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ELECROMAGNETIC THEORY FORMULA SHEET
CONTENTS
PART-A (CORE)
1. Electrostatics…………………………………………………………………..... (5-17)
1.1 Coulomb’s Law and Superposition Principle…….................................................. (5)
1.1.1 Electric field
1.2 Gauss’s law ……..................................................................................................... (6)
1.2.1 Field lines and Electric flux
1.2.2 Applications
1.3 Electric Potential……............................................................................................. (7)
1.3.1 Curl of Electric field
1.3.2 Potential of localized charges
1.4 Laplace’s and Poisson equations…….................................................................... (8)
1.5 Electrostatic boundary condition…….................................................................... (8)
1.6 Work and Energy in electrostatics…….................................................................. (9)
1.6.1 The energy of point charge distribution
1.6.2 Energy of continuous charge distribution
1.7 Basic properties Conductors ……........................................................................ (10)
1.8 Multiple Expansions……..................................................................................... (11)
1.8.1 The Electric Potential and Field of a Dipole
1.8.2 Approximate potential at large distances
1.9 Polarization……................................................................................................. (13)
1.9.1 The Field of a polarized Object (Bound Charges)
1.10 The Electric Displacement............................................................................. (13-14)
1.10.1 Gauss Law in the Presence of Dielectrics
1.10.2 Linear Dielectrics (Susceptibility, Permittivity, Dielectric Constant)
1.10.3 Boundary Condition
1.10.4 Energy in dielectric system
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1.12 Image problems.................................................................................... (15-17)
1.12.1 The Classic Image problem
1.12.2 Induced Surface Charge
1.12.3 Force and Energy
1.12.4 Other Image Problem
2. Magnetostatic.......................................................................................... (18-24)
2.1 Magnetic force on current element............................................................. (18)
2.1.1 Current in a wire
2.1.2 Surface current density
2.1.3 Volume current density
2.2 Continuity equation................................................................................... (19)
2.3 Biot-Savart law.......................................................................................... (19)
2.3.1 Magnetic field due to wire
2.3.2 Magnetic field due to Solenoid and Toroid
2.4 Ampere's Law........................................................................................... (20)
2.5 Magnetic Vector Potential........................................................................ (20)
2.6 Magnetostatic boundary condition........................................................... (21)
2.7 Multiple Expansion of Vector Potential................................................... (22)
2.8 Magnetisation........................................................................................... (23)
2.8.1 The Field of a magnetized Object (Bound Currents)
2.9 The Auxiliary field H.......................................................................... (23-24)
2.9.1 Ampere’s Law in in presence of magnetic materials
2.9.2 Magnetic Susceptibility and Permeability
2.9.3 Boundary Condition
3. Dynamics of charged particles in static and uniform electromagnetic fields
3.1 Charged particle in static electric field………………………..……. (25-26)
3.1.1 Charged particle enters in the direction of field (Linear motion)
3.1.2 Charged particle enters in the direction perpendicular to field
(Parabolic motion)
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3.2 Charged particle in static magnetic field.......................................................... (27)
3.2.1 Charged particle enters in the direction perpendicular to field
(Circular motion)
3.2.2 Charged particle enters in the direction making an angle with the field
(Helical motion)
3.3 Charged particle in uniform electric and magnetic field (Cycloid motion).... (28)
4. Electromagnetic induction…………………………………………………... (29)
4.1 Faraday’s Law
4.1.1 Lenz’s Law
4.1.2 Inductance
4.1.3 Energy stored in the field
5. Maxwell's equations…………………………………………………..…... (30)
5.1 Maxwell’s equation in free space
5.1.1 Electrodynamics before Maxwell’s
5.1.2 How Maxwell fixed Ampere’s Law
5.1.3 Paradox of charging capacitor
5.1.4 Maxwell’s equation in free space
5.2 Maxwell’s equation in linear isotropic media …………………………... (30)
5.3 Boundary conditions on the fields at interfaces ……………………….... (31)
6. Electromagnetic waves in free space…………………………….….... (31-34)
6.1 Poynting Theorem…………………………………………………..….... (31)
6.2 Waves in one dimension (Sinusoidal waves) ……………….................... (32)
6.2.1 The wave equation
6.2.2 Terminology
6.2.3 Complex notation
6.2.4 Polarization
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6.3 Electromagnetic waves in vacuum……………….................................... (35)


6.3.1 The wave equation for E and B
6.3.2 Monochromatic plane waves
6.3.3 Energy and momentum in Electromagnetic wave
6.4 Electromagnetic waves in matter………………..................................... (38)
6.5 Electromagnetic waves in conductors……………….............................. (39)
7. Applications of Electromagnetic waves………................................ (41-43)
7.1 Reflection and Refraction………………................................................ (41)
7.1.1 Normal incidence
7.1.2 Oblique incidence
7.1.3 Fresnel’s relation (Parallel and Perpendicular Polarization)
PART ‘B’ ADVANCED
8. Potential and Field formulation for Time Varying Fields……........ (44-45)
8.1 Scalar and vector potentials..................................................................... (44)
8.2 Gauge transformation.............................................................................. (44)
8.3 Coulomb and Lorentz gauge.................................................................... (45)
9. Waveguides……................................................................................... (45-49)
9.1 Rectangular Waveguides......................................................................... (45)
9.2.1 TE Waves in Rectangular Waveguide
9.2.2 TM Waves in Rectangular Waveguide
10. Radiation- from moving charges ....................................................... (50)
10.1 Radiation
10.1.1 Power radiated by a point charge
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1. Electrostatics
The electric field at any point due to stationary source charges is called as electrostatic
field.
1.1 Coulomb’s Law and Superposition Principle
The electric force on a test charge Q due to a single point charge q, which is at rest and a
distance R apart is given by Coulomb’s law
Q

1 Qq 
F
.
4 0 R 2 R
R
The constant  0 is called the permittivity of free space.
q
In mks units,  0  8.85 10
12
C2
N .m 2
1.1.1 Electric field
If we have many point charges q1 , q2 ,...... at distances R1 , R2 ,...... from test charge Q,
then according to the principle of superposition the total force on Q is


  

q2Q
1  q1Q
F  F1  F2  ............ 
 2 R1  2 R2  ........  ,
4 0  R1
R2



 F  QE

where E  P  
1
4 0
n
qi
R
i 1
2
i

Ri


E is called the electric field of the source charges. Physically E  P  is the force per unit
charge that would be exerted on a test charge placed at P.
If charge is distributed continuously over some region, then

E r  
1
4 0
1 ˆ
Rdq .
2
R
line

The electric field of a line charge is  dq   dl  

E r  
1
4 0
 ( r ) ˆ
Rdl  where  is charge per unit length.
R2
line

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For surface charge  dq   da 

E r  
1
4 0
 ( r ) ˆ
Rda where  is charge per unit area.
2

R
surface
For a volume charge  dq   d  

E r  
1
4 0
 ( r ) ˆ
Rd  where  is charge per unit volume.
2
R
volume

1.2 Gauss’s law
1.2.1 Field lines and Electric flux

Consider that a point charge q is situated at the origin: E 
1 q
4 0 r 2

r
This field is represented by the field line as shown in
figure below. The magnitude of the field is indicated by
the density of the field lines: it's strong near the center
E
where the field line are close together, and weak farther
out, where they are relatively far apart.
The field strength (E) is proportional to the number of
field lines per unit area (area perpendicular to the lines).


The flux of E through a surface S, E   E .d a is a measure of the “number of field
S
lines” passing through S.
A charge outside the surface would contribute nothing to the total flux, since its field
lines go in one side and out the other. It follows, then, that for any closed surface,


1
 E .d a  
Qenc
0
where Qenc is the total charge enclosed within the surface. This is Gauss’s law in integral
form.
 
Gauss’s law in differential form:
 .E 
1

0
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1.2.2 Applications of Gauss’s Law
Gauss's law is always true, but it is not always useful. Gauss's law is useful for only three
kinds of symmetry:
1. Spherical symmetry. Make your Gaussian surface a concentric sphere.
2. Cylindrical symmetry. Make your Gaussian surface a coaxial cylinder.
3. Plane symmetry. Make your Gaussian surface a “pillbox,” which extends equally
above and below the surface.
Gaussian
surface
r
Gaussian pillbox
R
Gaussian surface
1.3 Electric Potential
1.3.1 Curl of Electric field


Then integral around a closed path is zero i.e.

E .d l  0
This line integral is independent of path. It depends on two end points.
 
Applying stokes theorem, we get   E  0 . The electric field is not just any vector but
only those vector whose curl is zero.
P 

So, we can define a function V  r     E .d l

where  is some standard reference point. V then depends only on the point r. It is called
the electric potential.
Evidently, the potential difference between two points a and b is
b 

V  b   V  a     E .d l


 E  V
a
Potential obeys the superposition principle.
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1.3.2 Potential of localized charges
Potential of a point charge q is V 
1 q
where R is the distance from the charge.
4 0 R
n
1
4 0
R
For continuous volume charge distribution V ( r ) 
1
4 0

The
charges
The potential of a collection of point charge is V 
potential
and V ( r ) 
of
line
and
surface
i 1
qi
.
i
 (r )
d 
R
are
V (r ) 
1
 (r )
dl 

4 0
R
1
 ( r )
da .

4 0
R
1.4 Laplace’s and Poisson equations
 



Since E  V and .E 
  2V   .
0
0
This is known as Poisson's equation.
In regions where there is no charge, so that   0 , Poisson's equation reduces to Laplace's
equation,  2V  0 .
1.5 Electrostatic boundary condition
The boundary between two medium is a thin sheet of surface charge  .

above  below





||
||
and Eabove
 Ebelow
 E above  E below  nˆ
0
0
where n̂ is unit vector perpendicular to the surface, pointing upward.
Vabove  Vbelow
where

Vabove Vbelow



n
n
0
V
 V  nˆ denotes the normal derivative of V (that is the rate of change in the
n
direction perpendicular to the surface.)
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1.6 Work and Energy in electrostatics
The work done in moving a test charge Q in an external field E , from point a to b is
b 
 

W   F  dl   Q  E  dl  Q V  b   V  a  
b
a
a
q1 

a


q2  
If a   and b  r
 W  Q V  r   V      QV  r  since V     0

Q

qi 
b
In this sense potential is potential energy (the work it takes to create the system) per unit
charge (just as the field is the force per unit charge).
1.6.1 The energy of point charge distribution
q3
When the first charge q1 is placed, no work has been
done. When q2 is placed work done W2  q2V1
where V1
is
the
potential
due
to q1

R13
so,
 q 
1
W2 
q2  1  .
4 0  R12 
Similarly
W3 
when
third
charge q3
is
placed
r1
q1
r3


R23

R12
r
2
q2
 q
q 
1
q3  1  2  .
4 0  R13 R23 
The work necessary to assemble the first three charges is W 
In general, W 
1
4 0
n
n

i 1 j 1
j i
qi q j
Rij

1
8 0
n
n

i 1 j 1
j i
qi q j
Rij

1  q1q2 q1q3 q2 q3 



.
4 0  R12
R13
R23 
1 n
 qiV  ri  ,
2 i 1
where V(ri) is the potential at point ri (the position of qi) due to all other charges.
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1.6.2 Energy of continuous charge distribution
For a volume charge density W 
1
Vd and
2
W
0
2

E 2 d
all space
(i) Energy of a uniformly charged spherical shell of total charge q and radius R is
W
q2
.
80 R
(ii) Energy stored in a uniformly charged solid sphere of radius R and charge q is
W
1 3q 2
.
40 5R
1.7 Basic properties Conductors

1. E  0 inside a conductor.
2.   0 inside a conductor.
3. Any net charge resides on the surface.
4. A conductor is an equipotential.
5. E is perpendicular to the surface, just outside a conductor.
Because the field inside a conductor is zero, boundary


 

E above  E below  nˆ requires that the field immediately outside is E  nˆ .
0
0
In terms of potential equation
condition
Vabove Vbelow

V
yields    0


n
n
n
0
These equations enable us to calculate the surface charge on a conductor, if we can

determine E or V .

1 2
Force per unit area on the conductor is f 
 nˆ .
2 0
This amounts to an outwards electrostatic pressure on the surface, tending to draw the
conductor into the field, regardless the sign of  . Expressing the pressure in terms of the
field just outside the surface,
P
0 2
E .
2
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1.8 Multiple Expansions
1.8.1 The Electric Potential and Field of a Dipole

If we choose coordinates so that p (dipole moment) lies at the
z
origin and points in the z direction, then potential at (r , ) is:

rˆ. p
p cos 
Vdip  r ,  

.
2
4 o r
4 o r 2

 E dip  r ,   

p

r
y


1  
ˆ
ˆ
3
p

r
r

p

4 0 r 3 
 
x
Note:

(a) When a dipole is placed in a uniform electric field ( E ), net force on the dipole is zero
  


where p  qd .
and it experiences a torque   p  E
   
(b) In non-uniform field, dipoles have net force F  p  E and torque  p  E .


 

(c) Energy of an ideal dipole p in an electric field E is U   p.E .
(d) Interaction energy of two dipoles separated by a distance r is
U


 
1
p1  p 2  3 p1  rˆ p 2  rˆ .
4 0 r 3
1.8.2 Approximate potential at large distances
Approximate potential at large distances due to arbitrary localized charge distribution
d '
r ' 

V r 

R
P
r
1 1
1
1
 r ' d '  2  r ' cos  '  r ' d '  2


4 0  r
r
r
 
 
' 2
 r 
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
1
3
2 '
'
'
 2 cos   2   r d  ...



 
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1
The first term (n = 0) is the monopole contribution (it goes like ). The second term
r
(n = 1) is the dipole term (it goes like
1
). The third term is quadrupole; the fourth
r2
octopole and so on.
The lowest nonzero term in the expansion provides the approximate potential at large r
and the successive terms tell us how to improve the approximation if greater precision is
required.
The monopole and dipole terms
Ordinarily, the multipole expansion is dominated (at large r) by the monopole term:
Vmon  r  
1 Q
.
4 0 r
where Q   d is the total charge of the configuration.
If the total charge is zero, the dominant term in the potential will be the dipole (unless, of
course, it also vanishes):


1 1
1 1
1 rˆ. p
'
'
'
'
'
'
'
Vdip  r  
r cos   r d 
rˆ. r  r d 
,
4 0 r 2 
4 0 r 2 
4 0 r 2


'
'
'
where dipole moment p   r  r d .
 
 
 
The dipole moment is determined by the geometry (size, shape and density) of the charge
distribute. The dipole moment of a collection of point charge is
 n '
p   qi ri .
i 1
Note: Ordinarily, the dipole moment does change when we shift the origin, but there is
an important exception: If the total charge is zero, then the dipole moment is independent
of the choice of origin.
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1.9 Polarization

(Polarization) P  dipole moment per unit volume
1.9.1 The Field of a polarized Object (Bound Charges)
Suppose we have a piece of polarized material with

polarization vector P containing a lot of microscopic

1
4 0

S


p
dipoles lined up. Then

V r 

R
1  
1  1   
P.d a '
   '.P d ' .
R
4 0 V  R 



The first term looks like the potential of a surface bound charge  b  P.nˆ (where n̂ is
the normal unit vector).
 
The second term looks like the potential of a volume bound charge b  .P .

1
Thus V r 
4 0


S
b
1
b
da '
d ' , this means the potential (and hence also the

R
4 0 V R
field) of a polarized object is the same as that produced by a volume charge
 

density b  .P plus a surface charge density  b  P.nˆ .
1.10 The Electric Displacement
1.10.1 Gauss Law in the Presence of Dielectrics
Within the dielectric, the total charge density can be written as   b   f where b is
volume bound charge  f free charge density.
 

 
.D   f where D   0 E  P is known as the electric displacement.
 
.D   f
Thus Gauss’ law reads,
 
Or, in integral form
 D.d a  Q fenc , where Q fenc denotes the total free charge enclosed
in the volume.
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1.10.2 Linear Dielectrics (Susceptibility, Permittivity, Dielectric Constant)

For any substances, the polarization is proportional to the field provided  is not too
  

strong:
P  E  P   0 e E
(Materials that obey this relation are called linear dielectrics)
The constant of proportionality,  e is called the electric susceptibility of the medium.
The value of  e depends on the microscopic structure of the substance and also on
external conditions such as temperature.
In linear media we have

 




D   0 E  P   0 E   0 e E   0 E (1   e )   E , where    0 (1   e )
This new constant  is called the permittivity of the material.
Also  r 

(1   e ) is called relative permittivity or dielectric constant, of the material.
0

1.10.3 Boundary Condition on D
The boundary between two medium is a thin sheet of free surface charge  f .
||
||
||
||


Dabove
 Dbelow
  f and D above  D below  P above  P below
1.10.4 Energy in dielectric system
W
 
1
 D  E d .
2 all space


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1.12 Image problems
1.12.1 The Classic Image problem
Suppose a point charge q is held a distance d above an infinite grounded conducting
plane. We can find out what is the potential in the region above the plane.
z
z
q
 q
d
d
y
V 0
y
d
 q
x
x
Forget about the actual problem; we are going to study a complete different situation.
The new problem consists of two point charges +q at (0, 0, d) and –q at (0, 0,-d) and no
conducting plane.
For this configuration we can easily write down the potential:

1 
q
q
V  x, y, z  

2
4  0  x 2  y 2  z  d 2
x 2  y2   z  d 







(The denominators represent the distances from (x, y, z) to the charges +q and –q,
respectively.) It follows that
1. V  0 when z  0 and
2. V  0 for x 2  y 2  z 2  d 2 ,
and the only charge in the region z  0 is the point charge +q at (0, 0, d). Thus the
second configuration produces exactly the same potential as the first configuration, in the
upper region z  0 .
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1.12.2 Induced Surface Charge
The surface charge density  induced on the conductor surface is    0
V
z
z 0
qd
   x, y  

2 x 2  y 2  d 2

3
2
As expected, the induced charge is negative (assuming q is positive) and greatest at
x=y=0.
Q    da  q
The total induced charge
1.12.3 Force and Energy
The charge q is attracted towards the plane, because of the negative induced surface

1
q2
charge. The force: F  
zˆ .
4 0  2d 2
One can determine the energy by calculating the work required to bring q in from
infinity.
d
d
 
1
q2
1  q2 
1 q2
W   F .dl 
dz






4 0  4 z 2
4 0  4 z  
4 0 4d

d
1.12.4 Other Image Problem
The method just described is not limited to a single point charge; any stationary charge
distribution near a grounded conducting plane can be treated in the same way, by

introducing its mirror image.

R

R'
r
R

a
q


V 0
b

q'
q
a
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Let us examine the completely different configuration, consisting of the point charge q
together with another point charge q '  
2
R
q placed at a distance b  R to the right of
a
a
the centre of sphere. No conductor, now-just two point charges.
The potential of this configuration is V(r, ) 
V(r, ) 
1  q q' 
  
40  R R  


q 
1
1




2
2
2
2
40  r  a  2ra cos 

R

(ra
/
R)

2ra
cos


Clearly when r = R, V  0
Force
The force on q, due to the sphere, is the same as the force of the image charge q, thus:
F
1
q 2 Ra
40 (a 2  R 2 )2
Energy
To bring q in from infinity to a, we do work
W
1
q 2R
40 2(a 2  R 2 )
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2. MAGNETOSTATICS
The magnetic field at any point due to steady current is called as magnetostatic field.
2.1 Magnetic force on current element

The magnetic force on a charge Q, moving with velocity v in a magnetic field B is,
 
F mag  Q v  B . This is known as Lorentz force law. In the presence of both electric and
  
magnetic fields, the net force on Q would be: F  Q E  v  B
2.1.1 Current in a wire
A line charge λ traveling down a wire at a



speed v constitutes a current I   v .
v t
v
Magnetic force on a segment of current
 
carrying wire is, F mag  I  dl  B

P

2.1.2 Surface current density

When charge flows over a surface, we describe it by the surface current K .
K
d
dl 
is the current per unit widthFlow
perpendicular to flow.
dl 


Also K   v where is  surface charge

density and v is its velocity.

 
Magnetic force on surface current F mag   K  B da

K

2.1.3 Volume current density

 d 
J
is the current per unit area-perpendicular to
da
da


flow. Also J   v where  is volume charge density

and v is its velocity. Magnetic force on volume

 
 
current F mag   J  B d .Current crossing a surface S is    J  d a

J
Flow

S
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2.2 Continuity equation
General continuity equation   J  

t
This is the precise mathematical statements of local charge conservation.
Thus for magnetostatic fields

 0 and hence the continuity equation becomes:
t
 
 J  0.
2.3 Biot-Savart law (magnetic field of steady line current)
The magnetic field of a steady line current is given by



0 I  Rˆ
0
dl '  Rˆ
B r 
dl 
I
4  R 2
4  R 2
where 0  4 10 7
r
I

R
N
( permeability of free space)
A2
dI '
For surface and volume current Biot - Savart law becomes:




0 K ( r ')  Rˆ
0 J ( r ')  Rˆ
( r ) 
da ' and (r ) 
d ' .
4 
R2
4  R 2
2.3.1 Magnetic field due to wire
Let us find the magnetic field a distance d from a long straight wire carrying a steady
current I.
  I
B  0 (sin 2  sin 1 )ˆ
4d
For Infinite wire: 1  

1
  I


and 2   B  0 ˆ
2
2
2d
2
Note:
I
1. Force (per unit length) of attraction between two
Wire segment
long, parallel wires a distance d apart, carrying
currents I1 and I2 in same direction are: f 
 0 I1 I 2
.
2 d
2. If currents are in opposite direction they will repel with same magnitude.
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3. The magnetic field a distance d above the center of a circular loop of radius R, which
  I
R2
carries a steady current I is B  0
zˆ .
2 (R 2  d 2 )3/2

 I
At the center of the circle B(0)  0 zˆ
2R
2.3.2 Magnetic field due to Solenoid and Toroid
The magnetic field of a very long solenoid, consisting of n closely wound turns per unit
length of a cylinder of radius R and carrying a steady current I is:
  0 nI zˆ inside the solenoid
B
outside the solenoid.
0
^
  0 NI 
Magnetic field due to toroid is B   2 r
0
for points inside the coil
for points outside the coil
where N is the total number of turns.
2.4 Ampere's Law
In general we can write
 
 
B

dl


I
where
I

0
enc
enc

 J .d a is the total current enclosed
by the amperian loop.
 

   B  0 J
Right hand Rule
If the fingers of your right hand indicate the direction of integration around the boundary,
then your thumb defines the direction of a positive current.
 
  
2.5 Magnetic Vector Potential A : Since   B  0  B    A
 


For magnetostatic fields,   A  0
and  2 A   0 J


0 J  r '
If J goes to zero at infinity, A  r 
d ' for volume current.
4  R



0 I 1
0 K
For line and surface currents, A  r 
dl '; A  r 
da '
4  R
4  R

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
2.6 Magnetostatic boundary condition (Boundary is sheet of current, K )
Just as the electric field suffers a discontinuity at a surface charge, so the magnetic field
is discontinuous at a surface current. Only this time it is the tangential component that
changes.
 
 B.d a  0
Since


 Babove
 Bbelow
For tangential components
 


||
||
r
 B.dl  0 I enc  Babove

B


K
B
is
parallel
to
surface
but

to
K
below
0


  


||
||
B is parallel to surface and along K
  B.dl  0 I enc  Babove  Bbelow

Thus the component of B that is parallel to the surface but perpendicular to the current




 
is discontinuous in the amount 0 K . A similar amperian loop running parallel to the
current reveals that the parallel component is continuous. The result can be summarized
in a single formula:


B above  B below   0 K  nˆ ,
where n̂ is a unit vector perpendicular to the surface, pointing “upward”.
Like the scalar potential in electrostatics, the vector potential is continuous across, an


boundary:
Aabove  Abelow
 
  
For . A  0 guarantees that the normal component is continuous, and   A  B , in the
 
 
form
A
.
dl


 B.d a  
line
S




  Aabove  Abelow
But the derivative of A inherits the discontinuity of B :

  0 K
n
n
 
 
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2.7 Multiple Expansion of Vector Potential
1
We can always write the potential in the form of a power series in , where r is the
r
distance to the point in question. Thus we can always write
0 I
4
1

1
1
1
2 3
2
 r  d l '  r 2  r ' cos ' d l '  r 3  r '  2 cos  ' 2 d l '  .........





z
First term, monopole   dl   0 (no magnetic
Ar  
monopole)


0 m  rˆ
Second term, dipole  Adip  r  
4 r 2
where
m
is
the
magnetic
dipole


moment:
m
m  I  d a  I A where A is area vector
Thus A dip
r

y

 m sin  ˆ
r  0

4 r 2

x
Hence

B dip r    A 


 1
0m
ˆ  0  3 3 m  rˆ rˆ  m 
ˆ
2
cos

r

sin



4 r 
4 r 3

 

Note:

(a) When a magnetic dipole is placed in a uniform magnetic field ( B ), net force on the
  
dipole is zero and it experiences a torque  m  B .
   
(b) In non-uniform field, dipoles have net force F   m  B and torque  m  B .



 
(c) Energy of an ideal dipole m in an magnetic field B is U   m.B .
(d) Interaction energy of two dipoles separated by a distance r is
U




1
m1  m2  3 m1  rˆ m2  rˆ .
4 0 r 3
.
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 
2.8 Magnetisation M
 
Magnetization M is magnetic dipole moment per unit volume.
2.8.1 The Field of a magnetized Object (Bound Currents)
 
Consider a piece of magnetized material with magnetization M .
Then the vector potential is given by

 1  

A  r   0    M  r    d ' 0
4 R
4
v


1  
   d a .
M
r


R
This means the potential(and hence also the field) of a magnetized object is the same as
would be produced by a volume current J b    M throughout the material, plus a
surface current K b  M  nˆ , on the boundary. We first determine these bound currents,
and then find the field they produce.

2.9 The Auxiliary field ( H )
2.9.1 Ampere’s Law in in presence of magnetic materials
  

In a magnetized material the total current can be written as J  J b  J f where J b is


  
 B 
bound current and J f is free current. Thus   H  J f where H 
M
0
 
In integral form  H  dl  I fenc where I fenc is the total free current passing through the
amperian loop.



H plays a role in magnetostatic analogous to D in electrostatic: Just as D allowed us to

write Gauss's law in terms of the free charge alone, H permits us to express Ampere's
law in terms of the free current alone- and free current is what we control directly.
Note:


When we have to find B or H in a problem involving magnetic materials, first look for
symmetry. If the problem exhibits cylindrical, plane, solenoid, or toroidal symmetry, then

 
we can get H directly from the equation  H  dl  I fenc .
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2.9.2 Magnetic Susceptibility and Permeability
For most substances magnetization is proportional to the field H ,  M   m H ,
where  m is magnetic susceptibility of the material.


 B   0 H  M   0 1   m H  B   H
where
 0   0  r   0 1   m  ,
is
permeability of material.

2.9.3 Boundary Condition ( H )
The boundary between two medium is a thin sheet of free surface current K f .
The Ampere’s law states that



 H above  H below  K f  nˆ .
And





H above
 H below
  M above
 M below
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
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3. Dynamics of charged particles in static and uniform electromagnetic fields
(The Lorentz Force Law)

The magnetic force on a charge Q, moving with velocity v in a magnetic field B is,
 
F mag  Q v  B
This is known as Lorentz force law.
In the presence of both electric and magnetic fields, the net force on Q would be:


 
F  Q E  v  B 




3.1 Charged particle in static electric field
3.1.1 Charged particle enters in the direction of field (Linear motion)
Q

2
1

E
The force on the charge Q in electric field E is F  Q E .
Acceleration of the charge particle in the direction of the electric field is a 
If r is the position vector at any time t then
Let at t=0, r  r0  C1  r0  r 
If initially u  0, r0  0  r 
QE 2
t  ut  r0
2m

ut
F QE

.
m
m

QE 2
t
2m

2

r0

1
QE 2
QE
t and v 
t.
2m
m
The energy acquired by the charged particle in moving from point 1 to 2 is
2
2
2
W   F .d l  m  a.d l  m 
1
1
1
2
1
dv
.vdt m  v.d v  W  mv 22  v12 
2
dt
1
If the potential difference between points 1 to 2 is V then W  QV 
1
m v 22  v12
2


If the particle starts from rest i.e v1  0  and final velocity is v then
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W  QV 
1 2
mv  v 
2
2QV
m
1 2 1 Q2E2 2
QE 2
t  QE 
t  QEr
Kinetic energy of the particle K .E.  mv  m.
2
2
2
2m
m
3.1.2 Charged particle enters in the direction perpendicular to field (Parabolic
motion)
Let us consider a charge particle enters in an electric field region with velocity v x at t=0.
The electric field is in the y-direction and the field region has length l. After traversing a
distance l it strikes a point P on a screen which is placed at a distance L from the field
region.





 L
vx
P
y2
y1

Ey



y
x

l
QE y
1
Thus y  a y t 2 
2
2m
QE y  l

y1 
2m  v x
 x

 vx
2

 and which represents parabolic path.

2

 and y 2  L tan 

QE y  l

Thus distance of point P from the center of the screen is, y1  y 2 
2m  v x
Angle of deviation in the field region, tan  
2

  L tan 

dy QE y

x
dx mv x2
Angle of deviation in the field free region, tan  
QE y
mv x2
l
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3.2 Charged particle in static magnetic field

The magnetic force on a charge Q, moving with velocity v in a magnetic field B is,
 
F mag  Q v  B
This is known as Lorentz force law.
3.2.1 Charged particle enters in the direction perpendicular to field (Circular
motion)
If a charge particle enters in a magnetic field at angle of 90o, then motion will be circular
with the magnetic force providing the centripetal acceleration.
y
2
For uniform circular motion: QvB  m
v
mv
R
R
QB
v
R
where R is the radius of the circle and m is the mass of the charge

particle.
z
Momentum of the charged particle p  QBR
Kinetic energy (KE) 
Time period T 
F
Q
x
B
p 2 Q2 B2 R2

2m
2m
2R 2m

v
QB
3.2.2 Charged particle enters in the direction making an angle with the field
(Helical motion)
If the charge particle enters in a magnetic field making an angle θ,
then motion will be helical.
v  v sin 
and
v||  v cos  ,
mv
and the radius of helix is R  QB .
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B
||
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3.3 Charged particle in uniform electric and magnetic field (Cycloid motion)

  
F  Q E  v  B  Q  E zˆ  Bz yˆ  By zˆ 
z

 ma  m  
y yˆ  
z zˆ 


E

 
y   z , 
z     y 
B

QB
where  
(cylotron frequency )
m
E
Solving above differential equations, we get
o
E
E
y t  
t  sin t  , z  t   1  cos t 
B
B
E
2
2
  y  Rt    z  R   R 2
where R 
B
This is the formula for a circle, of radius R, x
whose center is
 0, Rt , R 
travels in the
a
c
b
B
y-direction at constant speed, v  R 
E
B
The curve generated in this way is called a cycloid.
 
 d l  Q v  B  vdt  0 .
Magnetic forces do not work because v  B is perpendicular to v , so
dWmag  F mag
Magnetic forces may alter the direction in which a particle moves, but they can not speed
up or slow down it.
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4. Electromagnetic induction
4.1 Faraday’s Law
v
v
I
I
B
I
B
B
b
a 
changing
magnetic field
c
Experiment 1: He pulled a loop of wire to the right through a magnetic field. A current
flowed in the loop (Figure a).
Experiment 2: He moved the magnet to the left, holding the loop still. Again, a current
flowed in the loop (Figure b).
Experiment 3: With both the loop and the magnet at rest, he changed the strength of the
field (he used an electromagnet, and varied the current in the coil). Once again current
flowed in the loop (Figure c).
Thus, universal flux rule is that, whenever (and for whatever reason) the magnetic flux
through a loop changes, an e.m.f.    will appear in the loop.
 
d
dt
 
(Where magnetic flux    B.d a )
In experiment 2,
A changing magnetic field induces an electric field.
 
d
It is this “induced” electric field that accounts for the e.m.f.    E  dl  
dt

 
B 
B
Then
 E  dl    t .d a    E   t
4.1.1 Lenz’s Law
In Faraday’s law negative sign represents the Lenz’s law. (The induced current will flow
in such a direction that the flux it produces tends to cancel the change).
For example if the magnetic flux is increasing then induced e.m.f will try to reduce and
vice versa.
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5. Maxwell's equations
5.1 Maxwell’s equation in free space
5.1.1 Electrodynamics before Maxwell’s
(i)    

0
(Gauss’ Law),
 
(ii)   B  0
(No name),

 
B
(iii)     
t
 

(iv)   B  0 J
(Farday’s Law),
(Ampere’s law).
5.1.2 How Maxwell fixed Ampere’s Law
From continuity equation and Gauss Law


   
 
   E 
  
E 
.J 
 ( 0 .E )  .   0
  .  J   0
  0 .
t
t

t

t





 

E
  B  0 J  0 0
Thus
t
A changing electric field induces a magnetic field.


E
Maxwell called this extra term the displacement current J d   0
.
t

 
 E  
Integral form of Ampere's law  B.dl 0 I enc  0 0  
 .d a
 t 
5.1.4 Maxwell’s equation in free space
(i)    

0
(Gauss’ Law),
 
(ii)   B  0
(No name),

 
B
(iii)     
t
(Farday’s Law),

 

E
(iv)   B  0 J  0 0
t
(Ampere’s law with Maxwell's correction).
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5.3 Boundary conditions on the fields at interfaces
||
||


(a ) Dabove
 Dbelow
  f 

(c) E above  E below











||
||
(b) Babove  Bbelow

(d ) H above  H below  K f  nˆ 
In particular, if there is no free charge or free current at the interface between medium1
and medium 2, then
(a )1 E1   2 E2  0 

(b) B1  B2

and
||  ||

(c) E1  E 2

.
1  || 1  ||
(d ) B1 
B 2  0
1
2

6. Electromagnetic waves in free space
6.1 Poynting Theorem (“work energy theorem of electrodynamics”)
The work necessary to assemble a static charge distribution is
We 
0
E 2 d , where E is the resulting electric field
2 
The work required to get currents going (against the back emf) is
Wm 
1
B 2 d , where B is the resulting magnetic field
2 0 
This suggests that the total energy in the electromagnetic field is
U em 
1 
B2 
2

E


 d .
0
2 
0 
Suppose we have some charge and current configuration which at time t, produces
 
fields E & B . In next instant, dt, the charges moves around a bit. The work is done by
electromagnetic forces acting on these charges in the interval dt.
According to Lorentz Force Law, the work done on a charge ‘q’ is
 
   
 
F  dl  q( E  v  B)  v dt  qE  v dt .
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 
Now q   d and  v  J , so the rate at which work is done on all the charges in a
volume V is
dW
d 1
1 2
1
   0E2 
B  d 
dt
dt V 2 
0 
0
 

  E  B d a ,
S
where S is the surface bounding V.
This is Poynting's theorem; it is the “work energy theorem” of electrodynamics.
The first integral on the right is the total energy stored in the fields, U em .
The second term evidently, represents the rate at which energy is carries out of V, across
its boundary surface, by the electromagnetic fields.
Poynting's theorem says, that, “the work done on the charges by the electromagnetic
force is equal to the decrease in energy stored in the field, less the energy that flowed out
through the surface”.
The energy per unit time, per unit area, transported by the fields is called the Poynting
vector
 1  
S
EB .
0



S  d a is the energy per unit time crossing the infinitesimal surface d a – the energy or
energy flux density.


Momentum density stored in the electromagnetic field is:dem  0 0 S
6.2 Waves in one dimension (Sinusoidal waves)
6.2.1 The wave equation
A wave propagating with speed v in z-direction can be expressed as:
2 f
1 2 f

z 2 v 2 t 2
The most general solution to the wave equation is the sum of a wave to the right
(+z direction) and a wave to the left (-z direction):
f  z, t   g  z  vt   h  z  vt  .
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6.2.2 Terminology
Let us consider a function f  z, t   A cos  k ( z  vt )   
A is the amplitude of the wave (it is positive, and represents the maximum displacement
from equilibrium).
The argument of the cosine is called the phase, and  is the phase constant (normally,
we use a value in the range 0    2 ).
Figure given below shows this function at time t  0 . Notice that at z  vt  
k
, the phase
is zero; let's call this the “central maximum.” If   0 , central maximum passes the origin

at time t  0 ; more generally k is the distance by which the central maximum (and
therefore the entire wave) is “delayed.”
Central
maximum
f  z, 0 

v
A
z
 /k

Finally k is the wave number; it is related to the wavelength  as  
advances by
2
, for when z
k
2
, the cosine executes one complete cycle.
k
As time passes, the entire wave train proceeds to the right, at speed v . Time period of
one complete cycle is T 
2
.
kv
The frequency  (number of oscillations per unit time) is 
1 kv v

 .
T 2 
The angular frequency   2  kv
In terms of angular frequency  , the sinusoidal wave can be represented as
f  z, t   A cos  kz   t    .
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A sinusoidal oscillation of wave number k and angular frequency  traveling to the left
would be written
f  z, t   A cos  kz  t    .
Comparing this with the wave traveling to the right reveals that, in effect, we could
simply switch the sign of k to produce a wave with the same amplitude, phase constant,
frequency, and wavelength, traveling in the opposite direction.
Central
maximum
f  z, 0 

v
 /k
z
6.2.3 Complex notation
In view of Euler's formula,
ei  cos   i sin  ,
the sinusoidal wave f  z, t   A cos  kz   t    can be written as
f  z, t   Re  Aei kz t    ,
where Re   denotes the real part of the complex number  . This invites us to introduce
the complex wave function
 i kz t 
f  z, t   Ae
with the complex amplitude A  Aei absorbing the phase constant.
The actual wave function is the real part of f :
f  z, t   Re  f  z , t   .
The advantage of the complex notation is that exponentials are much easier to manipulate
than sines and cosines.
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6.2.4 Polarization
In longitudinal wave, the displacement from the equilibrium is along the direction of
propagation. Sound waves, which are nothing but compression waves in air, are
longitudinal.
Electromagnetic waves are transverse in nature. In a transverse wave displacement is
perpendicular to the direction of propagation.
There are two dimensions perpendicular to any given line of propagation. Accordingly,
transverse waves occur in two independent state of polarization:
 i  kz t  xˆ ,
“Vertical” polarization fv  z , t   Ae
 i kz t  yˆ ,
“Horizontal” polarization fh  z , t   Ae
 i kz t  nˆ
or along any other direction in the xy plane f  z, t   Ae
The polarization vector n̂ defines the plane of vibration. Because the waves are
transverse, n̂ is perpendicular to the direction of propagation:
nˆ.zˆ  0
6.3 Electromagnetic waves in vacuum


6.3.1 The wave equation for E and B

Write Maxwell’s equations in free space (   0 and J  0 ) then,




2 E
2 B
2
2
  E  0 0 2
 B  0 0 2
and
t
t
Thus, E and B satisfy the wave equation  2 f 
1 2 f
.
v t 2
So, EM waves travels with a speed
v
where 0  4 10 7 N
1
0 0
A
2
 3 108 m / s  c(velocity of light in free space)
2
,  0  8.86  1012 C
Nm 2
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6.3.2 Monochromatic plane waves
Suppose waves are traveling in the z-direction and have no x or y dependence; these are
called plane waves because the fields
are
uniform
perpendicular
over
to
the
every
x
plane
direction
of
propagation.
The plane waves can be represented as:
~
~
~
~
E  z , t   E 0 e i kz t  , B z , t   B0 e i kz t 
c

0
z
0 / c
~
~
where E 0 and B0 are the (complex) y

~
~
amplitudes (the physical fields, of course are the real parts of E and B ).
That is, electromagnetic waves are transverse: the electric and magnetic fields are
perpendicular to the direction of propagation.


k
Also
B 0  ( zˆ  E 0 )

There is nothing special about the z direction; we can generalize the monochromatic

plane waves traveling in an arbitrary direction. The propagation vector or wave vector k
points in the direction of propagation, whose magnitude is the wave number k. The scalar

product k .r is the appropriate generalization of kz, so

 
 i k.r t
 

1  i k .r t  ˆ
1

ˆ
E  r , t   E 0 e
n,
B  r , t   E 0 e
k  nˆ  kˆ  E
c
c


where n̂ is polarization vector.
The actual (real) electric and magnetic fields in a monochromatic plane wave with

propagation vector k and polarization n̂ are

 
E  r , t   E0 cos k .r  t   nˆ ,



 
1
B  r , t   E0 cos k .r  t  
c

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  kˆ  nˆ 
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6.3.3 Energy and Momentum in Electromagnetic Wave
1
1 2
The energy per unit volume stored in electromagnetic field is u    0 E 2 
B 
2
0 
In case of monochromatic plane wave B 2 
E2
 0 0 E 2
2
c
1
1 2
So the electric and magnetic contributions are equal i.e. u E  u B   0 E 2 
B .
2
2 0
u  uE  u B   0 E 2 =  0 E0 2 cos 2 ( kz  wt   ) .
As the wave travels, it carries this energy along with it. The energy flux density (energy
per unit area, per unit time) transported by the fields is given by the Pointing vector
 1  
S
( E  B)
0
For monochromatic plane wave propagating in the z-direction,

S  c 0 E02 cos 2 ( kz  wt   ) zˆ  cu zˆ .
The energy per unit time, per unit area, transported by the wave is therefore uc.
Electromagnetic fields not only carry energy, they also carry momentum. The
momentum density stored in the field is
For monochromatic plane wave,
 1 
 2 S .
c
 1
1
  0 E0 2 cos 2 ( kz  wt   ) zˆ  u zˆ .
c
c
1
u    0 E02 ,
2
^
 1
Average of Poynting vector  S   c 0 E02 z ,
2

^
1
Average momentum density    0 E02 z .
2c
Average energy density
The average power per unit area transported by an electromagnetic wave is called the
intensity
1
I   S   c 0 E02 .
2
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Note:
(a) When light falls on perfect absorber it delivers its momentum to the surface. In a
p   Act ,
time t the momentum transfer is
A
so the radiation pressure (average force per unit area) is
P
c
1 p 1
I
  0 E02  .
A t 2
c
(b) When light falls on perfect reflector, the radiation pressure
P
ct
2I
c
because the momentum changes direction, instead of being absorbed.
6.4 Electromagnetic waves in matter
Inside matter, but in regions where there is no free charge or free current. If the medium


 1 
is linear and homogeneous, D   E and H= B .





2E
2B
2
Now the wave equation inside matter is  E   2 and  B   2 .
t
t
2
Thus EM waves propagate through a linear homogenous medium at a speed
v
1
c

 where n 
 0 0
 n
thus n   r is the index of refraction (since  r  1 for non-magnetic material).

1
1
1
The energy density u    E 2  B 2   u    E02
2
2
 
^
 1  
 1
The Poynting vector S  E  B   S   v E02 z

2


 1
Intensity I   S    vE02
2
Thus in a medium c→ v,  0   and  0  
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6.5 Electromagnetic waves in conductors
Any initial free charge density  f 0  given to conductor dissipate in a characteristic time
  / t
   /  where σ is conductivity and  f  t   e    f  0  .
This reflects the familiar fact that if we put some free charge on conductor, it will flow


out to the edges. The modified wave equation for E and B are,






2 E
E
2 B
B
2
2
 E   2  
and  B  
 
t
t
t
t
~
~ ~
~
~ ~
The admissible plane wave solution is E z , t   E 0 e i k z t  , B z , t   B 0 e i k z t 
“wave number” k is complex. Let k  k  i where k and  are real and imaginary part
of k .
1/2
2
2

 
 
 
 
 1  



k 

1
and



1


2 
 
2 
 










Thus,
E  z, t   E 0 e  z e i kz t  , B  z , t   B 0 e  z ei kz t 
The distance it takes to reduce the amplitude by a factor of
d
1/ 2




1
is called the skin depth (d)
e
1
;

it is a measure of how far the wave penetrates into the conductor.
The real part of k determines the wavelength, the propagation speed, and the index of
refraction:

2

, v ,
k
k
n
ck

Like any complex number, k can be expressed in terms of its modulus and phase:
k  Kei
2
 
where K  k  k 2   2    1  
and   tan 1 k


  
 
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The complex amplitudes E 0  E0 ei E and B 0  B0 ei B are related by
k
Kei
B 0  E 0  B0 ei B 
E0 ei E .


Evidently the electric and magnetic fields are no longer in phase; in fact  B   E   , the
magnetic field lags behind the electric fields.
1/2
2 
B0 K 
  
   1  

E0  
   


E  z , t   E0 e  z cos  kz  t   E  xˆ
Thus, 
B  z, t   B0 e  z cos  kz  t   E    yˆ
Note:
(a) In a poor conductor (    )


2

i.e independent of frequency.

(b) In a very good conductor (    )


,
2
d
1
2
1




 f 
(c) When an electromagnetic wave strikes a perfect conductor (    ) then all waves
are reflected back i.e. E 0 R   E 0 I
and
E 0T  0 .
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7. Applications of Electromagnetic waves
7.1 Reflection and Refraction
7.1.1 Normal incidence
At z  0 , the combined field on the left E I  E R and B I  B R , must join the fields on the right
ET & BT , in accordance with the boundary conditions.
 v v 
  0 R   2 1   0 I
 v2  v1 
 2v2  
 0T  
 0 I
 v1  v2 
,
1
I
x
2
T
v2
v1
Note:
T
I
Reflected wave is in phase if v2  v1 or n2  n1
z
y
and out of phase if v2  v1 or n2  n1 .
R
In terms of indices of refraction the real
v1
R
Interface
amplitudes are
0 R 
n1  n2
0I
n1  n2
,
 0T 
2n1
0 I .
n1  n2
1
Since Intensity I   vE02 , then the ratio of the reflected intensity to the incident intensity is
2
2
2
 E   n n 
I
the Reflection coefficient R  R   0 R    1 2  .
I I  E0 I   n1  n2 
The ratio of the transmitted intensity to the incident intensity is the Transmission coefficient
2
I
 v E 
4n1n2
T  T  2 2  0T  
.
I I  1v1  E0 I   n1  n2 2
 R T 1
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7.1.2 Oblique incidence
In oblique incidence an incoming wave meets the boundary at an arbitrary angle  I . Of course,
normal incidence is really just a special case of oblique incidence with  I  0 .
First Law (law of refraction)

kR
The incident, reflected and transmitted wave vectors

kT
form a plane (called the incidence plane),
R
which also includes normal to the surface.
T
Second law (law of reflection)
z
I
The angle of incidence is equal to the angle of reflection
i.e.
I  R
Plane of Incidence

kI
Third Law: (law of refraction, or Snell’s law)
2
1
sin T n1
 .
sin  I n2
7.1.3 Fresnel’s relation (Parallel and Perpendicular Polarization)
Case I: (Polarization in the plane of incidence)

kR
Reflected and transmitted amplitudes
 
E 0 R  
  

 E0 I and

cos T
where  
cos  I
and
 2
E 0T  
  
 
BR

 E0 I

v
n
 11  1 2
 2v2 2 n1
These are known as Fresnel’s equations.

ER

T
R
T

T

kT
z

I
Notice that transmitted wave is always in phase with
I

I

kI
1
2
the incident one; the reflected wave is either in phase ,
if    , or 180 0 out phase if    .
At Brewster’s angle (  B ) reflected light is completely extinguished when    , or
tan  B 
n2
n1
and  I  B  90o
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When light enters from denser to rarer medium ( n1  n2 ) then after a critical angle ( C )
there is total internal reflection.
sin 900 n1

sin  c
n2

sin  c 
n2
n1
at  C , T  90o .
   

Reflection and Transmission coefficients are R  
   
2
 2 
and T   

   
2
 R T 1
CaseII: (Polarization perpendicular to plane of incidence)

 1  
E0 R  
 1  

 E0 I


 2
E 0T  
 1  
and

 E0 I


kR

ER
R
In this case  B is not possible.
2
 1   
Thus R  
 and
 1   

 2 
T   

 1   

BR

T
T
2
z
I

I
R T 1
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
T

kT

kI

I
1
2
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8. Potential and Field formulation for Time Varying Fields
8.1 Scalar and vector potentials
  
B   A




A
E  V 
t
Now from first Maxwell’s equation (i)
  

 2V 
  A   ...............(1)
t
0


From fourth Maxwell’s equation


  2 
 2 A     
V 
   A  0 0 2       A  0 0



J
0 …............(2)
t 
t 


Equations (1) and (2) contain all the information of Maxwell’s equations. Thus we need

to calculate only four components (one for V and three for A ) instead of six components


(three for E and three for B ).
8.2 Gauge transformation


Suppose we have two sets of potentials, V , A and V , A , which correspond to the




same electric and magnetic fields.
  
Thus
A  A   and V   V   .
  
A  A   , 

It follows that
 
V   V  .
t 


Conclusion: For any old scalar function  , we can add  to A , provided we


from V. None of these will affect the physical quantities E
t


and B . Such changes in V and A are called gauge transformations.
simultaneously subtract
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8.3 Coulomb and Lorentz gauge
 
 A  0 .
Coulomb Gauge reads
 
V
Lorentz Gauge condition is   A   0 0
.
t



  2 


 2 A     
V 
2
Since   A  0 0 2       A  0 0
 A   ,
   0 J and  V 
t 
t 
t
0




Using Lorentz Gauge condition

 2
 2 

 2V

2 A

V



 .
  A  0 0 2   0 J and
0 0
2
t
0
t

The virtue of the Lorentz gauge is that it treats V and A on an equal footing: the same
 2
2





differential operator
0 0
t 2
2
(called the d' Alembertian) occurs in both
equations:
(i) 2 V  

,
0
(ii)
2


A   o J .
9.2 Waveguides
Electromagnetic waves confined to the interior of a hollow pipe or wave guide. The wave


guide is a perfect conductor, E  0 and B  0 inside the material itself, and hence the
boundary conditions at the inner wall are:
||

E  0 and B   0
Free charges and currents will be induced on the surface in such a way as to enforce these
constraints. Let us assume E.M. Waves that propagate inside the waveguide is
represented by:


E  x, y, z, t   E 0 ( x, y)ei  kz t  ,


B  x, y, z, t   B 0 ( x, y)ei  kz t  .
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These electric and magnetic field must satisfies Maxwell's equations in the interior of the
waveguide.Since confined waves are not (in general) transverse; in order to fit the
boundary conditions we shall have to include longitudinal components  Ez and Bz  :


E 0  Ex xˆ  E y yˆ  Ez zˆ, B 0  Bx xˆ  By yˆ  Bz zˆ .
Putting this into Maxwells equations(iii) and (iv) and compare R.H.S and L.H.S
E y Ex
Ez
E
 ikE y  i Bx , ikEx  z  i By ,

 i Bz
y
x
x
y
By Bx
Bz
i
B
i
i
 ikBy   2 Ex , ikBx  z   2 E y ,

  2 Ez
y
c
x
y
c
x
c
Let us rewrite these six equations
(i)
(ii)
E y
Ex
 i Bz ,
y
(iv)
Ez
 ikE y  i Bx ,
y
(v)
x

(iii) ikEx 
Ez
 i By ,
x
By
x

Bx
i
  2 Ez
y
c
Bz
i
 ikBy   2 Ex
y
c
(vi) ikBx 
Bz
i
  2 Ey
x
c
Equation (ii), (iii), (v), and (vi) can be solved for Ex , E y , Bx , and By :
(i) Ex 
 Ez
B 
k
 z 

y 
 k 2  x
i
 c 
(iii) Bx 
2
i
 c 
2
 Bz  Ez 
 2
k

2 

x
c y 
k
(ii) E y 
(iv) By 
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i
 c 
2
i
 c 
2
 Ez
B 
k
 z 

x 
 k 2  y
 Bz  Ez 
 2
k

2 

y
c x 
k
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Finally, we will get coupled differential equation
2
 2

2   
(i)  2  2     k 2  Ez  0
  x  y  c 

2
 2

2   
(ii)  2  2     k 2  Bz  0
  x  y  c 

If Ez  0 we call these TE (“transverse electric”) waves; if Bz  0 they are called TM
(“transverse magnetic”) waves; if both Ez  0 and Bz  0 , we call them TEM waves.
Note: It turns out that TEM waves can not occur in a hollow waveguide.
9.2.1 TE Waves in Rectangular Waveguide
Suppose we have a wave guide of rectangular shape with height a and with b  a  b  , and
we are interested in the propagation of TE waves.
x
a
z
b
The
problem
is
to
solve
y 2
2


2   
2
 2  2     k  Bz  0
  x  y  c 

(since Ez  0 )

subject to boundary condition B   0 .
 m x   n y 
Thus Bz  B0 cos
 cos
 where m  0,1,2..... and n  0,1,2......
 a   b 
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This solution is called the TE mn mode. (The first index is conventionally associated with
the larger dimension.)
2
 
Wave number k is obtaine from equation k x2  kx2     k 2  0 by putting k x and k y .
c
 c    m a   n b   where   2f .
2
So wave number k 
If
2
2
2
   
   
2
2
c  m 2 n 2
  c  m a  n b    mn or f mn 

b  ,


2  a
the wave number is imaginary, and instead of a traveling wave we have exponentially
attenuated fields. For this reason  mn or f mn is called cutoff frequency for the mode in
question.
The lowest cutoff frequency (fundamental mode) for the waveguide occurs for the
mode TE10 :
10 
c
a
c
.
2a
f10 
or
Frequencies less than this will not propagate at all.
Wave number can be written more simply in terms of the cutoff frequency
k
The wave velocity is v 
Group velocity vg 


k
c

1   mn 



2
c
f
1   mn 
f

2
1
dk
1
2
 2   mn
.
c
d
2
c.
2

f
 c 1   mn   c 1   mn   c .

f


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0
Wavelength inside the waveguide g 

1   mn 


377
Characteristic impedanceTE 
 
1   mn 
  
2
2
0

f
1   mn 
f

2
377

 f 
1   mn 
 f 
2
9.2.2 TM Waves in Rectangular Waveguide
The
problem
is
to
solve
2
 2

2   
2


 2
   k  Ez  0
2
  x  y  c 

(since Bz  0 )
||
subject to boundary condition E  0 .
We will get similar expression as we have derived in TE waves. Thus
 m x   n y 
E z  E0 sin 
 sin
 where m  1,2,3..... and n  1,2,3......
 a   b 
This solution is called the TM mn mode. (The first index is conventionally associated
with the larger dimension.)
Formula for cutoff frequency  mn , wave velocity v , group velocity vg and g are same as
TE waves.
w 
Characteristic impedance  TM  377 1   mn 
 w 
The fundamental mode is TM 11 and
f11 
c
2
2
2
 a  b
1
 1
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2
.
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10. Radiation- from moving charges
10.1 Radiation
Accelerating charges and changing current produce electromagnetic waves (radiation).
If the source is localized near the origin, the total power passing out through the spherical
shell of radius r is the integral of pointing vector:
  1
P  r    S .da 
0


r

  E  B .da
Source
The power radiated is the limit of this quantity as r goes to infinity:
Prad  lim P  r 
r 
This is the energy (per unit time) that is transported out to infinity, and never comes back.
10.1.1 Power radiated by a point charge

The electric field E and magnetic field B of an EM wave due to a point charge q, having


acceleration a at any point P (position vector r ) is given by
 qa sin 
 qa sin 
E 
, B
r
r
z
E
(where  is the angle which r makes with z-direction)
 q 2 a 2 sin 2   0 q 2 a 2
S


Thus
r2
16 2 c
 sin 2 
 2
 r

a

rˆ

The total power radiated is evidently
P
r
y
q
x
   q 2 a 2 sin 2 
 q 2a2
P   S .da  0 2  2 r 2 sin  d d  0
6c
16 c
r

The direction of E can be determined from the following rule:


(i) r and E is always perpendicular.

 
(ii) a , r and E lies in one plane.
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