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Chapter14_7th

# Chapter14_7th - .1 The dimensions of the outer conductor of...

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CHAPTER 14 14.1. The dimensions of the outer conductor of a coaxial cable are b and c , c > b . Assume σ = σ c and let µ = µ 0 . Find the magnetic energy stored per unit length in the region b < r < c for a uniformly distributed total current I ﬂowing in opposite directions in the inner and outer conductors: First, from the inner conductor, the magnetic field will be H 1 = I 2 πρ a φ The contribution from the outer conductor to the magnetic field within that conductor is found from Ampere’s circuital law to be: H 2 = I 2 πρ ρ 2 b 2 c 2 b 2 a φ The total magnetic field within the outer conductor will be the sum of the two fields, or H T = H 1 + H 2 = I 2 πρ c 2 ρ 2 c 2 b 2 a φ The energy density is w m = 1 2 µ 0 H 2 T = µ 0 I 2 8 π 2 c 2 ρ 2 c 2 b 2 2 J / m 3 The stored energy per unit length in the outer conductor is now W m = 1 0 2 π 0 c b µ 0 I 2 8 π 2 c 2 ρ 2 c 2 b 2 2 ρ dρ dφ dz = µ 0 I 2 4 π ( c 2 b 2 ) 2 c b c 4 ρ 2 c 2 ρ + ρ 3 = µ 0 I 2 4 π c 4 ( c 2 b 2 ) 2 ln c b + b 2 (3 / 4) c 2 ( c 2 b 2 ) J 14.2. The conductors of a coaxial transmission line are copper ( σ c = 5 . 8 × 10 7 S/m) and the dielectric is polyethylene ( r = 2 . 26, σ/ω = 0 . 0002). If the inner radius of the outer conductor is 4 mm, find the radius of the inner conductor so that (assuming a lossless line): a) Z 0 = 50Ω: Use Z 0 = 1 2 π µ ln b a = 50 ln b a = 2 π r (50) 377 = 1 . 25 Thus b/a = e 1 . 25 = 3 . 50, or a = 4 / 3 . 50 = 1 . 142mm 1

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14.2b. C = 100 pF/m: Begin with C = 2 π ln( b/a ) = 10 10 ln b a = 2 π (2 . 26)(8 . 854 × 10 2 ) = 1 . 257 So b/a = e 1 . 257 = 3 . 51, or a = 4 / 3 . 51 = 1 . 138mm . c) L = 0 . 2 µ H / m: Use L = µ 0 2 π ln b a = 0 . 2 × 10 6 ln b a = 2 π (0 . 2 × 10 6 ) 4 π × 10 7 = 1 Thus b/a = e 1 = 2 . 718, or a = b/ 2 . 718 = 1 . 472mm . 14.3. Two aluminum-clad steel conductors are used to construct a two-wire transmission line. Let σ Al = 3 . 8 × 10 7 S/m, σ St = 5 × 10 6 S/m, and µ St = 100 µ H / m. The radius of the steel wire is 0.5 in., and the aluminum coating is 0.05 in. thick. The dielectric is air, and the center-to-center wire separation is 4 in. Find C , L , G , and R for the line at 10 MHz: The first question is whether we are in the high frequency or low frequency regime. Calculation of the skin depth, δ , will tell us. We have, for aluminum, δ = 1 πfµ 0 σ Al = 1 π (10 7 )(4 π × 10 7 )(3 . 8 × 10 7 ) = 2 . 58 × 10 5 m so we are clearly in the high frequency regime, where uniform current distributions cannot be assumed. Furthermore, the skin depth is considerably less than the aluminum layer thickness, so the bulk of the current resides in the aluminum, and we may neglect the steel. Assuming solid aluminum wires of radius a = 0 . 5 + 0 . 05 = 0 . 55 in = 0 . 014 m, the resistance of the two-wire line is now R = 1 πaδσ Al = 1 π ( . 014)(2 . 58 × 10 5 )(3 . 8 × 10 7 ) = 0 . 023 Ω / m Next, since the dielectric is air, no leakage will occur from wire to wire, and so G = 0 S / m . Now the capacitance will be C = π 0 cosh 1 ( d/ 2 a ) = π × 8 . 85 × 10 12 cosh 1 (4 / (2 × 0 . 55)) = 1 . 42 × 10 11 F / m = 14 . 2 pF / m Finally, the inductance per unit length will be L = µ 0 π cosh( d/ 2 a ) = 4 π × 10 7 π cosh(4 / (2 × 0 . 55)) = 7 . 86 × 10 7 H / m = 0 . 786 µ H / m 2
14.4. Each conductor of a two-wire transmission line has a radius of 0.5mm; their center-to-center

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Chapter14_7th - .1 The dimensions of the outer conductor of...

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