ECE 6340 Intermediate EM Waves. Fall Prof. David R. Jackson Dept. of ECE. Notes 3

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1 C 634 Intermedate M Waves Fall 216 Prof. Davd R. akson Dept. of C Notes 3 1

2 Types of Current ρ v Note: The free-harge densty ρ v refers to those harge arrers (ether postve or negatve) that are free to move (usually eletrons or ons). It s zero for perfet nsulators. mpressed urrent (soure) onduton (ohm) urrent = ρ v v Lnear medum: = σ (Ohm s law) Note: The eletr feld s set up n response to the mpressed urrent soure. 2

3 Types of Current (ont.) ρ v Ampere s law: Η = + jωε Η = + σ + jωε Soure Conduton Dsplaement 3

4 ffetve Permttvty Η = + jωε = + σ + jωε = + σ + jωε ( ) σ = + jω ε + jω σ = + jω ε j ω 4

5 ffetve Permttvty (ont.) Defne: ε ε j σ ω Ths "effetve" permttvty aounts for the ondutvty. Note: If there s polarzaton loss (moleular or atom frton), than ε wll be omplex n addton to ε. Ampere s law beomes: Η = + jωε Ampere s law thus beomes n the same form as for free spae: H = + jωε 5

6 ffetve Permttvty (ont.) Note: ε s often alled ε for smplty n most books. However, be areful! D = ε D ε ven though the effetve permttvty appears n Ampere s law, t s the atual permttvty that relates the flux densty to the eletr feld. 6

7 ffetve Permttvty Prnple Ths prnple allows us to solve problems nvolvng a homogeneous (lossy) materal, as long as we know how to solve the orrespondng free-spae problems. H = + jωε (Free-spae problem) ε ε Η = + jωε (Materal problem) The formulas for the felds remans the same: we smply make ths smple substtuton. 7

8 xample Oean z εσ, x I = 1 y A dpole s embedded n an nfnte medum of oean water. What s the far-feld of the dpole? Frst examne problem n free spae (next slde). 8

9 xample (ont.) Dpole n free spae: z r ε, µ x θ I = 1 y As r jωµ jkr θ = sn θe, k = ω µε 4π r 9

10 xample (ont.) Dpole n oean: z Oean εσ, x I = 1 y As r jωµ jk1r = sn, = = 4π r θ θe k1 ω µε k1 jk1 ε = ε j σ ω ( j ) ε= εε = ε ε ε r r r 1

11 Loss Tangent ε = ε j σ ω Wrte ths as: ε = ε jε The loss tangent s defned as: ε tanδ ε Note: The loss tangent ombnes losses from atom and moleular frton together wth loss from ondutvty. ε = Re ( ε) σ ε = Im ( ε) + ω Note: In most books, the symbol ε s used to denote ε n the tme-harmon steady state. 11

12 Loss Tangent Some Common Materals f = 3 GHz Materal tanδ Water (pure).156 FR4.18 Durod board (typal).1 Polyethelene.31 Teflon.14 Quartz.61 Sapphre.2 12

13 Polarzaton Current H = + jωε = + σ + jωε = + σ + jωε + jω ε ε ( ) Soure Conduton Free-spae dsplaement Polarzaton p Four types of urrent densty (nonmagnet medum) ρ v p Note: The freespae dsplaement urrent s not an atual urrent flow. 13

14 Polarzaton Current (ont.) Model of polarzaton urrent: - + x q v N d dpoles per unt volume The dpoles streth rather than rotate. As the eletr feld hanges, we magne that the poston x of the postve harge hanges, wth the negatve harge beng statonary. P x = dp qx N ( ) d dx dt ( Nq) ( ) x = = d dt Nqv d x From the harge-urrent equaton: p x Hene = ρ v = qn v q v p x ( ) dp = x dt d 14

15 Polarzaton Current (ont.) In general, p d = P dt Tme-harmon steady state: ( 1) ( ) p = jω P= jωε χe = jωε ε r = jω ε ε p = jω( ε ε ) Ths agrees wth the onluson from Amperes law. 15

16 Polarzaton Current (ont.) If magnet materal s present: µ 1 B M = + jωε µ = + σ + jωε LHS s that of Ampere's law n free spae. 1 H B M = + σ + jωε + jω ε ε ( ) 1 H = B M µ 1 B = + σ + jωε ( ) + jω ε ε + M µ Polarzaton urrent from deletr propertes Polarzaton urrent from magnet propertes 16

17 Polarzaton Current (ont.) 1 B = + σ + jωε ( ) + jω ε ε + M µ Soure Note: The freespae dsplaement urrent s not an atual urrent flow. Conduton Free-spae dsplaement Fve types of urrent densty Polarzaton p Magnet polarzaton mp ρ v p mp 17

18 quvalent Current εσ, Body ε = ε j σ ω Insde the body, Nonmagnet body Defne: H = jωε = jωε + jω ε ε ( ) eq jω ε ( ε ) 18

19 quvalent Current (ont.) H = jωε + eq Interpretaton: ε, µ eq The body s replaed by ts equvalent urrent n free spae. Note: The equvalent urrent s unknown, sne the eletr feld nsde the body s unknown. 19

20 quvalent Current (ont.) The equvalent urrent ombnes the onduton urrent and the polarzaton urrent. eq jω ε ε ( ) = jω ε ε + σ ( ) ε = ε j σ ω so eq = jω ε ε + σ ( ) Polarzaton urrent Conduton urrent 2