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Unformatted text preview: CHAPTER 14 14.1. The dimensions of the outer conductor of a coaxial cable are b and c , c > b . Assume = c and let = . Find the magnetic energy stored per unit length in the region b < r < c for a uniformly distributed total current I owing in opposite directions in the inner and outer conductors: First, from the inner conductor, the magnetic eld will be H 1 = I 2 a The contribution from the outer conductor to the magnetic eld within that conductor is found from Amperes circuital law to be: H 2 = I 2 2 b 2 c 2 b 2 a The total magnetic eld within the outer conductor will be the sum of the two elds, or H T = H 1 + H 2 = I 2 c 2 2 c 2 b 2 a The energy density is w m = 1 2 H 2 T = I 2 8 2 c 2 2 c 2 b 2 2 J / m 3 The stored energy per unit length in the outer conductor is now W m = Z 1 Z 2 Z c b I 2 8 2 c 2 2 c 2 b 2 2 d d dz = I 2 4 ( c 2 b 2 ) 2 Z c b c 4 2 c 2 + 3 d = I 2 4 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 ( c = 5 . 8 10 7 S/m) and the dielectric is polyethylene ( r = 2 . 26, / = 0 . 0002). If the inner radius of the outer conductor is 4 mm, nd the radius of the inner conductor so that (assuming a lossless line): a) Z = 50: Use Z = 1 2 r ln b a = 50 ln b a = 2 p r (50) 377 = 1 . 25 Thus b/a = e 1 . 25 = 3 . 50, or a = 4 / 3 . 50 = 1 . 142mm 1 14.2b. C = 100 pF/m: Begin with C = 2 ln( b/a ) = 10 10 ln b a = 2 (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 H / m: Use L = 2 ln b a = 0 . 2 10 6 ln b a = 2 (0 . 2 10 6 ) 4 10 7 = 1 Thus b/a = e 1 = 2 . 718, or a = b/ 2 . 718 = 1 . 472mm . 14.3. Two aluminumclad steel conductors are used to construct a twowire transmission line. Let Al = 3 . 8 10 7 S/m, St = 5 10 6 S/m, and St = 100 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 centertocenter wire separation is 4 in. Find C , L , G , and R for the line at 10 MHz: The rst question is whether we are in the high frequency or low frequency regime. Calculation of the skin depth, , will tell us. We have, for aluminum, = 1 f Al = 1 p (10 7 )(4 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 twowire line...
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This note was uploaded on 09/09/2010 for the course ECE 3025 taught by Professor Citrin during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 CITRIN

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