App C_3 version 2 REVIEWED

This is the Title PhL 3.26.05 The plan of this doc has been installed. C.3 The DC inductance of a round wire We assume here that μi is the permeability of the wire, and μe of the region outside the wire. The magnetic field for an isolated round wire is easily computed from Ampere's Law H ds = ∫S J dS 2πr Hi = I(πr2/πa2) => Hi = (r/a2) => Bi = (r/a2) 2πr He = I => He = (1/r) => Be = (1/r) The internal and external inductances of the wire (per unit length) are defined as Lidz = [ total internal magnetic flux ] / I = ∫i B dS / I = ∫i (Bi/I) dS Ledz = [ total external magnetic flux ] / I = ∫e B dS / I = ∫e (Be/I) dS The inductance of either the internal or external retion We assume here that μi is the permeability of the wire, and μe of the region outside the wire. The energy density (joules/m3) stored in an electromagnetic field within a medium of negligible loss is given by uem = (ED + BH)/2 [ Jackson p 259 Eq. (6.106) ] . We are interested only in the portion of this energy density stored in the magnetic field, so u = (1/2) BH. Since μ and ε are assumed to be non-tensor in nature, we find that u = (1/2) μ H2. For any inductor of inductance L carrying current I and having potential difference V, we know that V = L dI/dt P = IV = I (L dI/dt) = d/dt [ (1/2)L I2 ]. But since the power fed into an ideal inductor goes into the magnetic field energy, P = dU/dt and so U = (1/2)L I2 For our straight wire, we now redefine symbol L to mean inductance per unit length, so then U = (1/2)[Ldz] I2 For either the internal or external region we have U = ∫dV u = dz ∫dS u where dV is a volume element and dS is a cross sectional area element. Thus, U = dz∫dS (1/2) μ H2 = (1/2)[Ldz] I2 and then L = μ ∫dS (H/I)2 In particular Li = μi ∫dS (Hi/I)2 Le = μe ∫dS (He/I)2 For the round wire it is a simple matter to compute Hi and He using Ampere's Law (1.1.37), H ds = ∫S J dS 2πr Hi = I(πr2/πa2) => Hi/I = (r/a2) . 2πr He = I => He/I = (1/r) We then compute the two inductances as follows: Li = μi ∫dS (Hi/I)2 = μi !Syntax Error, Irdr!Syntax Error, Idθ [(r/a2)] The total energy U stored in the full field of the inductor must then be U = (1/2)L I2 in order to obtain P = dU/dt.