagnetic version of Babinet’s principle. 2Solution The solution for small cylinders follows secs. 361-368 of [1]. The scattering cross section in cylindrical coordinates (r, φ, z), with the cylinder along the z-axis, is given by dσ power scattered into dφ Sscat(φ) = = r . (1) dφ incident power per unit area Sincident where S =(c/4π)E × B, (2) is the Poynting vector (in Gaussian units) and c is the speed of light in vacuum. Because the incident wavelength is large compared to the radius of the cylinder, and the wave is normally incident, the incident fields are essentially uniform over the cylinder, and we might suppose the induced fields near the cylinder are the same as the static fields of a conducting cylinder in an otherwise uniform electric and magnetic field. This approach was appropriate for the case of scatter by a small sphere [2], but there is no static solution for an external electric field along the axis of a conducting cylinder. Instead, we follow an approach perhaps first used by J.J. Thomson in sec. 359 of [1] in which we note that close to a small cylinder the incident plane wavefunction ei(kx−ωt) can be approximated as 1+ikx =1+ikr cos φ. We consider the total electric and magnetic fields to be the sum of the incident plane wave and a scattered wave, i(kx−ωt) −iωt i(kx−ωt) −iωt E = E0 e + Es(r, φ) e , B = xˆ × E0 e + Bs(r, φ) e , (3) 1 The electric and magnetic fields Es and Bs of angular frequency ω obey the vector Helmholtz equation (as do also the incident fields), 2 2 2 2 ∇ Es + k Es =0=∇ Bs + k Bs, (4) 2 but only the z-components ψ = Ez or Bz obey the scalar Helmholtz equation with ∇ in cylindrical coordinates, ∂2ψ 1 ∂ψ 1 ∂ψ + + + k2ψ =0, (5) ∂r2 r ∂r r2 ∂φ2 noting that the fields in this problem do not depend on z. This equation is separable, permitting solutions that are sums of terms Rn(r)cosnφ, where the symmetry of the incident wave in x implies that terms in sin nφ will not be present. The radial functions Rn obey Bessel’s equation, ∂2R ∂R n2 n 1 n k2 − R , ∂r2 + r ∂r + r2 n =0 (6) with solutions ⎧ ⎨ −i n+1 i/πkr eikr kr , (1) ( ) 2 ( large) Rn = Jn(kr)+iNn(kr)=Hn (kr) ≈ (7) ⎩ n n −i2 (n − 1)!/π(kr) (kr small,n>0). (1) (2) The Hankel functions Hn (kr) (rather than Hn (kr)) are appropriate in that for large r the −iωt solutions Rn e should be outgoing waves. A component ψ of E or B then has the form iωt ikx (1) ψe = ψ0 e + AnHn (kr)cosnφ. (8) n The Fourier coefficients An are to be determined by the conditions that the tangential com- ponent of the electric field, and the normal component of the magnetic field, vanish at the surface of the conducting cylinder. 2.1 Electric Field Polarized Parallel to the Wire In this case we identify ψ as Ez, and the condition is that Ez(a, φ)=ψ(a, φ) = 0. Close to the cylinder eq. (8) has the form iωt (1) ψe ≈ E0(1 + ikr cos φ)+ AnHn (kr)cosnφ. (9) n Clearly, An = 0 unless n = 0 or 1. Then, the condition ψ(a, φ) = 0 tells us that 2 2 A0 1 iπ A1 ika πk a A0 = − ≈ , = − ≈ , (10) E (1) C E (1) E 0 H0 (ka) 2 0 H1 (ka) 2 0 where (referring, for example, to sec. 3.7 of [3]) C =ln(2/ka) − 0.5772 < 2/ka, (11) 2 which becomes large for very small ka, but 1/C > ka. The electric field for r>ais, from eq. (8), 2 2 −iωt ikx iπ (1) πk a (1) E ≈ E0 e e + H (kr)+ H (kr)cosφ zˆ, (12) 2C 0 2 1 and the magnetic field follows from Faraday’s law as i iπka2 − ∇ × ≈ E e−iωt −eikx − H(1) kr φ B = k E 0 yˆ r 1 ( )sin ˆr (13) 2 (1) π 1 (1) (1) H (kr) + H (kr) − ik2a2 H (kr) − 1 cos φ φˆ . 2 C 1 0 kr For large r the electric and magnetic field are, neglecting terms in k2a2 compared to 1/C, given by −iωt ikx 1 iπ ikr E(kr 1) ≈ E0 e e + e zˆ, (14) C 2kr recalling eq. (7),1 and −iωt ikx 1 iπ ikr B(kr 1) ≈ E0 e −e yˆ − e φˆ . (15) C 2kr The time-average Poynting vector in the far zone is c S(kr 1) = Re(E × B) 8π cE2 sin[kr(1 − cos φ) − π/4] π ≈ 0 1 − xˆ 8π C 2kr π sin[kr(1 − cos φ) − π/4] π + − ˆr . (16) 2C2kr C 2kr The sine functions with arguments kr(1 − cos φ) − π/4 oscillate extremely rapidly in φ for large r, and average to zero. Hence, 2 cE0 π S(kr 1)≈ xˆ + ˆr ≡S0 xˆ + Sscat . (17) 8π 2C2kr At large r the cross terms in the Poynting vector involving the incident and scattered fields can be neglected, and the scattered Poynting vector Sscat is entirely due to the scattered fields. This appealing decomposition does not hold at small r. The differential scattering cross section is, according to eq. (1), dσ π ≈ , (18) dφ 2C2k 1This result appears at the top of p. 432 of [1]. 3 and the total scattering cross section is π2 π2 σ ≈ ≈ , (19) C2k k ln2(2/ka) which is much smaller than the wavelength λ, with a logarithmic dependence on the wire radius a. The result (19) gives only a faint hint of the fact that a grid of wires with a λ and spacing d such that a d λ is essentially totally reflecting for polarization parallel to the wires, as if the scattering cross section of each wire equals the spacing d λ [4]. 2.2 Electric Field Polarized Perpendicular to the Wire In this case the magnetic field is parallel to the z-axis, and we take ψ = Bz and the condition that the tangential electric Eφ(a, φ) field vanish at the surface of the wire becomes ∂B a, φ ∂ψ a, φ 1 z( ) 1 ( ) iE eika cos φ φ A H(1) ka nφ 0=k ∂r = k ∂r = 0 cos + n n ( )cos n 2 (1) ≈ iE0 cos φ − kaE0 cos φ + AnHn (ka)cosnφ n ka ka (1) ≈− E0 + iE0 cos φ − E0 cos 2φ + AnHn (ka)cosnφ. (20) 2 2 n Thus, An =0forn>2, A ka ka iπk2a2 0 ≈ − ≈− , (1) = (1) (21) E0 H ka 4 2H0 (ka) 2 1 ( ) 2 2 A1 i πk a ≈− ≈− , (22) E0 (1) 2 H1 (ka) A i πk3a3 2 ≈− ≈ ≈ , (1) 0 (23) E0 4 H2 (ka) (1) n n+1 noting that for n>0andsmallka, Hn (ka) ≈ iNn(ka) ≈ i2 n!/π(ka) . The magnetic field for r>ais, from eq. (8),2 2 2 2 2 −iωt ikx iπk a (1) πk a (1) B ≈ E0 e e − H (kr) − H (kr)cosφ zˆ, (24) 4 0 2 1 and the electric field follows from the fourth Maxwell equation as i iπka2 ∇ × ≈ E e−iωt eikx H(1) kr φ E = k B 0 yˆ + r 1 ( )sin rˆ (25) 2 2 2 (1) (1) πk a H (kr) (1) H (kr) + 1 + i H (kr) − 1 cos φ φˆ . 2 2 0 kr 2This result appears near the bottom of p. 434 of [1]. 4 For large r the electric and magnetic field are given by −iωt ikx 2 2 iπ ikr 1 B(kr 1) ≈ E0 e e − k a e − cos φ zˆ, (26) 2kr 2 and −iωt ikx 2 2 iπ ikr 1 E(kr 1) ≈ E0 e e yˆ − k a e − cos φ φˆ . (27) 2kr 2 The time-average Poynting vector in the far zone is, again neglecting the rapidly oscillating cross terms, 2 3 4 cE0 πk a 1 2 S(kr 1)≈ xˆ + − cos φ +cos φ ˆr ≡S0 xˆ + Sscat . (28) 8π 2r 4 The differential scattering cross section is, according to eq. (1), 3 4 dσ⊥ πk a 1 ≈ − cos φ +cos2 φ , (29) dφ 2 4 which is much larger in the backward hemisphere than in the forward, and the total scattering cross section is 3π2k3a4 σ⊥ ≈ , (30) 4 which is small compared to the geometric cross section 2a. The result (30) anticipates that a grid of fine wires is essentially transparent to fields polarized perpendicular to the wires [4]. 2.3 Surface Charge and Current on the Wire The surface charge and current densities σ and K on the wires of radius a are given by Er(a, φ) i ∂Bz(a, φ) c ic ς(φ)= = , K(φ)= ˆr×B(a, φ)=− rˆ×(∇×E(a, φ)).
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