old Jr argument REVIEWED

Old Jr argument PhL 10.12.13 This doc is now replaced by Section 3.7 (c) and (d). In retrospect, the Jr argument presented here is exactly right!! The only thing wrong is that the "sine" graph is reversed, but this does not affect anything since I just use the average of sine over a quarter wave which is invariant under flipping. You can identify the blue Gaussian box below with the left half of the green one in the Section 3.7(c) figure! This doc contains two other random items: Original Table 3 in Section 3.6 which I got rid of and stored here. Old cross section ratty fields picture from original doc (d) Jz and Jr in the conductor. In the table above for region 2 we have set Jz "large" and Jr "small". In this section, we show that the ratio Jr/Jz is quite small, that that is why Jr is small in the table. Consider the following drawing of a piece of a transmission line: In this picture the wavelength λ is highly distorted; it is intended to be much larger than the transverse dimensions of the transmission line. The picture is drawn at an instant in time when the total longitudinal current in the left conductor has its maximum value I at z = z1 and vanishes at z = z2. Thus, a total current of I is entering the interior of the left conductor between z1 and z2. This current has to go somewhere, and one can regard it as charging the capacitance of the quarter wave transmission line section between z1 and z2. In other words, this total current I is equal to the integral of the dielectric displacement current density over some area which divides the two conductors, such as the blue cylinder shown. As discussed above, the displacement current is fed by the radial current density Jr just inside the conductor. As a rough estimate, if the active perimeter of the left conductor is p, then p * λ/4 * Jr * (2/π) ≈ I => pλJr ≈ 2πI => Jr = 2πI/(pλ) (3.6.4) Here (2/π) represents the average value of the sine-shaped displacement current curve over the z region of interest, and p would be 2πa for a wire of radius a if the two conductors were widely separated. For an arbitrary conductor shape and position, p is some effective distance associated with the transverse geometry; it is the "active" perimeter discussed earlier. On the other hand, for a round conductor operating in the skin effect regime where δ < a, Jz ≈ I/(pδ) (3.6.5) where p is the same active perimeter just mentioned. So Jz ≈ I/(pδ) Jr ≈ 2πI/(pλ) Jr/Jz ≈ 2π (δ/λ) . (3.6.6) For δ we had δ ≡ (2.1.20) For λ one may write, λ = v/f = 2πv/ω (3.6.7) where v is the wave velocity. Then (δ/λ) = * = = . (3.6.8) Setting v ≈ c and μ = μ0 = 4π x 10-7 and σ = 5.81 x 107 (copper) and f = 109f(Ghz) we get (δ/λ) ≈ = = = 10-3 = 7 x 10-6 and so Jr/Jz ≈ (2π) (δ/λ) ≈ 4.4 x 10-5 (3.6.9) For f ≤ 10 GHz we then find Jr/Jz ≤ 1.4 x 10-4 . f ≤ 10 GHz (3.6.10) A round conductor is in the low-frequency non-skin-effect regime when δ >a, where a is the wire radius. This means > a => ω < 2/(μσa2) or ωa/2 < 1/(μσa) (3.6.11) In this low frequency regime we must replace (3.6.5) by Jz ≈ I/(πa2) (3.6.12) Since (3.6.4) is still valid, we find now that Jz ≈ I/(πa2) Jr ≈ 2πI/(pλ) ≈ 2πI/(2πaλ) ≈ I/(aλ) Jr/Jz ≈ π(a/λ) ≈ (πa)(ω/2πv) ≈ ωa/2v = (ωa/2)(1/v) (3.6.13) Using ** this says Jr/Jz < 1/(μσav) (3.6.14) With μ = μ0 = 4π x 10-7, σ = 5.81 x 107 (copper) and v = c = 3 x 108 we find for a wire of radius 1 mm, Jr/Jz < = = 4.6 x 10-8 (3.8.15) Therefore, our high frequency result of 1.4 x 10-4 for f ≤ 10 GHz represents a much worse case, so we will stick with that number in our work below. ****************************************************** We now display once again the table for a real transmission line, showing relative estimates of the sizes of things . A few primes have been added to remove ambiguities. [ huh? ] Table 3: E, B and J for a real transmission line Region 2. In the conductor, just under the surface charge layer, and in the current layer. Er' ≤ 10-6 Er (c) Br = 0 Jr ≤ 2 x 10-5 Jz (see later) Eφ = 0 Bφ ~ 20 μT Jφ = 0 Ez ≤ 2x10-4 Er(a) Bz ~ 0.25μT Jz = large [ ~107 A/m2 ] Region 3. In the dielectric, just outside the super-thin surface charge layer. Er = large [ 1500 v/m] Br = 0 J'r ≤ 2x10-4 Jr (d) Eφ = 0 Bφ ~ 20 μT Jφ = 0 Ez ≤ 2x10-4 Er (a) Bz ~ 0.25μT Jz ' ≈ 10-12 Jz (b) ******************************************************