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Chap 4 transmission line as inductor REVIEWED

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Phil's draft section dated 10.16.13, intended for Chapter 4 of his transmission line notes. It applies Stokes' theorem to B = curl A over a thin loop of width dz between the two conductors, neglecting transverse components of A. It equates the flux through the loop to the external inductance per unit length times the current, Le i(z), and notes that this excludes energy stored inside the conductors. The text breaks off partway through.

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Ch 4 Transmission line as inductor PhL 10.16.13 Something like this will get into Chapter 4. The Stokes theorem applied to B = curl A says curl A = B A ds = ∫S B dA (**) Consider the red loop shown in this top view of the two transmission line conductors. The loop is intended to have a tiny width dz, and the top view obscures the fact that each conductor has an arbitrary cross section. Since we neglect any transverse components of A, the Stokes theorem says [Az1(bottom) - Az2(top) ] dz = [ flux through red loop] We can imagine the entire transmission line as forming a long thin horizontal loop of wire as shown in blue. The long loop has some total external inductance which has this definition: (flux through blue loop) = Ltot,ext I where I is the loop current. This is an external inductance only because it does not account for the stored field energy inside the conductors, as we saw in the case of a round wire in Appendix C.3. The blue loop is a superposition of many red loops, and for the red loop shown we write [ flux through red loop] = (Ledz) i(z) where recall i(z) is the loop current. Now Le is the external inductance per unit length of the transmission line. Combining the above and cancelling the dz's we find from (4.4.1) that W(z) = Le i(z) Therefore, from (4.4.3) we have