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Fundamentals of Electromagnetics:Fundamentals of Electromagnetics:A Two-Week, 8-Day, Intensive Course for A Two-Week, 8-Day, Intensive Course for
Training Faculty in Electrical-, Electronics-, Training Faculty in Electrical-, Electronics-, Communication-, and Computer- Related Communication-, and Computer- Related
Engineering DepartmentsEngineering Departments
byby
Nannapaneni Narayana RaoNannapaneni Narayana RaoEdward C. Jordan Professor EmeritusEdward C. Jordan Professor Emeritus
of Electrical and Computer Engineeringof Electrical and Computer EngineeringUniversity of Illinois at Urbana-Champaign, USAUniversity of Illinois at Urbana-Champaign, USADistinguished Amrita Professor of EngineeringDistinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, IndiaAmrita Vishwa Vidyapeetham, India
Amrita Viswa Vidya Peetham, CoimbatoreAmrita Viswa Vidya Peetham, CoimbatoreAugust 11, 12, 13, 14, 18, 19, 20, and 21, 2008 August 11, 12, 13, 14, 18, 19, 20, and 21, 2008
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3-2
Module 3Maxwell’s EquationsIn Differential Form
Faraday’s law and Ampere’s Circuital LawGauss’ Laws and the Continuity EquationCurl and Divergence
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3-3
Instructional Objectives8. Determine if a given time-varying electric/magnetic
field satisfies Maxwell’s curl equations, and if so find the corresponding magnetic/electric field, and any required condition, if the field is incompletely specified
9. Find the electric/magnetic field due to one-dimensional static charge/current distribution using Maxwell’s divergence/curl equation for the electric/magnetic field
10. Establish the physical realizability of a static electric field by using Maxwell’s curl equation for the static case, and of a magnetic field by using the Maxwell’s divergence equation for the magnetic field
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3-4
Faraday’s Law andAmpère’s Circuital Law
(FEME, Secs. 3.1, 3.2; EEE6E, Sec. 3.1)
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3-5
Maxwell’s Equations in Differential Form
Why differential form?
Because for integral forms to be useful, an a priori knowledge of the behavior of the field to be computed is necessary.
The problem is similar to the following:
There is no unique solution to this.
If y(x) dx 2, what is y(x)?01
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3-6
However, if, e.g., y(x) = Cx, then we can find y(x), since then
On the other hand, suppose we have the following problem:
Then y(x) = 2x + C.
Thus the solution is unique to within a constant.
Cx dx 2 or Cx2
2
0
1
1
02 or C 4
y(x) 4x.
If dydx
2, what is y?
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3-7
FARADAY’S LAW
First consider the special case
and apply the integral form to the rectangular path shown, in the limit that the rectangle shrinks to a point.
E Ex (z,t) ax and H Hy (z, t) a y
(x, z)
x S C
z (x, z + z)
x(x + x, z) (x + x, z + z)
zy
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3-8
,zx x yz z x z
dE x E x B x zdt
00
Lim zx xz z
xz
E E x
x z
,
00
Limy x z
xz
d
dtB x z
x z
yxBE
z t
SC
dd ddt
E l B S
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3-9
General CaseE Ex (x, y, z,t)a x Ey (x, y, z,t)ay Ez (x, y,z, t)azH Hx (x, y, z,t)a x Hy (x, y, z,t)ay Hz (x, y, z, t)az
Lateral space derivatives of the components of E
Time derivatives of the components of B
– –
– –
– –
yz x
yzx
y zx
EE By z t
BEEz x tE BEx y t
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3-10
Combining into a single differential equation,
Differential formof Faraday’s Law
–
x y z
x y z
x y z tE E E
a a aB
–
tBE
x y zx y za a a
Del Cross or Curl of = – t
BE E
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3-11
AMPÈRE’S CIRCUITAL LAW
Consider the general case first. Then noting that
we obtain from analogy,E –
t(B)
E • dl –ddt
B • dSSC
H • dl J • dS ddt D • dSSSC
H J t (D)
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3-12
Thus
Special case:
Differential form of Ampère’s circuital law
E Ex (z,t)a x , H Hy (z,t)ay
0 0
0 0
J
x y z
y
z tH
a a aD
–
y xx
H DJ
z t
tDH J
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3-13
80 cos 6 10 yE t kz E × aEx. For
in free space
find the value(s) of k such that E satisfies both
of Maxwell’s curl equations.
Noting that E Ey (z,t)a y ,we have from
0 0, , , J = 0
– –
y xx
H DJ
z t
– ,
tBE
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3-14
80
80
cos 6 10
sin 6 10
yxEB
t z
E t kzzkE t kz
808 cos 6 10
6 10xkE
B t kz
– – 0 0
0 0
x y z
y
t zE
a a aB E
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3-15
Thus,
Then, noting that we have fromH Hx (z,t)ax ,
808
70
802
cos 6 106 10
4 10
cos 6 10240
x
x
kE t kz
kEt kz
B a
B BH
a
,
tDH
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3-16
0 0
0 0
x y z
x
t zH
a a aD × H
2
802 sin 6 10
240
y xHt z
k E t kz
D
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3-17
2
803 8 cos 6 10
1440 10yk E
D t kz
2
803 8 cos 6 10
1440 10 yk E t kz
D a
90
280
2
10 36
cos 6 104 yk E
t kz
D DE
a
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3-18
80 cos 6 10 2 yE t z E a
k 2
3 108(c) m s.
Comparing with the original given E, we have
20
0 24k EE
Sinusoidal traveling waves in free space, propagating in the z directions with velocity,
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3-19
Gauss’ Laws and the Continuity Equation
(FEME, Secs. 3.4, 3.5, 3.6; EEE6E, Sec. 3.2)
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3-20
GAUSS’ LAW FOR THE ELECTRIC FIELD
D • dS S dvVz
(x, y, z)y
x
z
y
x
x xx x x
y yy y y
z zz z z
D y z D y z
D z x D z x
D x y D x y
x y z
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3-21
000
Limxyz
x y zx y z
+Δ
000
Δ Δ
Lim
x xx x x
y yy y y
z zz z z
xyz
D D y z
D D z x
D D x y
x y z
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3-22
• D
Divergence of D =
Ex. Given that
Find D everywhere.
0 for – a x a0 otherwise
Longitudinal derivativesof the components of
y zx D DDx y z
D
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3-23
Noting that = (x) and hence D = D(x), we set
0
x=–a x=0 x=a
• •• •• •• •• •• •• •• •• •• •• •
• •• •• •• •• •• •• •• •• •• •• •
0 and 0, so that
y z
•
D y zx xD DD D
x y z x
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3-24
Thus,
which also means that D has only an x-component. Proceeding further, we have
where C is the constant of integration. Evaluating the integral graphically, we have the following:
• D = gives
Dx x dx C–x
( )
xD xx
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3-25
–a 0 a x
0
–a 0 a x
(x ) dx–x
20a
From symmetry considerations, the fields on the two sides of the charge distribution must be equal in magnitude and opposite in direction. Hence,
C = – 0a
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3-26
0a
–0a
–a a x
Dx
0
0
0
for for for
x
x
x
a x ax a x aa x a
aD a
a
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3-27
B • dS = 0 = 0 dvVS
• B 0
GAUSS’ LAW FOR THE MAGNETIC FIELD
From analogy
Solenoidal property of magnetic field lines. Provides test for physical realizability of a given vector field as a magnetic field.
D • dS = dvVS
• D
• B 0
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3-28
LAW OF CONSERVATION OF CHARGE
J • dS ddt dv 0VS
• J t ( ) 0
aaaa
• J t 0
ContinuityEquation
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3-29
SUMMARY
(4) is, however, not independent of (1), and (3) can be derived from (2) with the aid of (5).
(1)
(2)
(3)
(4)
(5)
–
•• 0
BE
DH J
DB
t
t
• 0
t
J
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3-30
Curl and Divergence (FEME, Secs. 3.3, 3.6; EEE6E, Sec. 3.3)
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3-31
Maxwell’s Equations in Differential Form
Curl
Divergence
=
=
t
t
B×E
D× H J
D
B
AA A= yx z
x y z
A
A A A
x y z
x y z
x y z
a a a
× Α
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3-32
Basic definition of curl
× A is the maximum value of circulation of A per unit area in the limit that the area shrinks to the point.
Direction of is the direction of the normalvector to the area in the limit that the area shrinksto the point, and in the right-hand sense.
× A
max
LimS 0 S
Cn
d
A l× A = a
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3-33
Curl Meteris a device to probe the field for studying the curl of thefield. It responds to the circulation of the field.
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3-343-34
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3-35
0
0
2 for 0
22
2 for 2
z
z
x av xa
x av x a
a
av
a
negative for 0
2
positive for 2
y
ax
ax a
× v
0
0
2
20 0
x y z
yz
y
yz
vv a
vx y z xav
a a aa
× v aa
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3-36
Basic definition of divergence
Divergence meter
is the outward flux of A per unit volume in the limit that the volume shrinks to the point.
is a device to probe the field for studying the divergence of the field. It responds to the closed surface integral of the vector field.
0d
vv
Lim A S
A
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3-37
Example:At the point (1, 1, 0)
Divergence zero
Divergence positive
Divergence negative
(a)
(b)
(c)
21 xx a
1 yy a
yxy
a
x
y
1
z 1
y
1
z 1
x
y
1
z 1
x
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3-38
Two Useful Theorems:
Stokes’ theorem
Divergence theorem
A useful identity
C S
d dA l = × A S
S V
d dvA S = A
× A
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3-39
x y z
x y z
x y zA A A
a a a
× Α
0
x y z
x y z
x y z
x y z
x y zA A A
× A = × A × A × A
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The End