10EC36
ENGINEERING ELECTROMAGNETICS
NOTES
10EC36
Introduction to vectors
The behavior of a physical device subjected to electric field can be studied either by Field
approach or by Circuit approach. The Circuit approach uses discrete circuit parameters like
RLCM, voltage and current sources. At higher frequencies (MHz or GHz) parameters would no
longer be discrete. They may become non linear also depending on material property and
strength of v and i associated. This makes circuit approach to be difficult and may not give very
accurate results.
Thus at high frequencies, Field approach is necessary to get a better understanding of
performance of the device.
The ‗Vector approach‘ provides better insight into the various as ects of Electromagnetic
phenomenon. Vector analysis is therefore an essential tool for the study of .
The ‗Vector Analysis‘ comprises of ‗Vector Algebra‘ and ‗Vect r Calculus‘.
Any physical quantity may be ‗Scalar quantity‘ or ‗Vector quantity‘. A ‗Scalar quantity‘ is
specified by magnitude only while for a ‗Vector quantity‘ requires both magnitude and direction
to be specified.
Examples :
Scalar quantity : Mass, Time, Charge, D nsity, Pote tial, Energy etc.,
Represented by alphab ts – A, B, q, t etc
Vector quantity : Electric field, force, velocity, acceleration, weight etc., represented by
alphabets with arrow on top.
A, B, E, B
etc.,
Vector algebra : If A, B, C
are vectors and m, n are scalars then
(1) Addition
A  B  B  A
Commutativ law
citystudentsgroup
A  ( B  C)  (A  B)  C Associative law
(2) Subtraction
A - B  A  (- B)
(3) Multiplication by a scalar
mA Am
Commutative law
m (n A)  n (m A)
Associative law
(m  n) A  m A  n A
Distributive law
m (A  B)  m A  m B
Distributive law
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A ‗vector‘ is represented graphically by a directed line segment.
A ‗Unit vector‘ is a vector of unit magnitude and directed along ‗that vector‘.
aˆ A
is a Unit vector along the direction of A .
Thus, the graphical representation of A and aˆ A are
A
Unit vector aˆ A
Vector A
Also aˆ A  A / A or A  aˆ A A
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Product of two or more vectors :
(1) Dot Product ( . )
A . B  A ( B COS θ OR {
A COS θ } B , 0  θ  π
B
B

A Cos θ

A
B Cos θ
A
A.B= B.A
(A Scalar quantity)
(2) CROSS PRODUCT (X)
C = A x B = A B SIN θ ˆ
Ex ,
where ' θ ' is angle between A and B ( 0
θπ)
and nˆ is unit vector perpendicular to plane of A and B
directed such that A B C form a right handed system of vectors
A x B 
-BxA
A x ( B  C)  A x B 
Department of ECE
AxC
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10EC36
CO-ORDINATE SYSTEMS :
For an explicit representation of a vector quantity, a ‗co-ordinate system‘ is essential.
Different systems used :
Sl.No.
1.
2.
3.
System
Rectangular
Cylindrical
Spherical
Co-ordinate variables
x, y, z
ρ, , z
r, , 
Unit vectors
ax , ay , az
aρ , a , az
ar , a , a
These are ‗ORTHOGONAL‗ i.e., unit vectors in such system of co-ordinates are mutually
perpendicular in the right circular way.
citystudentsgroup
i.e., x y z ,  z , r 
RECTANGULAR CO-ORDINATE SYSTEM :
Z
x=0 plane
az
p
y=0
plane
ax
X
Y
ay
z=0 plane
ax.ayay.azaz.ax0a
x
xayaz
ayxazax
azxaxay
az is in direction of ‗advance‘ of a right circular screw as is turned from ax to ay
Co-ordinate variable ‗x‘ is intersection of planes OYX and OXZ .e, z = 0 & y = 0
Location of point P :
If the point P is at a distance of r from O, then
If the components of r along X, Y, Z are x, y, z then
r  x a x  y a y  z az  r ar
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Equation of Vector AB :
If OA  A  Ax a x  Ay a y  Az a z
B
and OB  B  Bx a x  By a y  Bz a z then
A  AB  B or AB  B - A
0
where As , Ay & Az are components of
and Bs , By & Bz are components of
AB
B
A
A
A along X, Y and Z
B along X, Y and Z
Dot and Cross Products :
A . B  (Ax a x  Ay a y  Az a z ) . (Bx a x  By a y  Bz a z )  Ax Bx  Ay By  Az Cz
citystudentsgroup
A x B  (Ax a x 
A a
y
y

A
z
az ) x (Bx a x  By a y  Bz a z )
Taking 'Cross products' term by term and grouping, we get
ax ay
A x B 
Ax
Bx
A y Az
By Bz
Ax
A . (B x C ) 
az
Ay
Az
x
y
z
Cx
Cy
Cz
B
B
B
If A, B and C are non zero vectors,
(i) A . B  0 then Cos θ  0 .e., θ  90
0
A and B are perpendicular
A x B  0 then Sin θ  0
θ  0 A and B are parallel
(ii) A . ( B x C) represents the volume of a parallelopoid of sides A , B and C
Unit Vector along AB
a AB 
AB
AB
where
Vector length AB  AB
Department of ECE
 (AB . AB)
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Differential length, surface and volume elements in rectangular co-ordinate systems
r  x aˆ x  y aˆ y  z aˆ z
r
dr  x
r
r
dx  y dy  z
dz
dr  dx aˆ x  dy aˆ y  dz aˆ z
Differential length dr  [ dx
2
2
2 1/2
 dy  dz ]
-----1
Differential surface element, ds
1. 
r
to z : dxdy aˆ z
2.  to z : dxdy aˆ z
r
3.  to z : dxdy aˆ z
r
------ 2
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Differential Volume element
dv = dx dy dz
------ 3
z
p‘
dx
p
dy
r
dz
r dr
0
y
x
Other Co-ordinate sy tems :Depending on the geometry of problem it is easier if we use the appropriate co-ordinate system
than to use the Car esian co-ordinate system always. For problems having cylindrical symmetry
cylindrical -ordinate system is to be used while for applications having spherical symmetry
spherical o-ordinate system is preferred.
Cylindrical Co-ordiante systems :z
P(ρ, , z)
r
ρ
ap
r
az
x = ρ Cos 
y = ρ Sin 
z=z
2
2
ρ x y
0

y
-1
φ  tan y / x
zz
ρ
x
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10EC36
r  x aˆ x  y aˆ y  z aˆ z
r  ρ Cos  aˆ x  ρ Sin  aˆ y  z aˆ z
r
r
r
dr  ρ d ρ   d  z dz
------1
r
r
ρ  Cos  aˆ x  Sin  aˆ y  ρ aˆ   hρ aˆρ
r  - ρ Sin  aˆ  ρ Cos  aˆ   r


x
y
aˆ

h
;
ρ
r
ρ  1

 ρ aˆ ; h   r  ρ



r  aˆ
z
z
h   r  1
z
z
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Thus unit vectors in (ρ, , z) systems can be expressed in (x,y,z) system as
aρ  Cos  a x  Sin  a y
a x  Cos  a   Sin  a
a   - Sin  a x  Cos  a y
a y  Sin  a   Cos  a 
az az
;
a  , a and a z are orthogonal
Further , dr  d ρ aˆ   ρ d aˆ  dz aˆ z
2
and dr
------2
 d ρ  (ρ d)  (dz)
2
2
2
Differential areas :
ds aˆz  (d ρ) (ρ d) . aˆz
ds aˆ   (dz) (ρ d) . aˆ 
-------3
ds aˆ (d ρ dz) aˆ
Differential volume :
d  (d ρ) (ρ d) (dz)
or d  ρ d ρ  dz
Spherical Co-ordinate Systems :Z
----- 4
z
X = r Sin  Cos 
Y = r Sin  Sin 
Z = r Cos 
p
R

0
x
r
y

Y
r Sin
X
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R  r Sin  Cos  aˆ x  r Sin  Sin  aˆ y  r Cos  aˆ z
aˆ
r
 R / R  Sin  Cos  aˆ x  Sin  Sin  aˆ y  Cos  aˆ z
r
r

R
R
 Cos  Cos  aˆ x  Cos  Sin  aˆ y  Sin  aˆ z
/
aˆ 
 

R
R
 - Sin  aˆ x  Cos aˆ y
/
aˆ 
 
dR 
R
r
dr 
R
d 

R
d

dR  dr aˆ r  r d aˆ  r Sin  d aˆ
2
d S r  r Sin  d d
2
d S  r Sin  dr d

d S  r dr d
2
dvr
Sin  dr d d
General Orthogonal Curvilinear Co-ordinates :z u1 a3
u3
a1
u2
a2
y
x
Co-ordinate Variables : (u1 , u2, u3)
; Here
u1 is Intersection of surfaces u2 = C & u3 = C
u2 is Intersection of surfaces u1 = C & u3 = C
u3 is Intersection of surfaces u1 = C & u2 = C
aˆ1 , aˆ 2 , aˆ3 are ubnit vectors tangential to u1 , u2 & u3 System
is Orthogonal if aˆ1 . aˆ 2  0 , aˆ 2 . aˆ3  0 & aˆ3 . aˆ1  0
If R  x aˆ x  y aˆ y  z aˆ z & x, y, z are functions of u1 , u2 & u3
then d R   R du   R du   R du
u1 1 u2
 u3
2
3
 h1 du1 aˆ1  h2 du2 aˆ 2  h3 du3 aˆ3
where h1 , h2 , h3 are scale factors ;
h 1   R , h 2   R
 u1
 u2
,h
3
  R

u3
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Co-ordinate Variables, unit Vectors and Scale factors in different systems
Systems
Co-ordinate Variables Unit Vector
General
u1
Scale factors
u2
u3
a1
a2
a3
h1
h2
h3
Rectangular x
y
z
ax
ay
az
1
1
1
Cylindrical ρ

z
aρ
a
az
1
ρ
1
Spherical


ar
a
a
1
r
r sin 
r
Transformation equations (x,y,z interms of cylindrical and spherical co-ordinate system
variables)
Cylindrical : x = ρ Cos  , y = ρ Sin  , z = z ; ρ  0, 0  2 - < z < 
Spherical
x = r Sin Cos , y = r Sin Cos , z = r Sin
r 0 , 0  , 0  2
 V  1  v aˆ  1  v aˆ  1  v aˆ
1
2
h
2
h3 u 3
u 2
h 1 u 1
1  

. A 

(h 2 h 3 A 1 ) 
(h 1 h 3
h h
h 1 2 3   u1
u2
h
h
h 1 aˆ 1
2
3
aˆ 3
aˆ 2



1
 x A  h 1 h 2 h 3
u
u
h A1
1
h A2
1
where V  V ( u 1 , u 2 , u 3 )
2
2
3
A 2 ) 

u3

(h 1 h 2 A 3 ) 

u
h A3
3
a Scalar
3
field
& A  A 1 aˆ 1  A 2 aˆ 2  A 3 aˆ 3 is a Vector field where A1  A1 (u1 , u2 , u3 )
A2  A2 (u1 , u2 , u3 ) and A3  A3 (u1 , u2 , u3 )
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Vector Transformation from Rectangular to Spherical :
Rectangula r : AR  Ax aˆ x  Ay aˆ y  Az aˆ z
 ( AR aˆ r ) aˆ r  (AR aˆ ) aˆ   (AR . aˆ ) aˆ
Spherical : AS
 Ar aˆ r  A aˆ  A aˆ
where Ar , A , A are related to Ax , Ay , Az as
A r 

A
 


 

aˆ

x
aˆ
. aˆ
r
aˆ x . aˆ
aˆ
x
. aˆ
A
. aˆ
r
y
aˆ

. aˆ
y
aˆ y

A x 
aˆ z . aˆ 
aˆ . aˆ 



. aˆ

 
aˆ z . aˆ r 
z



A

y 
A 
 z 
A ‗field‘ is a region where any object experiences a force. The study of performance in the
presence of Electric field (E) , Magnetic field () is the essence of EM Theory.
P1 : Obtain the equation for the line between the points P(1,2,3) and Q (2,-2,1)
PQ ax- 4 ay- 2 az
P2 : Obtain unit vector from the origin to G (2, -2, 1)
Problems on Vector Analysis
Examples :1. Obtain the vector equation for the line PQ between the points P (1,2,3)m and Q (2, -2, 1)
m
Z
PQ
P (1,2,3)
Q(2,-2,-1)
0
Y
X
The vector PQ  (xq - xp ) aˆ x  (yq - yp ) aˆ y  (zq - zp ) aˆz
 (2 - 1) aˆ x  (-2 - 2) aˆ y  (-1 - 3) aˆz

 (aˆ x - 4 aˆ y - 2 aˆz )

2. Obtain unit vector from origin to G (2,-2,-1)
G
G
0
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The vector G  (xg - 0) aˆ x  (yg - 0) aˆ y  (zg - 0) aˆ z
 (2 aˆ x - 2 aˆ y - aˆ z )
ˆ

The unit vector , ag
2
G
G
2
G  2
2
 (-2)
 (-1)
3
 aˆg  (0.667 aˆ x - 0.667 aˆ y - 0.333 aˆ z )
3. Given
A  2 aˆ x - 3 aˆ y  aˆz
B  - 4 aˆ x - 2 aˆ y  5aˆz
find (1) A . B and (2) A x B
Solution :
(1) A . B  (2 ax - 3 a y  az ) . (-4 ax - 2 a y  5 az )
=-8+6+5= 3
Since ax . ax = ay . ay = az . az = 0 and ax ay = ay az = az ax = 0
(2) A x B 
ax
a y az
2
4
3 1
2 5
= (-13 ax -14 ay - 16 az)
4. Find the distance between A( 2, /6, 0) and B = ( 1, /2, 2)
Soln : The points are given in Cylindrical Co-ordinate (ρ,, z). To find the distance between
two points, the co-ordinates are to be in Cartesian (rectangular). The corresponding
rectangular co-ordinates are (ρ Cos, ρ Sin, z)
 A  2 Cos
& B  Cos


aˆ
6
aˆ
6
x
 Sin
x

 2 Sin

aˆ  1.73 aˆ  aˆ
6
y
x
aˆ  2 aˆ  aˆ  2 aˆ
2
y
z
y
y
z
 AB  (Bx - Ax ) aˆ x  (By - Ay ) aˆ y  (Bz - Az ) aˆ z

 -1.73 aˆ x  (1 -1) aˆ y  (2 - 0) aˆ z
 - 1.73 aˆ x  2 aˆ z
2
 (AB)  1.73
2
 2 2.64
5. Find the distance between A( 1, /4, 0) and B = ( 1, 3/4, )
Soln : The specified co-ordinates (r, , ) are spherical. Writing in rectangular, they are (r
Sin  Cos , r Sin  Sin , r Cos ).
Therefore, A & B in rectangular co-ordinates,
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A  (1 Sin

Cos 0 aˆ  1 Sin
4
 ( 0.707 aˆ x


Sin 0 aˆ y  1 Cos aˆ z )
x
4
4
 0.707 aˆ y )
B  ( Sin 3 Cos  aˆ x  Sin 3Sin  aˆ y  Cos 3 aˆ z )
4
4
 ( 0.707 aˆ x  0.707 aˆ y )
4
AB  (Bx - Ax ) aˆ x  (By - Ay ) aˆ y  (Bz - Az ) aˆ
z
 - 1.414 aˆ x  (- 0.707) aˆ y  (-0.707) aˆ z
AB  ( AB . AB )1/2
 (2  0.5  0.5)1/2  1.732

6. Find a unit vector along AB in Problem 5 above.
ˆ
AB
a
AB
 AB
1
= [ - 1.414 ax + (-0.707) ay + (-0.707) az] 1.732
= ( - 0.816 aˆx - 0.408 aˆ y  0.408 aˆz )
7. Transform F  (10 aˆx - 8 aˆ y  6 aˆz ) into F in Cylindrical Co - ordinates.
Soln :
FCyl  (F . aˆp ) aˆp  (F . aˆ ) aˆ  (F . aˆz ) aˆz
 [(10 aˆ x - 8 aˆ y  6 aˆ z ) . (Cos  aˆ x  Sin  aˆ y )] aˆ 
 [ (10 aˆ x - 8 aˆ y  6 aˆ z ) . (- Sin  aˆ x  Cos aˆ y )] aˆ
 [ (10 aˆ x - 8 aˆ y  6 aˆ z ) . (aˆ x )] aˆ x
ρ
 (10 Cos  - 8 Sin  ) aˆ   (-10 Sin  - 8 Cos  ) aˆ 6 aˆ z

x   Cos 
 x
y   Sin 
  tan
2
-1
2
y
 12.81
y
x
y
0
 - 38.66
x
FCyl  [ 10 Cos (- 38.66) - 8 Sin ( - 38.66) ] aˆp  [- 10 Sin (- 38.66) - 8 Cos (- 38.66)] aˆ
 6 aˆz  (12.8 aˆ   6 aˆz )
8. Transform B  y aˆx - x aˆ y  z aˆz into Cylindrical Co-ordinates.
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10EC36
x   Cos  , y   Sin 
B   Sin  aˆ x -  Cos  aˆ y  z aˆ z
BCyl  (B. aˆ  ) aˆ   (B. aˆ ) aˆ  (B . aˆ z ) aˆ z
 [ (  Sin  aˆ x -  Cos  aˆ y  z aˆ z ). (Cos  aˆ x  Sin  aˆ y )] aˆ 
 [ (  Sin  aˆ x -  Cos  aˆ y  z aˆ z ). (- Sin  aˆ x  Cos  aˆ y )] aˆ   z aˆ z
 [  Sin  Cos  -  Sin  Sin ] aˆ   [ -  Sin  -  Cos ] aˆ  z aˆ z
2
2
 -  aˆ  z aˆ z

9. Transform 5 aˆ
x
into Spherical Co-ordinates.
ASph  (A. aˆ r ) aˆ r  (A. aˆ ) aˆ  (A. aˆ ) aˆ
 [ 5 aˆ x . (Sin  Cos  aˆ x  Sin  Sin  aˆ y  Cos  aˆ z )] aˆ r 
[ 5 aˆ x . (Cos  Cos  aˆ x  Cos  Sin  aˆ y - Sin  aˆ z ] aˆ 
[ 5 aˆ x . (- Sin  aˆ x  Cos  aˆ y )] aˆ
 5 Sin  Cos  aˆ r  5 Cos  Sin  aˆ  5 Sin  aˆ

10. Transform to Cylindrical Co-ordinates G  (2 x  y) aˆx - (y - 4x) aˆ y at Q (, , z)
Soln :
GCyl  (G. a  ) aˆ   (G. a ) aˆ  (G. a z ) aˆ z
GCyl  [ (2 x  y) aˆ x - (y - 4x) aˆ y ] . [ Cos  aˆ x  Sin  aˆ  ] aˆ  
[ (2 x  y) aˆ x - (y - 4x) aˆ y ] . [ - Sin  aˆ x  Cos  aˆ y ] aˆ  0
 [ ( 2x  y) Cos  - (y - 4x) Sin ] aˆ 
 [ - (2 x  y) Sin  - (y - 4x ) Cos ] aˆ
x   Cos  , y   Sin 
GCyl  [ ( 2  Cos    Sin  ) Cos  - (  Sin  - 4  Cos ) Sin ] aˆ 

 [ - ( 2  Cos    Sin  ) Sin  - (  Sin  - 4  Cos ) Cos ] aˆ
 [ 2  Cos2    Sin  Cos  -  Sin 2   4  Sin  Cos ] aˆ 
 [ - 2  Sin  Cos  -  Sin 2  -  Sin  Cos   4 Cos2  ] aˆ

 ( 2  Cos2   5  Sin  Cos  -  Sin 2  ) aˆ 
 ( 4  Cos2    Sin 2  - 3  Sin  Cos ) aˆ

11. Find a unit vector from ( 10, 3/4, /6) to (5, /4, )
Soln :
A(r, , ) expressed in rectangular co-ordinates
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OA  r Sin  Cos  aˆ x  r Sin  Sin  aˆ y  r Cos  aˆ z 10 Cos
 A  10 Sin
B  5 Sin

3
Cos
4

aˆ
x
6
Cos  aˆ
4
x
 10 Sin
 5 Sin

4
A  6.12 aˆ x  3.53 aˆ y - 7.07 aˆ z
3
Sin

3
aˆ
4
6
y
Sin  aˆ  5 Cos
y

4
aˆ
4
aˆ
z
z
B  - 3.53 aˆ x  3.53 aˆ z
AB  B - A  - 9.65 aˆ x - 3.53 aˆ y  10.6 aˆ z
2
AB  9.65
aˆ AB
2
 3.53  10.6
 AB  (- 0.65 aˆx
AB
2
 14.77
- 0.24 aˆ y  0.72 aˆ z )
12. Transform F  10 aˆx - 8 aˆ y  6 aˆz into F in Spherical Co-orindates.
aˆ r  Sin  Cos  aˆ x  Sin  Sin  aˆ y  Cos  aˆ z
aˆ  Cos  Cos  aˆ x  Cos  Sin  aˆ y - Sin  aˆ z
aˆ  - Sin  aˆ x  Cos  aˆ y


FSph  (F . aˆ r ) aˆ r  (F . aˆ ) aˆ  (F . aˆ ) aˆ
 (10 Sin  Cos  - 8 Sin  Sin   6 Cos  ) aˆ r
 (10 Cos  Cos  - 8 Cos  Sin  - 6 Sin  ) aˆ
 (- 10 Sin  - 8 Cos  ) aˆ
r  102  82  62  200 ;   Cos
-1
z  Cos-1
r
6  64.890
200
-1
0
  tan - 8  - 38.66
10
Sin   Sin 64.69  0.9
Sin   Sin (-38.66)  - 0.625
Cos   Cos 64.69  0.42 Cos   Cos (-38.66)  0.781
F  (10 x 0.9 x 0.781 - 8 x 0.9 x (-0.625)) aˆ r
 ( 10 x 0.42 x 0.781 - 8 x 0.42 x (0.625)) aˆ
 (-10 x - 0.625 - 8 x 0.781) aˆ
F  (11.529 aˆ r  5.38 aˆ  0.783 aˆ )
Line Integrals
In general orthogonal Curvilinear Co-ordinate system
dl  h1 du1 aˆ1  h2 du2 aˆ 2  h3 du3 aˆ3
F  F1 aˆ1  F2 aˆ2  F3 aˆ3
3
CF . dl  h1C F1 du1  h2 FC2 du2  h3 F3 du
C
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
Conservative Field A field  is said to be conservative if it is such that  . dl  0
C
 . dl  
b
d (b) -  (a) (does not depend on the path !). If  is electrostatic flux, then
a
E  -  represent the electric field intensity and

b
 . dl represent the potential between b and a and is zero if it is taken around a closed contour.
a

i.e.,  . dl  0
Therefore ES flux field is ‗Conservative‘.
EXAMPLES :

I  a . dl
13. Evaluate
line
integral
2
along y  x from A (1,1) to B (4,2)
where
a  (x  y) aˆ x  (y - x) aˆ y
Soln : dl  dx aˆx  dy aˆ y
a . dl  (x  y) dx  (y - x )
2
dy y  x or 2 dy dy  dx
2
2
 a . dl   (y  y) 2y dy  (y - y ) dy
2
2
1

1

(2 y3  2 y2  y - y2 ) dy

(2 y
2
1

2
3
y
2
 y) dy
1
2 y
4



 4 2
2 4



3
2

y 2


3

 2
8

 8 

y
3
2
3

3

2


2 
 4 
 2  - 

2
2 1
 1
- 
1

2
3
14. Evaluate the Integral I  E . ds

1
3
3
-1
 3 
1 
2 
1
3
2
 12

 11
where E  x aˆx and S is hunisphere of radius a
S
Soln:
If S is hemisphere of radius a, then S is defined by
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10EC36
x
2
2
y
z
2
a
2
,z0;
ds  (a d ) (a Sin  ) d aˆ r
ds  a Sin  d d aˆ r
2
E  (E . aˆ r ) aˆ r  (E. aˆ ) aˆ  (E. aˆ ) a
Er  x Sin  Cos  aˆ r
; x  a Sin  Cos 
E . ds  E r . ds  a ( Sin  Cos  ) aˆ r . a Sin  d d
2
2
E . ds  a Sin  Cos  d d
0     / 2 , 0    2
3
/2
2 
E . ds  a Sin  d
3
2
3
0
0 Cos
2
 d  a
3
x
2
3
x2a
3
3
where r, r1 , r2 ….. rm are the vector distances of q, q1 , …… qm from origin, 0.
r - rm is distance between charge qm and q.
aˆ m is unit vector in the direction of line joining qm to q.
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Unit-1
a. Coulomb’s Law and electric field intensity: Experimental law of Coulomb
  Electric field intensity
  Field due to continuous volume charge distribution
  Field of a line charge
b. Electric flux density, Gauss’ law and divergence:
  Electric flux density
  Gauss‘ law
  Divergence, Maxwell‘s First equation(Electrostatics)
 vector operator and divergence theorem
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Electric field is the region or vicinity of a charged body where a test charge experiences a
force. It is expressed as a scalar function of co-ordinates variables. This can be illustrated by
drawing ‗force lines‘ and these may be termed as ‗Electric Flux‘ represented by  and unit
is coulomb (C).
Electric Flux Density (D) is the measure of cluster of ‗electric lines of force‘. It is the
number of lines of force per unit area of cross section.
ψ
2
i.e., D  A c/m
or ψ  D nˆ ds C
where nˆ is unit vector normal to surface
S
Electric Field Intensity (E) at any point is the electric force on a unit +ve charge at that
point.
q
i.e., E
F
1
 
2
q
4 r
aˆ1
N/c
0 1

1  q 

 aˆ
  4  r2 
0

1 
D

N / c 
1
1
N / c or D 
E C in vacuum
0
0
In any medium other than vacuum, the field Intensity at a point distant r m from + Q C is
Q
E  4   r2 aˆ r N / c ( or V / m)
0r
Q
and D 0 r E Cor D 
4 r
2
aˆr C
Thus D is independent of medium, while E depends on the property of medium.
r
+QC
E
q = 1 C (Test Charge)
Source charge
E
E 
0
r,m
Electric Field Intensity E for different charge configurations
1. E due to Array of Discrete charges
Let Q, Q1 , Q2 , ……… Qn be +ve charges at P, P1 , P2 , ……….. Pn . It is required to find
E at P.
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Q1r 1
En
P1
Q2
P
r2
E
2
E1
P2
Qn
r1
0
rn
Pr
Er
 4 
1
Qm
 
2
aˆm V / m
r-r
0
m
2. E due to continuous volume charge distribution
aˆ R
R
P
ρv C / m
3
The charge is uniformly distributed within in a closed surface with a volume charge density
dQ
3
of ρv C / m i.e, Q  V dvand V 
dv
V
Q
 E 
aˆ
4  R 2 R
0
Er
 V (r )

1

aˆ R
0
1
2
1
4   (r - r )
V
V  V
2
4  R
aˆ R N / C
0
aˆ R is unit vector directed from ‗source‘ to ‗filed point‘.
3. Electric field intensity E due to a line charge of infinite length with a line charge
ˆ
a
density of ρl C / m
R
P
R
ρl C / m
dl
L

Ep

4
1
  R
l
dl
2
aˆR N / C
0 L
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4. E due to a surface charge with density of ρS C / m
2
ˆ
a R
P (Field point)
ds
R
(Source charge)
Ep

1


4
S ds

R
2
aˆR
N/C
0 S
Electrical Potential (V) The work done in moving a unit +ve charge from Infinity to that is
called the Electric Potential at that point. Its unit is volt (V).
Electric Potential Difference (V12) is the work done in moving a unit +ve charge from one
point to (1) another (2) in an electric field.
Relation between E and V
If the electric potential at a point is expressed as a Scalar function of co-ordinate variables
(say x,y,z) then V = V(x,y,z)
dV  -
f
 - E . dl
dl
- - - - - - - - (1)
q
Also, dV  V dx  V dy  V dz
x
y
z
dV  V . dl
- - - - - - - - - (2)
From (1) and (2) E
-V
Determination of electric potential V at a point P due to a point charge of + Q C
aˆ l
R  dR

0
+Q
At point P, E

R
P
aˆ R
Q
2 aˆR N / C
4  R
0
Therefore, the force f on a unit charge at P.
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10EC36
Q
 4   R 2 aˆR N
 f  1 x Ep
0
The work done in moving a unit charge over a distance dl in the electric field is
dV - f . dl  - E . dl
R
Q dl
 Vp  -  4   R 2 (aˆ R . aˆl ) 
V
P
0
Q
 4   R 2 Volt
R

Q
4  

2
R dR
0
(a scalar field)
0
Electric Potential Difference between two points P & Q distant Rp and Rq from 0 is
V
pq
 1
R
Q
 (Vp - Vq ) 
4 0


p
-
1 
 volt
Rq 

Electric Potential at a point due to different charge configurations.
1. Discrete charges
. Q1
.
Q2
Qm
P
Rm
V
1P
1

4 
n

1
Q
m
V
R
m
0
2. Line charge
V 
xP
2P
1
4  
0

ll dl V
R
l
ρl C / m
3. Surface charge
V
xP
ρs C / m
3P
2

1
4 0

 ds
S
R
V
S
4. Volume charge
xP
3
ρv C/ m
V 
R
4P
V dv
1
4  

R
V
0 V
5. Combination of above V5P = V1P + V2P + V3P + V4P
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Equipotential Surface : All the points in space at which the potential has same value lie on a
surface called as ‗Equipotential Surface‘.
Thus for a point change Q at origin the spherical surface with the centre of sphere at the
origin, is the equipotential surface.
Sphere of
Radius , R
R
P
0
equipotential surfaces
+Q
Q
V
0
R
Potential at every point on the spherical surface is
V
R

Q
4 0 R
volt
VPQ is difference of potential two equipotential surface potential
Gauss’s law : The surface integral of normal component of D emerging from a closed
surface is equal to the charge contained in the space bounded by the surface.
i.e., D . nˆ ds  Q C
(1)
S
where ‗S‘ is called the ‗Gaussian Surface‘.
By Divergence Theorem,
 D . nˆ ds   . D dv
S
----------- (2)
V

Also, Q  V dv
---------- (3)
V
From 1, 2 & 3,
----------- (4) is point form (or differential form) of Gauss‘s law while
.D

equation (1) is Integral form of Gauss law.
Poisson’s equation and Laplace equation
In equation 4, D 0 E
 . E   / 0 or  . (-  V)   / 0
2

 V - Poisson equation
0
If   0,  V  0 Laplace equation
2
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Till now, we have discussed (1) Colulomb‘s law (2) Gauss law and (3) Laplace equation.
The determination of E and V can be carried out by using any one of the above relations.
However, the method of Coulomb‘s law is fundamental in approach while the other two use
the physical concepts involved in the problem.
(1) Coulomb’s law : Here E is found as force f per unit charge. Thus for the simple case of
point charge of Q C,
1  Q 
E 

 V/ M
2
4 0  R 
V  E dl
Volt
l
(2) Gauss’s law : An appropriate Gaussian surface S is chosen. The charge enclosed is
determined. Then
D nˆ ds  Q
enc
S
Then D and hence E are determined

Also V  E dl
volt
l
(3) Laplace equation : The Laplace equation 2 V  0
is solved subjecting to different
boundary conditions to get V. Then, E  -  V
Solutions to Problems on Electrostatics :1. Data : Q1 = 12 C , Q2 = 2 C , Q3 = 3 C at the corners of equilateral triangle d m.
To find : F on Q3
Solution :
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Let Q1 , Q2 and Q3 lie at P1 , P2 and P3 the corners of equilateral triangle of side d meter.
If P1 , P2 and P3 lie in YZ plane, with P1
at origin the n
Z
P1  (0,0,0) m
P3
P2  (0, d, 0) m
P3  (0, 0.5 d, 0.866 d) m
d
r1  0
d
d
r2  d aˆ y
P1
Y
P2
r3  0.5 d aˆ y  0.866 aˆ z
The force F3 is F3  F13 
Q
F
Q

3
aˆ13 
3
r
3
aˆ 23 
d
r - r
aˆ
2

4  0
3
1
Q
aˆ
1
13
2

F
23 
2
d

0.5 d aˆ y  0.866 d aˆ z
r3 - r2
 0.5 aˆ y  0.866 aˆ z
d
- r1
r3 - r2
X
23

 - 0.5 aˆ y  0.866 aˆ z
Substituting,
F3  (3 x 10

-6
27 x 10
-3
12 x 10
) 9 x 10 
2
 d
5 aˆ y  12.12 aˆz 

2
-6
9
2
2
2 x 10
( 0.5 aˆ y  0.866 aˆz ) 
2
d
-6
( - 0.5 aˆ y  0.866 aˆz

)

 13.11
 5  12.12 
d
F3  0.354 aˆF N
where aˆF  (0.38 aˆ y  0.924 aˆz )
2
2
2
2. Data : At the point P, the potential is V  (x  y  z ) V
p
To find :
(1) Ep (2) VPQ given P(1,0.2) and Q (1,1,2) (3) VPQ by using general expression for V
Solution :
(1) Ep  -  Vp
V
 - 
aˆ x 
p
V
p
aˆ y 
V
p

aˆ z 
 x
y
z

2
 - [ 2 x aˆ x  2 y aˆ y  3z aˆ z ] V /m
P
1
(2) VPQ  - Ep . dl
Q
1
0
 2x dx  2y dy
1
2
0y
0
2
 3z dz
2
2
 0  -1 V
(3) VPQ  VQ - VP  -1 V
Department of ECE
Page 26
10EC36
3. Data : Q = 64.4 nC at A (-4, 2, -3) m
To find : E at 0 (0,0,0) m
Solution :
A
E0
0
Q
E0  4   (AO)2 aˆ AO N / C
0
64.4 x 10-9
-9
10

[ aˆ AO ] N/ C
4  x 36  (AO)
2
AO  (0  4) aˆ x  (0 - 2) aˆ y  (0  3) aˆ z  4aˆ x - 2 aˆ y  3 aˆ z
aˆ
AO

AO

1
AO
E

29
aˆ
64.4 x 9
0
AO
29
(AO)  (0.743 aˆ x - 0.37 aˆ y  0.56 aˆ z )
 20 aˆN / C
AO
4. Q1 = 100 C at P1 (0.03 , 0.08 , - 0.02) m
Q2 = 0.12 C at P2 (- 0.03 , 0.01 , 0.04) m
F12 = Force on Q2 due to Q1 = ?
Solution :
F
12

Q1 Q2
2 aˆ 12
4  R
0
12
R12  R 2 - R1  (-0.03 aˆ x  0.01 aˆ y  0.04 aˆ z ) - (0.03 aˆ x  0.08 aˆ y - 0.02aˆ z )
 ( - 0.06 aˆ x - 0.07 aˆ y  0.06 aˆ z ) ;
R
12
 0.11 m
aˆ12  ( - 0.545 aˆ x - 0.636 aˆ y  0.545 aˆ z )
-6
F 100 x 10 x 0.121 x 10
2
12
0.11
F
12
-6
x 9 x 10
9
aˆ
12
 9 aˆ12 N
-9
-9
5. Q1 = 2 x 10 C , Q2 = - 0.5 x 10 C C
-2
(1) R12 = 4 x 10 m , F
?
12
-2
(2) Q1 & Q2 are brought in contact and separated by R12 = 4 x 10 m F12`  ?
Solution :
Department of ECE
Page 27
10EC36
-9
-5
2 x 10 -9 x - 0.5 x 10
 - 9 x 10 aˆ  5.63  N (attractive)
aˆ
-9
-2 2
16
12
12
12
4  x 10 x ( 4 x 10 )
36 
`
`
-9
(2) When brought into contact Q  Q  1 (Q  Q )  1.5 x 10 C
1
2
2
(1)
2 1
-9 2
2
-18  13
`
( 1.5 x 10 )
aˆ  12.66  N aˆ
 1.5 x 9 x 10
F 
aˆ
-9
16
-2 2
12
12
12
x ( 4 x 10 )
4  x 10
36 
`
F  12.66  N (repulsive)
F 
12
12
6.
Y
x
P3
x
x
P2
x
P1
0
X
Q1 = Q2 = Q3 = Q4 = 20  C
QP = 200  C at P(0,0,3) m
P1 = (0, 0 , 0) m P2 = (4, 0, 0) m
P3 = (4, 4, 0) m P4 = (0, 4, 0) m
FP = ?
Solution :
F
F
Fp  1p  F2p F3p  4p
R
R
 3 m aˆ1p  aˆ z
1p  3 aˆ z
1p
R
R
2p  - 4 aˆ x  3 aˆ z ;
2p
 5 m aˆ 2p  - 0.8 aˆ x  0.6 aˆ z
R
R
3p  - 4 aˆ x - 4 aˆ y  3 aˆ z ;
3p
 6.4 m ; aˆ3p  - 0.625 aˆ x - 0.625 aˆ y  0.47 aˆ z
R
R
 5 m ; aˆ 4p  - 0.8 aˆ y  0.6 aˆ z
4p  - 4 aˆ y  3 aˆ z ;
4p
Fp 
 Q
Qp
10
4
-9

1
2
R1

p
aˆ1 p

Q
2
2
R2
aˆ 2 p
p

Q
R3
3
3
p
aˆ3 p 
Q4
R4
2
aˆ
p

4 p 


36 
1
1
1

2
2
2
3 aˆz  5 ( - 0.8 aˆ x 0.6 aˆ z ) 6.4 (-0.625 aˆ x - 0.625 aˆ y  0.47 aˆ z ) 20 x 10 -6
 200 x 10 -6 x 9 x 10 9 


 1 ( - 0.8 aˆ  0.6 aˆ )

 2

y
z
 5

Department of ECE
Page 28
10EC36
100
100
100

aˆ

(0.8
aˆ

0.6
aˆ
)

z
x
z
-6
9
 200x10 x9x10 x10 9 x10 -6 x 10-2  9
25
40.96
 100

 25




( - 0.8 aˆ y  0.6 aˆz )
1
 0.36 (3.2 1.526) aˆ x 

(-0.625 aˆ x - 0.625 aˆ y  0.47 aˆz )
(-1.526 - 3.2) aˆ y  (11.11  2.4 1.15 
2
6.4

 (- 1.7 aˆ x - 1.7 aˆ y 17 aˆz ) N  17.23 aˆp N


2.4) aˆz ) 

7. Data : Q1 , Q2 & Q3 at the corners of equilateral triangle of side 1 m.
Q1 = - 1C, Q2 = -2 C , Q3 = - 3 C
To find : E at the bisecting point between Q2 & Q3 .
Solution :
Z
P1 Q1
P1 : (0, 0.5, 0.866) m
P2 : (0, 0, 0) m
P3 : (0, 1, 0) m
P : (0, 0.5, 0) m
Q2
P E1P
Q3
Y
P2
E2P
E3P
P3
E  E  E
2P
3P
EP  1P
1 Q
Q2
Q3


 2 aˆ1P  2 aˆ 2P  2 aˆ3P 
R
R
4 R
0  1P
2P
3P

R
R
aˆ1P  - aˆ z
1P  - 0.866 aˆ z
1P
 0.866
1
R
2P
 0.5 aˆ y
R3P  - 0.5 aˆ y
EP 
1
10 -9
R
2P
R
aˆ 2P  aˆ y
 0.5
aˆ3 P  - aˆ y
-6
-6
- 2 x 10
- 3 x 10
 - 1 x 10
 0.866 2 ( - aˆ z )  0.5 2 ( - aˆ y )  0.5 2

3P
 0.5
-6

( - aˆ y )

4 36 
ˆ
ˆ
ˆ
3
9 x 10 1.33 a z - 8 a y  12 a y 
ˆ
ˆ
ˆ
3
9 x 10
ˆ
4 a y  1.33 a z   36 a y 12 a z   0
0
3
V / m  37.9 18
k V/m
Z
E1P
EP
( EP ) = 37.9 k V / m
Y
E2P
Department of ECE
(E3P – E2P)
E3P
Page 29
10EC36
8. Data Pl = 25 n C /m on (-3, y, 4) line in free space and P : (2,15,3) m
To find : EP
Solution :
Z
ρl = 25 n C / m
A
R
P
ρ (2, 15, 3) m
Y
X
The line charge is parallel to Y axis. Therefore EPY = 0
 (2 - (-3)) aˆ x  (3 - 4) aˆ z  (5 aˆ x - aˆ z ) ;  R  5.1 m
R  AP
aˆ R
 R
R
 (0.834 aˆ x - 0.167 aˆ z
l
EP  2   R aˆ R 
0
)
25 x
10-9
aˆ R
2  36  x 5.1
EP  88.23 aˆ R V / m
9. Data : P1 (2, 2, 0) m ; P2 (0, 1, 2) m ; P3 (1, 0, 2) m
Q2 = 10 C ; Q3 = - 10 C
To find : E1 , V1
Solution :
1  Q 2
Q3

E E E
1
21
21
   R  2 aˆ 21  R 2 aˆ31 
4 0  21
31

R
R 21  (2 aˆ x  aˆ y - 2 aˆ z)
aˆ 21  0.67 aˆ x  0.33 aˆ y - 0.67 aˆ z
21  3
R
R31  aˆ x  2 aˆ y  2 aˆ z
aˆ3 1  0.33 aˆ x  0.67 aˆ y  0.67 aˆ z
31  3
-6
-6
10

9 10
(0.67 aˆ x  0.33 aˆ y - 0.67 aˆ z ) 
(0.33 aˆ x  0.67 aˆ y  0.67 aˆ z )
E1  9 x 10 
 9
9

3
 10 [ aˆ x  aˆ y ]  14.14 (0.707 aˆ x  0.707 aˆ y ) V / m
V
1

Q 2

1
4 R
0
E1

-6
-6
10
10 
Q 
 3   9 x 109 

  3000 V
R
 14.14 V / m

 3
V1  3000 V
31
21
3 
10. Data : Q1 = 10 C at P1 (0, 1, 2) m ; Q2 = - 5 C at P2 (-1, 1, 3) m
P3 (0, 2, 0) m
To find : (1) E3
Solution :
Department of ECE
(2) Q at (0, 0, 0) for E3x  0
Page 30
10EC36
Q2
 Q
 2 aˆ1 3 
R 2
 R
1
(1) E3 
1
aˆ

2 3 
0  13
23

4
R1 3  (2 - 1) aˆ y  (0 - 2) aˆ z  aˆ y - 2 aˆ z
R1 3  5
R 2 3  (0  1) aˆ x  (2 - 1) aˆ y  (0 - 3) aˆ z  aˆ x  aˆ y - 3 aˆ z
ˆ
a
13
R
ˆ
13


R
ˆ
( 0.447 a y
R 2 3  11
)
- 0.894 a z
13
ˆ
a
23
R
23

 0.3 a x
R
ˆ
ˆ
 0.3 a y
ˆ
- 0.9 a z
23
 10 x 10
E3  9 x 10 
2
 ( 5)
ˆ
ˆ
-6
9
  (8 a y
ˆ
 (-1.23 a x
- 16 a z )
ˆ

(0.447 aˆ y - 0.894 aˆ z ) 
ˆ 
- 1.23 a y
ˆ
ˆ

- 5 x 10
-6
2
ˆ
(0.3 aˆ x  0.3 aˆ y
( 11)

- 0.9 aˆ z )

3.68 a z ) 
3
 - 1.23 a x  6.77 a y - 12.32 a z 10 V / m
Q2
Q aˆ  ; R
 Q
9
(2) E3  9 x 10 
aˆ

aˆ

03 
03
 2 aˆ y
2
2
2
23
13 R
R
R 03
 13
23

1
E3x  - 1.23 aˆ x
E3x cannot be zero
-9
11. Data : Q2 = 121 x 10 C at P2 (-0.02, 0.01, 0.04) m Q1
-9
= 110 x 10 C at P1 (0.03, 0.08, 0.02) m
P3 (0, 2, 0) m
To find : F12
Solution :
QQ
1
F
12
 4   R
2
2
12
0
R
aˆ12 N ;
-9
12
 - 0.05 aˆ x - 0.07 aˆ y  0.02 aˆ z
-9
F 121 x 10 x 110 x 10 [aˆ ]
10-9
12
12
x 7.8 x 10-3
4 
36 
F
12
 0.015 aˆ12 N
2
 0.088
R
12
2
12. Given V = (50 x yz + 20y ) volt in free space
Find VP , EP and aˆnp at P (1, 2, - 3) m
Solution :
Department of ECE
Page 31
10EC36

2
VP  50 (1) (2) (-3)  20 (2)
E-V-
  - 220 V
2
V aˆ


- V aˆ - V aˆ
x
y
z
x
y
2
z
2
E  - 100 x y z aˆ x - 50 x z aˆ y - 50 x y aˆ z
EP  - 100 (2) (-3) aˆ x - 50 (-3) aˆ y - 50 (2) aˆ z
 600 aˆ x 150 aˆ y - 100 aˆ z

 62 6.5 aˆ P V / m ; aˆ P  0.957 aˆ x  0.234 aˆ y - 0.16 aˆ z
Additional Problems
A1. Find the electric field intensity E at P (0, -h, 0) due to an infinite line charge of density
ρl C / m along Z axis.
+
Z
A dz
R
AP
z
dEPy

P
dEPz
Y
h
0
d EP
aˆ P
X
- 
Solution :
Source : Line charge ρl C / m. Field point : P (0, -h, 0)
dEP

4
dQ
R2
aˆ R 
ρl dz
4   R 2 aˆ R
0
V / m ; R  AP  - z aˆ z - h aˆ y
0
2
2
R  AP  z  h
ˆ
R
a R 
R
dEP
dEPy


1
- h a y
ˆ
-zaz
ˆ 
R
ρl dz
- h
2 
aˆ y -
4  0 R  R
ρl dz h aˆ
4  R 2 R y
0

aˆ z  d EPy aˆ y  d EPz aˆ z
R 
z aˆ
dEPz - ρl dz
z
4 R2R
z
0
Expressing all distances in terms of fixed distance h,
2
h = R Cos  or R = h Sec  ; z = h tan  , dz = h sec  d
Department of ECE
Page 32
10EC36
dEPy -
ρl h Sec  d
x Cos   2
2
4   h Sec 
ρ
2
l
4
h
0

E
Py
dE
Pz
ρl
-
0
[  Sin   ]/ 2
4  0 h
2
ρl h Sec  d

2
2
4   h Sec 
-/2

Pz
E
-
ρl
4  0 h
[ Cos ]
/2
-/2

-
ρl
x2-
4  0 h
ρl
2  0 h
aˆ
y
ρl
tan 
Sin  d
x h
 
h
h Sec 
4
0
E
Cos  d
0
0
ρl
aˆy V / m
2 π 0 h
An alternate approach uses cylindrical co-ordinate system since this yields a more general
insight into the problem.
Z + 
A dz
z
R

0
P (ρ ,  / 2, 0)
Y
P
/2
AP
X
- 
Department of ECE
Page 33
10EC36
dQ  ρl dz is the elemental change at Z.
The field intensity dEP due to dQ is
dQ
dEP 
2 aˆ R V / m
4  0 R
and aˆ R  1 ( ρ aˆ ρ - z aˆ z )
R
where R  ρ aˆ ρ - z aˆ z
dQ  ρl dz C
ρl dz  ρ
z aˆ 
dEP 
aˆρ z   dEPρ aˆρ  dEP z aˆ z
2 
4  0 R R
R

ρ
ρ
l
l
(i) dEPρ  4  R 2 ρ dz ; (ii) dEP z  - 4  R 2 z dz
0
0
ˆ
Taking   OPA as integration variable, and expressing all distances in terms of ρ and 
ρ
2
z  ρ tan  , dz  ρ Sec  d
and R 
 ρ Sec 
Cos 
(i) dE
 ρ l x ρ x 3ρ Sec 3  d 
Pρ
4ρ
2
ρ
Sec 
4ρ
0
ρ
0
ρ
/2
l
Pρ
(ii) dEP z 
-/2
l
2 
0
/2
l
 EP 
ρ
x
0
[ Cos ]

Pz

4  0 ρ
2  0 ρ
4 0 ρ
2
 ρ l x ρ tan  x ρ Sec 
ρ l
(- Sin  ) d
d 
3
3
4   ρ Sec 

 ρ
4
ρ
E
Cos  d
l
[ Sin  ]

E
l
4 0 ρ
ρl
2   ρ aˆρ V / m
/2
0
0
Thus, E is radial
in direction
A2. Find the electric field intensity E at (0, -h, 0) due to a line charge of finite length along Z
axis between A (0, 0, z1) and B(0, 0, z 2)
Z
B (0, 0, z2)
2
1
P
dz
A(0, 0, z1)
Y
X
Solution :
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ρl dz  h
dEP 
z1
0


EP
aˆ y -
4  0 R  R
ρl
z2
 d EP  - 4   h

EP
2 
ρl
z
R
2

1

z 

Cos  d aˆ y -
2
(- Sin  )
4  0 h
- Sin 2 ) a y
4
ρl
h
0

aˆ
1
ρl
4   h (Sin 1
aˆ
y
ˆ
ρl
2
Sin  d aˆ
1
( Cos )
2
4  0 h
(Cos 1
z
aˆ
1
z
V/m
ˆ
- Cos 2 ) a z
0
is extending from -  to  ,
If the line
 ,   -

2

1
2
2
- ρl
 2   h aˆy V / m
EP
0
A3. Two wires AB and CD each 1 m length carry a total charge of 0.2  C and are disposed
as shown. Given BC = 1 m, find E at P, midpoint of BC.
A
B
P
.
C
1m
1m
D
Solution :
(1)
A
B
1 = 180
P
0
2 = 180
0
1m
EPAB

ρl
ˆ
 Cos 2 - Cos 1
 ˆ 0
4  0 h
(Indeterminate)
0
az
(2)
Pay
C
1
1 = - tan
2 = 0
-1
1
0.5
0
= - 63.43
D
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E

PCD

E
 - (Sin 2
ρl
4  0 h -6
ˆ
 (Cos 2 - Cos 1 ) a z 
 - (Sin (-63.43)) a y
-9
4 10 0.5
36 
ˆ
 (Cos 0 - Cos 63.43) a z
ˆ
ˆ
- 0.894 a y  (1 - 0.447) a z   (-3218 a y
 3.6 x 10
ˆ
ˆ
0.2 x 10
3
PCD
- Sin 1 ) a y

ˆ
ˆ
 1989.75 a z )
Since Eρ AB is indeterminate, an alternate method is to be used as under :
Z
dEPz
d
dy
y
B
Y
P
dEPy
A
L
R
ρ dy
dEP  l
2 aˆ R V / m
4R
1
; aˆ  (-aˆ )
0
R  (L  d - y)aˆ
R
dEPy 
R
y
R
 ρl aˆ y
2 dy
4   (L  d - y)
0
Let L  d - y  - t ; - dy  - dt ; y  0 , t  1
Ld
y  L ; t 1
d
- ρl
dEP 
dt
  t2
4
0
1
 ρ l d
EP  

4  0 t 

1
Ld

E
PA B
EP


1
ρl

4 0 d
0.2 x 10
-6
-9
10
4  36 
ρ l 1
1 
 
L  d 
4 0 d
1 
-
V / m
L  d 
 1 1 

 aˆ y
1.5 
0.5
EPA B  1800 [ 2 - 0.67] aˆ y  2400 aˆ y V/ m
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 EP  EP
 EP
AB
 2400 aˆ y - 3218 aˆ y  1990 aˆ z
CD
 (-820 aˆ y  1990 aˆ z )
 2152 aˆ P V / m
where
aˆ P  (- 0.381 aˆ y  0.925 aˆ z )
A4. Develop an expression for E due to a charge uniformly distributed over an infinite plane
2
with a surface charge density of ρS C / m .
Solution :
Let the plane be perpendicular to Z axis and we shall use Cylindrical Co-ordinates. The
2
source charge is an infinite plane charge with ρS C / m .
dEP Z

AP  R
z
P
0
Y
d

X
A
ρ
AP  AO  OP  - OA  OP
R  ( - ρ aˆρ  z aˆ z )
aˆ
R
 1 ( - ρ aˆ ρ  z aˆ z )
R
The field intensity dEP due to dQ = ρS ds = ρS (d dρ) is along AP and given by
ρS
( - ρ aˆ  z aˆ ) d ρ dρ
dEP  ρS ρ d dρ aˆ 
R
4R
2
4R
3
0
ρ
z
0
Since radial components cancel because of symmetry, only z components exist
ρ
z
 dEP  4   R3 d ρ d ρ
S
0
EP


ρS
ρS
zρdρ
zρ
2 
d
x 2 
 dEP 

3

4   
 R
 R3
4 
S
0
0
‗z‘ is fixed height of ρ above plane and let OPA  
0
are expressed in terms of z and 
0
dρ
0
ˆ
be integration variable. All distances
2
ρ = z tan  , d ρ = z Sec  d ; R = z Sec  ; ρ = 0,  = 0 ; ρ =  ,  =  / 2
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
E
P
ρS

z z tan 
2   z3 Sec 
3
0
2
z Sec
ρS
 d 
/2
2 
0
0

Sin  d

ρS
2 
0
[- Cos  ]
/2
0
aˆ
z
0

ρS
aˆ (normal to p lane)
2  z
0
A5. Find the force on a point charge of 50 C at P (0, 0, 5) m due to a charge of 500 C that
is uniformly distributed over the circular disc of radius 5 m.
Z
P
h =5 m
0
Y
ρ
X
Solution :
Given : ρ = 5 m, h = 5 m and Q = 500 C
To find : fp & qp = 50 C
ρS
fP  EP x qP where EP  2 0 aˆ z
Q
A
 2  aˆ z 
0
500 x 10
-9
10 aˆ z
2
2 ( 5 ) x 36 
-6


 2
500
x25 x 36  x 10 aˆ z
3

3
 1131 x 10 aˆ z N / C
3
fP  1131 x 10 aˆ z
x 50 x 10-6
fP  56.55 aˆ z N
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Unit-2






a. Energy and potential : Energy expended in moving a point charge in an
electric field
 The line integral
 Definition of potential difference and Potential
 The potential field of a point charge and system of charges
 Potential gradient, Energy density in an electrostatic field
b. Conductors, dielectrics and capacitance: Current and current density
 Continuity of current metallic conductors
 Conductor properties and boundary conditions
 boundary conditions for perfect Dielectrics, capacitance
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An electric field surrounds electrically charged particles and time-varying magnetic fields. The
electric field depicts the force exerted on other electrically charged objects by the electrically
charged particle the field is surrounding. The concept of an electric field was introduced by
Michael Faraday.
−1
The electric field is a vector field with SI units of newtons per coulomb (N C ) or, equivalently, volts
−1
−3 −1
per metre (V m ). The SI base units of the electric field are kg⋅m⋅s ⋅A . The strength or magnitude of
the field at a given point is defined as the force that would be exerted on a positive test charge of 1
coulomb placed at that point; the direction of the field is given by the direction of that force. Electric
fields contain electrical energy withenergy density proportional to the square of the field amplitude. The
electric field is to charge as gravitational acceleration is to mass and force densityis to volume.
An electric field that changes with time, such as due to the motion of charged particles in the
field, influences the local magnetic field. That is, the electric and magnetic fields are not
completely separate phenomena; what one observer perceives as an electric field, another
observer in a differentframe of reference perceives as a mixture of electric and magnetic fields.
For this reason, one speaks of "electromagnetism" or "electromagnetic fields". In quantum
electrodynamics, disturbances in the electromagnetic fields are called photons, and the energy of
photons is quantized.
Consider a point charge q with position (x,y,z). Now suppose the charge is subject to a force Fon
q due to other charges. Since this force varies with the position of the charge and by Coloumb's
Law it is defined at all points in space, Fon q is a continuous function of the charge's position
(x,y,z). This suggests that there is some property of the space that causes the force which is
exerted on the charge q. This property is called the electric field and it is defined by
Notice that the magnitude of the electric field has units of Force/Charge. Mathematically, the
E field can be thought of as a function that associates a vector with every point in space.
Each such vector's magnitude is proportional to how much force a charge at that point would
"feel" if it were present and this force would have the same direction as the electric field
vector at that point. It is also important to note that the electric field defined above is caused
by a configuration of other electric charges. This means that the charge q in the equation
above is not the charge that is creating the electric field, but rather, being acted upon by it.
This definition does not give a means of computing the electric field caused by a group of
charges.
From the definition, the direction of the electric field is the same as the direction of the force
it would exert on a positively charged particle, and opposite the direction of the force on a
negatively charged particle. Since like charges repel and opposites attract, the electric field is
directed away from positive charges and towards negative charges.
Array of discrete point charges
Electric fields satisfy the superposition principle. If more than one charge is present, the total
electric field at any point is equal to the vector sum of the separate electric fields that each point
charge would create in the absence of the others.
The total E-field due to N point charges is simply the superposition of the E-fields due to
each point charge:
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where ri is the position of charge Qi,
the corresponding unit vector.
Continuum of charges
The superposition principle holds for an infinite number of infinitesimally small elements of
charges – i.e. a continuous distribution of charge. The limit of the above sum is the integral:
where ρ is the charge density (the amount of charge per unit volume), and dV is the differential
volume element. This integral is a volume integral over the region of the charge distribution.
The electric field at a point is equal to the negative gradient of the electric
potentialthere,
Coulomb's law is actually a special case of Gauss's Law, a more fundamental description of the
relationship between the distribution of electric charge in space and the resulting electric field.
While
Columb's law (as given above) is only true for stationary point charges, Gauss's law is true for all
charges
either
in
static
or
in
motion.
Gauss's
equations governing electromagnetism.
law
is
one
of Maxwell's
Gauss's law allows the E-field to be calculated in terms of a continuous distribution of charge
density
where ∇⋅ is the divergence operator, ρ is the total charge density, including free and bound charge, in other words all the
charge present in the system (per unit volume).
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where ∇ is the gradient. This is equivalent to the force definition above, since electric potential Φ is defined by the electric potential energy U per unit (test) positive charge:
and force is the negative of potential energy gradient:
If several spatially distributed charges generate such an electric potential, e.g. in a solid, an
electric field gradient may also be defined
where ϕ is the potential difference between the plates and d is the distance separating the plates. The negative
sign arises as positive charges repel, so a positive charge will experience a force away from the positively
charged plate, in the opposite direction to that in which the voltage increases. In micro- and nanoapplications,
for instance in relation to semiconductors, a typical magnitude of an electric field is in the order of 1 volt/µm
achieved by applying a voltage of the order of 1 volt between conductors spaced 1 µm apart.
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Parallels between electrostatic and gravitational fields
is similar to Newton's law of universal gravitation:
.
This suggests similarities between the electric field E and the gravitational field g, so sometimes
mass is called "gravitational charge".
Similarities between electrostatic and gravitational forces:
Both act in a vacuum.
Both are central and conservative.
Both obey an inverse-square law (both are inversely proportional to square of r).
Differences between electrostatic and gravitational forces:
Electrostatic forces are much greater than gravitational forces for natural values of charge and
mass. For instance, the ratio of the electrostatic force to the gravitational force between two
42
electrons is about 10 .
Gravitational forces are attractive for like charges, whereas electrostatic forces are repulsive for
like charges.
There are not negative gravitational charges (no negative mass) while there are both positive and
negative electric charges. This difference, combined with the previous two, implies that
gravitational forces are always attractive, while electrostatic forces may be either attractive or
repulsive.
Electrodynamic fields are E-fields which do change with time, when charges are in motion.
An electric field can be produced, not only by a static charge, but also by a changing magnetic
field. The electric field is given by:
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in which B satisfies
and ∇× denotes the curl. The vector field B is the magnetic flux density and the
vector A is
the magnetic vector potential. Taking the curl of the electric field equation we obtain,
which is Faraday's law of induction, another one of Maxwell's equations.
Energy in the electric field
The electrostatic field stores energy. The energy density u (energy per unit volume) is given by
where ε is the permittivity of the medium in which the field exists, and E is the electric field
vector.
The total energy U stored in the electric field in a given volume V is therefore
Line integral
Line integral of a scalar field
Definition
n
For some scalar field f : U ⊆ R → R, the line integral along a piecewise smooth curve C ⊂ U is
defined as
where r: [a, b] → C is an arbitrary bijective parametrization of the curve C such that r(a) and
r(b) give the endpoints of Cand
.
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The function f is called the integrand, the curve C is the domain of integration, and the
symbol ds may be intuitively interpreted as an elementary arc length. Line integrals of scalar
fields over a curve C do not depend on the chosen parametrization r of C.
Geometrically, when the scalar field f is defined over a plane (n=2), its graph is a surface
z=f(x,y) in space, and the line integral gives the (signed) cross-sectional area bounded by the
curve C and the graph of f. See the animation to the right.
Derivation
For a line integral over a scalar field, the integral can be constructed from a Riemann sum
using the above definitions off, C and a parametrization r of C. This can be done by
partitioning the interval [a,b] into n sub-intervals [ti-1, ti] of length t = (b − a)/n, then r(ti)
denotes some point, call it a sample point, on the curve C. We can use the set of sample
points {r(ti) : 1 ≤ i ≤ n} to approximate the curve C by a polygonal path by introducing a
straight line piece between each of the sample points r(ti-1) and r(ti). We then label the
distance between each of the sample points on the curve as si. The product of f(r(ti)) and
si can be associated with the signed area of a rectangle with a height and width of f(r(ti))
and si respectively. Taking the limit of the sum of the terms as the length of the partitions
approaches zero gives us
We note that, by the mean value theorem, the distance between subsequent points on the
curve, is
Substituting this in the above Riemann sum yields
which is the Riemann sum for the integral
Line integral of a vector field
n
n
For a vector field F : U ⊆ R → R , the line integral along a piecewise smooth curve C ⊂ U, in the direction of r, is defined as
where · is the dot product and r: [a, b] → C is a bijective parametrization of the curve C such
that r(a) and r(b) give the endpoints of C.
A line integral of a scalar field is thus a line integral of a vector field where the vectors are
always tangential to the line.
Line integrals of vector fields are independent of the parametrization r in absolute value, but they
do depend on its orientation. Specifically, a reversal in the orientation of the parametrization
changes the sign of the line integral.
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Electric Potential Difference
In the previous section of Lesson 1, the concept of electric potential was introduced. Electric
potential is a location-dependent quantity that expresses the amount of potential energy per unit
of charge at a specified location. When a Coulomb of charge (or any given amount of charge)
possesses a relatively large quantity of potential energy at a given location, then that location is
said to be a location of high electric potential. And similarly, if a Coulomb of charge (or any
given amount of charge) possesses a relatively small quantity of potential energy at a given
location, then that location is said to be a location of low electric potential. As we begin to apply
our concepts of potential energy and electric potential to circuits, we will begin to refer to the
difference in electric potential between two points. This part of Lesson 1 will be devoted to an
understanding of electric potential difference and its application to the movement of charge in
electric circuits.
Consider the task of moving a positive test charge within a uniform electric field
from location A to location B as shown in the diagram at the right. In moving the
charge against the electric field from location A to location B, work will have to
be done on the charge by an external force. The work done on the charge changes
its potential energy to a higher value; and the amount of work that is done is equal to the change
in the potential energy. As a result of this change in potential energy, there is also a difference in
electric potential between locations A and B. This difference in electric potential is represented
by the symbol V and is formally referred to as the electric potential difference. By definition,
the electric potential difference is the difference in electric potential (V) between the final
and the initial location when work is done upon a charge to change its potential energy. In
equation form, the electric potential difference is
The standard metric unit on electric potential difference is the volt, abbreviated V and named in
honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb. If the electric
potential difference between two locations is 1 volt, then one Coulomb of charge will gain 1
joule of potential energy when moved between those two locations. If the electric potential
difference between two locations is 3 volts, then one coulomb of charge will gain 3 joules of
potential energy when moved between those two locations. And finally, if the electric potential
difference between two locations is 12 volts, then one coulomb of charge will gain 12 joules of
potential energy when moved between those two locations. Because electric potential difference
is expressed in units of volts, it is sometimes referred to as the voltage.
Electric Potential Difference and Simple Circuits
Electric circuits, as we shall see, are all about the movement of charge between varying locations
and the corresponding loss and gain of energy that accompanies this movement. In the previous
part of Lesson 1, the concept of electric potential was applied to a simple battery-powered
electric circuit. In thatdiscussion, it was explained that work must be done on a positive test
charge to move it through the cells from the negative terminal to the positive terminal. This work
would increase the potential energy of the charge and thus increase its electric potential. As the
positive test charge moves through the external circuit from the positive terminal to the negative
terminal, it decreases its electric potential energy and thus is at low potential by the time it
returns to the negative terminal. If a 12 volt battery is used in the circuit, then every coulomb of
charge is gaining 12 joules of potential energy as it moves through the battery. And similarly,
every coulomb of charge loses 12 joules of electric potential energy as it passes through the
external circuit. The loss of this electric potential energy in the external circuit results in a gain in
light energy, thermal energy and other forms of non-electrical energy.
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With a clear understanding of electric potential difference, the role of an electrochemical cell or
collection of cells (i.e., a battery) in a simple circuit can be correctly understood. The cells
simply supply the energy to do work upon the charge to move it from the negative terminal to the
positive terminal. By providing energy to the charge, the cell is capable of maintaining an
electric potential difference across the two ends of the external circuit. Once the charge has
reached the high potential terminal, it will naturally flow through the wires to the low potential
terminal. The movement of charge through an electric circuit is analogous to the movement of
water at a water park or the movement of roller coaster cars at an amusement park. In each
analogy, work must be done on the water or the roller coaster cars to move it from a location of
low gravitational potential to a location of high gravitational potential. Once the water or the
roller coaster cars reach high gravitational potential, they naturally move downward back to the
low potential location. For a water ride or a roller coaster ride, the task of lifting the water or
coaster cars to high potential requires energy. The energy is supplied by a motor-driven water
pump or a motor-driven chain. In a battery-powered electric circuit, the cells serve the role of the
charge pump to supply energy to the charge to lift it from the low potential position through the
cell to the high potential position.
It is often convenient to speak of an electric circuit such as the simple circuit discussed here as
having two parts - an internal circuit and an external circuit. The internal circuit is the part of
the circuit where energy is being supplied to the charge. For the simple battery-powered circuit
that we have been referring to, the portion of the circuit containing the electrochemical cells is
the internal circuit. The external circuit is the part of the circuit where charge is moving outside
the cells through the wires on its path from the high potential terminal to the low potential
terminal. The movement of charge through the internal circuit requires energy since it is an
uphill movement in a direction that isagainst the electric field. The movement of charge through
the external circuit is natural since it is a movement in the direction of the electric field. When at
the positive terminal of an electrochemical cell, a positive test charge is at a highelectric
pressure in the same manner that water at a water park is at a high water pressure after being
pumped to the top of a water slide. Being under high electric pressure, a positive test charge
spontaneously and naturally moves through the external circuit to the low pressure, low potential
location.
As a positive test charge moves through the external circuit, it encounters a variety of types of
circuit elements. Each circuit element serves as an energy-transforming device. Light bulbs,
motors, and heating elements (such as in toasters and hair dryers) are examples of energytransforming devices. In each of these devices, the electrical potential energy of the charge is
transformed into other useful (and non-useful) forms. For instance, in a light bulb, the electric
potential energy of the charge is transformed into light energy (a useful form) and thermal
energy (a non-useful form). The moving charge is doing work upon the light bulb to produce two
different forms of energy. By doing so, the moving charge is losing its electric potential energy.
Upon leaving the circuit element, the charge is less energized. The location just prior to entering
the light bulb (or any circuit element) is a high electric potential location; and the location just
after leaving the light bulb (or any circuit element) is a low electric potential location. Referring
to the diagram above, locations A and B are high potential locations and locations C and D are
low potential locations. The loss in electric potential while passing through a circuit element is
often referred to as a voltage drop. By the time that the positive test charge has returned to the
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negative terminal, it is at 0 volts and is ready to be re-energized andpumped back up to the high
voltage, positive terminal.
Electric Potential Diagrams
An electric potential diagram is a convenient tool for representing the electric potential
differences between various locations in an electric circuit. Two simple circuits and their
corresponding electric potential diagrams are shown below.
In Circuit A, there is a 1.5-volt D-cell and a single light bulb. In Circuit B, there is a 6-volt
battery (four 1.5-volt D-cells) and two light bulbs. In each case, the negative terminal of the
battery is the 0 volt location. The positive terminal of the battery has an electric potential that is
equal to the voltage rating of the battery. The battery energizes the charge to pump it from the
low voltage terminal to the high voltage terminal. By so doing the battery establishes an electric
potential difference across the two ends of the external circuit. Being under electric pressure, the
charge will now move through the external circuit. As its electric potential energy is transformed
into light energy and heat energy at the light bulb locations, the charge decreases its electric
potential. The total voltage drop across the external circuit equals the battery voltage as the
charge moves from the positive terminal back to 0 volts at the negative terminal. In the case of
Circuit B, there are two voltage drops in the external circuit, one for each light bulb. While the
amount of voltage drop in an individual bulb depends upon various factors (to be discussed
later), the cumulative amount of drop must equal the 6 volts gained when moving through the
battery.
Electric Potential from a Point Charge
The potential a distance r from a point charge Q is given by:
V = kQ/r
As with electric field, potential can be represented by a picture. We draw equipotential surfaces
that connect points of the same potential, although in two dimensions these surfaces just look
like lines.
For a 2-D representation of the equipotentials from a point charge, the equipotentials are circles
centered on the charge. The difference in potential between neighboring equipotentials is
constant, so the equipotentials get further apart as you go further from the charge. In 3-D the
equipotentials are actually spherical shells.
Potential energy in a uniform field is U = qEd, so potential is:
V = U/q = Ed
d here is some distance moved parallel to the field, and is measured from some convenient
reference point.
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More important is the potential difference, which increases as you move in the opposite direction
of the field:
DV = EDd
Even more generally, DV = -E · Dr
Equipotentials

Equipotential surfaces are always perpendicular to field lines.

No work is required to move a charge along an equipotential.

Equipotentials connect points of the same potential. They are similar to contour lines on a
topographical map, which connect points of the same elevation, and to isotherms (lines of
constant temperature) on a weather map.


The calculation of potential is inherently simpler than the vector sum required to calculate the
electric field.
The electric field outside a spherically
symmetric charge distribution is identical to that of a point charge as can be shown byGauss'
Law. So the potential outside a spherical charge distribution is identical to that of a point charge.
Potential due to a System of Charges
Potential at a point due to a system of charges is the sum of potentials due to individual charges
Consider a system of charges q1, q2, q3, …qn with positive vectors r1, r2, r3,…..rn relative to the
origin P.
The potential V1 at P due to charge 'q1' is given by
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Similarly, the potential at P due to charge 'q2' is
We know the potential at 'P' is the due to total charge configuration and is the algebraic sum of
potentials due to individual charges (By superposition principle).
V = V1 + V2 + v3 + ….. + Vn
From our previous chapter, we learnt that for a uniformly charged spherical shell, the electric
field outside the shell is as if the entire charges are concentrated at the centre. Thus, the potential
outside the shell is given by
where 'q' is the total charge on the shell and 'R' the radius.
Potential gradient
Fundamentally - the expression for a potential gradient F in one dimension takes the form:
[1]
where ϕ is some type of potential, and x is displacement (not distance), in the x direction. In the limit of infinitesimal displacements, the ratio of differences becomes a ratio of differentials:
In three dimensions, the resultant potential gradient is the sum of the potential gradients in each
direction, in Cartesian coordinates:
where ex, ey, ez are unit vectors in the x, y, z directions, which can be compactly and neatly written in terms of the gradient operator ∇,
Vector calculus
The mathematical nature of a potential gradient arises from vector calculus, which directly has
application to physical flows, fluxes and gradients over space. For any conservative vector
field F, there exists a scalar field ϕ, such
the vector field;
that the gradient ∇ of the scalar field is equal to
[2]
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using Stoke's theorem, this is equivalently stated as
meaning the curl ∇× of the vector field vanishes.
In physical problems, the scalar field is the potential, and the vector field is a force field, or
flux/current density describing the flow some property.
Current and current density
In electromagnetism,
and
related
fields
in solid
state
physics, condensed
matter
physics etc. current density is the electric current per unit area of cross section. It is defined as a
vector whose magnitude is the electric current per cross-sectional area. In SI units, the electric
current density is measured in amperes per square metre
Charge carriers which are free to move constitute a free current density, which are given by
expressions such as those in this section.
Electric current is a coarse, average quantity that tells what is happening in an entire wire. At
position r at time t, the distribution of charge flowing is described by the current density:[5]
where J(r, t) is the current density vector, vd(r, t) is the particles' average drift velocity (SI unit:
m∙s−1), and
is the charge density (SI unit: coulombs per cubic metre), in which n(r, t) is the number of
particles per unit volume ("number density") (SI unit: m−3), q is the charge of the individual
particles with density n (SI unit: coulombs).
A common approximation to the current density assumes the current simply is proportional to the
electric field, as expressed by:
where E is the electric field and σ is the electrical conductivity.
Conductivity σ is the reciprocal (inverse)
of electrical resistivity and has the
SI units
of siemens per metre (S m−1), and E has
the SI units of newtons per coulomb (N
C−1) or,
equivalently, volts permetre (V m−1).
A more fundamental approach to calculation of current density is based upon:
indicating the lag in response by the time dependence of σ, and the non-local nature of response
to the field by the spatial dependence of σ, both calculated in principle from an underlying
microscopic analysis, for example, in the case of small enough fields, the linear response
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function for the conductive behaviour in the material. See, for example, Giuliani or
Rammer.[6][7] The integral extends over the entire past history up to the present time.
The above conductivity and its associated current density reflect the fundamental mechanisms
underlying charge transport in the medium, both in time and over distance.
A Fourier transform in space and time then results in:
where σ(k, ω) is now a complex function.
In many materials, for example, in crystalline materials, the conductivity is a tensor, and the
current is not necessarily in the same direction as the applied field. Aside from the material
properties themselves, the application of magnetic fields can alter conductive behaviour.
Continuity equation
Since charge is conserved, current density must satisfy a continuity equation. Here is a derivation
from first principles.[11]
The net flow out of some volume V (which can have an arbitrary shape but fixed for the
calculation) must equal the net change in charge held inside the volume:
where ρ is the charge density, and dA is a surface element of the surface S enclosing the volume
V. The surface integral on the left expresses the current outflow from the volume, and the
negatively signed volume integral on the right expresses the decrease in the total charge inside
the volume. From the divergence theorem:
Hence:
This relation is valid for any volume, independent of size or location, which implies that:
and this relation is called the continuity equation
Boundary conditions on the electric field
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Figure 44:
What are the most general boundary conditions satisfied by the electric field at the interface
between two media: e.g., the interface between a vacuum and a conductor? Consider an
interface
between two media
and
. Let us, first of all, apply Gauss' law,
(632)
to a Gaussian pill-box
of cross-sectional area
whose two ends are locally parallel to the
interface (see Fig. 44). The ends of the box can be made arbitrarily close together. In this limit,
the flux of the electric field out of the sides of the box is obviously negligible. The only
contribution to the flux comes from the two ends. In fact,
(633)
where is the perpendicular (to the interface) electric field in medium
at the interface, etc. The
charge enclosed by the pill-box is simply
, where
is the sheet charge density on the
interface. Note that any volume distribution of charge gives rise to a negligible contribution to
the right-hand side of the above equation, in the limit where the two ends of the pill-box are very
closely spaced. Thus, Gauss' law yields
(634)
at the interface: i.e., the presence of a charge sheet on an interface causes a discontinuity in the
perpendicular component of the electric field. What about the parallel electric field? Let us apply
Faraday's law to a rectangular loop
whose long sides, length , run parallel to the interface,
(635)
The length of the short sides is assumed to be arbitrarily small. The dominant contribution to the
loop integral comes from the long sides:
(636)
where
is the parallel (to the interface) electric field in medium
at the interface, etc. The
flux of the magnetic field through the loop is approximately
, where
is the
component of the magnetic field which is normal to the loop, and
is the area of the loop.
But,
as the short sides of the loop are shrunk to zero. So, unless the magnetic field
becomes infinite (we shall assume that it does not), the flux also tends to zero. Thus,
(637)
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i.e., there can be no discontinuity in the parallel component of the electric field across
an interface.
Boundary Conditions at Dielectric Surfaces
Boundary conditions at dielectric surfaces state how the electric vectors E and D change at the
interface between two different media (e.g. vacuum and a dielectric, two different dielectrics
etc).
Consider a plane interface between two media 1 and 2 (these can be vacuum, or insulators). They
have dielectric constants k1 and k2 (permittivity ε1 and ε2 respectively). For generality it is
assumed that there is a free charge density σf due to charge q at the interface. E1 and E2 are
electric field intensity vectors making angles θ1 and θ2 with the normal to the interface.
Corresponding displacement vectors D1 and D2 will make angles θ1 and θ2 with normal (shown
in the figure 2.6).
A) Consider the Gaussian surface at the interface in the from of a cylinder. If its height is made
infinitesimally small, then the electric flux through only the two flat end (top and bottom)
surfaces needs to be considered.Applying Gauss‘s law (Integral form) over closed surface for D.
n1 and n2 are unit vectors normal to the surfaces S1 and S2 respectively.
n1 = − n2;dS1 = dS2 = dS; in the limit
D1n − D2n = q
where Dn is the component of D normal to the surface. This result shows that the normal
component of the vector D is discontinuous if charge q is present across the interface.
If charge is absent then q = 0 = > D1n = D2n
The normal component of electric displacement vector is same on the two sides of the boundary
i.e. D is continuous at the interface.
B) Consider the closed loop at the interface in the form of a rectangular path. The electric field is
conservative, the work done (given by the line integral) around the closed path is zero. If the
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height of the rectangle is made infinitesimally small then the only contribution to the line integral
is from the top and bottom edges of the rectangle.
Where Et is the tangential component of E.
Hence E1t = E2t
The tangential component of E is continuous across any interface.
C) Now consider a general case where an E-field (and hence D-field) passes from a first
dielectric to a second. If there is no surface charge at the interface then
E1sinθ1 = E2sinθ2(E − tangential)
D1cosθ1 = D2cosθ2(D − normal)
but D1 = ε0k1E1 = εr1E1 and D2 = ε0k2E2 = εr2E2
E1εr1cosθ1 = E2εr2cosθ2
Finally dividing expressions for D and E in terms of E
Or Tanθ1 / Tanθ2 = k1 / k2
It can be said that while crossing the interface between two dielectrics, the electric field incident
at an angle θ1 in the first medium having dielectric constant k1 gets refracted at an angle θ2 in the
second medium having dielectric constant k2.
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Fig. 2.7 Boundary Conditions for Field Vectors D and E (at the boundary between two
media)
st
We have the phasor form of the 1 Maxwell‘s curl eqn.
 H   E  j  E  J c  Jdisp
2
where J c   E  conduction current density ( A/m )
2
J disp  j  E  displacement current density ( A/m )

J
cond
J
disp
 

We can choose a demarcation between dielectrics and conductors;
  1


8
*   1 is conductor.
*
Cu: 3.5*10 @ 30 GHz
  1 is dielectric.

Mica: 0.0002 @ audio and RF
* For good conductors,
 &  are independent of freq.
* For most dialectics,
 &  are function of freq.
*
 is relatively constant over frequency range of interest

Therefore dielectric ― constant ―
*
 dissipation factor D

if D is small, dissipation factor is practically as the power factor of the
dielectric. PF = sin
-1
 = tan D
PF & D difference by <1% when their values are less than 0.15.
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Example
Express
E y  100 cos 210 t  0.5 z  30
6
0

j 2 10 t  0.5 z 30 
E
8
 R 100 e

jwt
Drop Re and suppress e term to get phasor
e
y
0
v / m as a phasor

Therefore phasor form of Eys = 100e0.5 z300
Whereas Ey is real, Eys is in general complex.
Note: 0.5z is in radians; 300 in degrees.
Example 11.2
Given
Es  100 30 axˆ  20  50 ayˆ  40 210 azˆ , V / m
0
0
0
find its time varying form representation
Let us rewrite Es as
E s  100e
ER
j 300
axˆ  20e
E
e

s

 j 500
ayˆ  40e
j 2100
azˆ . V / m
e jt 


j  t  30
0
 Re 100e

E  100 cos  t  30
0
0

j  t  50 
 20e
 40e
j t 210
0
 

V/m
20 cos  t  50 0  40 cos t  210 0 V / m
None of the amplitudes or phase angles in this are expressed as a function of x,y or
z. Even if so, the procedure is still effective.
a) Consider
 20e   0.1 j 20z axˆ A / m
jt


H t   Re 20e 0.1j20 zaxˆ e

H
s
 20e
0.1z

cos t  20 z axˆ
E x  E x  x , y , z 
Note :

consider
A/m

e  x 
Ex

t  t R  E x , y , z
 R  j E
e jt 

e 
x

e jt 
Therefore taking the partial derivative of any field quantity wrt time is equivalent to multiplying
the corresponding phasor by j .
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Example
Given
E
0s
  500 40 ayˆ   200  j 600 azˆ  e
0
Find a  



 
b E at
c E at
 j 0.4 x
V/m

2, 3,1 at t  0


2, 3,1 at t  10 ns.
 d  E at 3, 4, 2 at t  20 ns.



a) From given data,
00
  0.4  
8
0.4  3 10

4  10
7
 120 106
9

10
9
36
6
f  19.1 10 Hz
b) Given,
E s   500 40
 500e
 j 40
500e 
e
 j 0.4 x
ayˆ
 632.456e  j 71.565 e  j 0.4 x azˆ
0
    ayˆ 632.456e j0.4x 71.565 azˆ
j 0.4 x
E t   500 Re  e
 j 0.4 x
ayˆ   200  j 600azˆ e
0

40
0


  j t

 j 0.4 x  40
0
e
 500 cos t  0.4 x  40

0

ayˆ  632.456 e

 j 0.4 x71.565
j t
e
0


azˆ 

ayˆ  632.456 cos t  0.4 x  71.565 azˆ
0
E at 2, 3,1t  0  500 cos 0.4 x  40  ayˆ  632.456 0.4 x  71.565 azˆ
0
 36.297 ayˆ  291.076 azˆ V / m
c)
E at t  10 ns  at 2, 3,1


 500 cos 120  10  10  10
6
9
 632.456 cos 120  10  10  10
6
0
 0.4  2  40
9
 ayˆ
0
 0.4  2  71.565
 azˆ
 477.823 ayˆ  417.473 azˆ V / m
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d)
at t = 20 ns,
E at 2, 3,1
 438.736 ayˆ  631.644 azˆ V / m
D 11.2:
Given H s  2 400 axˆ  320 ayˆ e  j 0.07 z A / m for a UPW traveling in free space. Find
(a) 
(b) Hx at p(1,2,3) at t = 31 ns.
(a) we have
(c) H at t=0 at the origin.
(e  j  z term)
p = 0.07
   0.07
0.07

8
  0.07  3  10  21.0 10
 21.0 10
6
6
rad / sec
rad / sec
(b)
H  t   Re
 2 e
 j 40 0
e  j 0.07 z axˆ  3 e j 20
 2 cos  t  0.07 z  40
0
H x (t )  2 cos t  0.07 z  40
0
e
 j 0.07 z
ayˆ  e
j t

axˆ  3 cos t  0.07 z  200 ayˆ
0
 H x (t ) at p 1, 2, 3
6
0
 2 cos 2.1  10 t  0.21  40 
6
9
0
At t  31n sec;  2 cos 2.1  10  31  10  0.21  40 
3
0
 2 cos 651 10  0.21  40 
 1.9333
A/m
(c)
H t  at t  0  2 cos 0.07 z  0.7 axˆ  3 cos 0.7 z  0.35ayˆ
H t   2 cos 0.7
axˆ  3 cos 0.3ayˆ
 1.53axˆ  2.82ayˆ
 3.20666
A/m
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In free space,
E  z , t   120 sin  t   z  ayˆ
V /m
H  z , t 
find
Ey
Hx
we have
 120
Ey
120
 H x 
120
120

sin  t   z  ayˆ
1
   sin  t   z 

1
 H  z , t    sin  t   z  axˆ
Problem 3
Non uniform plans waves also can exist under special conditions. Show that the function
Fe
 z

sin  x t 
2
satisfies the wave equation  F 
1 2 F
2
c t
2
provided the wave velocity is given by
e


1

 2 c2 


2


Ans:
From the given eqn. for F, we note that F is a function of x and z,
2
 F 

F
x
2
 F
x
F
2
z
2
 F
z
2
 F
 F
2 
2
x
y
2
2

 e  z  cos   x t 
 z
 e
 e
 z
 e
2
Department of ECE


2  z
 x  t   e


 sin
2
 
   




sin   x t 


   2 F
  
 z
sin
F
xt
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 

2
2
 F  
 F
2
 

dF  e  z     cos   x t 


 

dt
2
d F
 z  
 e

 2    sin  x t 
dt
 
2
 F
2
The given wave equation is
2
2
 F 


2

1  F
2
c 2 2t
 
2  F 
c
 
2
2


2
 
2
2 
c


2
2
2
2
c
1
2
 2  F
2
 2

2
2 2
2
c
 
2

2c2
 2 c2   2
2
c

2 2

c
1 
2

c
or  
1 
Department of ECE
 2 c2
2
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Unit-3
Poisson’s and Laplace’s equations



Derivations of Poisson‘s and Laplace‘s Equations

Uniqueness theorem

Examples of the solutions of Laplace‘s and Poisson‘s equations
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Deriving Poisson's and LaPlace's equations
First we will derive the Divergence Theorem (Gauss's Theorem)
Divergence Theorem and Gauss's Theorem
Consider flux (flow) of material (force lines) through an infinitesimal box:
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Relationship to Green's Theorem
The flux out of a volume V equals the divergence throughout volume V
Poisson's and Laplace's Equations
For gravity,
Consider the net flux out of (or into) a closed volume:
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If M is inside the volume, the surface surrounding the mass takes up the entire "field of view",
which is another way of saying that the total solid angle subtended by the surrounding surface is
4psteradians (a steradian is the 3D equivalent of a radian; the circumference of a unit circle is
2p, hence 2p radians in a circle. Similarly, the surface area of a unit sphere is 4p steradians).
Derivation of Laplace Equation from Maxwell's Equations
Consider the first of Maxwell's Equations
.
Remember that
. Since, for static electric fields, we also recognise
that
.
Putting these two relationships together yields
This is Poisson's Equation, which, together with suitable boundary conditions, yields the
potential in terms of the distribution of charge density ρ.
In regions where the charge density is zero, Poisson's Equation becomes the Laplace
Equation
.
The Laplace Equation is very useful problems where potentials are defined on boundaries and
one wishes to compute the field in a source-free region. The Poisson Equation is convenient for
problems where charge distributions are used to compute fields.
General solution of the Poisson Equation
Without sacrificing any generality, we can solve for the "impulse response" of the electrostatic
system by deriving a solution for
.
By integrating this equation over a spherical volume that encloses the origin and applying the
Divergence Theorem, we see that
.
If the sphere is "small", the surface integral on the left-hand-side becomes
. Rearranging yields
.
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This is easily integrable in r and we see that the final result for the impulse response potential is
.
is a constant that describes the reference potential (usually zero). Given the spherical
symmetry of the delta function source, we know that this is the complete solution.
By shifting the origin to
, the impulse response can be generalised to any point in space:
.
It is left to the reader to show that this is the solution of
.
(Hint: use the Divergence Theorem on the
operator when
The next stage is to show that any distribution of charge
constant value, a potential everywhere in space.
.)
uniquely defines, within a
By using Green's Theorem,
,
we can illustrate what the expression for potential must look like. Assume
is the unknown
potential and
is the "impulse response" function that we found in the first part of this
chapter. All we know is that the function
satisfies Poisson's Equation as well as any necessary
boundary conditions on conductors as well as at infinity (where it must go to zero). Sources
(charges) must be defined over a finite volume.
Inserting what we know and changing the integration to primed (source) coordinates yields
.
Carrying out the volume integration on the left hand side and the surface integration on the right
gives
.
The surface charge density
in the final surface integral comes about if we recognise that
The first integral on the left did not change. However, the integration of the product of the delta
function and the unknown potential over the problem volume just yields back the potential at the
obervation point in the second term. The surface integral is a bit trickier to understand. At
infinity, contributions to the surface integral vanish. Only surfaces (like conductors, dielectrics)
at finite distances in the problem domain contribute to the potential (since we assume a finite
charge distribution). The first surface integral term (over surfaces where potential is fixed)
actually collapses to a constant value on metallic boundaries, because the conducting boundary
must be an equipotential. As a result, the surface integral
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is
if (the observation point) lies on a surface source and 0 if the observation point is not
on the surface.
Hence, we find that
,
where the first surface integral term vanishes for observation points away from surface sources.
If we rearrange, the potential anywhere in space can be written as
.
In source-free regions (where Laplace's equation is valid) only the surface integral term is
needed. Many useful numerical methods have been developed using
as the basis for solving many complicated problems. Furthermore, in source-free regions, we can
make the philosophically important observation that knowing the field along a surface is enough
to know the field everywhere in space!
By finding the "impulse response" (known as a "Green's function") of the potential function, we
can use what signal-processing people will recognise as convolution to generate solutions for
general potentials based on a known distribution of charge. Special techniques can also be used
to reconstruct the charge distribution from a known potential distribution. These form the basis
of a wide class of computer simulation techniques known as boundary element models.
Laplace's equation
Laplace's equation is a second-order partial differential equation named after Pierre-Simon
Laplace who first studied its properties. This is often written as:
where ∆ = ∇² is the Laplace operator and
is a scalar function. In general, ∆ = ∇² is the Laplace–Beltrami or Laplace–de Rham operator.
Laplace's equation and Poisson's equation are the simplest examples of elliptic partial differential
equations. Solutions of Laplace's equation are called harmonic functions.
The general theory of solutions to Laplace's equation is known as potential theory. The solutions
of Laplace's equation are the harmonic functions, which are important in many fields of science,
notably the fields of electromagnetism, astronomy, and fluid dynamics, because they can be used
to accurately describe the behavior of electric, gravitational, and fluid potentials. In the study
ofheat conduction, the Laplace equation is the steady-state heat equation.
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n three dimensions, the problem is to find twice-differentiable real-valued functions , of real
variables x, y, and z, such that
In Cartesian coordinates
In cylindrical coordinates,
In spherical coordinates,
In Curvilinear coordinates,
or
This is often written as
or, especially in more general contexts,
where ∆ = ∇² is the Laplace operator or "Laplacian"
where ∇ ⋅ = div is the divergence, and ∇ = grad is the gradient.
If the right-hand side is specified as a given function, h(x, y, z), i.e., if the whole equation is
written as
then it is called "Poisson's equation".
The Laplace equation is also a special case of the Helmholtz equation.
According to Maxwell's equations, an electric field (u,v) in two space dimensions that is
independent of time satisfies
and
where ρ is the charge density. The first Maxwell equation is the integrability condition for the
differential
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so the electric potential φ may be constructed to satisfy
The second of Maxwell's equations then implies that
which is the Poisson equation.
It is important to note that the Laplace equation can be used in three-dimensional problems in
electrostatics and fluid flow just as in two dimensions.
Uniqueness theorem
The uniqueness theorem for Poisson's equation states that the equation has a unique gradient of
the solution for a large class of boundary conditions. In the case of electrostatics, this means that
if an electric field satisfying the boundary conditions is found, then it is the complete electric
field.
In Gaussian units, the general expression for Poisson's equation in electrostatics is
Here
is the electric potential and
is the electric field.
The uniqueness of the gradient of the solution (the uniqueness of the electric field) can be proven
for a large class of boundary conditions in the following way.
Suppose that there are two solutions
and
. One can then define
which is
the difference of the two solutions. Given that both
and
satisfy Poisson's Equation, must
satisfy
Using the identity
And noticing that the second term is zero one can rewrite this as
Taking the volume integral over all space specified by the boundary conditions gives
Applying the divergence theorem, the expression can be rewritten as
Where
are boundary surfaces specified by boundary conditions.
Since
and
, then
when the surface integral vanishes.
must be zero everywhere (and so
)
This means that the gradient of the solution is unique when
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The boundary conditions for which the above is true are:
Dirichlet boundary condition: is well defined at
all of the boundary surfaces. As
such
so at the boundary
and correspondingly the surface integral vanishes.
Neumann boundary condition:
is well defined
at all of the boundary suaces. As
such
so at the boundary
and correspondingly the surface integral
vanishes.
Modified Neumann boundary condition (where boundaries are specified as conductors with
known charges):
is also well defined by applying locally Gauss's Law. As such, the surface
integral also vanishes.
Mixed boundary conditions (a combination of Dirichlet, Neumann, and modified Neumann
boundary conditions): the uniqueness theorem will still hold.
Example
The electric field intensity of a uniform plane wave in air has a magnitude of 754 V/m and is
in the z direction. If the wave has a wave length  = 2m and propagating in the y direction.
Find
(i)
Frequency and  when the field has the form A cos  t   z .
(ii)
Find an expression for H .
In air or free space,
  c  3108 m / sec
(i)
f 
e 3 108



2 
8
m / sec  1.5  10 Hz  150MHz
2m
2
  2 m  3.14 rad / m
 E z  754 cos 2  150  10 t  y 
6
(ii)
For a wave propagating in the +y direction,
E
z

Hz
Ex
Hz
For the given wave,
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E z  754 V / m; Ex  0
H
 754    754  754 A / m
120
x
377
 H  2 cos 2  150  10 t  y axˆ
6
A/m
Example
7
find  for copper having  = 5.8*10 (/m) at 50Hz, 3MHz, 30GHz.
2


1
   f 
1 
1
4  10

7

 1
7
f
5.8 10
1
66 10
1
1

 1 

2
2
4 5.8
f
23.2 f
f
3
66 10
3
 9.3459 10 m
(i ) 
50
(ii ) 
66 10
(iii ) 
3
6
3 10
3
66 10
3 10
6
Department of ECE
 3.8105 10
5
 3.8105 10
3
m
7
m
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Unit-4:
The steady magnetic field






Biot-Savarts law
Ampere circuital law
stokes‘ theorem
magnetic flux and flux density
scalar and vector magnetic potential
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Biot–Savart law
In physics, particularly electromagnetism, the Biot–Savart law (is an equation that describes
themagnetic field generated by an electric current. It relates the magnetic field to the magnitude,
direction, length, and proximity of the electric current. The law is valid in the magnetostatic
approximation, and is consistent with both Ampère's circuital law and Gauss's law for magnetism
The Biot–Savart law is used to compute the resultant magnetic field B at position r generated by
a steady current I (for example due to a wire): a continual flow of charges which is constant in
time and the charge neither accumulates nor depletes at any point. The law is a physical example
of a line integral: evaluated over the path C the electric currents flow. The equation in SI units is
where r is the full displacement vector from the wire element to the point at which the field is
being computed and r is the unit vector of r. Using this the equation can be equivalently written
where dl is a vector whose magnitude is the length of the differential element of the wire, in the
direction of conventional current, and μ0 is the magnetic constant. The symbols in boldface
denotevector quantities.
The integral is usually around a closed curve, since electric currents can only flow around closed
paths. An infinitely long wire (as used in the definition of the SI unit of electric current theAmpere) is a counter-example.
To apply the equation, the point in space where the magnetic field is to be calculated is chosen.
Holding that point fixed, the line integral over the path of the electric currents is calculated to
find the total magnetic field at that point. The application of this law implicitly relies on the
superposition principle for magnetic fields, i.e. the fact that the magnetic field is a vector sum of
the field created by each infinitesimal section of the wire individually
he formulations given above work well when the current can be approximated as running
through an infinitely-narrow wire. If the current has some thickness, the proper formulation of
the Biot–Savart law (again in SI units) is:
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or equivalently
where dV is the differential element of volume and J is the current density vector in that volume.
In this case the integral is over the volume of the conductor.
The Biot–Savart law is fundamental to magnetostatics, playing a similar role to Coulomb's
law in electrostatics. When magnetostatics does not apply, the
Biot–Savart law should be
replaced byJefimenko's equations.
In the special case of a steady constant current I, the magnetic field B is
i.e. the current can be taken out the integral.
In the case of a point charged particle q moving at a constant velocity v, then Maxwell's
equations give the following expression for the electric field and magnetic field:
[5]
where r is the vector pointing from the current (non-retarded) position of the particle to the point
at which the field is being measured, and θ is the angle between v and r.
2
2
When v ≪ c , the electric field and magnetic field can be approximated as
[5]
These equations are called the "Biot–Savart law for a point charge
Ampère's circuital law
In classical electromagnetism, Ampère's circuital law, discovered by André-Marie Ampère in
1826, relates the integrated magnetic field around a closed loop to the electric current passing
through the loop. James Clerk Maxwell derived it again using hydrodynamics in his 1861 paper
On Physical Lines of Force and it is now one of the Maxwell equations, which form the basis of
classical electromagnetism.
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It relates magnetic fields to electric currents that produce them. Using Ampere's law, one can
determine the magnetic field associated with a given current or current associated with a given
magnetic field, providing there is no time changing electric field present. In its historically
original form, Ampère's Circuital Law relates the magnetic field to its electric current source.
The law can be written in two forms, the "integral form" and the "differential form". The forms
are equivalent, and related by the Kelvin–Stokes theorem. It can also be written in terms of either
the B or H magnetic fields. Again, the two forms are equivalent (see the "proof" section below).
Ampère's circuital law is now known to be a correct law of physics in a magnetostatic situation:
The system is static except possibly for continuous steady currents within closed loops. In all
other cases the law is incorrect unless Maxwell's correction is included (see below).
Integral form
In SI units (cgs units are later), the "integral form" of the original Ampère's circuital law is a line
integral of the magnetic field around some closed curveC (arbitrary but must be closed). The
curve C in turn bounds both a surface S which the electric current passes through (again arbitrary
but not closed—since no three-dimensional volume is enclosed by S), and encloses the current.
The mathematical statement of the law is a relation between the total amount of magnetic field
around some path (line integral) due to the current which passes through that enclosed path
(surface integral). It can be written in a number of forms.
In terms of total current, which includes
the magnetic B-field (in tesla, T) around
[2][3]
both free and bound current, the line integral of
closed curve C is proportional to
the total
current Ienc passing through a surface S (enclosed by C):
−2
where J is the total current density (in ampere per square metre, Am ).
Alternatively in terms of free current, the line integral of the magnetic H-field (in ampere per
−1
metre, Am ) around closed curve C equals the free current If, enc through a surface S:
where Jf is the free current density only. Furthermore

is the closed line integral around the closed curve C,

denotes a 2d surface integral over S enclosed by C
 • is the vector dot product,

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 dℓ is an infinitesimal element (a differential) of the curve C (i.e. a vector with
magnitude equal to the length of the infinitesimal line element, and direction given
by the tangent to the curve C)
 dS is the vector area of an infinitesimal element of surface S (that is, a vector with
magnitude equal to the area of the infinitesimal surface element, and direction
normal to surface S. The direction of the normal must correspond with the
orientation of C by the right hand rule), see below for further explanation of the
curve C and surface S.

The B and H fields are related by the constitutive equation



where μ0 is the magnetic constant.

There are a number of ambiguities in the above definitions that require clarification and a choice
of convention.

First, three of these terms are associated with sign ambiguities: the line integral
could go
around the loop in either direction (clockwise or counterclockwise); the vector area dS could
point in either of the two directions normal to the surface; and Ienc is the net current passing
through the surface S, meaning the current passing through in one direction, minus the current in
the other direction—but either direction could be chosen as positive. These ambiguities are
resolved by the right-hand rule: With the palm of the right-hand toward the area of integration,
and the index-finger pointing along the direction of line-integration, the outstretched thumb
points in the direction that must be chosen for the vector area dS. Also the current passing in the
same direction as dS must be counted as positive. The right hand grip rule can also be used to
determine the signs.
Second, there are infinitely many possible surfaces S that have the curve C as their border.
(Imagine a soap film on a wire loop, which can be deformed by moving the wire). Which of
those surfaces is to be chosen? If the loop does not lie in a single plane, for example, there is no
one obvious choice. The answer is that it does not matter; it can be proven that any surface with
boundary C can be chosen.
Differential form
By the Kelvin–Stokes theorem, this equation can also be written in a "differential form". Again,
this equation only applies in the case where the electric field is constant in time, meaning the
currents are steady (time-independent, else the magnetic field would change with time); see
below for the more general form. In SI units, the equation states for total current:
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and for free current
where ∇× is the curl operator.
The electric current that arises in the simplest textbook situations would be classified as "free
current"—for example, the current that passes through a wire or battery. In contrast, "bound
current" arises in the context of bulk materials that can be magnetized and/or polarized. (All
materials can to some extent.)
When a material is magnetized (for example, by placing it in an external magnetic field), the
electrons remain bound to their respective atoms, but behave as if they were orbiting the nucleus
in a particular direction, creating a microscopic current. When the currents from all these atoms
are put together, they create the same effect as a macroscopic current, circulating perpetually
around the magnetized object. This magnetization current JM is one contribution to "bound
current".
The other source of bound current is bound charge. When an electric field is applied, the positive
and negative bound charges can separate over atomic distances in polarizable materials, and
when the bound charges move, the polarization changes, creating another contribution to the
"bound current", the polarization current JP.
The total current density J due to free and bound charges is then:
with Jf the "free" or "conduction" current density.
All current is fundamentally the same, microscopically. Nevertheless, there are often practical
reasons for wanting to treat bound current differently from free current. For example, the bound
current usually originates over atomic dimensions, and one may wish to take advantage of a
simpler theory intended for larger dimensions. The result is that the more microscopic Ampère's
law, expressed in terms of B and the microscopic current (which includes free, magnetization
and polarization currents), is sometimes put into the equivalent form below in terms of H and the
free current only. For a detailed definition of free current and bound current, and the proof that
the two formulations are equivalent, see the "proof" section below.
Shortcomings of the original formulation of Ampère's circuital law
There are two important issues regarding Ampère's law that require closer scrutiny. First, there is
an issue regarding the continuity equation for electrical charge. There is a theorem in vector
calculus that states the divergence of a curl must always be zero. Hence
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and so the original Ampère's law implies that
But in general
which is non-zero for a time-varying charge density. An example occurs in a capacitor circuit
where time-varying charge densities exist on the plates
Second, there is an issue regarding the propagation of electromagnetic waves. For example, in
free space, where
Ampère's law implies that
but instead
To treat these situations, the contribution of displacement current must be added to the current
term in Ampère's law.
James Clerk Maxwell conceived of displacement current as a polarization current in the
dielectric vortex sea, which he used to model the magnetic field hydrodynamically and
mechanically.
[9]
He added this displacement current to Ampère's circuital law at equation (112)
in his 1861 paper On Physical Lines of Force
Curl Theorem
A special case of Stokes' theorem in which
embedded 2-manifold with boundary in
is a vector field and
is an oriented, compact
, and a generalization of Green's theorem from the
plane into three-dimensional space. The curl theorem states
(1)
where the left side is a surface integral and the right side is a line integral.
There are also alternate forms of the theorem. If
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(2) then
(3)
and if
(4) then
Stokes' theorem
In differential geometry, Stokes' theorem (also called the generalized Stokes' theorem) is a
statement about the integration of differential forms onmanifolds, which both simplifies and
generalizes several theorems from vector calculus. Stokes' theorem says that the integral of a
differential form ωover the boundary of some orientable manifold Ω is equal to the integral of its
exterior derivative dω over the whole of Ω, i.e.
This modern form of Stokes' theorem is a vast generalization of a classical result first discovered by
Lord Kelvin, who communicated it to George Stokes in July 1850.
[1][2]
Stokes set the theorem as a
question on the 1854 Smith's Prize exam, which led to the result bearing his name.
[2]
This classical
Kelvin–Stokes theorem relates the surface integral of the curl of a vector field F over a surface Σ in
Euclidean three-space to the line integralof the vector field over its boundary ∂Σ:
This classical statement, as well as the classical Divergence theorem, fundamental theorem of
calculus, and Green's Theorem are simply special cases of the general formulation stated above.
he fundamental theorem of calculus states that the integral of a function f over the interval [a, b]
can be calculated by finding an antiderivative F of f:
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Stokes' theorem is a vast generalization of this theorem in the following sense.

By the choice of F,
. In the parlance of differential forms, this is saying
that f(x) dx is the exterior derivative of the 0-form, i.e. function, F: in other words, that
dF = f dx. The general Stokes theorem applies to higher differential forms
instead of
F.
 A closed interval [a, b] is a simple example of a one-dimensional manifold with
boundary. Its boundary is the set consisting of the two points a and b. Integrating f over
the interval may be generalized to integrating forms on a higher-dimensional manifold.
Two technical conditions are needed: the manifold has to be orientable, and the form has
to be compactly supported in order to give a well-defined integral.

The two points a and b form the boundary of the open interval. More generally, Stokes'
theorem applies to oriented manifolds M with boundary. The boundary ∂M of M is itself
a manifold and inherits a natural orientation from that of the manifold. For example, the
natural orientation of the interval gives an orientation of the two boundary points.
Intuitively, a inherits the opposite orientation as b, as they are at opposite ends of the
interval. So, "integrating" F over two boundary points a, b is taking the difference F(b) −
F(a).
In even simpler terms, one can consider that points can be thought of as the boundaries of curves,
that is as 0-dimensional boundaries of 1-dimensional manifolds. So, just as one can find the
value of an integral (f dx = dF) over a 1-dimensional manifolds ([a,b]) by considering the antiderivative (F) at the 0-dimensional boundaries ([a,b]), one can generalize the fundamental
theorem of calculus, with a few additional caveats, to deal with the value of integrals (dω) over
n-dimensional manifolds (Ω) by considering the anti-derivative (ω) at the (n-1)-dimensional
boundaries (dΩ) of the manifold.
So the fundamental theorem reads:
Let
be an oriented smooth manifold of dimension n and let
is compactly supported on
be an n-differential form that
. First, suppose that α is compactly supported in the domain of a
single,oriented coordinate chart {U, φ}. In this case, we define the integral of
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over
as
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n
i.e., via the pullback of α to R .
More generally, the integral of
over
is defined as follows: Let {ψi} be a partition of unity
associated with a locally finite cover {Ui, φi} of (consistently oriented) coordinate charts, then
define the integral
n
where each term in the sum is evaluated by pulling back to R as described above. This quantity
is well-defined; that is, it does not depend on the choice of the coordinate charts, nor the partition
of unity.
Stokes' theorem reads: If
the boundary of
is an (n − 1)-form with compact support on
denotes
with its induced orientation, then
A "normal" integration manifold (here called D instead of
Here
and
) for the special case n=2
is the exterior derivative, which is defined using the manifold structure only. On the
r.h.s., a circle is sometimes used within the integral sign to stress the fact that the (n-1)-manifold
[3]
is closed.
The r.h.s. of the equation is often used to formulate integral laws; the l.h.s. then
leads to equivalent differentialformulations (see below).
The theorem is often used in situations where
bigger manifold on which the form
is an embedded oriented submanifold of some
is defined.
A proof becomes particularly simple if the submanifold
is a so-called "normal manifold", as in
the figure on the r.h.s., which can be segmented into vertical stripes (e.g. parallel to the xn
direction), such that after a partial integration concerning this variable, nontrivial contributions
come only from the upper and lower boundary surfaces (coloured in yellow and red,
respectively), where the complementary mutual orientations are visible through the arrows.
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Kelvin–Stokes theorem
An illustration of the Kelvin–Stokes theorem, with surface
, its boundary
and the "normal"
vector n.
This is a (dualized) 1+1 dimensional case, for a 1-form (dualized because it is a statement about
vector fields). This special case is often just referred to as the Stokes' theorem in many
introductory university vector calculus courses and as used in physics and engineering. It is also
sometimes known as the curl theorem.
The classical Kelvin–Stokes theorem:
which relates the surface integral of the curl of a vector field over a surface Σ in Euclidean threespace to the line integral of the vector field over its boundary, is a special case of the general
Stokes theorem (with n = 2) once we identify a vector field with a 1 form using the metric on
Euclidean three-space. The curve of the line integral, ∂Σ, must have positive orientation,
meaning that dr points counterclockwise when the surface normal, dΣ, points toward the viewer,
following the right-hand rule.
One consequence of the formula is that the field lines of a vector field with zero curl cannot be
closed contours.
The formula can be rewritten as:
where P, Q and R are the components of F.
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These variants are frequently used:
Magnetic flux
In physics, specifically electromagnetism, the magnetic flux (often denoted Φ or ΦB) through a
surface is the component of the B field passing through that surface. The SI unit of magnetic flux
is the weber (Wb) (in derived units: volt-seconds), and the CGS unit is the maxwell. Magnetic
flux is usually measured with a fluxmeter, which contains measuring coils and electronics that
evaluates the change of voltage in the measuring coils to calculate the magnetic flux.
The magnetic interaction is described in terms of a vector field, where each point in space (and
time) is associated with a vector that determines what force a moving charge would experience at
that point (see Lorentz force). Since a vector field is quite difficult to visualize at first, in
elementary physics one may instead visualize this field with field lines. The magnetic flux
through some surface, in this simplified picture, is proportional to the number of field lines
passing through that surface (in some contexts, the flux may be defined to be precisely the
number of field lines passing through that surface; although technically misleading, this
distinction is not important). Note that the magnetic flux is the net number of field lines passing
through that surface; that is, the number passing through in one direction minus the number
passing through in the other direction (see below for deciding in which direction the field lines
carry a positive sign and in which they carry a negative sign). In more advanced physics, the
field line analogy is dropped and the magnetic flux is properly defined as the component of the
magnetic field passing through a surface. If the magnetic field is constant, the magnetic flux
passing through a surface of vector area S is
where B is the magnitude of the magnetic field (the magnetic flux density) having the unit of
2
Wb/m (Tesla), S is the area of the surface, and θ is the angle between the magnetic field lines
and the normal (perpendicular) to S. For a varying magnetic field, we first consider the magnetic
flux through an infinitesimal area element dS, where we may consider the field to be constant:
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A generic surface, S, can then be broken into infinitesimal elements and the total magnetic flux
through the surface is then the surface integral
From the definition of the magnetic vector potential A and the fundamental theorem of the curl
the magnetic flux may also be defined as:
where the line integral is taken over the boundary of the surface S, which is denoted ∂S.
Gauss's law for magnetism, which is one of the four Maxwell's equations, states that the total
magnetic flux through a closed surface is equal to zero. (A "closed surface" is a surface that
completely encloses a volume(s) with no holes.) This law is a consequence of the empirical
observation that magnetic monopoles have never been found.
In other words, Gauss's law for magnetism is the statement:
for any closed surface S.
While the magnetic flux through a closed surface is always zero, the magnetic flux through
an open surface need not be zero and is an important quantity in electromagnetism. For example,
a change in the magnetic flux passing through a loop of conductive wire will cause an
electromotive force, and therefore an electric current, in the loop. The relationship is given by
Faraday's law:
where
is the EMF,
ΦB is the magnetic flux through the open surface Σ,
∂Σ is the boundary of the open surface Σ; note that the surface, in general, may be in
motion and deforming, and so is generally a function of time. The electromotive force is
induced along this boundary.
dℓ is an infinitesimal vector element of the contour ∂Σ,
v is the velocity of the boundary ∂Σ,
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E is the electric field,
B is the magnetic field.
The two equations for the EMF are, firstly, the work per unit charge done against the Lorentz
force in moving a test charge around the (possibly moving) surface boundary ∂Σ and, secondly,
as the change of magnetic flux through the open surface Σ. This equation is the principle behind
an electrical
generator.
The magnetic vector potential
Electric fields generated by stationary charges obey
(315)
This immediately allows us to write
(316)
since the curl of a gradient is automatically zero. In fact, whenever we come across an
irrotational vector field in physics we can always write it as the gradient of some scalar field.
This is clearly a useful thing to do, since it enables us to replace a vector field by a much simpler
scalar field. The quantity
in the above equation is known as the electric scalar potential.
Magnetic fields generated by steady currents (and unsteady currents, for that matter) satisfy
(317)
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This immediately allows us to write
(318)
since the divergence of a curl is automatically zero. In fact, whenever we come across a
solenoidal vector field in physics we can always write it as the curl of some other vector field.
This is not an obviously useful thing to do, however, since it only allows us to replace one vector
field by another. Nevertheless, Eq. (318) is one of the most useful equations we shall come
across in this lecture course. The quantity
is known as the magnetic vector potential.
We know from Helmholtz's theorem that a vector field is fully specified by its divergence and its
curl. The curl of the vector potential gives us the magnetic field via Eq. (318). However, the
divergence of
has no physical significance. In fact, we are completely free to choose
to be whatever we like. Note that, according to Eq. (318), the magnetic field is invariant under
the transformation
(319)
In other words, the vector potential is undetermined to the gradient of a scalar field. This is just
another way of saying that we are free to choose
. Recall that the electric scalar potential
is undetermined to an arbitrary additive constant, since the transformation
(320)
leaves the electric field invariant in Eq. (316). The transformations (319) and (320) are examples
of what mathematicians call gauge transformations. The choice of a particular function
particular constant
or a
is referred to as a choice of the gauge. We are free to fix the gauge to be
whatever we like. The most sensible choice is the one which makes our equations as simple as
possible. The usual gauge for the scalar potential
gauge for
is such that
at infinity. The usual
is such that
(321)
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This particular choice is known as the Coulomb gauge.
It is obvious that we can always add a constant to
so as to make it zero at infinity. But it is not
at all obvious that we can always perform a gauge transformation such as to make
Suppose that we have found some vector field
zero.
whose curl gives the magnetic field but whose
divergence in non-zero. Let
(322)
The question is, can we find a scalar field
are left with
function
such that after we perform the gauge transformation we
. Taking the divergence of Eq. it is clear that we need to find a
which satisfies
But this is just Poisson's equation. We know that we can always find a unique solution of this
equation (see Sect. 3.11). This proves that, in practice, we can always set the divergence of
equal to zero.
Let us again consider an infinite straight wire directed along the
-axis and carrying a current
. The magnetic field generated by such a wire is written
We wish to find a vector potential
whose curl is equal to the above magnetic field, and whose
divergence is zero. It is not difficult to see that
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fits the bill. Note that the vector potential is parallel to the direction of the current. This would
seem to suggest that there is a more direct relationship between the vector potential and the
current than there is between the magnetic field and the current. The potential is not very wellbehaved on the
-axis, but this is just because we are dealing with an infinitely thin current.
Let us take the curl of Eq. We find that
where use has been made of the Coulomb gauge condition We can combine the above relation
with the field to give
Writing this in component form, we obtain
But, this is just Poisson's equation three times over. We can immediately write the unique
solutions to the above equations:
These solutions can be recombined to form a single vector solution
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Of course, we have seen a equation like this before:
Equations are the unique solutions (given the arbitrary choice of gauge) to the field equations
(they specify the magnetic vector and electric scalar potentials generated by a set of stationary
charges, of charge density
, and a set of steady currents, of current density
Incidentally, we can prove that Eq. satisfies the gauge condition
analysis of Eqs. (with
and
.
by repeating the
), and using the fact that
for
steady currents.
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Part-B
Unit-5: Magnetic forces:



 forces on moving charge
 differential current element
 force between differential current element
 force and torque on closed circuit.
Magnetic material and inductance:



 magnetization and permeability
 magnetic boundary condition
 magnetic circuit
 inductance and mutual inductance
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Magnetic Force on a Moving Charge:
 The force on a moving charge in a magnetic field is equal to the cross product of the
particles velocity with the magnetic field times the magnitude of the charge.
 The direction of the Magnetic Force is always at right angle to the plane formed by the
velocity vector v and the magnetic field B. (Right-hand rule)







The Magnetic Force depends upon the size of the charge q, the magnitude of the
magnetic field B, and how fast the charge is moving v perpendicular to the magnetic
field (equivalently, the size of v and the component of B perpendicular to the direction of
the moving charge).





 Also see the motion of a free charge moving in a magnetic field.
Properties of the Magnetic Field acting on a Moving Charge at one moment:


When q <
0, F is
in
the
opposite
direction.
If the sign of the charge on the particle is inverted, then the direction of the magnetic
force will be opposite that of a positive charge. The magnitude of the magnetic force
remains the same, only its direction is inverted.











When
v is
parallel
to B,
then F =
0.

There is one direction in space where the moving particle will experience no magnetic
force acting on it; this direction is along the direction of the magnetic field.


o
The direction of the magnetic force is always at 90 to the direction of motion of the
particle.


o
The direction of the magnetic force is always at 90 to the direction of the magnetic
field. (See Right-hand rule)

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
Plane
formed
by v and B
All positive particles which have v-B planes that are identical (or parallel) will
experience a magnetic force in the same
direction independent of the angle
between v and B. Changing the angle between v and Bwithout changing the orientation
of the v-B plane will not change the direction of the magnetic force on the charge. This
direction is perpendicular to the v-B plane and determined by the right-hand rule.

If the direction of the motion of the particle is changed, then the magnitude of the
magnetic force may or may not change.

When

If a particle moves in a plane that is perpendicular to B, then the magnitude of the
magnetic force will always have the same size (but necessarily the same direction) no
matter which direction the particle moves in the plane perpendicular to B. In this plane
the magnetic force will be maximum.

F is maximum, when = 90 . The magnitude of the magnetic force is greatest when the
charge is moving at right angles to the magnetic field.


o
 If the charge is not constrained by other forces then magnetic force will cause the free
charge to change its direction each moment. The end results is that the free charge will
move in a helix around the axis of the magnetic field. See Free Charge in Magnetic Field.
Forces between differential current elements
•
We need to find the force between two conductors without calculating the magnetic field
H
•
•
Assume two current loops
2
1
The magnetic field at point P due to P is
dH 2 
•
I dl  aˆ
1
1
12
4 R122
The magnetic field at P2 due to loop 1 is
H 
2

l
1
• The force on P2 due to loop 1 is
dF2  I 2 dl 2 B2
F2   I 2 dl 2  B 2
• This leads to:
l
2
o I 1 I 2
F2  4  dl
l 2
2
 dl 1 2aˆ12
 R
l 1
12
I dl 1  aˆ
1
4 R12
2
12
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• Finally we get
F2 
•
 I 1I
o
aˆ dl
2
4
12
l
2
R
l
2
1
dl 2
12
1
We can notice that this form is more complicated than calculating H2 first, then
calculating F2
•
Force and torque on closed circuit






A magnetic field exerts on a force on a wire (or other conductor) when a current passes
through it.
The general formula for determining the size and direction of the magnetic force on a
current-carrying wire involves a complicated integral
There are several cases in which the solution is relatively easy: (note that each one
involves a cross product of two vectors)
o Straight wire in uniform magnetic field:
o
F = I (L x B)
o
o
Curved wire in uniform magnetic field:
F = I (L' x B)
where L' is a straight vector from the starting point of the wire to its end point
o
o









Closed circuit loop in a magnetic field:
F=0
Although the force on a closed loop of current in a uniform magnetic field is zero, the
torque is not.
If one defines a special "area vector" A, the magnitude of which is the area of the closed
loop, and the direction of which is perpendicular to the plane of the loop (as given by
right-hand rule), then the torque on the loop is
tau = I (A x B)
The torque acts to make the "area vector" parallel to the direction of the magnetic field 
The magnetic moment "mu" of a closed circuit loop is a vector quantity, the product of
its current and "area vector". One can express the torque on a loop as
tau = mu x B
Permeability (electromagnetism)
In electromagnetism, permeability is the measure of the ability of a material to support the
formation of a magnetic field within itself. In other words, it is the degree of magnetization that a
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material obtains in response to an applied magnetic field. Magnetic permeability is typically
represented by the Greek letter μ. The term was coined in September, 1885 by Oliver Heaviside.
The reciprocal of magnetic permeability is magnetic reluctivity.
−1
(H·m ),
In SI units,
permeability
is
measured
in henrys per
meter
−2
or newtons per ampere squared (N·A ). The permeability constant (μ0), also known as
themagnetic constant or the permeability of free space, is a measure of the amount of resistance
encountered when forming a magnetic field in a classicalvacuum. The magnetic constant has the
[1]
−7
−6
−1
−2
exact (defined) value µ0 = 4π×10 ≈ 1.2566370614…×10 H·m or N·A ).
A closely related property of materials is magnetic susceptibility, which is a measure of the
magnetization of a material in addition to the magnetization of the space occupied by the
material.
In electromagnetism, the auxiliary magnetic field H represents how a magnetic field B influences
the organization of magnetic dipoles in a given medium, including dipole migration and
magnetic dipole reorientation. Its relation to permeability is
where the permeability, μ, is a scalar if the medium is isotropic or a second rank tensor for an
anisotropic medium.
In general, permeability is not a constant, as it can vary with the position in the medium, the
frequency of the field applied, humidity, temperature, and other parameters. In a nonlinear
medium, the permeability can depend on the strength of the magnetic field. Permeability as a
function of frequency can take on real or complex values. In ferromagnetic materials, the
relationship between B and H exhibits both non-linearity and hysteresis: B is not a single-valued
function of H, but depends also on the history of the material. For these materials it is sometimes
useful to consider the incremental permeability defined as
This definition is useful in local linearizations of non-linear material behavior, for example in a
Newton–Raphson iterative solution scheme that computes the changing saturation of a magnetic
circuit.
Permeability is the inductance per unit length. In SI units, permeability is measured in henrys per
−1
2
−2
metre (H·m = J/(A ·m) = N A ). The auxiliary magnetic field H has dimensions current per
−1
unit length and is measured in units of amperes per metre (A m ). The product μH thus has
2
dimensions inductance times current per unit area (H·A/m ). But inductance is magnetic flux per
unit current, so the product has dimensions magnetic flux per unit area. This is just the magnetic
2
field B, which is measured in webers (volt-seconds) per square-metre (V·s/m ), or teslas (T).
B is related to the Lorentz force on a moving charge q:
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The charge q is given in coulombs (C), the velocity v in meters per second (m/s), so that the
force F is in newtons (N):
H is related to the magnetic dipole density. A magnetic dipole is a closed circulation of electric
current. The dipole moment has dimensions current times area, units ampere square-metre
2
(A·m ), and magnitude equal to the current around the loop times the area of the loop. The H
field at a distance from a dipole has magnitude proportional to the dipole moment divided by
[4]
distance cubed, which has dimensions current per unit length.
Relative permeability and magnetic susceptibility
Relative permeability, sometimes denoted by the symbol μr, is the ratio of the permeability of a
specific medium to the permeability of free space, μ0:
where μ0 = 4π × 10
−7
−2
N A . In terms of relative permeability, the magnetic susceptibility is
χm, a dimensionless quantity, is sometimes called volumetric or bulk susceptibility, to distinguish
it from χp (magnetic mass or specific susceptibility) and χM (molar or molar mass susceptibility).
Magnetic circuit
A magnetic circuit is made up of one or more closed loop paths containing a magnetic flux. The
flux is usually generated by permanent magnets orelectromagnets and confined to the path by
magnetic cores consisting of ferromagnetic materials like iron, although there may be air gaps or
other materials in the path. Magnetic circuits are employed to efficiently channel magnetic fields
in many devices such as electric motors, generators, transformers, relays,
lifting electromagnets, SQUIDs, galvanometers, and magnetic recording heads.
The concept of a "magnetic circuit" exploits a one-to-one correspondence between the equations
of the magnetic field in an unsaturated ferromagnetic material to that of an electrical circuit.
Using this concept the magnetic fields of complex devices such as transformers can be quickly
solved using the methods and techniques developed for electrical circuits.
Some examples of magnetic circuits are:

horseshoe magnet with iron keeper (low-reluctance circuit)
horseshoe magnet with no keeper (high-reluctance circuit)
electric motor (variable-reluctance circuit)


Similar to the way that EMF drives a current of electrical charge in electrical circuits,
magnetomotive force (MMF) 'drives' magnetic flux through magnetic circuits. The term
'magnetomotive force', though, is a misnomer since it is not a force nor is anything moving. It is
perhaps better to call it simply MMF. In analogy to the definition of EMF, the magnetomotive
force around a closed loop is defined as:

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The MMF represents the potential that a hypothetical magnetic charge would gain by completing
the loop. The magnetic flux that is driven is not a current of magnetic charge; it merely has the
same relationship to MMF that electric current has to EMF. (See microscopic origins of
reluctance below for a further description.)
The unit of magnetomotive force is the ampere-turn (At), represented by a steady, direct electric
current of one ampere flowing in a single-turn loop of electrically conducting material in a
vacuum. The gilbert (Gi), established by the IEC in 1930 [1], is the CGS unit of magnetomotive
force and is a slightly smaller unit than the ampere-turn. The unit is named after William Gilbert
(1544–1603) English physician and natural philosopher.

The magnetomotive force can often be quickly calculated using Ampère's law. For example, the
magnetomotive force
of long coil is:

,
where N is the number of turns and I is the current in the coil. In practice this equation is used for
the MMF of real inductors with N being the winding number of the inducting coil.
In electronic circuits, Ohm's law is an empirical relation between the EMF
element and the current I it generates through that element. It is written as:
applied across an

where R is the electrical resistance of that material. Hopkinson's law is a counterpart to Ohm's
law used in magnetic circuits. The law is named after the British electrical engineer, John
[1][2]
Hopkinson. It states that

where is the magnetomotive force
(MMF) across a magnetic element, is the magnetic
flux through the magnetic element, and
is the magnetic reluctance of that element. (It shall
be shown later that this relationship is due to the empirical relationship between the H-field and
the magnetic field B, B=μH, where μ is the permeability of the material.) Like Ohm's law,
Hopkinson's law can be interpreted either as an empirical equation that works for some materials,
or it may serve as a definition of reluctance.

Magnetic reluctance, or magnetic resistance, is analogous to resistance in
an electrical circuit (although it does not dissipate magnetic energy). In likeness to the
way an electric field causes an electric current to follow the path of least resistance, a
magnetic field causes magnetic flux to follow the path of least magnetic reluctance. It is a
scalar, extensive quantity, akin to electrical resistance.
 The total reluctance is equal to the ratio of the (MMF) in a passive magnetic circuit and
the magnetic flux in this circuit. In an AC field, the reluctance is the ratio of the
amplitude values for a sinusoidal MMF and magnetic flux. (see phasors)




where
The definition can be expressed as:

is the reluctance in ampere-turns per weber (a unit that is equivalent to turns per henry).
Magnetic flux always forms a closed loop, as described by Maxwell's equations, but the path of
the loop depends on the reluctance of the surrounding materials. It is concentrated around the
path of least reluctance. Air and vacuum have high reluctance, while easily magnetized materials
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such as soft iron have low reluctance. The concentration of flux in low-reluctance materials
forms strong temporary poles and causes mechanical forces that tend to move the materials
towards regions of higher flux so it is always an attractive force(pull).
The inverse of reluctance is called permeance.

Its SI derived unit is the henry (the same as the unit of inductance, although the two concepts are
distinct).
Inductance
In electromagnetism and electronics, inductance is the property of a conductor by which a
change in current in the conductor "induces" (creates) avoltage (electromotive force) in both the
[1][2][3]
[4][5]
conductor itself (self-inductance)
and any nearby conductors (mutual inductance).
This
effect derives from two fundamental observations of physics: First, that a steady current creates a
[6]
steady magnetic field (Oersted's law) and second, that a time-varying magnetic field induces a
[7]
[8]
voltage in a nearby conductor (Faraday's law of induction). From Lenz's law, in an electric
circuit, a changing electric current through a circuit that has inductance induces a proportional
voltage which opposes the change in current (self inductance). The varying field in this circuit
may also induce an e.m.f. in a neighbouring circuit (mutual inductance).
[9]
The term 'inductance' was coined by Oliver Heaviside in February 1886. It is customary to use the
[10][11]
symbol L for inductance, in honour of the physicist Heinrich Lenz.
In the SI system the
unit of inductance is the henry, named in honor of the scientist who discovered
inductance,Joseph Henry.
To add inductance to a circuit, electronic components called inductors are used, typically
consisting of coils of wire to concentrate the magnetic field and so that the magnetic field is
linked into the circuit more than once.
The relationship between the self inductance L of an electrical circuit in henries, voltage, and
current is
where v denotes the voltage in volts and i the current in amperes. The voltage across an inductor
is equal to the product of its inductance and the time rate of change of the current through it.
All practical circuits have some inductance, which may provide either beneficial or detrimental
effects. In a tuned circuit inductance is used to provide a frequency selective circuit. Practical
inductors may be used to provide filtering or energy storage in a system. The inductance of a
transmission lineis one of the properties that determines its characteristic impedance; balancing
the inductance and capacitance of cables is important for distortion-free telegraphy and
telephony. The inductance of long power transmission lines limits the AC power that can be sent
over them. Sensitive circuits such as microphone and computer network cables may use special
cable constructions to limit the mutual inductance between signal circuits.
The generalization to the case of K electrical circuits with currents im and voltages vm reads
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Inductance here is a symmetric matrix. The diagonal coefficients Lm,m are called coefficients of
self inductance, the off-diagonal elements are called coefficients of mutual inductance. The
coefficients of inductance are constant as long as no magnetizable material with nonlinear
characteristics is involved. This is a direct consequence of the linearity of Maxwell's equations in
the fields and the current density. The coefficients of inductance become functions of the
currents in the nonlinear case, see nonlinear inductance.
The inductance equations above are a consequence of Maxwell's equations. There is a
straightforward derivation in the important case of electrical circuits consisting of thin wires.
Consider a system of K wire loops, each with one or several wire turns. The flux linkage of loop
m is given by
Here Nm denotes the number of turns in loop m, Φm the magnetic flux through this loop, and
Lm,n are some constants. This equation follows from Ampere's law - magnetic fields and fluxes
are linear functions of the currents. By Faraday's law of induction we have
where vm denotes the voltage induced in circuit m. This agrees with the definition of inductance
above if the coefficients Lm,n are identified with the coefficients of inductance. Because the total
currents Nnin contribute to Φm it also follows that Lm,n is proportional to the product of turns NmNn
Mutual Inductance of Two Coils
In the previous tutorial we saw that an inductor generates an induced emf within itself as a result
of the changing magnetic field around its own turns, and when this emf is induced in the same
circuit in which the current is changing this effect is called Self-induction, ( L ). However, when
the emf is induced into an adjacent coil situated within the same magnetic field, the emf is said to
be induced magnetically, inductively or by Mutual induction, symbol ( M ). Then when two or
more coils are magnetically linked together by a common magnetic flux they are said to have the
property of Mutual Inductance.
Mutual Inductance is the basic operating principal of the transformer, motors, generators and
any other electrical component that interacts with another magnetic field. Then we can define
mutual induction as the current flowing in one coil that induces an voltage in an adjacent coil.
But mutual inductance can also be a bad thing as "stray" or "leakage" inductance from a coil can
interfere with the operation of another adjacent component by means of electromagnetic
induction, so some form of electrical screening to a ground potential may be required.
The amount of mutual inductance that links one coil to another depends very much on the
relative positioning of the two coils. If one coil is positioned next to the other coil so that their
physical distance apart is small, then nearly nearly all of the magnetic flux generated by the first
coil will interact with the coil turns of the second coil inducing a relatively large emf and
therefore producing a large mutual inductance value.
Likewise, if the two coils are farther apart from each other or at different angles, the amount of
induced magnetic flux from the first coil into the second will be weaker producing a much
smaller induced emf and therefore a much smaller mutual inductance value. So the effect of
mutual inductance is very much dependant upon the relative positions or spacing, ( S ) of the two
coils and this is demonstrated below.
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Mutual Inductance between Coils
The mutual inductance that exists between the two coils can be greatly increased by positioning
them on a common soft iron core or by increasing the number of turns of either coil as would be
found in a transformer. If the two coils are tightly wound one on top of the other over a common
soft iron core unity coupling is said to exist between them as any losses due to the leakage of
flux will be extremely small. Then assuming a perfect flux linkage between the two coils the
mutual inductance that exists between them can be given as.
Where:
-7
µo is the permeability of free space (4.π.10 )
µr is the relative permeability of the soft iron core
N is in the number of coil turns
A is in the cross-sectional area in m
2
l is the coils length in meters
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Mutual Induction
Here the current flowing in coil one, L1 sets up a magnetic field around itself with some of these
magnetic field lines passing through coil two, L2 giving us mutual inductance. Coil one has a
current ofI1 and N1 turns while, coil two has N2 turns. Therefore, the mutual inductance, M12 of
coil two that exists with respect to coil one depends on their position with respect to each other
and is given as:
Likewise, the flux linking coil one, L1 when a current flows around coil two, L2 is exactly the
same as the flux linking coil two when the same current flows around coil one above, then the
mutual inductance of coil one with respect of coil two is defined as M21. This mutual inductance
is true irrespective of the size, number of turns, relative position or orientation of the two coils.
Because of this, we can write the mutual inductance between the two coils as: M12 = M21 = M.
Hopefully we remember from our tutorials on Electromagnets that the self inductance of each
individual coil is given as:
and
Then by cross-multiplying the two equations above, the mutual inductance that exists between
the two coils can be expressed in terms of the self inductance of each coil.
giving us a final and more common expression for the mutual inductance between two coils as:
Mutual Inductance Between Coils
However, the above equation assumes zero flux leakage and 100% magnetic coupling between
the two coils, L 1 and L 2. In reality there will always be some loss due to leakage and position,
so the magnetic coupling between the two coils can never reach or exceed 100%, but can become
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very close to this value in some special inductive coils. If some of the total magnetic flux links
with the two coils, this amount of flux linkage can be defined as a fraction of the total possible
flux linkage between the coils. This fractional value is called the coefficient of coupling and is
given the letter k.
Coupling Coefficient
Generally, the amount of inductive coupling that exists between the two coils is expressed as a
fractional number between 0 and 1 instead of a percentage (%) value, where 0 indicates zero or
no inductive coupling, and 1 indicating full or maximum inductive coupling. In other words, if k
= 1 the two coils are perfectly coupled, if k > 0.5 the two coils are said to be tightly coupled and
if k < 0.5 the two coils are said to be loosely coupled. Then the equation above which assumes a
perfect coupling can be modified to take into account this coefficient of coupling, k and is given
as:
Coupling Factor Between Coils
or
When the coefficient of coupling, k is equal to 1, (unity) such that all the lines of flux of one coil
cuts all of the turns of the other, the mutual inductance is equal to the geometric mean of the two
individual inductances of the coils. So when the two inductances are equal and L 1 is equal to L
2, the mutual inductance that exists between the two coils can be defined as:
Example No1
Two inductors whose self-inductances are given as 75mH and 55mH respectively, are positioned
next to each other on a common magnetic core so that 75% of the lines of flux from the first coil
are cutting the second coil. Calculate the total mutual inductance that exists between them.
In the next tutorial about Inductors, we look at connecting together Inductors in Series and the
affect this combination has on the circuits mutual inductance, total inductance and their induced
voltages.
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Unit-6
Time varing field and maxwell’s equation





Faraday‘s law
displacement current
maxwell‘s equation in point and integral form
retarded potentials.
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Faraday's law of induction
Faraday's law of induction is a basic law of electromagnetism that predicts how a magnetic
field will interact with an electric circuit to produce anelectromotive force (EMF). It is the
fundamental
operating
principle
of transformers, inductors,
and
many
types
of electrical motors and generators.
Faraday's law of induction makes use of the magnetic flux ΦB through a hypothetical surface Σ
whose boundary is a wire loop. Since the wire loop may be moving, we write Σ(t) for the surface.
The magnetic flux is defined by a surface integral:
where dA is
an element of surface area of the moving surface Σ(t), B is the magnetic field,
and B·dA is
a vector dot product (the infinitesimal amount of magnetic flux). In more visual
terms, the magnetic flux through the wire loop is proportional to the number of magnetic flux
lines that pass through the loop.
When the flux changes—because B changes, or because the wire loop is moved or deformed, or
both—Faraday's law of induction says that the wire loop acquires an EMF , defined as the
energy available per unit charge that travels once around the wire loop (the unit of EMF is the
[2][15][16][17]
volt).
Equivalently, it is the voltage that would be measured by cutting the wire to
create an open circuit, and attaching a voltmeter to the leads. According to the Lorentz force law
(in SI units),
the EMF on a wire loop is:
where E is the electric field, B is the magnetic field (aka magnetic flux density, magnetic
induction), dℓ is an infinitesimal arc length along the wire, and the line integral is evaluated
along the wire (along the curve the conincident with the shape of the wire).
The EMF is also given by the rate of change of the magnetic flux:
where
is the magnitude of the electromotive force (EMF) in volts and ΦB is the magnetic flux
in webers. The direction of the electromotive force is given by Lenz's law.
For a tightly wound coil of wire, composed of N identical loops, each with the same ΦB,
Faraday's law of induction states that
where N is the number of turns of wire and ΦB is the magnetic flux in webers through a single
loop.
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The Maxwell–Faraday equation states that a time-varying magnetic field is always accompanied
by a spatially-varying, non-conservative electric field, and vice-versa. The Maxwell–Faraday
equation is
(in SI units) where
is the curl operator and again E(r, t) is the electric field and B(r, t) is
the magnetic field. These fields can generally be functions of position r and time t.
The Maxwell–Faraday equation is one of the four Maxwell's equations, and therefore plays a
fundamental role in the theory of classical electromagnetism. It can also be written in an integral
[20]
form by the Kelvin-Stokes theorem:
where, as indicated in the figure:
Σ is a surface bounded by the closed contour ∂Σ,
E is the electric field, B is the magnetic field.
dℓ is an infinitesimal vector element of the contour ∂Σ,
dA is an infinitesimal vector element of surface Σ. If its direction is orthogonal to that
surface patch, the magnitude is the area of an infinitesimal patch of surface.
Both dℓ and dA have a sign ambiguity; to get the correct sign, the right-hand rule is used, as
explained in the article Kelvin-Stokes theorem. For a planar surface Σ, a positive path element dℓ
of curve ∂Σ is defined by the right-hand rule as one that points with the fingers of the right hand
when the thumb points in the direction of the normal n to the surface Σ.
The integral around ∂Σ is called a path integral or line integral.
Notice that a nonzero path integral for E is different from the behavior of the electric field
generated by charges. A charge-generated E-field can be expressed as the gradient of a scalar
field that is a solution to Poisson's equation, and has a zero path integral. See gradient theorem.
The integral equation is true for any path ∂Σ through space, and any surface Σ for which that
path is a boundary.
If the path Σ is not changing in time, the equation can be rewritten:
The surface integral at the right-hand side is the explicit expression for the magnetic flux ΦB
through Σ.
Displacement current
In electromagnetism, displacement current is a quantity appearing in Maxwell's equations that
is defined in terms of the rate of change of electric displacement field. Displacement current has
the units of electric current density, and it has an associated magnetic field just as actual currents
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do. However it is not an electric current of moving charges, but a time-varying electric field. In
materials, there is also a contribution from the slight motion of charges bound in atoms,
dielectric polarization.
The idea was conceived by James Clerk Maxwell in his 1861 paper On Physical Lines of Force
in connection with the displacement of electric particles in a dielectric medium. Maxwell added
displacement current to the electric current term in Ampère's Circuital Law. In his 1865 paperA
Dynamical Theory of the Electromagnetic Field Maxwell used this amended version of Ampère's
Circuital Law to derive the electromagnetic wave equation. This derivation is now generally
accepted as a historical landmark in physics by virtue of uniting electricity, magnetism and
optics into one single unified theory. The displacement current term is now seen as a crucial
addition that completed Maxwell's equations and is necessary to explain many phenomena, most
particularly the existence of electromagnetic waves.
The electric displacement field is defined as:
where:
ε0 is the permittivity of free space
E is the electric field intensity
P is the polarization of the medium
Differentiating this equation with respect to time defines the displacement current density, which
[1]
therefore has two components in a dielectric:
The first term on the right hand side is present in material media and in free space. It doesn't
necessarily involve any actual movement of charge, but it does have an associated magnetic
field, just as does a current due to charge motion. Some authors apply the name displacement
current to only this contribution.
The second term on the right hand side is associated with the polarization of the individual
molecules of the dielectric material. Polarization results when the charges in molecules move a
little under the influence of an applied electric field. The positive and negative charges in
molecules separate, causing an increase in the state of polarization P. A changing state of
polarization corresponds to charge movement and so is equivalent to a current.
This polarization is the displacement current as it was originally conceived by Maxwell.
Maxwell made no special treatment of the vacuum, treating it as a material medium. For
Maxwell, the effect of P was simply to change the relative permittivity εr in the relation D = εrε0
E.
The modern justification of displacement current is explained below.
Isotropic dielectric case
In the case of a very simple dielectric material the constitutive relation holds:
where the permittivity ε = ε0 εr,

εr is the relative permittivity of the dielectric and
 ε0 is the electric constant.

In this equation the use of ε, accounts for the polarization of the dielectric.
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The scalar value of displacement current may also be expressed in terms of electric flux:
The forms in terms of ε are correct only for linear isotropic materials. More generally ε may be
replaced by a tensor, may depend upon the electric field itself, and may exhibit time dependence
(dispersion).
For a linear isotropic dielectric, the polarization P is given by:
where χe is known as the electric susceptibility of the dielectric. Note that:
Maxwell‘s Equations (Integral Form)
It is sometimes easier to understand Maxwell’s equations in their integral form; the version
we outlined last time is the differential form.
For Gauss’ law and Gauss’ law for magnetism, we‘ve actually already done this. First, we
write them in differential form:
We pick any region
we want and integrate both sides of each equation over that region:
On the left-hand sides we can use the divergence theorem, while the right sides can simply be
evaluated:where
is the total charge contained within the region . Gauss‘ law tells us
that the flux of the electric field out through a closed surface is (basically) equal to the charge
contained inside the surface, while Gauss‘ law for magnetism tells us that there is no such
thing as a magnetic charge.
Faraday‘s law was basically given to us in integral form, but we can get it back from the
differential form:
We pick any surface
and integrate the flux of both sides through it:
On the left we can use Stokes’ theorem, while on the right we can pull the derivative
outside the integral:
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where
is the flux of the magnetic field through the surface . Faraday‘s law tells
us that a changing magnetic field induces a current around a circuit.
A similar analysis helps with Ampère‘s law:
We pick a surface and integrate:
Then we simplify each side.
where
is the flux of the electric field
through the surface , and
is the total
current flowing through the surface . Ampère‘s law tells us that a flowing current induces
a magnetic field around the current, and Maxwell‘s correction tells us that a changing
electric field behaves just like a current made of moving charges.
We collect these together into the integral form of Maxwell‘s equations:
Retarded potentials
We are now in a position to solve Maxwell's equations. Recall that in steady-state, Maxwell's
equations reduce to
(502)
(503)
The solutions to these equations are easily found using the Green's function for Poisson's
equation (480):
(504)
(505)
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The time-dependent Maxwell equations reduce to
(506)
(507)
We can solve these equations using the time-dependent Green's function (499). From Eq. (486)
we find that
(508)
with a similar equation for
equations reduce to
. Using the well-known property of delta-functions, these
(509)
(510)
These are the general solutions to Maxwell's equations. Note that the time-dependent solutions,
(509) and (510), are the same as the steady-state solutions, (504) and (505), apart from the weird
way in which time appears in the former. According to Eqs. (509) and (510), if we want to work
out the potentials at position
and time
then we have to perform integrals of the charge
density and current density over all space (just like in the steady-state situation). However, when
we calculate the contribution of charges and currents at position
to these integrals we do not
use the values at time , instead we use the values at some earlier time
. What is
this earlier time? It is simply the latest time at which a light signal emitted from position
would be received at position before time . This is called the retarded time. Likewise, the
potentials (509) and (510) are called retarded potentials. It is often useful to adopt the following
notation
(511)
The square brackets denote retardation (i.e., using the retarded time instead of the real time).
Using this notation Eqs. (509) and (510), become
(512)
(513)
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The time dependence in the above equations is taken as read.
We are now in a position to understand electromagnetism at its most fundamental level. A charge
distribution
can thought of as built up out of a collection, or series, of charges which
instantaneously come into existence, at some point
and some time , and then disappear
again. Mathematically, this is written
(514)
Likewise, we can think of a current distribution
as built up out of a collection or series of
currents which instantaneously appear and then disappear:
(515)
Each of these ephemeral charges and currents excites a spherical wave in the appropriate
potential. Thus, the charge density at
and
sends out a wave in the scalar potential:
(516)
Likewise, the current density at
and
sends out a wave in the vector potential:
(517)
These waves can be thought of as messengers which inform other charges and currents about the
charges and currents present at position
and time . However, these messengers travel at a
finite speed: i.e., the speed of light. So, by the time they reach other charges and currents their
message is a little out of date. Every charge and every current in the Universe emits these
spherical waves. The resultant scalar and vector potential fields are given by Eqs. (512) and
(513). Of course, we can turn these fields into electric and magnetic fields using Eqs. (421) and
(422). We can then evaluate the force exerted on charges using the Lorentz formula. We can see
that we have now escaped from the apparent action at a distance nature of Coulomb's law and the
Biot-Savart law. Electromagnetic information is carried by spherical waves in the vector and
scalar potentials, and, therefore, travels at the velocity of light. Thus, if we change the position of
a charge then a distant charge can only respond after a time delay sufficient for a spherical wave
to propagate from the former to the latter charge.
Let us compare the steady-state law
(518)
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with the corresponding time-dependent law
(519)
These two formulae look very similar indeed, but there is an important difference. We can
imagine (rather pictorially) that every charge in the Universe is continuously performing the
integral (519), and is also performing a similar integral to find the vector potential. After
evaluating both potentials, the charge can calculate the fields, and, using the Lorentz force law, it
can then work out its equation of motion. The problem is that the information the charge receives
from the rest of the Universe is carried by our spherical waves, and is always slightly out of date
(because the waves travel at a finite speed). As the charge considers more and more distant
charges or currents, its information gets more and more out of date. (Similarly, when
astronomers look out to more and more distant galaxies in the Universe, they are also looking
backwards in time. In fact, the light we receive from the most distant observable galaxies was
emitted when the Universe was only about one third of its present age.) So, what does our
electron do? It simply uses the most up to date information about distant charges and currents
which it possesses. So, instead of incorporating the charge density
electron uses the retarded charge density
time). This is effectively what Eq. (519) says.
(i.e., the density evaluated at the retarded
Consider a thought experiment in which a charge
for a while, and then disappears at time
charge? Using Eq. (519), we find that
in its integral, the
appears at position
at time
, persists
. What is the electric field generated by such a
(520)
Now,
so
(since there are no currents, and therefore no vector potential is generated),
(521)
This solution is shown pictorially in Fig. 37. We can see that the charge effectively emits a
Coulomb electric field which propagates radially away from the charge at the speed of light.
Likewise, it is easy to show that a current carrying wire effectively emits an Ampèrian magnetic
field at the speed of light.
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Figure 37:
We can now appreciate the essential difference between time-dependent electromagnetism and
the action at a distance laws of Coulomb and Biot & Savart. In the latter theories, the field-lines
act rather like rigid wires attached to charges (or circulating around currents). If the charges (or
currents) move then so do the field-lines, leading inevitably to unphysical action at a distance
type behaviour. In the time-dependent theory, charges act rather like water sprinklers: i.e., they
spray out the Coulomb field in all directions at the speed of light. Similarly, current carrying
wires throw out magnetic field loops at the speed of light. If we move a charge (or current) then
field-lines emitted beforehand are not affected, so the field at a distant charge (or current) only
responds to the change in position after a time delay sufficient for the field to propagate between
the two charges (or currents) at the speed of light.
In Coulomb's law and the Biot-Savart law, it is not entirely obvious that the electric and
magnetic fields have a real existence. After all, the only measurable quantities are the forces
acting between charges and currents. We can describe the force acting on a given charge or
current, due to the other charges and currents in the Universe, in terms of the local electric and
magnetic fields, but we have no way of knowing whether these fields persist when the charge or
current is not present (i.e., we could argue that electric and magnetic fields are just a convenient
way of calculating forces, but, in reality, the forces are transmitted directly between charges and
currents by some form of magic). However, it is patently obvious that electric and magnetic
fields have a real existence in the time-dependent theory. Consider the following thought
experiment. Suppose that a charge
comes into existence for a period of time, emits a
Coulomb field, and then disappears. Suppose that a distant charge
interacts with this field,
but is sufficiently far from the first charge that by the time the field arrives the first charge has
already disappeared. The force exerted on the second charge is only ascribable to the electric
field: it cannot be ascribed to the first charge, because this charge no longer exists by the time the
force is exerted. The electric field clearly transmits energy and momentum between the two
charges. Anything which possesses energy and momentum is ``real'' in a physical sense. Later on
in this course, we shall demonstrate that electric and magnetic fields conserve energy and
momentum.
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Figure 38:
Let us now consider a moving charge. Such a charge is continually emitting spherical waves in
the scalar potential, and the resulting wavefront pattern is sketched in Fig. 38. Clearly, the
wavefronts are more closely spaced in front of the charge than they are behind it, suggesting that
the electric field in front is larger than the field behind. In a medium, such as water or air, where
waves travel at a finite speed, (say), it is possible to get a very interesting effect if the wave
source travels at some velocity which exceeds the wave speed. This is illustrated in Fig. 39.
Figure 39:
The locus of the outermost wave front is now a cone instead of a sphere. The wave intensity on
the cone is extremely large: this is a shock wave! The half-angle of the shock wave cone is
simply
. In water, shock waves are produced by fast moving boats. We call these
bow waves. In air, shock waves are produced by speeding bullets and supersonic jets. In the latter
case, we call these sonic booms. Is there any such thing as an electromagnetic shock wave? At
first sight, the answer to this question would appear to be, no. After all, electromagnetic waves
travel at the speed of light, and no wave source (i.e., an electrically charged particle) can travel
faster than this velocity. This is a rather disappointing conclusion. However, when an
electromagnetic wave travels through matter a remarkable thing happens. The oscillating electric
field of the wave induces a slight separation of the positive and negative charges in the atoms
which make up the material. We call separated positive and negative charges an electric dipole.
Of course, the atomic dipoles oscillate in sympathy with the field which induces them. However,
an oscillating electric dipole radiates electromagnetic waves. Amazingly, when we add the
original wave to these induced waves, it is exactly as if the original wave propagates through the
material in question at a velocity which is slower than the velocity of light in vacuum. Suppose,
now, that we shoot a charged particle through the material faster than the slowed down velocity
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of electromagnetic waves. This is possible since the waves are traveling slower than the velocity
of light in vacuum. In practice, the particle has to be traveling pretty close to the velocity of light
in vacuum (i.e., it has to be relativistic), but modern particle accelerators produce copious
amounts of such particles. Now, we can get an electromagnetic shock wave. We expect an
intense cone of emission, just like the bow wave produced by a fast ship. In fact, this type of
radiation has been observed. It is calledCherenkov radiation, and it is very useful in high energy
physics. Cherenkov radiation is typically produced by surrounding a particle accelerator with
perspex blocks. Relativistic charged particles emanating from the accelerator pass through the
perspex traveling faster than the local velocity of light, and therefore emit Cherenkov radiation.
We know the velocity of light ( , say) in perspex (this can be worked out from the refractive
index), so if we can measure the half angle of the radiation cone emitted by each particle then
we can evaluate the speed of the particle
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via the geometric relation
.
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Unit- -7
Uniform plane wave:





wave propagation in free space and dielectric

Poynting theorem

wave power

propagation in good conductors

skin effect.
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UNIFORM PLANE WAVES
In free space ( source-less regions where
D0
EE0
  J    0 ), the gauss law is
or
 E  0 ________ (1)
The wave equation for electric field, in free-space is,
 E
2
2 E
t
________ (2)
2
The wave equation (2) is a composition of these equations, one each component
wise, ie,
2
2
 Ex
 Ey
2

2

x
2
 Ey
y
2
t
2
 Ey
t
_______(2) b
2
2
 Ez
z
_______(2) a
2
 Ez

2
t
_______(2) c
2
Further, eqn. (1) may be written as
Ey Ez
Ex 

x
y
z
 0 ________ (1) a
For the UPW, E is independent of two coordinate axes; x and y axes, as we have assumed.

Ez

x


y
0
Therefore eqn. (1) reduces to
 0 ______ (3)
z
ie., there is no variation of Ez in the z direction.
2
 Ez
Also we find from 2 (a) that
t2= 0 ____(4)
These two conditions (3) and (4) require that Ez can be
(i)
Zero
(ii)
Constant in time or
(iii)
Increasing uniformly with time.
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A field satisfying the last two of the above three conditions cannot be a part of wave motion.
Therefore Ez can be put equal to zero, (the first condition).
Ez = 0
The uniform plane wave (traveling in z direction) does not have any field components of E & H
in its direction of travel.
Therefore the UPWs are transverse., having field components (of E & H ) only in directions
perpendicular to the direction of propagation does not have any field component only the
direction of travel.
RELATION BETWEEN E & H in a uniform plane wave.
We have, from our previous discussions that, for a UPW traveling in z direction, both E & H are
independent of x and y; and E & H have no z component. For such a UPW, we have,
 E 
 H 
ˆ
ˆ
i
j
ˆ
k
 (  0)  (  0) 
x
y
z
Ez(  0)
Ex
Ey
ˆ
ˆ
i
j
ˆ  Ey  ˆ  Ex 
i

  j


z


_____ (5)

 z 
ˆ
k
 (  0)  (  0) 
x
y
z
Hx
Hy Hz(  0)
i
ˆ  H y  ˆ  H x 

  j



z


_____ (6)
 z 
Then Maxwell‘s curl equations (1) and (2), using (5) and (6), (2) becomes,
Ex ˆ
 H E 
t
t
and
i
Ey ˆ

j
t
 E  H  Hx
t
t
Hy   ˆ  Hx 

ˆ  z

j
 i 
 
ˆ Hy ˆ
i
t
j


 i 
Ey 



______ (7)
z 
ˆ  Ex 
j
 z

Thus, rewriting (7) and (8) we get

______ (8)
z 
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 Hy ˆ  Hx ˆ
z
z
i
j
Ex ˆ
 t


 Ey ˆ  Ex ˆ
z
z
i
Hx ˆ
 t
j

i
 Ey ˆ 
t

______ (7)
i
 Hy ˆ
t

______ (8)

j

j 
ˆ
Equating i th and j th terms, we get
Hy
z
 Ex
t
______ 9 ( a)
Hx
Ey
z
t
Ey ˆ
Hx
 z
i  t
and
______ 9 ( b)

Ex
z
Let
  Hy
______ 9( c)
______ 9 ( d )
t

Ey  f1  z   0 t  ;
Ey
t
1
0

E
.
Then,
 f1  z   0 t  0 .  0 f1.
 From eqn. 9( c ), we get ,
Hx
t

0 f '


 Hx 
Department of ECE
0

0
0

f
'
1
f1 dz  c.
'
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Now
f
  z  0t 
'
'
z
1
f
z
1
H
z


f
1
f


'

1
C
z
Now
f
'
1
z
  z  0t 
'
f
z
f1 dz  c 
1



1


z

Hx 
'
f
f c
1
 Ey  c

The constant C indicates that a field independent of Z could be present. Evidently this is not a
part of the wave motion and hence is rejected.
Thus the relation between HX and EY becomes,
H x   E


Ey

Hx
y


__________ (10)
Similarly it can be shown that
E

x
H
y

 _____________ (11)
ˆ
ˆ
In our UPW, E  E x i  E y j
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 E  
2

E 
t 
t 
 E

2
 E
E
 E  
 2 _______ ( xi)
t
t
2
But  E 


0
E
DERIVATION OF WAVE EQUATION FOR A CONDUCTING MEDIUM:
In a conducting medium,  = 0,  = 0. Surface charges and hence surface currents exist, static
fields or charges do not exist.
For the case of conduction media, the point form of maxwells equations are:
 D
 E


E

 t
 t
B
H
 H  J 
________ (i )

_________ (ii)
t
t
 D  E  E  0 _________ (iii)

 B   H   H  0 _________ (iv)
 E 
Taking curl on both sides of equation (i ), we get

   H   E 
E 

t



t  E ________ ( v)
   E  
substituting eqn. (ii ) in eqn. (v ), we get

H 

t 

 H  


2


 H 


2  _________ ( vi)
t


2
ButH H  H_________ (vii)
eqn. ( vi ) becomes
2
 H
H
2
  H  H  
t  t
_________ (viii)
B  1  B  1 0  0
But  H 


2



eqn. ( viii ) becomes,
 H  
2
2
 H
H

2  0 ________ (ix)
t
t
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This is the wave equation for the magnetic field H in a conducting medium.
Next we consider the second Maxwell‘s curl equation (ii)
H ________ (ii)
t
 E 
Taking curl on both sides of equation (ii) we get
H
 E  


t
t
But  E 
  H  ________ ( x)
  E 2 E ;
Vector identity and substituting eqn. (1) in eqn (2), we get

E 

  E 
t 
t 
2
 E _______ ( xi)
E


t
t 2
  E  2 E  

But  E 
 
0
(Point form of Gauss law) However, in a conductor,  = 0, since there is no net charge within a
conductor,
 E0
Therefore we get
Therefore eqn. (xi) becomes,
 E  
2
E

t
2 E
____________ (xii)
2
t
This is the wave equation for electric field E in a conducting medium.
Wave equations for a conducting medium:


Regions where conductivity is non-zero.

Conduction currents may exist.
For such regions, for time varying fields
The Maxwell‘s eqn. Are:
 E
 t _________ (1)
 H
  E   t __________ (2)
  H  J 

JE
:Conductivity(/m)
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= conduction current density.
Therefore eqn. (1) becomes,
E
t
 H   E
_________ (3)
Taking curl of both sides of eqn. (2), we get

  E   

t  H 
 

2E
 
E
________ (4)
2
t t
But


2
 E   E  E ( vector identity ) u sin g this eqn. (4)
becomes vector identity,


E
t
  E  E  
 But  D  
2
 is cons tan t ,E

1
2 E
2
t
_______ (5)
D

Since there is no net charge within a conductor the charge density is zero ( there can be charge
on the surface ), we get.
1
E D0
Therefore using this result in eqn. (5)
we get
 E  
2
E
t

2 E
 0 ________(6)
2
t
This is the wave eqn. For the electric field E in a conducting medium.
This is the wave eqn. for E . The wave eqn. for H is obtained in a similar manner.
Taking curl of both sides of (1), we get
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 H 
E
   E ________ (7)
t
But  E 
(1) becomes,
 H   
H
t
________ (2)
2
 H 
2
t
H
t
________ (8)
As before, we make use of the vector identity.
 H 
  H  2 H
in eqn. (8) and get


2 H
  H  H    H 
2
t
t
2
________ (9)
But
 H  B  1  B  1 0  0



eqn.(9)becomes

2
2
 H
H
H  
 2 ________ (10)
t
t
This is the wave eqn. for H in a conducting medium.
Sinusoidal Time Variations:
In practice, most generators produce voltage and currents and hence electric and magnetic fields
which vary sinusoidally with time. Further, any periodic variation can be represented as a weight
sum of fundamental and harmonic frequencies.
Therefore we consider fields having sinusoidal time variations, for
example, E = Em cos t
E = Em sin t
Here, w = 2f, f = frequency of the variation.
Therefore every field or field component varies sinusoidally, mathematically by an additional
term. Representing sinusoidal variation. For example, the electric field E can be represented as
E  x , y , z , t  as
ie., E r , t ; r  x , y , z 
Where E is the time varying field.
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The time varying electric field can be equivalently represented, in terms of corresponding phasor
quantity E (r) as

jt

E r , t   R e  E r e
 ________ (11)
The symbol ‗tilda‘ placed above the E vector represents that E is time – varying quantity.
The phasor notation:
We consider only one component at a time, say Ex.
The phasor Ex is defined by
Ex r , t   Re Ex r 
e jt  ________ (12)
| Ex |
| Ex |
t

E
x
E x r  denotes Ex as a function of space (x,y,z). In general E x r is complex and hence can be
represented as a point in a complex and hence can be represented as a point in a complex plane.
(see fig) Multiplication by e jwt results in a rotation through an angle wt measured from the angle
. At t increases, the point Ex e jwt traces out a circle with center at the origin. Its projection on
the real axis varies sinusoidally with time & we get the time-harmonically varying electric field
Ex (varying sinusoidally with time). We note that the phase of the sinusoid is determined by ,
the argument of the complex number Ex.
Therefore the time varying quantity may be expressed as
E x  Re  E x e
j
e
j t
 ________ (13)
 E x cos( t ) ________ (14)
Maxwell’s eqn. in phasor notation:
In time – harmonic form, the Maxwell‘s first curl eqn. is:
 H  J 
 D
 t _______ (15)
using phasor notation, this eqn. becomes,
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
j t
 Re He
 
 t Re  De

j t
 R


j t

 Je
e
 ________ (16)
The diff. Operator  & Re part operator may be interchanged to get,

j t
Re  He
  Re


t

 Re  j D e

e 
 j D  J
R   H
De
j t
  Re
j t
Je





j t 


j t 

 Re  Je

e
jt
 0
This relation is valid for all t. Thus we get
 H  J  j D ________ (17)
This phasor form can be obtained from time-varying form by replacing each time derivative by

jw  ie.,



is to be replaced by  
t

For the sinusoidal time variations, the Maxwell‘s equation may be expressed in phasor form as:
(17)
LH
 H  J  j D

dL   J  j D
 ds
S
(18)
 E  j B
(19)
 D  
 L E dl    Sj B ds
D ds   d
S

V
V
V
B ds  0

The continuity eqn., contained within these is,
(20)
B0
S
 S J ds   j dv _______ (21)
 J  j
vol
The constitutive eqn. retain their forms:
D  E
BH
JE
____ (22)
For sinusoidal time variations, the wave equations become
 2 E  2   E
( for electric field ) 
 H    H
( for electric field )
2
2

_________ (23)

Vector Helmholtz eqn.
In a conducting medium, these become
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 E      j  E  0
2
2


 j
 H
2
________ (24)
H0
2
Wave propagation in a loss less medium:
In phasor form, the wave eqn. for VPW is

2
 E    E   Ey
2
2
2
x
; x  E y
2
 E

E C

j

x
e
C
e j  x _______ (26)
2
2

y
1
_______ (25)
2
C1 & C2 are arbitrary constants.
The corresponding
time varying field is
e
y

E


y



x , t  R  E x e jt 
 R  C e j  t   z   C
e
1
2
______ (27)
e j  t  z  

 C1 cos  t   z   C 2 cos  t   z  ______ (28)
When C1 and C2 are real.
Therefore we note that, in a homogeneous, lossless medium, the assumption of sinusoidal time
variations results in a space variation which is also sinusoidal.
Eqn. (27) and (28) represent sum of two waves traveling in opposite directions.
If C1 = C2 , the two traveling waves combine to form a simple standing wave which does not
progress.
If we rewrite eqn. (28) with Ey as a fn of (xt), we get  =


Let us identify some point in the waveform and observe its velocity; this point is


 t   x  a constant
  dx
Then

dt  
 ' a 't 


x

t 




 
This velocity is called phase velocity, the velocity of a phase point in the wave.
 is called the phase shift constant of the wave.
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Wavelength: These distance over which the sinusoidal waveform passes through a full cycle of
2 radians
ie.,
  2
2


But

2

or

2



 

f
or
  f ;
:

f in H Z

1



0
Wave propagation in a conducting medium
We have,
 E E0
2
Where
2
2  2   j
 j
  j 
Department of ECE


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 is called the propagation constant is, in general,
complex. Therefore,  =  + j
 = Attenuation constant

 = phase shift constant.
The eqn. for UPW of electric field strength is
2
E
2
2
E x
One possible solution is
E x   E0 e x
Therefore in time varying form, we get
 
e 

 x jt
e 
E x,t REe
jwt
 e  x R  E e 
e0

This eqn. shown that a up wave traveling in the +x direction and attenuated by a factor e x
. The phase shift factor
2 


and velocity  f  


 = Real part of  = RP j   jt 


=


2
 2 2 1 
 
2 




2


 1 2 2
2 
 


1

1


THE UNIFORM PLANE WAVE:
Topics dealt:




Principles of EM wave propagation

Physical process determining the speed of em waves; extent to which attenuation may
occur.

Energy flow in EM waves; power carried by em waves. Pointing theorem.
 Wave polarization.


1. Wave propagation in free space
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We have the generalized Maxwell‘s equations.
Point form
Integral form
Differential form
Macroscopic form
Microscopic form
 H  J 
D
t
 H dL   J  s
D
s
B
 E  t
 E dL  s 
 D  v
 D dS   enc
B0
S B dS  0
L
ds
t
B ds
t

   v dv
S
vol
In free space   0    ( source less v  0  J ) these equations become
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 H 0 
 E 0







E
H
t
 H dL   s
L
t
D  0  E
E ds ( I )
s

E dL  
s
0
t
H ds ( II )
t
L
S D dS   E
dS  0( III )
B  0  H
J  0
 B dS   H
dS  0( IV )
S
J
dS  0  continuityequ .  (V )
S
The Constituent equations, in free space, are,
D 0 E ___________________(VI )
B  0 H ___________________(VII )
JE
Concept of wave motion:
Eqn (1) states that if the electric field E changes with time, at some point, this change produces a
rotating curling magnetic field at that point; H varying spatially in a direction normal to its
orientation. Further, if E changes with time, in general, so does H although not necessarily in the
same way.
Next, from eqn. (2), we note that a time varying H generates a rotating E , ( curl E ), and this
E varies spatially in a direction normal to its orientation. Because H varies with time, so does
E but need not be in the same way therefore we once again have a time changing electric field
Department of ECE
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10EC36
( our original hypothesis from (1) ), but this field is present a small distance away from the point
of original disturbance. The velocity with which the effect moves away from the original point is
the velocity of light as we are going to see later.
Let us rewrite the point form of Maxwell‘s equations in ( source free ) free space
    J  0:








 D
 t ________(1)
 B
  E  B   t ______(2)
 H  D 
 D  0 _______(3)
 B  0 _______(4)
Taking curl on both sides of equation ( 1 ), we get
D

 H  t  t  E

 H  t
D  E;
B H;
 E
and  and  are independent of time.
But from ( 2 ),
 B
 E  B   t ______(2)
Next we take curl on both sides of eqn (2) and get
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H
 E  
But
 H 
D
t
 H
t
t
 E
t
 
______(1)
2
 E
2
t
 E   
But
2
 E  E  E
2
 E
2
t
 E  E 
2
But
E0
 weget
2
 E
2 _________(6)
t
 E   
2
Equations (5) and (6) are known as ― Wave Equations‖.
The first condition on either E or H is that it must satisfy the wave equation ( Although E & H
obey the same law E  H ).
Wave Propagation:
Consider the special case where E and H are independent of two dimensions, say x and y.
Then we get

2
E
2
E
2
E
 
2
2
2
y z x
2
E
2

E
x
2
Therefore eqn. (6) becomes
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2
 E
 E   
2
z
2
( E independent of x & y ) ______ (7)
This is a set of 3 scalar equations, one for each of the scalar components of E .
Let us consider one of them, the Ey component for which the wave equation (6) is :
2
 Ey
z

2
nd
This is a 2
t
__________ 7( a)
2
order PDE having a standard solution of the form
 f1 Z   0 t   f 2 Z 0t  ________(8)
Ey
Here
2
 Ey
1
0
;
00
f1,f2 : any functions of x 0t  and x 0t 
respectively. Examples of such functions are
Acos  x 0t 
h
ce
x 0t 
x  0 t
etc.,
All these equations represent a wave.
The Wave motion :
If a physical phenomenon that occurs at one place at a given time is reproduced at later time, the
time delay being proportional to the space separation from the fixed location, then the group of
phenomena constitutes a wave. ( A wave not necessarily be a repetitive phenomenon in time)
The functions f1 x 0t and f2 x 0t
 describe such a wave mathematically. Here the wave
varies in space as a function of only one dimension.
f1 x 0 t1 
Department of EEE, SJBIT
t = t1
Page 132
10EC36
Z
f1 x 0 t2 
t = t2
Z
v0 (t2 – t1 )
Figure shows the function f
x 0t  at two different instances of time t
1
f
since t gets fixed here. 1
f
1 and t2 .
x 0t 
1
becomes a function of z only
at t = t1 is shown in figure above as
f1 z 0 t1 . At another time t ( t > t ) we get another function of z namely f1 z 0 t2 . This is
2 2
2
nothing but time shifted version of
f1 z 0 t1  , shifted along + z axis by a distance ‗z‘ = 0 t 2  t1 .
This means that the function f1 x 0t
 has traveled along + z axis with a velocity 0 . This is
called a traveling wave.
On the other hand f 2 z 0t  represents a wave traveling along – z axis with a velocity 0 and
is called a reflected wave, as we shall further seen in the next semester, in the topic transmission
line.
This shows that the wave equation has two solutions ( as expected, since the wave eqn. is a
second order PDE ) a traveling wave ( or forward wave ) along + z direction represented by
f1 z 0t  and the other a reverse traveling wave ( reflected wave ) along – z axis. If there is no
reflecting surface, the second term of eqn. (8) is zero, resulting is
E = f1 z 0t  _________(9)
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Remember that eqn. (9) is a solution of the wave equation and is only for the particular case
where the electric field E is independent of x and y directions; and is a function of z and t only.
Such a wave is called also the equation does not indicate the specific shape of the wave
(amplitude variation) and hence is applicable to any arbitrary waveform.
In free space ( source-less regions where
D0
EE0
  J    0 ), the gauss law is
or
 D  0 ________ (1)
The wave equation for electric field, in free-space is,
 E
2
2 E
t
________ (2)
2
The wave equation (2) is a composition of these equations, one each component
wise, ie,
2
2
 Ex
2

2

x
2
 Ey
y
 Ey
2
z
t
_______(2) b
2
2
 Ez
2
_______(2) a
2
t
2
 Ey
 Ez

t
_______(2) c
2
Further, eqn. (1) may be written as
Ey Ez
Ex 

x
y
z
 0 ________ (1) a
For the UPW, E is independent of two coordinate axes; x and y axes, as we have assumed.

Ez

x


y
0
Therefore eqn. (1) reduces to
 0 ______ (3)
z
ie., there is no variation of Ez in the z direction.
2
 Ez
Also we find from 2 (a) that
Department of ECE
t 2= 0 ____(4)
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These two conditions (3) and (4) require that Ez can be
(iv)
Zero
(v)
Constant in time or
(vi)
Increasing uniformly with time.
A field satisfying the last two of the above three conditions cannot be a part of wave motion.
Therefore Ez can be put equal to zero, (the first condition).
Ez = 0
The uniform plane wave (traveling in z direction) does not have any field components of E & H
in its direction of travel.
Therefore the UPWs are transverse., having field components (of E & H ) only in directions
perpendicular to the direction of propagation does not have any field component only the
direction of travel.
RELATION BETWEEN E & H in a uniform plane wave.
We have, from our previous discussions that, for a UPW traveling in z direction, both E & H are
independent of x and y; and E & H have no z component. For such a UPW, we have,
 E 
 H 
ˆ
ˆ
i
j
ˆ
k
 (  0)  (  0) 
x
y
z
Ez(  0)
Ex
Ey
ˆ
ˆ
i
j
ˆ  Ey  ˆ  Ex 
i

  j


z


_____ (5)

 z 
ˆ
k
 (  0)  (  0) 
x
y
z
Hx
Hy Hz(  0)
i
ˆ  H y  ˆ  H x 


  j


z


_____ (6)
 z 
Then Maxwell‘s curl equations (1) and (2), using (5) and (6), (2) becomes,
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Ex ˆ
Ey ˆ
 H E 

i
j
t
t
t
and
Hx
 E  H 
t
t
ˆ 
 i 

Hy 

ˆj  Hx ______
(7)



z 

 z 
 Ex 
 ˆ

 Ey
j
______ (8)

i
z 


 z 
ˆi Hy ˆj

t
Thus, rewriting (7) and (8) we get
 Hy ˆ  Hx
z
z
i
 Ey ˆ
z
ˆ
i
 Ex ˆ
z
ˆ
j
j
Equating i th and j th terms, we get
Ex ˆ
 t

Hx ˆ
 t

i
i


Ey
t
Hy
ˆ 
j 

ˆ 
j
t
______ (7)


______ (8)
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 Hy

Hx
z
 Ey
z
Ey

ˆ
Ex
t
i  
z
______ 9 ( a)
______ 9 ( b)
t
Hx
t
______ 9( c)
and

Ex
z
Let
Hy
t
______ 9 ( d )
Ey  f1  z   0 t 
;0
1

.Then,
E
 f1  z   0 t  0 .  0 f1.
Ey
t
 From eqn. 9( c ), we get ,
Hx
t


0
f
'
 Hx 
0
'
1
z
H


'
f


Department of ECE
 f
'
1
f1

1
'
z

'
f1 dz  c.
 z  0t 
1
z

0
Now
f
0


f



z
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Now
f
'
1
z

  z  0t 
'
f
z
f1
1


'
f
1
dz  c  

 z
Hx 
f c
1
 Ey  c

The constant C indicates that a field independent of Z could be present. Evidently this is not a
part of the wave motion and hence is reflected.
Thus the relation between HX and EY becomes,
H x   E


Ey

Hx
y


__________ (10)
Similarly it can be shown that
Ex

Hy

 _____________ (11)
ˆ
ˆ
In our UPW, E  E x i  E y j
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

2
 E     E E 
t 
t 
2
 E
E
2
 E  
 2 _______ ( xi)
t
t
But  E 


0
E
DERIVATION OF WAVE EQUATION FOR A CONDUCTING MEDIUM:
In a conducting medium,  = 0,  = 0. Surface charges and hence surface currents exist, static
fields or charges do not exist.
For the case of conduction media, the point form of maxwells equations are:
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 H  J 
D
  E  E ________ (i )
t
t
B
 H _________ (ii)
t
t
 D  E  E  0 _________ (iii)

 B   H   H  0 _________ (iv)
 E 
Taking curl on both sides of equation (i ), we get

E 
  H   E 

t


  E ________ ( v)
t
   E 
substituting eqn. (ii ) in eqn. (v ), we get




2


 H     
  H2  _________ ( vi)
H 


t



t

2
ButH H  H_________ (vii)
eqn. ( vi ) becomes

  H   2 H  
But  H 

t
B  1  B 

2
 H
H

t 2
_________ (viii)
1 00

eqn. ( viii ) becomes,
 H  
2
2
 H
H

2  0 ________ (ix)
t
t
This is the wave equation for the magnetic field H in a conducting medium.
Next we consider the second Maxwell‘s curl equation (ii)
 E 
H
t
________ (ii)
Taking curl on both sides of equation (ii) we get
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H
 E  
t
 
  H  ________ ( x)
t
  E 2 E ;
But  E 
Vector identity and substituting eqn. (1) in eqn (2), we get

E 

  E 
t 
t 
2
 E _______ ( xi)
E


t
t 2
  E  2 E  

But  E 
 

0
(Point form of Gauss law) However, in a conductor,  = 0, since there is no net charge within a
conductor,
 E0
Therefore we get
Therefore eqn. (xi) becomes,
 E  
2
E

t
2 E
____________ (xii)
2
t
This is the wave equation for electric field E in a conducting medium.
Wave equations for a conducting medium:


Regions where conductivity is non-zero.

Conduction currents may exist.
For such regions, for time varying fields
The Maxwell‘s eqn. Are:
 E
 t _________ (1)
 H
  E   t __________ (2)
  H  J 

JE
:Conductivity(/m)
= conduction current density.
Therefore eqn. (1) becomes,
 H   E
Department of ECE
E
t
_________ (3)
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Taking curl of both sides of eqn. (2), we get
  E   


t  H 
 

2E
 
E
________ (4)
2
t t
But


2
 E   E  E ( vector identity ) u sin g this eqn. (4)
becomes vector identity,


E
t
  E  E  
 But  D  
2
2 E
2
t

 is cons tan t ,E
1
_______ (5)
D

Since there is no net charge within a conductor the charge density is zero ( there can be charge
on the surface ), we get.
1
E D0
Therefore using this result in eqn. (5)
we get
 E  
2
E
2 E

t
 0 ________(6)
2
t
This is the wave eqn. For the electric field E in a conducting medium.
This is the wave eqn. for E . The wave eqn. for H is obtained in a similar manner.
Taking curl of both sides of (1), we get
 E
 t    E ________ (7)
H
 H 
But  E 
________ (2)
t
(1) becomes,
 H   
2H
 
2
t
H
________ (8)
t
As before, we make use of the vector identity.
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 H 
  H  2 H
in eqn. (8) and get


2 H
  H  H    H 
2
t
t
2
________ (9)
But
B
1
1
 H      B   0  0

eqn.(9)becomes
H
2 H
2

H





________ (10)

tt2
This is the wave eqn. for H in a conducting medium.
Sinusoidal Time Variations:
In practice, most generators produce voltage and currents and hence electric and magnetic fields
which vary sinusoidally with time. Further, any periodic variation can be represented as a weight
sum of fundamental and harmonic frequencies.
Therefore we consider fields having sinusoidal time variations, for
example, E = Em cos t
E = Em sin t
Here, w = 2f, f = frequency of the variation.
Therefore every field or field component varies sinusoidally, mathematically by an additional
term. Representing sinusoidal variation. For example, the electric field E can be represented as
E  x , y , z , t  as
ie., E r , t ; r  x , y , z 
Where E is the time varying field.
The time varying electric field can be equivalently represented, in terms of corresponding phasor
quantity E (r) as

jt

E r , t   R e  E r e
 ________ (11)
The symbol ‗tilda‘ placed above the E vector represents that E is time – varying quantity.
The phasor notation:
We consider only one component at a time, say Ex.
The phasor Ex is defined by
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e jt  ________ (12)
Ex r , t   Re Ex r 
E x r  denotes Ex as a function of space (x,y,z). In general E x r is complex and hence can be
represented as a point in a complex and hence can be represented as a point in a complex plane.
(see fig) Multiplication by e jwt results in a rotation through an angle wt measured from the angle
. At t increases, the point Ex e jwt traces out a circle with center at the origin. Its projection on
the real axis varies sinusoidally with time & we get the time-harmonically varying electric field
Ex (varying sinusoidally with time). We note that the phase of the sinusoid is determined by ,
the argument of the complex number Ex.
Therefore the time varying quantity may be expressed as
E x  Re  E x e j e jt  ________ (13)
 E x cos( t ) ________ (14)
Maxwell’s eqn. in phasor notation:
In time – harmonic form, the Maxwell‘s first curl eqn. is:
D
t
 H  J 
_______ (15)
using phasor notation, this eqn. becomes,

 Re He
j t

 
R De
t e 
j t



jt

 ________ (16)
 Re  Je
The diff. Operator  & Re part operator may be interchanged to get,
Re  He
j t

  Re


t

 Re  j D e
e 
R   H  j D  J
j t
De
  Re


j t 
 Re

j t
Je
 Je



j t 


e
jt
 0
This relation is valid for all t. Thus we get
 H  J  j D ________ (17)
This phasor form can be obtained from time-varying form by replacing each time derivative by
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

jw  ie.,


is to be replaced by  
t

For the sinusoidal time variations, the Maxwell‘s equation may be expressed in phasor form as:
LH
 H  J  j D
(17)

dL   J  j D
 ds
S
(18)
 E  j B
(19)
 D  
 L E dl    Sj B ds
D ds   d
S

V
V
V
B ds  0

The continuity eqn., contained within these is,
B0
(20)
S
 S J ds   j dv _______ (21)
 J  j
vol
The constitutive eqn. retain their forms:
D  E
BH
JE
____ (22)
For sinusoidal time variations, the wave equations become
 2 E  2   E
( for electric field ) 
 H    H
( for electric field )
2
2

_________ (23)

Vector Helmholtz eqn.
In a conducting medium, these become
 E      j  E  0
2
2

 H
2

 j
2
H0
________ (24)
Wave propagation in a loss less medium:
In phasor form, the wave eqn. for VPW is

2
E    E   Ey
2
2
2
x
 ; x  E y
_______ (25)
2
 E

E C
e  j  x  C e j  x _______ (26)
2
2

y
1
2
C1 & C2 are arbitrary constants.
The corresponding time varying field is
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x,t R E
E
y

x
e jt 
 y  

 R  1C e j  t   z   C

e
e 
2
e j  t  z 

______ (27)

 C1 cos  t   z   C 2 cos  t   z  ______ (28)
When C1 and C2 are real.
Therefore we note that, in a homogeneous, lossless medium, the assumption of sinusoidal time
variations results in a space variation which is also sinusoidal.
Eqn. (27) and (28) represent sum of two waves traveling in opposite directions.
If C1 = C2 , the two traveling waves combine to form a simple standing wave which does not
progress.
If we rewrite eqn. (28) with Ey as a fn of (xt), we get  =


Let us identify some point in the waveform and observe its velocity; this point is


 t   x  a constant
 ' a 't 
  dx

dt  
Then
x
t 






 
This velocity is called phase velocity, the velocity of a phase point in the wave.
 is called the phase shift constant of the wave.
Wavelength: These distance over which the sinusoidal waveform passes through a full cycle of
2 radians
ie.,
  2
2


But

2

or

2



 

f
or
  f ;
:

f in H Z

1



0
Wave propagation in a conducting medium
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We have,
 E E0
2
2
2  2   j
 j
  j 
Where


 is called the propagation constant is, in general,
complex. Therefore,  =  + j
 = Attenuation constant

 = phase shift constant.
The eqn. for UPW of electric field strength is
2
E
2
2
E x
One possible solution is
E x   E0 e x
Therefore in time varying form, we get
 
e 

 x jt
e 
E x,t REe
jwt
 e  x R  E e 
e0

This eqn. shown that a up wave traveling in the +x direction and attenuated by a factor e x
. The phase shift factor
2 


and velocity  f  


 = Real part of  = RP j   jt 


=


2
 2 2 1 
 
2 




2


 1
2 
2 2


1
Department of ECE

1


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Conductors and dielectrics:
st
We have the phasor form of the 1 Maxwell‘s curl eqn.
 H   E  j  E  J c  Jdisp
2
where J c   E  conduction current density ( A/m )
2
J disp  j  E  displacement current density ( A/m )

J
cond
J
disp
 

We can choose a demarcation between dielectrics and conductors;
  1


8
*   1 is conductor.
*
Cu: 3.5*10 @ 30 GHz
  1 is dielectric.

Mica: 0.0002 @ audio and RF
* For good conductors,
 &  are independent of freq.
* For most dialectics,
 &  are function of freq.
*
 is relatively constant over frequency range of interest

Therefore dielectric ― constant ―
*
 dissipation factor D

if D is small, dissipation factor is practically as the power factor of the dielectric.
PF = sin
-1
 = tan D
PF & D difference by <1% when their values are less than 0.15.
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b) Express
E y  100 cos 210 t  0.5 z  30
6
0

j 2 10 t  0.5 z 30 
E
8
 R 100 e
y
e
0
v / m as a phasor


Drop Re and suppress e
jwt
term to get phasor
Therefore phasor form of Eys = 100e0.5 z300
Whereas Ey is real, Eys is in general complex.
Note: 0.5z is in radians; 300 in degrees.
c) Given
Es  100 300 axˆ  20  500 ayˆ  40 210 0 azˆ , V / m
find its time varying form representation
Let us rewrite Es as
E s  100e
ER
j 300
axˆ  20e
E
e

ayˆ  40e
j 2100
azˆ . V / m
e jt 
s

 j 500


j  t  30
0
 Re 100e

E  100 cos  t  30
0
0

j  t  50 
 20e
 40e
j t 210
0
 

V/m
20 cos  t  50 0  40 cos t  210 0 V / m
None of the amplitudes or phase angles in this are expressed as a function of x,y or
z. Even if so, the procedure is still effective.
d) Consider
 20e   0.1 j 20z axˆ A / m
jt


H t   Re 20e 0.1j20 zaxˆ e

H
s
 20e
0.1z

cos t  20 z axˆ
E x  E x  x , y , z 
Note :

consider
A/m

e  x 
Ex

t  t R  E x , y , z
 R  j E
e jt 

e 
x

e jt 
Therefore taking the partial derivative of any field quantity wrt time is equivalent to multiplying
the corresponding phasor by j .
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Next, the wave equation in free space is:


E
2
2
E
2
 k0 
2
2
 k0  

t

  E
2
2
 E
s
2
2
2
 Es   E s   E
2
2
2
x
y
z
for E x component ,
2
2
k
s
sx
2
s
E
0
2
 Esx   E
2
2
x
y
  E
2
s
  E sx
2
z
s
k
2
E
0
xs
For a UPW traveling along z axis,
We get
2
 E
k
sx
2
0
2
x
E
xs
One solution:
E xs  E x 0 e
 jk
0
z
  E x  z , t   E x 0 cos t  k 0 z  E x  z , t
  E x 0 cos t  k 0 z 
These two are called the real instantaneous forms of the electric field.
1
 
0
k
0
x
E
3 10
0


1
0


 e
z,t E
x0
 3  108  c
00
8
 
cos  t  z / c



We can visualize wave propagation by putting t-0
E x z , 0
Ex0
 z 
cos 
  E x 0 cos  z   E x 0 cos k 0 z 
 e 
This is a simple periodic fn that repeats every incremental distance , known as wavelength. The
requirement is that k0 = 2
Department of ECE
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8
ie., x  2  c
k0
3 10 in f ree space
f
f
Given
E0 s   500 40 ayˆ   200  j 600 azˆ  e
0
Find a  


V/m


at t  0
b E at 2, 3,1
c E at
 j 0.4 x

2, 3,1 at t  10 ns.
 d  E at 3, 4, 2 at t  20 ns.



c) From given data,
 
  0.4  
0
0
8
0.4  3 10

4  10
7
 120 106
9
10

9
36
6
f  19.1 10 Hz
d) Given,
E s   500 40
 500e
 j 40
500e 
e
 j 0.4 x
ayˆ
 632.456e  j 71.565 e  j 0.4 x azˆ
0
    ayˆ 632.456e j0.4x 71.565 azˆ
j 0.4 x
E t   500 Re  e
 j 0.4 x
ayˆ   200  j 600azˆ e
0

40
0
  j t


e

 j 0.4 x  40
 500 cos t  0.4 x  40
0

0

ayˆ  632.456 e
j t
e

 j 0.4 x71.565
0
 
azˆ 

ayˆ  632.456 cos t  0.4 x  71.565 azˆ
0
E at 2, 3,1t  0  500 cos 0.4 x  40  ayˆ  632.456 0.4 x  71.565 azˆ
0
 36.297 ayˆ  291.076 azˆ V / m
c)
E at t  10 ns  at 2, 3,1


 500 cos 120  10  10  10
6
9
 632.456 cos 120  10  10  10
6
0
 0.4  2  40
9
 ayˆ
0
 0.4  2  71.565
 azˆ
 477.823 ayˆ  417.473 azˆ V / m
Department of ECE
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d)
at t = 20 ns,
E at 2, 3,1
 438.736 ayˆ  631.644 azˆ V / m
D 11.2:
Given H s  2 400 axˆ  320 ayˆ e  j 0.07 z A / m for a UPW traveling in free space. Find
(a) 
(b) Hx at p(1,2,3) at t = 31 ns.
(a) we have
(c) H at t=0 at the origin.
(e  j  z term)
p = 0.07
   0.07
0.07

8
  0.07  3  10  21.0 10
 21.0 10
6
6
rad / sec
rad / sec
(b)
H  t   Re
 2 e
 j 40 0
e  j 0.07 z axˆ  3 e j 20
 2 cos  t  0.07 z  40
0
H x (t )  2 cos t  0.07 z  40
0
e
 j 0.07 z
ayˆ  e
j t

axˆ  3 cos t  0.07 z  200 ayˆ
0
 H x (t ) at p 1, 2, 3
6
0
 2 cos 2.1  10 t  0.21  40 
6
9
0
At t  31n sec;  2 cos 2.1  10  31  10  0.21  40 
3
0
 2 cos 651 10  0.21  40 
 1.9333
A/m
(c)
H t  at t  0  2 cos 0.07 z  0.7 axˆ  3 cos 0.7 z  0.35ayˆ
H t   2 cos 0.7
axˆ  3 cos 0.3ayˆ
 1.53axˆ  2.82ayˆ
 3.20666
A/m
Department of ECE
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In free space,
E  z , t   120 sin  t   z  ayˆ
V /m
H  z , t 
find
Ey
Hx
we have
 120
Ey
120
 H x 
120
120

sin  t   z  ayˆ
1
   sin  t   z 

1
 H  z , t    sin  t   z  axˆ
Problem 3. J&B
Non uniform plans waves also can exist under special conditions. Show that the function
Fe
 z

sin  x t 
2
satisfies the wave equation  F 
1 2 F
2
c t
2
provided the wave velocity is given by
e


1

 2 c2 


2


Ans:
From the given eqn. for F, we note that F is a function of x and z,
2
 F 

F
x
2
 F
x
F
2
z
2
 F
z
2
 F
 F
2 
2
x
y
2
2

 e  z  cos   x t 
 z
 e
 e
 z
 e
2
Department of ECE


2  z
 x  t   e


 sin
2
 
   




sin   x t 


   2 F
  
 z
sin
F
xt
Page 153
10EC36
 

2
2
 F  
 F
2
 

dF  e  z     cos   x t 


 

dt
2
d F
 z  
 e

 2    sin  x t 
dt
 
2
 F
2
The given wave equation is
2
2
 F 


2

1  F
2
c 2 2t
 
2  F 
c
 
2
2


2
 
2
2 
c


2
2
2
2
c
1
2
 2  F
2
 2

2
2 2
2
c
 
2

2c2
 2 c2   2
2
c

2 2

c
1 
2

c
or  
1 
 2 c2
2
The electric field intensity of a uniform plane wave in air has a magnitude of 754 V/m and is
in the z direction. If the wave has a wave length  = 2m and propagating in the y direction.
Find
(iii)
Frequency and  when the field has the form A cos  t   z .
(iv)
Find an expression for H .
In air or free space,
  c  3108 m / sec
(i)
Department of ECE
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10EC36
f 
e 3 108
m / sec  1.5  10 Hz  150MHz
2m


8

2 
2
  2 m  3.14 rad / m
 E z  754 cos 2  150  10 t  y 
6
(ii)
For a wave propagating in the +y direction,
Ez

Hz
Ex
Hz
For the given wave,
E z  754 V / m; Ex  0
 H  754   
754
120
x
 754 A / m
377
 H  2 cos 2  150  10 t  y axˆ
6
A/m
7
find  for copper having  = 5.8*10 (/m) at 50Hz, 3MHz, 30GHz.
2


1
   f 
1 
1
4  10

7

 1
7
f
5.8 10
1
66 10
1

 1 

2
4 5.8
f
23.22 f
f
3
66 10
3
 9.3459 10 m
(i ) 
50
1
(ii ) 
66 10
(iii ) 
3
6
3 10
3
66 10
3 10
6
 3.8105 10
5
 3.8105 10
3
m
7
m
Wave Propagation in a loss less medium:
Definition of uniform plane wave in Phasor form:
Department of ECE
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In phasor form, the uniform plane wave is defined as one for which the equiphase surface is
also an equiamplitude surface, it is a uniform plane wave.
For a uniform plane wave having no variations in x and y directions, the wave equation in
phasor form may be expressed as
2
2
 E
 E
   E0 r
2
Z
2
 E ________ (i )
2
Z
2
where    . Let us consider eqn.(i) for, the Eycomponent, we get
2
 E
y
 Ey
2

Z2
Ey has a solution of the form,
E y  C1e
jz
jz
 C2 e
________ (2)
Where C1 and C2 are arbitrary complex constants. The corresponding time varying form of
E
y
is
E y z , t   Re E y z  e
 R  C
jt

ejz C
j t
e j  z  e  _______ (3)


2
  1
If C1 and C2 are real, the result of real part extraction operation is,
e
 E y z , t   C1 cos  t   z  C2 cos  t   z  _______ (4)
From (3) we note that, in a homogeneous lossless medium, sinusoidal time variation results
in space variations which is also sinusoidal.
Equations (3) and (4) represent sum of two waves traveling in opposite directions.
If C1 = C2, the two wave combine to form a standing wave which does not progress.
Phase velocity and wavelength:
The wave velocity can easily obtained when we rewrite Ey as a function and z t  , as in
eqn. (4). This shows that



________(5)
In phasor form, identifying a some reference point on the waveform and observing its
velocity may obtain the same result. For a wave traveling in the +Z direction, this point is
given by  t   z  a constant.
 
dz

dt   , as in eqn. (5)
Department of ECE
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This velocity of some point on the sinusoidal waveform is called the phase velocity.  is
called the phase-shift constant and is a measure of phase shift in radians per unit length.
Wavelength: Wavelength is defined as that distance over which the sinusoidal waveform
passes through a full cycle of 2 radius.
ie.,
  2
 2 
2 
2
 ;   1
________(7)
 2 f  f

f in Hz
________(8)

  f  ,
For the value of  given in eqn. (1), the phase velocity is,




 

0  C

1
0

_______(9)
8
C  3 10 m / sec
;
Wave propagation in conducting medium:
The wave eqn. written in the form of Helmholtz eqn. is

 E   E  0 _______(10)
2
2

where 
2

   j
2


 j  j
_______(11)
.
, the propagation constant is complex =  + j _________(12)
We have, for the uniform plane wave traveling in the z direction, the electric field E must
satisfy
2
 E
  E _______(13)
2
2
Z
This equation has a possible solution
E Z   E0 e
 Z
_______(14)
In time varying form this is becomes
E z , t   R e  E 0 e
=e
 z
Re
E0
 Z
e
e
j t
j  t  z
 _______(15)
________(16)

This is the equation of a wave traveling in the +Z direction and attenuated by a factor eZ .
The phase shift factor and the wavelength phase, velocity, as in the lossless case, are given
by
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 2

  f   

The propagation constant
We have,

2


j  j 


   j 
2
2

________(11)
 2 j    j     ________(17)
2
2
      ;        ________(18)

2
2

2
2
2
2

________(19)
2
Therefore (19) in (18) gives:
2
4

2


 4
4
 
2

 4      
2
2
2
2
2
0
 2  2 2  0
   
4
2
2
4
2
        
2
4
2
2
2
2
2
2

22

1

2

2

2




2


2 2 
    1  1 
2
2 
2
  



 
 2 
2


1

2

2
2

1 _________(20)
  
and




2
1 

2
22


  
1 ___________(21)
We choose some reference point on the wave, the cosine function,(say a rest). The value of
the wave ie., the cosine is an integer multiple of 2 at erest.
th
 k z  2m at m erest.
0
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Now let us fix our position on the wave as this m
th
erest and observe time variation at this
position, nothing that the entire cosine argument is the same multiple of 2 for all time in
order to keep track of the point.
 t  k0  0 z  2m   t  z / c
ie.,
Thus at t increases, position z must also increase to satisfy eqn. ( ). Thus the wave erest (and
the entire wave moves in a +ve direction) with a speed given by the above eqn.
Similarly, eqn. ( ) having a cosine argument  t  0 z  describes a wave that moves in the
negative direction (as + increases z must decrease to keep the argument constant). These two
waves are called the traveling waves.
Let us further consider only +ve z traveling wave:
We have
ˆ
ˆ
i
j
ˆ
k
  0
x
  0
y

z
Ex
Ey
0
 E s  j H s
 Ey 
z 

Exs
z

Ex
  j
i  
z
 j H
 H oy 
z,tEx0
E
 

H
y
0y
1
j E z 0
Hy
x
ˆ  j iH x  j  b 
0
y
k0
e

0
 0
;
 jk
0
z
Ex0
0

0
e
j z
0
cos  t  0 z 
 0  377  120

Ey and Hx are in phase in time and space. The UPW is called so because  is uniform
thought any plane Z = constant.
Energy flow is in +Z direction.
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E and H are perpendicular to the direction of propagation; both lie in a plane that is
transverse to the direction of propagation. Therefore also called a TEM wave.
11.1. The electric field amplitude of a UPW in the azˆ direction is 250 V/m. If E = E x axˆ and
 = 1m rad/sec, find (i) f (ii) 
(iii) period (iv) amplitude of H .
6
10
f    2 f 
 159.155 KHz
2
2
2
 C
 1.88495 km
f
1  6.283  s
f
period 
amplitude of H
 H y 
y

Ex
   120
Hy
Ex
250

 0.6631 A / m
120
120
0
0
 j 0.07 z
A / m for a certain UPW traveling in free
11.2. Given H s   2 40 axˆ  320 ayˆ e
space.
Find (i), (ii)Hx at p(1,2,3) at t = 31ns and (iii) H at t = 0 at the orign.
Wave propagation in dielectrics:
For an isotopic and homogeneous medium, the wave equation becomes
2
 Es  k
2
s
k   k0
rr0
rr
For Ex component
We have
2
d Exs k 2 E
xs for E
2
dz
x
comp. Of electric field wave traveling in Z – direction.
k can be complex one of the solutions of this eqn. is,
jk    j 
E
xs
E
x0
e  z e j  z
Therefore its time varying part becomes,
E
xs
 Ex 0e
 z
Department of ECE
cos  t   z 
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This is UPW that propagates in the +Z direction with phase constant  but losing its amplitude
with increasing Z e z . Thus the general effect of a complex valued k is to yield a traveling
wave that changes its amplitude with distance.
If  is +ve
 = attenuation coefficient if  is +ve wave decays
If  is -ve
 = gain coefficient
In passive media,  is +ve
wave grows
 is measured in repers per meter
In amplifiers (lasers)  is –ve.
Wave propagation in a conducting medium for medium for time-harmonic fields:
(Fields with sinusoidal time variations)
For sinusoidal time variations, the electric field for lossless medium ( = 0) becomes
 E    E
2
2
In a conducting medium, the wave eqn. becomes for sinusoidal time variations:
 E     j  E  0
2
2
Problem:
Using Maxwell‘s eqn. (1) show that
.D  0
in a conductor
if ohm‘s law and sinusoidal time variations are assumed. When ohm‘s law and sinusoidal time
variations are assumed, the first Maxwell‘s curl equation is
 H   E  j  E
Taking divergence on both sides, we get,

  H     E  j  E 
0  E   j   0
or  D
 

   j 
 0 
 , & are
constants and of finite values and  0
D0
Wave propagation in free space:
The Maxwell‘s equation in free space, ie., source free medium are,
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 H  E H _________(1)
0 t
 E H _________(2)
t
 D  0 _________(3)
 B  0 _________(4)
Note that wave motion can be inferred from the above equation.
How? Let us see,
Eqn. (1) states that if electric field
E is changing with time at some, point then magnetic field
H has a curl at that point; thus
H varies spatially in a direction normal to its orientation
direction. Further, if E varies with time,
then H will, in general, also change with
time;
H generates E ; this electric field, having a
curl,
although not necessarily in the same way.
Next
From (2) we note that a time varying
therefore varies spatially in a direction normal to its orientation direction.
We thus have once more a time changing electric field, our original hypothesis, but this field is
present a small distance away from the point of the original disturbance.
The velocity with which the effect has moved away from the original disturbance is the
velocity of light as we are going to prove later.
Uniform plane wave is defined as a wave in which (1) both fields E and H lie in the transverse
plane. Ie., the plane whose normal is the direction of propagation; and (2) both E and H are of
constant magnitude in the transverse plane.
Therefore we call such a wave as transverse electro magnetic wave or TEM wave.
The spatial variation of both E and H fields in the direction normal to their orientation (travel)
ie., in the direction normal to the transverse plane.
Differentiating eqn. (7) with respect to Z1 we get
2
 Ex
Z
2
  0
2
 Hy 
 H
Z  t 
t Z

Department of ECE
0
________(9)
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Differentiating (8) with respect to t1 we get
2
2
 H   E x _________(10)
2
t
t Z
0
Therefore substituting (10) into (9) gives,
2
2
 E x    E x
2
0t2
t
0
_________(11)
This eqn.(11) is the wave equation for the x-polarized TEM electric field in free space.
1
The constant


0
is the velocity of the wave in free space, denoted c and has a value
0
 4  10
3 108 m / sec , on substituting the values, 
7
H / m and 
109
.
0
0
36
Differentiating (10) with respect to Z and differentiating (9) with respect to ‗t‘ and following the
similar procedure as above, we get
2
2
 H y    H y _________(13)
2
2
Z
00 t
eqn. (11 and (13) are the second order partial differential eqn. and have solution of the form, for
instance,
Ex Z , t   f1 t  Z /    f 2 t  Z /  ________(14)
Let E  E x axˆ
(ie., the electric field is polarized (!) in the x- direction !) traveling along Z
direction. Therefore variations of E occurs only in Z direction.
Form (2) in this case, we get
aˆ x

aˆ y

 E  x  0  y  0 
Ex
aˆz

z
0
0
Ex ˆ
H
 z j 0 t 0
H ˆ
t
j _________(5)
Note that the direction of the electric field E determines the direction of H , we is now along the
y direction.
Therefore in a UPW, E and H are mutually orthogonal. (ie., perpendicular to each other). This in
a UPW .
(i) E and H are perpendicular to each other (mutually orthogonal and
(ii) E and H are also perpendicular to the direction of travel.
Form eqn. (1), for the UPW, we get
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 H 
H y axˆ  E  t
0
t
Z
0
Ex axˆ
t
(using the mutually orthogonal property) _______________(6)
Therefore we have obtained so far,
H y
E x

Z
t
0
H y
 E x
Z
t
0
________(7)
________(8)
f1 and f2 can be any functions who se argument is of the form t  Z / .
The first term on RHS represents a forward propagating wave ie., a wave traveling along positive
Z direction.
The second term on RHS represents a reverse propagating wave ie., a wave traveling along
negative Z direction.
(Real instantaneous form and phaser forms).
The expression for Ex (z,t) can be of the form
E x z , t   E x z , t   E 1x z , t 




1
  t  Z /  p  1   E x 0 cos   t  Z / p 2 
 E x 0 cos  t  k 0 z  1   E 1x0 cos  t  k 0 z 2  _______ 15
 E x 0 cos


 p is called the phase velocity = c in free space k0 is called the wave number in free space =

c
rad/m _________(16)
eqn. (15) is the real instantaneous forms of the electric (field) wave. ( experimentally
measurable)
0t and k0z have the units of angle usually in radians.
 : radian time frequency, phase shift per unit time in
rad/sec. k0 : spatial frequency, phase shift per unit distance in
rad/m. k0 is the phase constant for lossless propagation.
Wavelength in free space is the distance over which the spatial phase shifts by 2 radians, (time
fixed)
ie.,
k 0 z  k0  2
2
or   k0
(in free space) _________(17)
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Let us consider some point, for instance, the crest or trough or zero crossing (either –ve to +ve or
+ve to –ve). Having chosen such a reference, say the crest, on the forward-propagating cosine
function, ie., the function cos  t  k0 z 1 . For a erest to occur, the argument of the cosine
th
must be an integer multiple of 2. Consider the m erest of the wave from our reference point,
the condition becomes,
K0z = 2m, m an integer.
This point on the cosine wave we have chosen, let us see what happens as time increases.
The entire cosine argument must have the same multiple of 2 for all times, in order to keep
track of the chosen point.
Therefore we get,
tk0ztZ/2m_______(18)
As time increases, the position Z must also increase to satisfy (18). The wave erest, and the entire
wave, moves in the positive Z-direction with a phase velocity C (in free space).
Using the same reasoning, the second term on the RHS of eqn. (15) having the cosine argument
t  k 0 z represents a wave propagating in the Z direction, with a phase velocity C, since as
time t increases, Z must decrease to keep the argument constant.
POLARISATION:
It shows the time varying behavior of the electric field strength vector at some point in space.
Consider of a UPW traveling along Z direction with E and H vectors lying in the x-y plane.
 If Ey  0 and only Ex is present, the wave is said to be polarized in the x-direction.

If Ex = 0 and only Ey is present, the wave is said to be polarized in the y-direction.
Therefore the direction of E is the direction of polarization

If both Ex and Ey are present and are in phase, then the resultant electric field E has a
direction that depends on the relative magnitudes of Ex and Ey .
The angle which this resultant direction makes with the x axis is tan
-1
Ey
Ex ; and this angle will be
constant with time.
(a) Linear polarization:
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In all the above three cases, the direction of the resultant vector is constant with time and the
wave is said to be linearly polarized.
If Ex and Ey are not in phase ie., they reach their maxima at different instances of time, then the
direction of the resultant electric vector will vary with time. In this case it can be shown that the
locus of the end point of the resultant E will be an ellipse and the wave is said to be elliptically
polarized.
0
In the particular case where Ex and Ey have equal magnitudes and a 90 phase difference, the
locus of the resultant E is a circle and the wave is circularly polarized.
Linear Polarisation:
Consider the phasor form of the electric field of a UPW traveling in the Z-direction:
E   Z   E0 e
jz
Its time varying or instanious time form is
E Z , t   Re E0 e
jz
e
jt

The wave is traveling in Z-direction.
Therefore Ez lies in the x-y plane. In general, E0 is a complex vector ie., a vector whose
components are complex numbers.
Therefore we can write E0 as,
E0  Er  jE0i
Where E0 and E0i are real vectors having, in general, different directions. At
some point in space, (say z = 0) the resultant time varying electric field is
E 0, t   Re
 E
0r
 j E0i
 e jt 
 E0 r cos  t  E0i sin t
Therefore E not only changes its magnitude but also changes its direction as time varies.
Circular Polarisation:
Here the x and y components of the electric field vector are equal in magnitude.
0
If Ey leads Ex by 90 and Ex and Ey have the same amplitudes,
Ie., E x  Ey , we have, E   axˆ  j ayˆ E0
The corresponding time varying version is,
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E 0, t    axˆ cos  t  ayˆ sin t E0
 E x  E0 cost
and E y  E0 sin t
2
2
2
E E E
x
y
0
Which shows that the end point of E0 0, t  traces a circle of radius E0 as time progresses.
Therefore the wave is said to the circularly polarized. Further we see that the sense or direction
of rotation is that of a left handed screw advancing in the Z-direction ( ie., in the direction of
propagation). Then this wave is said to be left circularly polarized.
Similar remarks hold for a right-circularly polarized wave represented by the complex vector,
E   axˆ  j ayˆ E0
0
It is apparent that a reversal of the sense of rotation may be obtained by a 180 phase shift
applied either to the x component of the electric field.
Elliptical Polarisation:
Here x and y components of the electric field differ in amplitudes E x  Ey .
0
Assume that Ey leads Ex by 90 .
Then,
E0 axˆ A  j ayˆ B
Where A and B are +ve real constants.
Its time varying form is
E 0, t   axAˆ cos  t  ayBˆ sin t
E

x
 A cost
E y B sin t

E
A
x 2
2

E
y
B 22
 1
Thus the end point of the
E 0, t  vector traces out an ellipse and the wave is elliptically
polarized; the sense of polarization is left-handed.
Elliptical polarization is a more general form of polarization. The polarization is completely
specified by the orientation and axial ratio of the polarization ellipse and by the sense in which
the end point of the electric field moves around the ellipse.
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Unit-8
Plane waves at boundaries and in dispersive media:



reflection of uniform plane wave at normal incidence
SWR
Plane wave propagation in general direction
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When an em wave traveling in one medium impinges upon a second medium having a different
,  or  , then the wave will be partially transmitted, and partially reflected.
When a plane wave in air is incident normally on the surface of a perfect conductor the wave is
for fields that vary with time, neither E nor H can exist within a conductor., therefore no energy
of the incident wave is transmitted.
As there can be no loss within a perfect conductor; therefore none of the energy is obsorbed.
Therefore, the amplitudes of E and H in the reflected wave are the same as in the incident wave;
the only difference is in the direction of power flow.
Let Ei e j  x ________(1) be the incident wave.
Let the boundary, the surface of the perfect conductor be at x = 0.
The reflected wave is Er e j  x __________(2)
Er must be determined from the boundary
conditions. With respect to,
(i)
Etan is continuous across the boundary
(ii)
E is zero within the conductor.
Therefore at the boundary, ie., at x = 0, the electric field is zero. This requires that, the sum of the
electric field strengths in the initial and reflected waves add to give zero resultant field strength
in the plane x = 0.
 Er Ei _______(3)
The amplitude of the reflected electric field strength is equal to that of the incident electric field
strength but its phase has been reversed on reflection.
The resultant electric field strength at any point at any point a distance –x from the x = 0 plane is
the sum of the field strengths of the incident and reflected wave at that point, given by
ET x   Ei e
jx
 2 jEi e

jx
 Er e
jx
jx
e

 2 jEi sin  x _______ 4
Its time varying version is
E
T

x , t  R 2 jE sin  x e
Department of ECE
e 
i
jt


 2 Ei sin  x sin t , if Ei real _______ 5
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1. Eqn. (3) shows that (1) the incident and reflected waves combine to produce a standing
wave, which does not progress.
2. The magnitude of the electric field varies sinusoidally with distance from the reflecting
plane.
3. It is zero at the surface and at multiples of half wave lengths from the surface.
4. It has a maximum value of twice the electric field strength of the incident wave at
distances from the surface that are odd multiples of a quarter wavelength.
In as much as the BCs require that the electric field is reversed in phase on reflection to produce
zero resultant field at the boundary surface.
Therefore if follows that H must be reflected without phase reversal. (otherwise if both are
reversed, on reversal of direction of energy propagation), which is required in this case).
Therefore the phase of the mag field strength is the same as that of the incident mag field
strength Hi at the surface of reflection.
H T x   H i e
jx
 2H i e

Hre
jx
e
jx
jx

 2 H i cos  x _______ 6
Hi is real since it is in phase with Ei
H T x , t   R e  H T x e
Further,
jt

 2 H i cos  x cos t ______ 7
The resultant magnetic field strength H also has a standing was distribution. This SWD has
maximum value at the surface of the conductor and at multiples of a half  from the surface,
where as the zero points occur at odd multiples of a quarter wavelength from the surface. From
the boundary conditions for H its follows that there must be a surface current of Js amperes per
such that JS = HT (at x = 0).
Since Ei and Hi were in phase in the incident plane wave, eqns. (6) and (7) show that ET and HT
0
are 90 out of time phase because of the factor j in eqn. (4).
This is as it should be, for it indicates no average flow of power. This is the case when the energy
transmitted in the forward direction is equaled by that reflected back.
Let us rewrite eqns. (4)e and (6)
T




i
E x , t  R  2 E sin  x e

i

 
 j / 2
e jt  2 E sin  x cos  t  / 2 _______ 8
H T x , t   2 H i cos  x cos t _______ 9
0
Eqns. (8) and (9) show that ET and HT differ in time phase by 90 .
Department of ECE
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REFLECTION BY A PERFECT CONDUCTOR – OBLIQUE INCIDENCE:
TWO SPACIAL CASES:
1. Horizontal Polarisation: (also called perpendicular polarization) Here the electric field
vector is parallel to the boundary surface, or perpendicular to the plane of incidence.
( Transverse electric TE)
2. Vertical Polarisation: (also called parallel polarization) Here the magnetic field vector is
parallel to the boundary surface, and the electric field vector is parallel to the plane of
incidence. (Transverse magnetic TM)
TE or TM are used to indicate that the electric or magnetic vector respectively is parallel to the
boundary surface/plane.
When a wave is incident on a perfect conductor, the wave is totally reflected with the angle of
incidence equal to the angle of reflection.
Case 1: E perpendicular to the plane of incidence: (perpendicular Polarisation)
The incident and reflected waves have equal wavelengths and opposite directions along the Z
axis, the incident and reflected waves form a standing wave distribution pattern along this axis.
In the y direction, both the incident and reflected waves progress to the right (+y direction) with
the same velocity and wavelength and so there will be a traveling wave along the +y direction.
The expression for reflected wave, using the above fig, is
E
refelected
and E
incident
i

 Er
e j   y sin  z cos  ______ 8
 Ei
e j   y sin  z cos  ______ 9
E  E  e  j   y sin   z cos    e j   y sin  z cos  

 j  y sin
 2 jEi sin  z cos  e

jyy
 2 jEi sin z z  e

_______ 10
From the BCs we have,
Er = - Ei
Therefore total electric field strength E is given by
Department of ECE
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j

Erefelected  Er e
and E
 y sin   z cos 
______ 8
 Ei e j   y sin  z cos  ______ 9
incident
E  E  e  j   y sin   z cos    e j   y sin   z cos  
i


 j  y sin
 2 jEi sin  z cos  e

jyy
 2 jEi sin z z  e

_______ 10
Where,


 2 
    Phase shift constant of the incident wave,
 z   cos = Phase shift constant in the Z direction.

 y   sin = Phase shift constant in the y direction.
 2  2  
z

 cos 
cos :
wavelength: distance twice between modal points of the
standing wave distribution.
The planes of zero electric field strength occur at multiples of

2
z
The planes of max electric field strength occurs at odd multiples of
from the reflecting surface.

4
z
from the surface.
The whole standing wave distribution of electric field strength is seen from eqn. (10) above to be
traveling in the y direction with a velocity,
      
y  sin  sin
y
This is the velocity with which a erest of the incident wave moves along the y axis. The
wavelength in this direction is,

g
 
sin 
Case 2: E parallel to the plane of incidence: (parallel polarization)
Here, Ei and Er will have the instantaneous directions shown above, because the components
parallel to the perfectly conducting boundary must be equal and opposite.
The magnetic field strength vector H will be reflected without phase reversal.
Department of ECE
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10EC36
The magnitudes of E and H are related by
E
E
 r
i
Hi
Hr
For the incident wave, the wave expression for the magnetic field strength would be
j
H incident  H i e
 y sin  z cos 
and for the reflected wave,
 j   y sin  z cos 
H
reflected
Hre
Therefore Hi = Hr
The total magnetic field H is,
H  2 H i cos z z e
where
jyy
z   cos and
y   sin
The magnetic field strength has a standing wave distribution in the Z-direction with the planes of
z
maximum H located at the conducting surface and at multiples of
from the surface. The
2
planes of zero magnetic field strength occur at odd multiples of

4
z
from the surface.
For the incident wave,
Ei   H i ,
Ez   sin  H i ;
E y  cos Hi
For the reflected wave,
Hr Hi,
Ez   sin  H r ;
E y  cos Hr
The total z component of the electric field strength is,
Ex  2 sin 
j
H i cos z z e
y
y
The total y component of the electric field strength is,
E y  2 j cos 
H i sin z z e
jyy
Both Ey and Ez have a standing wave distribution above the reflecting surface. However, for the
normal or z components of E , the maxima occur at the plane and multiples of
z
2 from the
plane, whereas for the component E parallel to the reflecting surface the minima occur at the
plane and at multiples of
Department of ECE
z
2 from the plane.
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10EC36
REFLECTION BY PERFECT DIELECTRICS
Normal incidence:
In this case part of the energy is transmitted and part of the energy is reflected.
Perfect dielectric: = 0. no absorption or loss of power in propagation through the
dielectric. Boundary is parallel to the x = 0 plane.
Plane wave traveling in +x direction is incident on it.
Ei  1H r
E  H
We have,
r
1
r
Et  2 Ht
BC: Tang comp. Of E or H is continuous across the boundary.
ie.,
Hi Hr Hz
Ei  E r  Et
 H i  H r 
1 E  E   H
1 E  E 
i
r
z 
i
r
2

1
 2 Ei  E r   1 Ei  Er 
Ei  2  1   Er  2 1 

E
r
E
 21
 
i
2
1
E
Et  Ei  Er  1  r  21
c

E
E
i
i
1
H
E
r
Further,
 r  12
H

E
t
2
1
E
H t  1 t  21
H
E

Also,
i
2 i
1
2
2
The permeabilities of all known insulators do not differ appreciably from that of free space, so
that,
Department of ECE
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10EC36
12
 0 / 2 
 Er 
Ei
Et 
E
0
r

2
Ht 
Hi

2
12
2
2  1
2 
Hi

1

1


1
 
i
H
0 / 1 
/ 2  0 / 1
1
2
2
1  2
REFLECTION BY PERFECT DIELECTRIC:
OBLIQUE INCIDENCE:
1. There is a transmitted wave, reflected wave and the incident wave.
2. The transmitted wave is refracted 9direction of propagation is altered)
1 ______ vel. Of wave in medium (1)
2 ______ vel. Of wave in medium (2)
Then from figure, we get
CB 

1
AD 2
Now CB = AB sin1 and AD = AB sin2.

 sin1  1  22
sin 2 2
11
 2
1
In addition,
AE = CB
sin1 = sin3
or 1= 3
2
The power transmitted = E
E and H are perpendicular to each other.

Incident power striking AB

1
E
1
2
cos
1
1
Department of ECE
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1 E22 cos1
Reflected power leaving AB 
2
The power transmitted =
1 E 2 cos .
t
2
 2
Therefore by conservation of energy we get
1 E 2 cos   1 E 2 cos   1

t

1
1
E

2  1 
E


1
E cos
2
E cos
2
2
1
t
2
t
2
2
t
i
E 2 cos
2
1
r
 1 
t
1
2
E cos
1
2
Ei
2
t
2
cos1
Case 1:
Perpendicular polarization (HP):
( E is perpendicular to the plane of incidence parallel to the reflecting surface)
Let Ei propagate along +x direction, so as the direction of Er and Et.
According to BCs. Etan and Htan are continuous. Across the boundary.
 Ei  Er  Et
E
EE
 Et  1  r i
i
But we have,
E
2
r
 1 

r
 1 
2
i
1
E2
E2
1 
Er
E
r

i

 
1 
 

 
1
 
i
1
E r 
Ei

2
1 


2
E r 

E 
2
i

2
2
1
1 
cos
t
Ei cos1
2
E r  cos
 
1 

 
E  cos
2
Ei
E 2
E 2
2
2 E
i

i

E r 

E 
2
1
cos 2
cos
cos2
cos
1
1
1 cos1 
2 cos 2
 cos   cos
1
1
2
2
But we have,
Department of ECE
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
sin1
2

sin

2

2
1
1  sin 2 2 
cos 2 2
2 1 sin 1
2
 E r  1 cos 12 12 sin 1
Ei
 cos  sin 
2
1

1
2
1
1
2
cos  2  sin 
1
1
1
2
cos  2  sin 
1
1
1
This equation gives the ratio of the reflected to incident electric field strength for the case of a
perpendicular polarized wave.
.
Case II:
Parallel Polarisation:
Here E is parallel to the plane of incidence.
H is parallel to the reflecting surface.
The BCs on tangential components give
Htan = Etan is continuous across the boundary.
Therefore this BC when applied, we get
Ei

E
 Er cos1  Et cos2

t 1 
E r cos 1


Ei 

E
i
 cos2
But we already have
Department of ECE
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
E2
Ei
2
E
2
 1 
r
2
cos 2
2
cos1
t
 Ei
1

E2
2
Ei


E


E2


1
2

Er
Ei
E r 
1
E i 

2

2
2
cos1

 cos 2
Er 




i
1 
 2 cos
2
Ei  cos
1
2
E  cos 1
E2

 


r
r

2
E  cos
1
i
1 
1
1 
2 
2
r
1 
2 
 1 
r
1
cos

Ei  cos 2

 cos 1 
 cos
 
 cos 
 cos
1


2
1

E
r
2

2
cos1 


Ei 
cos1 
But from Snell‘s law we get
1 1
1
2

cos 2
1

2
2
1
2

cos1 
cos 2 
cos1 
2
2
sin   /  sin 
2
2
1
2

1

1
1  sin 2  2 
1  sin 2  2 
1
Therefore we get
 /  cos 
E r 
E
i

2
 sin 2 
1
1
 /  cos   sin 2 
2
1
1
1
1
2
1
1
2
This equation gives the reflection coefficient for parallel or vertical polarization, ie., the ratio of
reflected to incident electric field strength when E is parallel to the plane of incidence.
BRESNSTER ANGLE:
We have
 /  cos
2  sin 
1
2
1
1
1
E r 
Ei
 /  cos 2  sin 2 
2
1
1
2

1
1
When Nr = 0, Er = 0.
Therefore no reflection at all.
Therefore for zero reflection condition, we have,
Department of ECE
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2
2  sin 2 
cos 1 
1
1
1
2
2
2

1
2 
2
2

1

1
2
sin   2  sin 
2
1


2
2
2
1
1
 sin  1  sin
2
 1
2
 sin 2  1

cos 2 1 
2
2

2
2
2
sin
2
2
1
2
2
1
2
1

2
1
1
1 2 1 2
 sin 2   
2
1
1
2
2
1
2
2
 sin 1 
1 2
2
2
 cos  1 
1
1 2
2
tan1 

1
At this angle, which is called the Bresoster angle, there is no reflected wave when the incident
wave is parallel (or vertically) polarized. If the incident wave is not entirely parallel polarized,
there will he some reflection, but the reflected wave is entirely of perpendicular (or horizontal)
polarization.
Note:1
For perpendicular paolarisation, we have
cos 
E
 / sin 
2
1
2
Ei cos  
1
putting
cos 
2
1
or
1
/ sin 
2
1
1
N r  0, we get
cos 
1
2
1
 / sin
2
1
2 
 / sin 
2
1
2
1
1
21
ie., there is no corresponding Bresvster angle for this polarization.
Note 2:
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For parallel polarization,
We can show that
E
r
tan1
Ei  tan1

2 
2
and for perpendicular polarization, we can show that,
E
r

1 
sin2 1
Ei  sin2
TOTAL INTERNAL REFLECTION:
If 1 2 , then, both the reflection coefficients given by equations,
Er

Ei
2
2
 sin 1
1

2
 sin 1

1
cos1 
cos1
( perpendicular polarization )
2
and
 /  cos
2
1
Er

Ei
 /  cos
2
1

1

1

2
 sin 
2
1
 sin 2 
1
1
1
( parallel polarization )
2
2
1
become complex numbers when, sin1 
 a  jb 
Both coefficients take the form


 and thus have a unit magnitude. In other words, the

 a 
jb 
reflection is total provided that 1 is great enough and also provided that medium (1) is denser
than medium. (2) but total reflection does not imply that there is no field in medium (2). In
medium (2), the fields have the form, e
Snell‘s law gives the y variation as, e
j2 y sin2 Z cos2 
 j2 y 1/2
And the Z variation as,
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e
 j Z cos
2
2
e

j2 Z

1sin

2
2


j

e
 
Z   j
1


1 
2
 sin 

1
 2 

Z
2


2
e


   2

 sin 
  
 2 


1
1
1
In the above expression, the lower sign must be chosen such that the fields decrease
exponentially as Z becomes increasingly negative.
  2
 


sin
 1
 j


sin 2 


  
 
2
1
1
 2 
 2 
2
Therefore under the condition of TIR, a field does exist in the rarer medium. However, this field
ie.,
cos
 j
1
1
2
has a phase progression along the boundary and decreases exponentially away from it. If is thus
the example of a non-uniform plane wave.
The phase velocity along the interface is given by ,

2



1
sin 
1
2
Which, under the conditions of TIR is less than the phase velocity  of a UPW in medium (2).
2
Consequently, the non-uniform plane wave in medium (2) is a slow wave. Also, since some kind
of a surface between two media is necessary to support this wave, it is called a surface wave.
.
Maxwell’s Equations
In static electric and magnetic fields, the Maxwell’s equations obtained so far are:
Differential form
 E  0
 D  ρ
 H  J
 B
Department of ECE
Controlling principle
Integral form
Potential around a closed path is zero
Gauss‘s Law
Ampere‘s Circuital Law
E  dL
0
D  dS  ρ dv
(1)
(2)
H  dL  J  dS
Non-existence of isolated magnetic poles B  dS  0
(3)
(4)
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Contained in the above equations is the equation of continuity for steady currents,
J  dS  0
 J  0
(5)
Modification of Maxwell‘s equations for the case of time varying fields

 E  0 ; E  dL  0; ----(1)
First Modification of the first Maxwell‘s equation
To discuss magnetic induction and energy, it is necessary to include time-varying fields, but
only to the extent of introducing the Faraday‘s law.
Faraday‘s law states that the voltage around a closed path can be generated by

a time changing magnetic flux through a fixed path ( transformer action)

by a time-varying path in a steady magnetic field (electric generator action)
or
Faraday‘s law: The electromotive force around a closed path is equal to the negative of the
time rate of change of magnetic flux enclosed by the path.
φ

 t
 t B  dS
S
In our study of electromagnetics, interest centers on the relation between the time changing
electric and magnetic fields and a fixed path of integration.
E  dL
For this case the Faraday‘s law reduces to,

t B  dS
The partial derivative w.r.t time indicates that only variations of magnetic flux through a fixed
E  dL

S
closed path or a fixed region in space are being considered.
Thus, for time varying fields, equation (1) gets modified to
 E  
B
 E  dL   B
 dS
(6)
t
t
S
Second Modification: Modification of the Continuity equation for time varying fields:
Current is charges in motion. The total current flowing out of some volume must be equal to
the rate of decrease of charge within the volume [charge cannot be created or destroyed- law of
conservation of charges]. This concept is needed in order for understanding why current flows
between the capacitor plates. The explanation is that the current flow is accompanied by charge
build up on the plates. In the form of equation, the law of conservation of charge is
J  dS
d
 -
ρdv
dt
If the region of integration is stationary, this relation becomes,
J  dS
Department of ECE

 -
ρ
t
dv
---- (7)
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Applying divergence theorem to this equation, we get,
ρ
 ( J)dv   t dv
If this relation is to hold for any arbitrary volume, then, it must be true that,
 J  ρ
---- ( 8
t
)
This is the time-varying form of the equation of continuity that replaces equation (5).
Third Modification: Modification of the Maxwell‘s equation for the Ampere‘s Law:
Taking the divergence of equation (3) we get the equation of continuity as,
 H  J  0
(Divergence of curl is zero- vector identity).
Thus Ampere‘s law is inconsistent with the time varying fields for which the equation of
ρ . To resolve this inconsistency, James Clerk Maxwell in the mid
t
1860‘s suggested modification of the Ampere‘s law to include the validity for time varying
fields also; He suggested substitution of Gauss‘s law (2) into the equation of continuity (8)
giving,
continuity is  J 
 J 
ρ
.
t
But we know that  D  ρ .
Therefore we get ,
 J 
  ( D) 

D
t
t
, on interchanging the
time and space differentiation.
D
D
 J  t  0
Therefore
or
 (J  t )  0
----- (9)
This equation may be put into integral form by integrating over a volume and then applying
the divergence theorem:
D
(J 
Equations 9 and 10 suggest that (J 
t )  dS  0
------ (10)
D
t
) may be regarded as a total current density for time
varying fields. Since D is the displacement density,
D
is known as the displacement current
t
density.
Consider now a capacitor connected to an ac source as shown in figure.
I

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When applied as shown in figure to a surface enclosing one plate of a two-plate capacitor,
equation ( 10 ) shows that during charge or discharge, the conduction current in the wire
attached to the plates is equal to the displacement current passing between the plates.
Maxwell reasoned that the total current density should replace J in Ampere‘s law with the
result that
D
 H  J  t
----- (11)
Taking the divergence of this equation gives equation (9) and thus the inconsistency has been
removed. Note that the Equation (11) has not been derived from the preceding equations but
rather suggested by them. Therefore when Maxwell proposed it, it was a postulate whose
validity had to be established by experiment.
Integrating equation (11) over a surface and application of Stoke‘s theorem gives the integral
form of the equation:
D
---- (12)
H  dL  (J 
t )  dS
This equation states that the mmf around a closed path is equal to the total current enclosed by
the path. Thus equations 11 and 12 replace the static form of Ampere‘s law (3).
Maxwell’s equations:
In summary, the Maxwell’s equations are as follows:
Differential form
Controlling principle
 H  D  J
Integral form
Ampere‘s Circuital Law
 E B
Potential around a closed path is zero
 D  ρ
Gauss‘s Law
 B
 J  dS (I)
H  dL  D
E
D  dS
Non-existence of isolated magnetic poles B
 dL B  dS(II)
 ρ dv
 dS  0
(III)
(IV)
Contained in the above equations is the equation of continuity,
ρ
ρ
 J 
J  dS  - dv
t
t
In all the cases the region of integration is assumed to be stationary.
WORD STATEMENT FORM OF FIELD EQUATIONS:
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The word statements of the field equations may readily be obtained from the integral form of
the Maxwell‘s equations:
I.
The mmf around a closed path is equal to the conduction current plus the time
derivative of the electric displacement through any surface bounded by the path.
II.
The emf around a closed path is equal to the time derivative of the magnetic
displacement through any surface bounded by the path.
III.
The total electric displacement through the surface enclosing a volume is equal to the
total charge within the volume.
IV.
The net magnetic flux emerging through any closed surface is zero.
Alternate way of stating the first two equations:
1.
The magnetic voltage around a closed path is equal to the electric current through the
path.
2.
The electric voltage around a closed path is equal to the magnetic current through the
path
Boundary Conditions using Maxwell’s equations:
The integral form of Maxwell‘s equations can be used to determine what happens at the
boundary surface between two different media.( Find out why not the differential form?)
The boundary conditions for the electric and magnetic fields at any surface of discontinuity
are:
1.
The tangential component of E is continuous at the surface. i.e., it is the same just outside
the
surface as it is at the inside the surface.
2.
The tangential component of H is discontinuous across the surface except at the surface
of a perfect conductor. At the surface of a perfect conductor, the tangential component of
H is discontinuous by an amount equal to the surface current per unit width.
3.
The normal component of B is continuous at the surface of discontinuity.
4.
The normal component of D is continuous if there is no surface charge density.
Otherwise D is discontinuous by an amount equal to the surface charge density.
y
X
Proof:
Ex1
1, 1 , 1
( medium 1)
x/2
Ex2
2, 2, 2
( medium 2)
x/2
EY1
Department of ECE
EY2
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Ex3 Ex4
X
Fig 2 A boundary between two media
Let the surface of discontinuity be the plane x=0 as shown in fig 2.
Conditions on the tangential components of E and H
1. Condition for Etan at the surface of the boundary:
Consider a small rectangle of width x and length y enclosing a small portion of each media
(1) and (2).
The integral form of the second Maxwell equation ( II ) is,
E  dL

 B  dS
S
For the elemental rectangle of fig 2, we apply this equation and get
Δx
Δx
Δx
E
Δx
E
x4
Ey2Δy  Ex2 2
2  Ey1Δy  Ex3 2
2 Bz Δx Δy ----(13)
where Bz is the average magnetic flux density through the rectangle Δx Δy . Now, as this area of
the rectangle is made to approach to zero, always keeping the surface of discontinuity between
the sides of the triangle. If Bz is finite, then as x  o, the RHS of equation 13 will approach
zero. If E is also assumed to be finite everywhere, then, x/2 terms of the LHS of equation 13
will reduce to zero, leaving
x1
Ey2 y - Ey1 y = 0
for x = 0. Therefore
Ey2
=
That is, the tangential component of E is continuous.
Ey1
2. Condition for Htan at the surface of the boundary:
Now the integral form of the first Maxwell‘s equation ( I ) is
 H  dL

(D  J)  dS
S
For the elemental rectangle this equation becomes,
Δx
Δx
Δx
Δx
Hy2Δy  Hx2 2  Hx1 2  Hy1Δy  Hx3 2  Hx4 2  (Dz  J z ) Δx Δy
Department of ECE
----(14)
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If the rate of change of electric displacement
and the current density J are both considered to
D
be finite, then, as before, equation (14) reduces to
Hy2 y - Hy1 y = 0 for
x = 0. Therefore
Hy2
=
Hy1
That is, the tangential component of H is continuous (for finite current densities i.e., for any
actual case).
In case of a perfect conductor: A perfect conductor has infinite conductivity. In such a conductor,
the electric field strength E is zero for any finite current density. However, the actual
conductivity may be very large and for many practical applications, it is useful to assume it to be
infinite. Such an approximation will lead to difficulties because of indeterminacy in formulating
the boundary conditions unless care is taken in setting them up. The depth of penetration of an ac
field into a conductor decreases as the conductivity increases. Thus in a good conductor a hf
current will flow in a thin sheet near the surface, the depth of the sheet approaching zero as the
conductivity approaches infinity. This gives to the useful concept of a current sheet. In a current
sheet a finite current per unit width, Jz amperes per meter flows in a sheet of vanishingly small
depth x, but with the required infinitely large current density J, such that
limx  0, Jx = Jz
Now consider again the above example the mmf around the small rectangle. If the current
density Jz becomes infinite as x0, the RHS of equation 14 will not become zero. Let J z
amperes per metre be the actual current per unit width flowing along the surface. Then as x0,
equation 14 for H becomes,
Hy2 y - Hy1 y = Jsz y.
Hence
Hy1 = Hy2 - Jsz
---- (15)
Now if the electric field is zero inside a perfect conductor, the magnetic field must also be zero,
for alternating fields, as the second Maxwell‘s equation shows. Then, in equation 15, H y2 must
be zero.
So,
Hy1 = - Jsz
----(16)
This equation states that the current per unit width along the surface of a perfect conductor is
equal to the magnetic field strength H just outside the conductor. The magnetic field and the
surface current will be parallel to the surface, but perpendicular to each other. In vector notation,
this is written as,
J  nˆ  H
where nˆ is the unit vector along the outward normal to the surface.
Conditions on the normal component of B and D
3.Condition on the normal component of D (Dnor):
The integral form of the third Maxwell‘s equation is
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 D  dS
ρ dV

S
----( III )
vol
Consider the pill-box structure shown in fig 3. Applying the third Maxwell‘s equation to this pillbox structure, we get
Dn1dS  Dn2dS  ψedge  ρ x dS
----(17)
DN1
dS
1 11
X
222
DN2
Fig 3 A pill-box volume enclosing a portion of a boundary surface
In this expression, dS is the area of each of the flat surfaces of the pillbox, x is their separation,
and  is the average charge density within the volume x dS. edge is the outward electric flux
through the curved edge surface of the pillbox. As x  0, that is, as the flat surfaces of the
pillbox are squeezed together, always keeping the boundary surface between them, edge  0,
for finite values of displacement density. Also, for finite values of average density ρ , the RHS of
equation (17) reduces to
Dn1dS  Dn2dS

0
for x = 0.Then for the case of no surface charge condition on the normal components of D
Dn1 = Dn2
---- (18)
That is, if there is no surface charge, the normal component of D is continuous across the
surface.
In the case of a metallic surface: In the case of a metallic surface, the charge is considered to
reside on ‗the surface‘. If this layer of surface charge has a surface charge density S Coulombs
per square meter, the charge density  of the surface layer is given by
ρ
S
C/m3
ρ  x
layer. As x approaches zero, the charge density
where x is the thickness of the surface
approaches infinity in such a manner that
limx0 ρx  ρS
Then in fig 3, if the surface charge is always kept between the two flat surfaces as the seperation
between them is decreased, the RHS of equation (17) approaches S dS as x approaches zero.
Equation 17 then reduces to
D D ρ
n1
Department of ECE
n2
S
----( 19 )
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When there is a surface charge density S, the normal component of displacement density
is discontinuous across the surface by the amount of the surface charge density.
For any metallic conductor the displacement density, D = E within the conductor will be a small
quantity( it will be zero in the electrostatic case, or in the case of a perfect conductor). Then if
the medium (2) is a metallic conductor, Dn2 = 0 ; and the equation (19) becomes
Dn1 = S
----(20)
The normal component of the displacement density in the dielectric is equal to the surface
charge density on the conductor.
4 Condition on the normal component of B (Bnor):
The integral form of the fourth Maxwell‘s equation is
 B  dS

----( IV )
0
S
The pill-box structure is again shown in fig 4 for magnetic flux density. Applying the fourth
Maxwell‘s equation to this pill-box structure, we get
Bn1dS  Bn2dS  ψedge  0
----(21)
BN1
dS
111
X
222
BN2
Fig 4 A pill-box volume enclosing a portion of a boundary surface
In this expression, dS is the area of each of the flat surfaces of the pillbox, x is their separation,
and  is the average charge density within the volume x dS. edge is the outward electric flux
through the curved edge surface of the pillbox. As x  0, that is, as the flat surfaces of the
pillbox are squeezed together, always keeping the boundary surface between them, edge  0,
for finite values of magnetic flux density. The RHS of equation (21) reduces to
Bn1dS  Bn2dS

for x = 0.Then the condition on the normal components of B
magnetic charges,
Bn1 = Bn2
0
since there are no isolated
---- (22)
i.e., The normal component of magnetic flux density is always continuous across the
boundary.
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