ELECTROMAGNETIC FIELDS AT BOUNDARIES

Transcription

1 ELECTROMGNETIC FIELDS T BOUNDRIES Differential Form of Maxwell s Equations: E = B t, H = J + D t D = ρ, B = 0 Gauss s Divergene Theorem: ( ) v Ddv = D nda ˆ s Stoke s Theorem: ( G) nda ˆ = î G ds S da Integral Form of Maxwell s Equations: dv V da ds C s î î D nda ˆ = ρ dv, B nda ˆ = 0 Gauss's Laws v s E ds = d B nda ˆ Faraday's Law dt H ds = ˆ + d J nda D nda ˆ mpere's Law dt L8-1

2 FIELDS PERPENDICULR TO BOUNDRIES Using Gauss s Law (s D nda ˆ = v ρdv ) : ˆ s D nda = (D 1n D 2n ) (lim δ 0) = v ρdv = σ s D 1 surfae harge density σ s D2 surfae S Therefore: D 1n D 2n = σ s yields: ( 1 2 ) s D D = σ ˆ s B nda = (B 1n B 2n ) = 0 yields: ( 1 B B = 0 2 ) L8-2

3 BOUNDRY CONDITIONS FOR PRLLEL FIELDS Using Faraday s Law: î E ds = - d ˆ dt B nda : î C E ds = (E 1// E 2 // )L (lim δ 0) = d dt B da 0 Therefore E 1// = E 2// and ( E 1 E2 ) = 0 Using mpere s Law: î H ds = J da + dt D da î H ds = (H 1// H 2 // )L (lim δ 0) = J da + d D da (J s a )L dt Therefore: ˆn ( H 1 H = J s 2 ) E 1// E 1 E 2 da, a d L -E 2// δ ds L8-3

4 FIELDS INSIDE PERFECT CONDUCTORS Eletri Fields Inside Perfet Condutors: If σ and E 0 : Then J = σe But if J : Then H sine H = J + D t But if H : But w m annot : Therefore E = 0 Then W m = µh 2 2 and w m inside perfet ondutors Sine E = 0 inside: Therefore ρ= 0 inside sine εe = ρ Magneti Fields Inside Perfet Condutors: Sine E = 0 and E = B t, therefore B t = 0 nd therefore: B = 0 inside perfet ondutors if σ = 0 first Superondutors: B 0 inside Superondutivity fails above a I 2 I B B ritial B rit = f(t emperature ) outside 1 I 1 Thus urrents in superonduting wires are limited (use ribbons) Cryotrons (Prof. Dudley Buk ~ 58) I 1 turns off I 2 (swithes and logi) L8-4

5 BOUNDRY CONDITIONS, PERFECT CONDUCTORS General Boundary Conditions: ( ( ˆn D D = σ 1 2 s ˆn B B = ) ) ˆn ( E 1 E 2 ) = 0 ( 1 2 ) s ˆn H H = J E,D 1 1 H 2,B 2 D2 = B2 = E2 = 0 inside σ = : H,B 1 1 E 2,D 2 σ σ s ˆn D = σ 1 s ˆn B = 0 1 ˆn E = 0 1 ˆn H = J 1 s B is parallel to perfet ondutors E is perpendiular to perfet ondutors L8-5

6 REFLECTIONS FROM PERFECT CONDUCTORS Solution Method for Boundary Value Problems: 1) Write fields in terms of unknown oeffiients (no boundaries); typially a sum of terms 2) Write equations for fields that satisfy boundary onditions 3) Solve for unknowns and hek answer with Maxwell Equations Example Plane Wave Perpendiular to σ = : x 1) Inident: E i = ŷe o e jkz E( z,to ) σ= z 0 Refleted: E r = ŷe r e + jkz Transmitted: E t = ŷe t e jkz = 0 here 2) Math B.B.: E (0) = 0 : // E i (z 0) + E r (z = 0) = E t (z 0) = 0 = = 3) Solve: yˆ E e jk0 + E r e + jk0 o = 0 E r = E o Standing wave L8-6

7 PURE STNDING WVES Waves Refleted by Perfet Condutor: Inident: E i = ŷe o e jkz Refleted: E r = ŷe o e + jkz H i = xˆ (E o η o )e jkz Total: E = ye ˆ o ( e jkz e + jkz ) = 2jŷEo sinkz )( H= x ˆ (E o η o e jkz + e + jkz ) = 2xˆ (E o ηo )oskz H r = xˆ (E o η o )e jkz Time Domain: E(t,z) ω = R e { Ee j t } = 2ŷEo (sinkz)sin ωt H(t,z) ω = R e { He j t } = 2xˆ (E o η o )(oskz)os ωt E y 0 x z Surfae Charge: σ = D n = 0 Cm 2 s Surfae Current: J s = H 1 = 2ŷ (E η o o )os ωt m 1 L8-7

8 POWER ND ENERGY IN STNDING WVES Waves Refleted by a Perfet Condutor: E = 2jyE ˆ sinkz ω H= 2xˆ E η Sine: E(t,z) = R e { Ee j t } = 2ŷEo (sinkz)sin ωt H(t,z) ω = R e { He j t } = 2xˆ (E o η o )(oskz)os ωt Then: S(t,z) = E H = zˆ ( E 2 η )(sin2kz)sin2ωt o o ( o o o )oskz nd: S(z) = E W e (t,z) = ε H = 2jẑ (E 2 o η o )sin2kz 2 E(t,z) 2 = 2εE o 2 ( sin 2 kz ) sin 2 ωt Jm 3 W e y E 0 z W m (t,z) = µ ( ) ( H(t,z) 2 = 2µ E o ηo os kz) os 2 ωt Jm 3 W m x H 0 z L8-8