Section 4 6 TRƯỜNG ĐIỆN TỪ

4 6 Boundary Conditions Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edw[.]

Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jordan Professor of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, Urbana, Illinois, USA Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India 4.6 Boundary Conditions 4.6-3 Why boundary conditions? Medium Inc wave Ref wave Medium Trans wave 4.6-4 Maxwell’s equations in integral form must be satisfied regardless of where the contours, surfaces, and volumes are Example: C3 C1 Medium C2 Medium 4.6-5 Example of derivation of boundary conditions d C E  d l  dt S B  d S Medium an Lim ad  bc   abcda as a b d c E  d l  Lim ad  bc  Medium d area B  d S  dt abcd 4.6-6 Eab ab   Ecd cd  0 Eab  Edc 0 aab  E1  E2  0 as × an  E1  E2  0 as  an × E1  E2  0 or, an × E1  E2  0 Et1  Et 0 4.6-7 Summary of boundary conditions an × E1  E2  0 or Et1  Et 0 an × H1  H2  J S or Ht1  H t  J S an  D1  D2  S or Dn1  Dn S an  B1  B2  0 or Bn1  Bn 0 4.6-8 Perfect Conductor Surface (No time-varying fields inside a perfect conductor Also no static electric field; may be a static magnetic field.) Assuming both E and H to be zero inside, on the surface, an × E = or Et 0 an × H = J S or Ht  J S an  D   S or Dn S an  B 0 or Bn 0 4.6-9 an E + - E   an JS H   H JS 4.6-10 Dielectric-Dielectric Interface S 0, J S 0 an × E1  E2  0 or Et1 Et an × H1  H2  0 or Ht1 Et an  D1  D2  0 or Dn1 Dn an  B1  B2  0 or Bn1 Bn 4.6-11 an Medium 1, e0 Dn1 En1 Dn2 En2 Bn1 Hn1 Bn2 Hn2 Et1 Et2 Medium 2, 3e0 an Medium 1, m0 Ht1 Ht2 Medium 2, 2m0 4.6-12 Example: D4.11 At a point on a perfect conductor surface,   (a) D D0 ax  2a y  2az and pointing away from the surface Find S D0 is positive D D0 ax  2a y  2az  an   D D0 ax  2a y  2az D D S an  D = D= D D  D D0 ax  2a y  2az 3D0 4.6-13   (b) D D0 0.6 ax  0.8 a y and pointing toward the surface D0 is positive D an  D D D S an  D =  D =  D D  D  D0 0.6 ax  0.8 a y  D0 4.6-14 Example: E1 E0 az for r  a (a) At 0, 0, a , an az E1 is entirely normal  D2 D1 2  E1 D2 E2  2E1 2 E0 az 0 z r>a e =e (0, 0, a) a a   0, ,   2   (0, a, 0) r