Phys. 506 Electricity and Magnetism Winter 2004 Prof. G. Raithel Problem Set 5 Total 40 Points. 1. Problem Points

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1 Phys. 56 Electricity and Magnetism Winter 4 Prof. G. Raithel Problem Set 5 Total 4 Points. Problem. Points The partial-wave analysis presented in Chapter.4 applied to the case of a perfectly conducting sphere with radius ka leads to the result stated in Eq..7, which applies to incident electric fields of either ɛ + (upper sign or ɛ polarization (lower sign, dω = π a (ka 4 X,± iˆn X,± ( The scattering cross section equals the radiated power per solid angle divided by the incident intensity, dω = dp sc dω /I inc = r E sc E sc/e E where E sc and E are the scattered and incident electric fields, respectively. Thus, up to a pre-factor including exp(ikr/r the term X,± iˆn X,± represents the scattered electric field in the radiation zone for the case of either clean ɛ + or ɛ polarizations. Based on the superposition principle, for an incident field with a unit polarization vector ɛ = + r (ɛ + + r exp(iαɛ ( the scattered electric field is obtained via a corresponding coherent superposition of the scattered fields of ɛ + and ɛ polarizations. Thus, for the incident polarization of Eq. the scattering cross section is dω = π a (ka 4 + r X, iˆn X, ] + r exp(iα X, + iˆn X, ] = : π a (ka 4 + r F Using that X l,m = ˆL = i l(l + ˆLYlm ( ˆφ θ ˆθ sin θ φ Y,± = sin θ exp(±iφ 8π it is found that

2 ( ˆφ X,± = cos θ ˆθ exp(±iφ 6π i Inserting into Eq. we find, with ˆn ˆθ = ˆφ and ˆn ˆφ = ˆθ, that the components of the transverse field ˆF are F θ = F φ = exp(iφ( cos θ + r exp( iφ + iα( cos θ] 6π i exp(iφ(cos θ + ir exp( iφ + iα( cos θ] 6π and dω = π a (ka 4 + r (F θf θ + F φ F φ = = k 4 a 6 8( + r k 4 a 6 8( + r ( cos θ ( + r + r cos(π α + ( cos θ ( + r r cos(φ α ] ( + r (5( + cos θ 8 cos θ + r cos(φ α( cos θ ] = k 4 a ( + cos θ cos θ ] r 4( + r sin θ cos(φ α q.e.d. A more basic but cumbersome approach is to calculate the scattered electric field in the far zone, ( E sc = k exp(ikr (ˆn p + Z m ˆn 4π r ɛ for the electric and magnetic dipoles p = 4πɛ a E in and m = π µ a B in induced by an incident field with k-vector k = kẑ E in = E + r (ɛ + + r exp(iαɛ B in = c ẑ E in. The scattering cross section follows from an elementary calculation of dω = r E sc E sc E in E in

3 . Problem. Points a: Since the skin depth is much smaller than the radius, δ R, the magnetic field does not penetrate significantly into the sphere. Also, since kr, the magnetic field in the vicinity of the sphere is essentially static (near-field limit. Because of both these facts, the H-field in the vicinity of the sphere is obtained by considering a sphere with radius R and µ = in an external homogeneous magnetic field H. Due to the absence of free currents in this model, the magnetostatic potential may be used. We first assume a magnetic field H in the z-direction. The magnetic potentials inside and outside the sphere are then of the form Φ i = l Φ o = l a l r l P l (cos θ b l r l P l (cos θ H rp (cos θ where the second term in the last equation is added to match the boundary condition H(r = Φ o (r = H = H ẑ The radial boundary condition on the surface is µh r,i = = µ H r,o, i.e. yielding b l = for l and = µ b l ( l R l P l H P l b = R H. The θ-boundary condition on the surface is H r,i = H r,o, i.e. l a l R l P l = l b l R l P l H P. With the previous result for the b l, we find a l = for l, and a = H. Result inside: Φ i = H r cos θ H r,i = H cos θ H θ,i = H sin θ B r,i = B θ,i =

4 Result outside: ( R Φ o = r + r H cos θ H r,o = ( R r + H cos θ ( R H θ,o = r + H sin θ ( B r,o = µ R r + H cos θ ( R B θ,o = µ r + H sin θ Immediately outside the surface, H r,s = and H θ,s = H sin θ. These results hold for the specialized case of H = H ẑ; in this case, all φ-components of the fields are zero. Also, from the form of Φ o it is seen that the outside field equals H plus that of a magnetic dipole with moment m = πr H. Next, we consider the field for general polarization. The outside field in the near zone equals H plus the field of a magnetic dipole m = πr H. Based on the superposition principle, we can - without further analysis - state that for a magnetic field H of the general form H = H ɛ H = H ˆk ɛ the induced magnetic moment is m = πr H ɛ H There, ɛ H is the polarization vector of the magnetic field of the incident wave, ɛ the (usual polarization vector of the electric field, and ˆk a unit vector in the direction of propagation of the incident wave. In a linear-polarization basis, a general polarization state is characterized by the equivalent forms ɛ = c ɛ + c ɛ ɛ H = c ɛ c ɛ, where c c + c c =. The magnetic field in the near-zone outside the sphere then is H = ] ˆn(ˆn m m 4πr + H ɛ H = R r (ˆn(ˆn ɛ H ɛ H + ɛ H H where ˆn is a radial unit vector. The surface field H s is obtained by setting r = R, H s = H ɛ H ˆn(ˆn ɛ H ] As a test, we can verify that this field is entirely tangential by seeing that ˆn H s =.

5 b: The absorbed power equals, by Eq. 8.5 of Jackson, P abs = σδ = R σδ ˆn H da = R σδ 9 H 4 = H R π σδ Since the incident intensity H s H sd cos θdφ ɛ H ˆn(ˆn ɛ H ] ɛ H ˆn(ˆn ɛ H] dω ɛ H ɛ H ˆn ɛ H ] dω 4π c c ˆn ɛ dω c c 4π c c 4π 4π c c 4π 4π (c c + c c 4π 4π ] + Re (c c ] ( ] ˆn ɛ dω + Re c c (ˆn ɛ (ˆn ɛ dω ] I in = Z E E = Z H H = Z H ɛ H ɛ H = Z H the absorption cross section is σ abs = P abs I in = 6R π Z σδ = 6R π µ ω Z σ = ɛ ω 6R π σ ω We find that P abs and σ abs are independent of the polarization state of the incident wave. Therefore, the results also apply for unpolarized light.

6 . Problem.8 Points a: According to Eq. 8. of Jackson, the tangential electric and magnetic fields on the surface of a non-ideal conductor with µ = µ follow E tan = µ ω σ ( i(ˆn H µ ω tan = ( i(ˆn H σ This is of the form of Eq..64, E tan = Z s (ˆn H with surface impedance µc ω Z s = σ ( i = Z kδ ( i, where we have used the skin depth δ = µ σω. b: The long-wavelength limit for l = is obtained from Eq..69 (set x = ka: α ± ( i(ka ( ka i kδ ( i ka + i kδ ( i = i(ka ( ( δ a i δ a ( + δ a + i δ a q.e.d. β ± ( i(ka ( ka i kδ( i ka + i kδ( i Since kδ < ka and, consequently, kδ, we can drop the ka in the large parentheses, and β ± ( i(ka ( i kδ( i = 4i(ka i kδ( i q.e.d. c: With X,± = 6π ( ˆφ cos θ ˆθ exp(±iφ i t := ( δ a i δ a ( + δ a + i δ a and ˆn ˆθ = ˆφ and ˆn ˆφ = ˆθ, from Eq..6 it follows for the given case

7 dω = π k αx,± ± iβˆn X,± = π 4(ka 6 k 9 = π 4(ka 6 k 9 = (ka6 8k tx,± ± iβˆn X,± ] 6π ± ˆφ t cos θ + i + i ˆθ t + cos θ] (t cos θ (t cos θ + (t cos θ(t cos θ] = (ka6 8k (tt + 4( + cos θ 4(t + t cos θ ] In first order of δ a, we find tt = δ a and t + t = Re(t = δ a. Thus, dω = (ka6 8k (5 δ a ( + cos θ 4( δ ] a cos θ As a quick test we note that the result agrees with Eq..7 in the limit δ. d: According to Eq..6, in the limit that only l = is important, as in the given case, the total absorption cross section is σ abs = π k ( αα ββ Re(α + β The terms αα and ββ are of order (ka 6. The term Re(β =, and the term Re(α is of order (ka. Thus, the only term of importance is Re(α, Re(α = (ka (ka ( δ a ( + δ a δ a ( δ a ( + δ ( δ a a + δ 4a = (ka δ a ( where the last line is valid for δ a. Thus, in first order of δ it is σ abs = π k (ka δ a = πkδa q.e.d. For δ = a we use the first line of Eq. to find Re(α = 5 (ka, and σ abs (δ = a = πka 5 This is only 4% of the result that would follow from the equation valid for δ a. Also, note that one cannot expect the underlying analysis of Chapter 8. to be very accurate for δ = a.

8 4. Problem.9a Points For ɛ r close to, we can use the Born approximation. For the given case, the normalized polarization-resolved scattering amplitude in Born approximation is, ɛ A sc D = k 4π (ɛ r ɛ ɛ exp(iq x d x r<a The integral r<a exp(iq x d x = a = 4π r a exp(iqr cos θ dω ] dr = π r sin(qr qr dr = 4π q qa z sin zdz = 4π q (sin(qa qa cos(qa = j (qa 4πa qa a r exp(iqr cos θ d cos θ ] dr Thus, ɛ A sc D = k (ɛ r ɛ ɛ a j (qa qa, and dω (ɛ, ɛ = ɛ A sc D = k 4 a 6 (ɛ r j (qa qa ɛ ɛ Averaging over the incident and summing over the exit polarizations leads to dω = k4 a 6 (ɛ r j (qa qa ( + cos θ (see Chapter. and lecture. We note that by the law of cosines it is q = k ( cos θ, where θ is the scattering angle. Trends. For ka, qa = ka ( cos θ also tends to be. This limit applies in all cases except for θ less than a critical value ka. The critical angle θ c = ka corresponds to a small forward-scattering cone π π with solid angle k a, in which the limit qa does not apply. However, for ka the solid angle k a will be negligibly small, and can be ignored. For an estimate, we may thus assume qa for all θ. Since for large qa = x it is j (x x sin(x π/, the following approximate scaling applies: j (qa qa (qa = (qa 4 Thus, the scattering cross section approximately scales as dω a ( 4 k q

9 and is clearly peaked at small q, corresponding to scattering in forward directions. Consequently, in the integral for the total scattering cross section we may set cos θ = and get σ = dω dω = k4 a 6 (ɛ r j (qa qa ( + cos θdω k 4 a 6 (ɛ r j (qa qa dω Also, with q = k ( cos θ it is σ πk 4 a 6 (ɛ r j (qa qa k q dq d cos θ = k q and k dq = πk a 4 (ɛ r k j(qa ka dq = πk a 4 (ɛ r q j(x x d(x Since for large x the scaling of j (x x range to infinity, is x, for the purpose of an estimate we may extend the integration σ πk a 4 (ɛ r σ π k a 4 (ɛ r q.e.d j (x x d(x = πk a 4 (ɛ r 4