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
1
Inc.
wave
Ref.
wave
Medium
2
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 1
C2
Medium 2
4.6-5
Example of derivation of boundary conditions
d
C E d l dt S B d S
Medium 1
an
Lim
ad 0
bc 0
abcda
as
a
b
d
c
E d l
Lim
ad 0
bc 0
Medium 2
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 2 0
4.6-7
Summary of boundary conditions
an × E1 E2 0
or
Et1 Et 2 0
an × H1 H2 J S or
Ht1 H t 2 J S
an D1 D2 S or
Dn1 Dn 2 S
an B1 B2 0
or
Bn1 Bn 2 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 = 0
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 2
an × H1 H2 0
or
Ht1 Et 2
an D1 D2 0
or
Dn1 Dn 2
an B1 B2 0
or
Bn1 Bn 2
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
2
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
2
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 0 E1
D2
E2 2E1 2 E0 az
0
z
r>a
e 2 =e 0
(0, 0, a)
a a
0,
,
2
2
(0, a, 0)
r
e 1 = 2e 0
y
4.6-15
(b) At 0, a, 0 ,
an a y
E1 is entirely tangential
E2 E1 E0 az
a a
,
(c) At 0,
,
2 2
1
an a y az
2
an × E2 E1 0
Solve.
an D2 D1 0