Physics 506 Winter 2006 Homework Assignment #2 Solutions

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Berniece Harrington
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1 Physics 506 Winter 006 Homework Assignment # Solutions Textbook problems: Ch. 8: 8.6, A resonant cavity of copper consists of a hollow, right circular cylinder of inner radius R and length L, with flat end faces. a) Determine the resonant frequencies of the cavity for all types of waves. With (1/ µɛr) as a unit of frequency, plot the lowest four resonant frequencies of each type as a function of R/L for 0 < R/L <. Does the same mode have the lowest frequency for all R/L? This cavity is essentially covered in Section 8.7 of the textbook. It is also similar to the waveguide problem 8., but with endcaps to form a resonant cavity. The normal modes are either TM or TE modes. The TM modes are given by ψ(ρ, φ) = E 0 J m (γ mn ρ)e ±imφ, γ mn = x mn R (1) where x mn are the zeros of the Bessel functions J m. The resonant frequencies are thus (TM) ω mnp = 1 µɛr x mn + with ( ) pπr (p 0) x 01 =.05, x 11 = 3.83, x 1 = 5.136, x 0 = 5.50 The lowest four resonant frequncies are plotted as follows ω 6 L Note that the p = 0 modes are independent of R/L. Clearly the same mode does not always have the lowest frequency. The cross-over points are accidental degeneracies, and are not related to any particular symmetry of the cylinder. R/L

2 The TE modes are given by ψ(ρ, φ) = H 0 J m (γ mn ρ)e ±imφ, γ mn = x mn R where x mn are the zeros of J m. The TE resonant frequencies are (TE) ω mnp = 1 µɛr x mn + with ( ) pπr (p > 0) x 11 = 1.81, x 1 = 3.05, x 01 = 3.83, x 31 =.01 In this case, the lowest four resonant frequencies are ω 7 L R/L b) If R = cm, L = 3 cm, and the cavity is made of pure copper, what is the numerical value of Q for the lowest resonant mode? For this geometry, it turns out the lowest mode is the TM 010 mode. We thus calculate the Q factor for TM mn0 modes. We start with the stored energy U = Lɛ ψ da () where ψ is given by (1). Note that this is double the p 0 result. The power loss expression for TM mn0 modes is P loss = ɛ ( ) CL 1 + ξ mn ψ da (3) σδµ A where we have taken p = 0 into account. Here C = πr and A = πr are the circumference and cross-sectional area of the cylinder. The geometrical factor ξ mn is the same as the waveguide result, which was obtained in Problem 8. as ξ mn = 1. More directly, we may start with the definition C dψ dn A A dl = ξ mn γmn C ψ da A A

3 Using (1), this statement is equivalent to C ( ( γmnj m(x mn ) ) ) = ξ mn γmn C R π J m (x mn ρ/r) ρ dρ A 0 = ξ mn γmn C ( πr J m+1 (x mn ) ) A where the second line is obtained by the Bessel function normalization condition. This results in ( ) J ξ mn = m (x mn ) J m+1 (x mn ) However, using the Bessel recursion relation J m+1 (ξ) = (m/ξ)j m (ξ) J m(ξ) and letting ξ = x mn be a zero of J m, we obtain simply J m+1 (x mn ) = J m(x mn ). This proves that the geometrical factor is simply ξ mn = 1. Finally, using () and (3) with ξ mn = 1 gives U Q mn0 = ω mn = µ L P loss µ c δ ( 1 + CL ) = µ ( L 1 + L ) A µ c δ R Since copper is non-ferromagnetic, we may take µ c = µ 0. Furthermore, we assume the interior of the cavity has µ = µ 0 and ɛ = ɛ 0. Substituting in R = cm, L = 3 cm then yields Q mn0 = m δ We calculate the lowest resonant frequency to be ω 010 = x 01c R =.05c R = s or ν 010 = 5.9 GHz, where we have used R = 3 cm. At this frequency, the skin depth for copper is δ = m νmnp (Hz) = m This gives a cavity Q of Q 010 = For the Schumann resonances of Section 8.9 calculate the Q values on the assumption that the earth has a conductivity σ e and the ionosphere has a conductivity σ i, with corresponding skin depths δ e and δ i. a) Show that to lowest order in h/a the Q value is given by Q = Nh/(δ e + δ i ) and determine the numerical factor N for all l.

4 The long-wavelength Schumann resonances have electric and magnetic fields approximately given by E r i ɛ 0 ω l a l(l + 1)H 0P l (cos θ), H φ H 0 Pl 1 (cos θ) () where ω l l(l + 1) c a To calculate the Q value, we begin with the stored energy. [ ɛ0 U = V E + µ 0 H ] d 3 x [ ha ɛ0 dω E r + µ 0 H φ ] = µ 0ha [ ] c H 0 dω ωl a l (l + 1) P l (cos θ) + Pl 1 (cos θ) = µ 0ha H 0 π d cos θ[l(l + 1)P l (cos θ) + P 1 l (cos θ) ] Using the (associated) Legendre polynomial normalization P l (x)p l (x) dx = l + 1 δ ll, P m l (x)pl m (l + m)! (x) dx = l + 1 (l m)! δ ll gives U = µ 0 hπa l(l + 1) H 0 l + 1 For the power lost, there are two contributions, one due to the conductivity at the surface of the Earth, and the other due to the conductivity at the ionosphere boundary. For a uniform tangential magnetic field at either boundary, we approximate P lost = 1 ˆn H σδ da 1 S σδ H 0 R = πr H 0 σδ = πr H 0 σδ dωp 1 l (cos θ) d cos θp 1 l (cos θ) l(l + 1) l + 1 where R is the radius of the sphere. Since h a, we use R = a for the Earth as well as R = a + h a for the ionosphere. Furthermore, we note that 1 σδ = 1 µ cωδ (5)

5 We may use µ c µ 0. Hence the sum of the power lost at the surface of the Earth and the ionosphere is given by Combining this with (5) gives P lost µ 0 ω(δ e + δ i )πa l(l + 1) H 0 l + 1 Q = ω U P lost h δ e + δ i This demonstrates that the numerical factor N is simply N = for all l. b) For the lowest Schumann resonance evaluate the Q value assuming σ e = 0.1 (Ωm), σ i = 10 5 (Ωm), h = 10 km. For the lowest Schumann resonance, we use ν 1 = 10.6 Hz or ω 1 = 66.6 s. The skin depths are δ e = 500 m µ 0 σ e ω 1 δ i = 5 10 m µ 0 σ i ω 1 This gives a Q of Q 10 5 m 5 10 m m c) Discuss the validity of the approximations used in part a) for the range of parameters used in part b). There are several issues to worry about, mainly related to the poor conductivity of the ionosphere. Firstly, the power loss expression dp loss da = 1 σδ K eff depended on extrapolating the tangential magnetic field H (which gives rise to the effective surface current density K eff = ˆn H ) into the conductor. The expressions that was used were made in the excellent conductor approximation, σ ωɛ 0. Using the above parameters, we compute σ ωɛ so this is actually a very good approximation, despite the poor conductivity of the ionosphere. In reality, of course, a constant conductivity σ i (independent of height) is not particularly realistic. But that is outside the scope of this problem.

6 The more important issue, however, is that the calculation of part a) involved using a perfect conductor approximation for the resonant mode. The perfect conductor fields () are then substituted in to the energy and power loss expressions. So long as the the skin-depth is small compared to the size of the cavity, this is a reasonable approximation. However, here, we see that the skin-depth of the ionosphere is about 50 km, while the height of the ionosphere is 100 km. Thus the penetration of the fields into the ionosphere is hardly small compared to the (vertical) size of the cavity. Since δ i /h 1/, we can expect such corrections to be on the order of 50%. Note, however, that the correction to the Schumann modes is not as severe as it could have been, as these modes are essentially independent of height. The penetration of the fields into the ionosphere give rise to an increased effective height, but otherwise do not drastically modify the resonances. This increased effective height, however, indicates that the fields penetrate quite a bit into the ionosphere. In particular, this means there is a substantial amount of field energy that was unaccounted for in the calculation of the stored energy U of part a). From this point of view, an improved expression for Q may be of the form Q h eff δ c + δ i where h eff = h+δ i /. Nevertheless, whatever expression we use for Q is essentially a reasonable result up to a factor of order unity.

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