An Off-Center Coaxial Cable

Transcription

1 1 Problem An Off-Center Coxil Cble Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ Nov. 21, 1999 A coxil trnsmission line hs inner conductor of rdius nd outer conductor of rdius b, but the xes of these two cylinders re offset by smll distnce δ b. Deduce the cpcitnce nd inductnce per unit length, nd the impednce Z, ccurte to order δ 2 /b 2. The reltive dielectric constnt nd permebility of the medium between the two conductors both unity. The relevnt frequencies nd conductivities re so lrge tht the skin depth is smll compred to δ. 1

2 2 Solution We use Gussin units, nd convert the impednce Z = L/C to MKSA units by noting tht 1/c = 30Ω, where c is the speed of light. We don t need to clculte both the cpcitnce C per unit length nd the inductnce L per unit length, since in the cse of perfectly conducting trnsmission line they re relted by LC = ɛµ c, 1 2 where the dielectric constnt ɛ nd the permebility µ re unity in the present cse. The ssumed smllness of the skin depth permits us to pproximte the present trnsmission line s perfectly conducting. We first present two clcultions of the cpcitnce secs. 3 nd 4, nd then clcultion of the inductnce sec. 5 s illustrtions of vrious possible techniques. 3 The Cpcitnce Vi the Imge Method It is expedient to use the imge method for 2-dimensionl cylindricl geometries. Recll tht in the cse of wire of chrge q per unit length t distnce b from ground conducting cylinder of rdius, s shown in the figure, one cn think of n imge wire of chrge q t rdius 2 /b. To pply this to the present problem, sketched in the figure below, note tht the imge wires of chrge ±q per unit length re both locted to the left of the center of the inner conductor, sy t distnces r nd r b. 2

3 For the inner cylinder to be n equipotentil, we must hve nd the outer cylinder is lso n equipotentil provided r b = 2 r b, 2 r b + δ = b2 r + δ, 3 noting the offset by δ between the inner nd outer cylinder. Combining eqs. 2 nd 3, nd noting tht r 0 s δ 0, we find r = b2 2 δ 2 b 2 2 δ δ δ The cpcitnce is relted by C = q/ V, where V = V b V is the potentil difference between the two cylinders. Recll tht the potentil t distnce r from wire of chrge q per unit length is 2q ln r + constnt. We evlute the potentils t the points where the cylinders re closest to one nother: using eq. 2, nd V = 2q ln r 2q lnr b = 2q ln r 2 /r = 2q ln r, 5 b r δ V b = 2q lnb δ r 2q lnr b b + δ = 2q ln b 2 /r + δ b = 2q ln r + δ, 6 b using eq. 3. Then, V = 2q ln [ ] 1 + δr. 7 b When combined with eq. 4, this is n exct solution for ny δ < b. In prticulr, s δ b, then r, nd the cylinders touch with the result tht V = 0. Here, we suppose tht δ b, nd expnd δ/r to second order: so tht δ = b2 2 δ 2 + b 2 2 δ δ 2 b2 2 b 2 δ 2 r b 2 2, 8 The cpcitnce per unit length is therefore, using eq δ b2 1 δ2. 9 r 2 b 2 2 C = q V 1 2 ln b, 10 δ2 b 2 2 3

4 The inductnce per unit length now follows from eq. 1: L = 2 c 2 ln b δ2 b 2 2, 11 nd the impednce is Z = L C 2 ln b c δ2 b 2 2 = 60 δ2 ln b b 2 2 Ω. 12 Remrk: The exct expression 7 is often written in different fshion, which is convenient for lrge δ, but perhps less useful for smll δ. The exct version of 8 leds to 1 + δ = b2 + 2 δ 2 + b δ b 2, 13 r 2 2 which in turn leds to C = q V = 1 2 ln b2 + 2 δ 2 + b δ b 2 2b = 1 2 cosh 1 2 +b 2 δ 2 2b Cpcitnce Vi Series Expnsion of the Potentil The imge method cn be deduced by n ppliction of series expnsion techniques for the electrosttic potentil. In this section, we explore direct use of such techniques. A full solution is long, nd when we leve off some steps t the end, we get n nswer tht is not quite correct. We define the electrosttic potentil φ to be zero on the inner conductor, φr = = 0, 15 nd V on the outer conductor whose surfce is pproximtely given by r = b + δ cos θ, The potentil is symmetric bout θ = 0: φr = b + δ cos θ = V. 16 φ θ = φθ, 17 so terms in sin nθ cnnot pper in the series expnsion of the potentil: φr, θ = A 0 ln r + n=1 A n r n + B n r n cos nθ. 18 The cpcitnce C per unit length is, of course, given by C = Q/V, where the chrge Q per unit length on the inner conductor is given by 2π 2π E r, θ Q = 2π σθ dθ = 2π 0 0 4π dθ = 2 2π 0 φ, θ r dθ = A

5 Thus, C = A 0 2V. 20 Applying the boundry condition 15 to the generl form 18, we hve 0 = A 0 ln + n=1 Likewise, the boundry condition 16 yields V = A 0 lnb + δ cos θ + n=1 A n n + B n n A n b + δ cos θ n + cos nθ. 21 B n cos nθ. 22 b + δ cos θ n With considerble effort, the terms in eq. 22 of the form cos l θ cos mθ cn be expressed s sums of terms in the orthogonl set of functions cos nθ. Then, eqs. 21 nd 22 cn be combined to yield the Fourier coefficients A n nd B n. Thus, subtrcting eq. 21 from 22 nd using the pproximtion 34, we hve V = A 0 ln b + δ cos θ δ2 cos 2 θ + F A b 2b 2 n, B n, θ 23 IF the integrl of F with respect to θ vnished, then integrting eq. 23 yields V = A 0 ln b δ2, 24 4b 2 nd the cpcitnce would be C = A 0 2V 1 2 ln b δ2 4b However, we the presence of terms like A 1 cos 2 θ in F mens tht we cnnot expect its integrl to vnish, nd eq. 25 is not quite correct. 5 Clcultion of the Inductnce The clcultion of the inductnce is complicted by the fct tht the currents in this problem re distributed over surfces, rther thn flowing in filmentry wires. We would like to use the reltion, Φ = cli, 26 where I is the totl stedy current flowing down the inner conductor nd bck up the outer conductor, nd Φ is the mgnetic flux per unit length linked by the circuit. From Ampere s lw, with the ssumption tht the currents re uniformly distributed on the inner nd outer conductors, the zimuthl component B θ of the mgnetic field in the region between the two conductors is given by B θ r = 2I cr. 27 5

6 If the cble were truly coxil, the flux would be simply Φ 0 = b nd the corresponding inductnce would be Then, from eq. 1 the cpcitnce would be B θ dr = 2I c ln b, 28 L 0 = 2 c 2 ln b. 29 C 0 = 1 2 lnb/, 30 s is redily verified by n electrosttic nlysis, nd the trnsmission line impednce would be L0 Z 0 = = 2 C 0 c ln b = 60 ln b Ω. 31 However, becuse the outer conductor is off center with respect to the inner, we cnnot simply use eq. 28. We cn segment the currents on the conductors into filments of zimuthl extent dθ, nd clculte the flux Φθ linked the circuit element defined by the segments centered on ngle θ on the inner nd outer conductors. Then, the effective inductnce of the whole cble cn be estimted from eq. 26 using the verge of Φθ: L = 1 2π Φθ dθ = 1 2π rmxθ dθ B θ rdr = 1 2π 2πcI 0 2πcI 0 πc 2 0 ln r mxθ dθ, 32 using 27 nd 28. The result holds only to the extent tht the current distribution is independent of zimuth, s discussed in sec. 6. However, there will be smll zimuthl dependence to the current in this problem, so we will not obtin completely correct result. 6

7 To complete the nlysis, we need r mx θ, the mximum rdius bout the center of the inner conductor of mgnetic field lines tht re linked by the segment of the outer conductor t zimuthl ngle θ. Assuming the currents is uniformly distributed over the inner nd outer conductors, the mgnetic field between the two conductors is entirely due to the current in the inner conductor, nd the field is purely zimuthl bout the xis of the inner conductor s given by eq. 27. Then, the geometry shown in the figure tells us tht r mx θ = b + δ cos θ. 33 This reltion is exct to the extent tht the currents re uniformly distributed; however, this is not ctully the cse in the present problem. To use reltion 33 in eq. 32, we pproximte ln r mxθ which leds to = ln b + δ cos θ = ln b + ln 1 + δ cos θ b ln b + δ cos θ b δ2 cos 2 θ 2b 2, 34 L 2 ln b c 2 δ b 2 This result hppens to gree with the result implied by sec. 4, but differs somewht from the more ccurte result of sec The Mgnetic Flux Linked by Distributed Circuit The mgnetic flux through filmentry circuit one in which the conductors re idelized s wires is well defined s Φ = B ds, 36 where the integrl is tken over ny surfce bounded by the circuit. However, when the conductors of the circuit re distributed nd hve finite cross sectionl re A, then eq. 36 is not well defined. We wish to show tht consistent definition of the flux through distributed circuit is obtined by segmenting the conductors into lrge number of circuits ech with very smll cross sectionl re A i, nd defining Φ = 1 A i Φ i, 37 A i where the mgnetic flux through subcircuit i is given by eq. 36. We re interested in definition of flux tht gives consistency to the reltion 26 in the context of circuit nlysis. In prticulr, if the circuit hs totl resistnce R, nd the mgnetic flux is chnging, then we desire Frdy s lw to be written s IR = E = 1 c dφ dt, 38 7

8 which is the sme form s holds for ech of the filmentry subcircuits: I i R i = E i = 1 c dφ i dt. 39 We suppose tht the current flowing in subcircuit i is relted to the totl current ccording to I i = A i I, 40 A in which cse the resistnce of subcircuit i is given by Then, we cn combine eqs s R i = A A i R. 41 I = i I i = 1 c i 1 R i dφ i dt = 1 cra i A i dφ i dt. 42 Hence, the definition 37 leds to the desired reltion 38 for the distributed circuit. 8