1. ELECTROMAGNETIC WAVES
Total number of questions in this chapter are :
(i) In chapter Examples ....................... 11
(ii) Solved Examples ....................... 06
Total no. of questions ....................... 17

2. 1. BASIC EQUATIONS OF ELECTRICITY AND
MAGNETISM
The whole concept of electricity and magnetism
can be explained by the four basic equations we
have deal so far.
(1)  ds
.
E =
0
Q
 (Gauss law for electrostatic)
(2)  ds
.
B = 0 (Gauss law for magnetism)
(3)  dl
.
B = µ0i (Ampere’s law for Magnetism)
(4)  dl
.
E = 0 (Ampere’s law for electrostaties)
The above stated equation are true for non-time
varying fields
2. FARADAYS LAW FOR TIME VARYING
MAGNETIC FIELD
To understand the concept of faradays law we
consider a circular conducting loop placed in a
region where time dependent magnetic field is
present
x
x
x
x
x
x
x
x
x
x
x
E
E
E
E




e
e
e e
Conducting
loop
Time dependent magnetic
field is switched on at t = 0
From the earlier concept we know that an induced
emf will be produced in the conducting loop due
to which current will flow in the loop.
For current to flow a force must act on the
electron which will move then from static state.
This force cannot be due to magnetic field (since
magnetic force does not act on stationary charge).
Hence this force must be due to an electric field
which has been generated due to changing
Magnetic field.
Note :- This electric field is non conservative in
nature.
Faraday stated this fact in his equation
 dl
.
E = – 




 
dt
d B
3. CONCEPT OF DISPLACEMENT CURRENT
(MODIFIED AMPERE’S LAW)
Maxwell tried to generalis the concept of faradays
law that if changing magnetic field can produce
changing electric field then the reverse should
also be true i.e. changing electric field must
produce magnetic fied.
To understand the concept of displacement
current let us try to understand this experiment
when the switch was closed at t = 0 both the
needles deflected.
V
t = 0
Magnetic Magnetic Needle
(2)
Parallel plate
capacitor
Needle
(1)
•
Deflection of needle (1) is under stood as M.F. is
produced due to current flowing in the wire.
But why did needle 2 deflect? It is lying in between
the two plates of capacitor where there is no
current. This magnetic field between the plates
is due to the changing electric field between the
plates (During charging of capacitor). Hence
maxwell conducted that changing electric field
produces a magnetic field
For Needle (1) Amper’s law
 dl
.
B = µ0ic ..... (1)
For needle (2) Amper’s law
 dl
.
B = µ00 dt
d E

..... (2)
Hence there are two methods of producing M.F.
(a) Due to flow of electron which is known as
conduction current
(b) Due to changing electric field
combining eq. (1) and eq. (2)
 dl
.
B = µ0 










 


dt
d
i E
0
C
Modifield ampere’s law
Note :
dt
d E
0

 is known as displacement current)

3. Displacement current
Ex. 1 A parallel plate capacitor with plate area A
and seperation between the plated d, is
charged by a constant current i, consider a
plane surface of area A/4 parallel to the plates
and drawn symetrically between the plates
what is the displacement current through this
area.
(A) i (B) 2i
(C) i/4 (D) i/2
Sol.(C) Electric field between the plates of the
capacitor is given by
E =
0


or
0
A
q

Flux through the area considered
 =
0
A
q

×
4
A
=
0
4
q

Displacement corrent id = 0 dt
d E

= 0 ×
dt
d








0
4
q
=
4
i
Ex. 2 The charging current for a capacitor is 1 A
then what is the displacement current ?
(A) 1 A (B) 0.5 A
(C) 0 (D) 2 A
Sol. (A) Electric field between the plates is
E = A
Q
0

 E = E. A or
A
Q
0

× A
A
 id = 0 dt
d E

or 0 dt
d








0
Q
 id =
dt
dQ
= i (charging current)
Hence id = 1A Ans.
4. FINAL FORM OF MAXWELL’S EQUATION
(a)  ds
.
E =
0
q

(b)  ds
.
B = 0
(c)  dl
.
E = –
dt
d B

(d)  dl
.
B = µ0 




 


dt
d E
0
I
The above equation is known as maxwell’s
equation for time varying form.
Howover for free space there are no charges and
no conduction current the equations that are
significant.








dt
d
µ
dl
.
B
dt
d
dl
.
E
E
0
0
B
Solving these two differential equation the
equation of electric field and magnetic field that
satisfies these differential equations are obtained
Ey = E0 sin (t – x/c)
Bz = B0 sin (t – x/c)
Here
0
0
B
E
= C and
0
0
µ
1
 = C
5. TRANSEVERSE NATURE OF ELECTRO
MAGNETIC WAVE AND ITS PROPERTIES
5.1 Electromagnetic waves :
The idea of electromagnetic waves was given by
Maxwell and experimental verification was
provided by Hertz and other scientists. A brief
history of electromagnetic waves is as follows:
On the basis of experimental study of
electromagnetic induction, Faraday concluded
that a magnetic field changing with time at a
point produces a time varying electric field at
that point. Maxwell in 1864 pointed out an electric
field changing with time at a point also produces
a time varying magnetic field. The two fields are
mutually perpendicular to each other. This idea
led Maxwell to conclude that the mutually
perpendicular time varying electric and magnetic
fields produce electromagnetic disturbances in
space. These disturbances have the properties
of wave which are called as electromagnetic
waves.
According to Maxwell, the electromagnetic
waves are those waves in which there are
sinusoidal variation of electric and magnetic
field vectors at right angle to each other as
well as right angles to the direction of wave
propagation. An electromagnetic wave is shown
in fig.

4. Z
Y
X
Direction of
propagation
E
B
E
B
E
B
Envelope of magnetic
induction vector
Envelope of electric
intensity vector
The velocity of electromagnetic wave in free space
is given by
c =
)
µ
(
1
0
0 
where µ0 (1 = 1.257 × 10–6 T mA–1) and 0 (=
8.854 × 10–12 C2N–1m–2) are permeability and
permittivity of free space respectively. The velocity
of electromagentic waves in free space is equal
to the velocity of light. Therefore, light is
electromagnetic waves. The electromagnetic
waves are transverse in nature.
5.2 Transverse nature of electromagnetic waves:
We have seen that electromagnetic waves
consists of a sinusoidally verying electric and
magnetic field. These fields act right angles to
each other as well as right angles to the direction
of propagation of waves. These fields are
represented by
E = E0 sin (t – x/c)
and B = B0 sin (t – x/c)
respectively. The two fields combine to constitute
electromagnetic wave. The electromagnetic wave
propagates in space in a direction perpendicular
to the directions of both fields as shown in fig.
The electric field vectors (E) is along Y-axis and
magnetic field vector (B) along Z-axis while the
wave propagation direction is along X-axis. As
both the fields are perpendicular to the direction
of propagation of electromagnetic wave and hence
the electromagnetic waves are transverse in
nature.
6. POYNTING VECTOR (DERIVATION NOT
REQUIRED)
Poynting vector is a vector that describes the
magnitude and direction of energy flow rate.
0
µ
B
E
S


 


S  Poynting vector
The magnitude of poynting vector represents the
rate at which energy flows through a unit surface
area perpendicular to the direction of wave
propogation SI unit J/sm2 or w/m2
Properties of E.M. wave
Ex.3 If E and B be the electric and magnetic field
of E.M. wave then the direction of propogation
of E.M. wave is along the direction.
(A) E (B) B
(C) E × B (D) B × E
Sol. (C)
Correct option is (C)
Ex.4 Which of the following pairs of space and
time varying E and B fields would generate a
plane electromagnetic wave travelling in (–Z)
direction-
(A) Ex, By (B) Ey, Bx
(C) Ex, By (D) Ey, Bx
Sol. (B)
Since the direction of propogation of EM wave
is given by E × B hence correct option is (b)
(
^
j ×
^
i = –
^
k )
Ex.5 Choose the wrong statement for E.M. wave.
They-
(A) Are Transverse
(B) Travels in free space with the speed of
light
(C) Are produced by accelerated charges
(D) Travels with same speed in all medium
Sol. (D)
Speed of E.M. wave =
r
r
0 µ
µ
1


in
medium hence it will travell with different
speed in different medium.

5. 7. ENERGY DENSITY AND INTENSITY
We know that electric and magnetic field have
energy and since EM wave have both these
components hence it carries energy with it.
We know that energy density associated with
E.F. =
2
1
0E2
We know that energy density associated with
M.F. =
0
2
µ
2
B
Thus total energy of EM wave is given by
U = UE + UB or
2
1
0E2 +
0
2
µ
2
B
putting the values of E.F. and M.F.
U =
2
1
0E0
2
sin2 (t – x/c) +
0
2
2
0
µ
2
)
c
/
x
t
(
sin
B 

If we take the average value over a long period of
time
Av
2
c
x
t
sin 












 =
2
1

2
0
0
2
0
av B
4µ
1
E
4
1
U 


The above equation can also be written as
Uav =
2
E2
0
0

or Uav =
0
2
0
µ
2
B
Intensity :
Energy crossing per unit area per unit time
perpendicular to the direction of propogation is
called intensity of wave.
Energy contianed in the volume :
x
C T

A
E.M. wave
U = Uav × vol =
2
1
0E2 (ACT)
intensity =
T
A
U

=
2
C
E2
0


2
0
0
2
0 B
2µ
1
or
CE
2
1
0
I 

Energy density and intensity
Ex.6 The intensity of light from a source is
500/W/m2. Find the amplitude of electric
field in this wave-
(A) 3 × 102 N/C (B) 2 3 × 102 N/C
(C)
2
3
× 102 N/C (D) 2 3 ×101 N/C
Sol. (B)
I =
2
1
0CE0
2
E0 = C
I
2
0
 or
3
10
36
10
500
2
8
9









E0 = 2 3 × 102 N/C
Hence correct Ans is (B)
Ex.7 A point source of 2 watts is radiating
uniformly in all direction in vaccum. Find the
amplitude of electric field at a distance 2m
from it-
(A) 3 × 10–4 (B) 3 × 10–2
(C) 3 × 10–4 (D) 3 × 10–2
Sol. (C)
I =
A
P
or
4
4
2


=

8
1
W/m2
I =
2
1
0 E0
2
c
E0 = C
I
2
0
 or 8
10
3
8
36
1
2






3 × 10–4 N/C
Hence correct ans in (C)
Ex.8 In a EM wave the amplitude of electric field
is 10 V/m. The frequency of wave is
5 × 104 Hz. The wave is propogating along Z-
axis. Then the average energy density of
Magnetic field is-
(A) 2.21 × 10–10 J/m3
(B) 2.21 × 10–8 J/m3
(C) 2 × 10–8 J/m3
(D) 2 × 10–10 J/m3
Sol. (A)
UB =
0
2
0
µ
4
B
Also
0
0
B
E
= C  B0 =
C
E0

6. Hence B0 =
0
0
0
µ
1
E

  UB =
0
0
0
2
0
µ
4
µ
E 
or
4
10
84
.
8
100 12



UB = 2.21 × 10–10 J/m3
Hence correct ans is (A)
8. ELECTROMAGNETIC SPECTRUM
Orderly arrangement of electromagnetic
radiations according to their wavelength or
frequency is called as electromagnetic
spectrum.
The electromagnetic spectrum encompasses
radiowaves, microwaves, infrared rays, visible
light, ultraviolet rays, X-rays and gamma rays fig.
shown the most commonly enecountered.
102
107
10
108
01
109
10–1
1010
10–2
1011
10–3
750 700 650 600 550 500 450 400 nm
1012
10–4
1013
10–5
1014
10–6
1015
10–7
1016
10–8
1017
10–9
1018
10–10
1019
10–11
1020
10–12
1021
10–13
V.H.
Radio
wave
Radio
wave
Micro wave Infra red
Ultra
violet Y rays
X rays
Red Orange Yellow Green Blue Violet
Visible light Frequency in Hz
Wavelength in nm
Name Frequency range (Hz) Wavelength range (m)
-ray 5 × 1020 – 3 × 1019 6 × 10–13 – 1 × 10–10
X-ray 3 × 1019 – 1 × 1016 1 × 10–10 – 3 × 10–8
Ultraviolet 1 ×1016 – 8 × 1014 3 × 10–8 – 4 × 10–7
Visible light 8 × 1014 – 4 × 1014 4 × 10–7 – 8 × 10–7
Infra-red 4 × 1014 – 1 × 1013 8 × 10–7 – 3 × 10–5
Micro-waves 3 × 1011 – 1 × 109 1 × 10–3 – 3 × 10–1
Ultra high radio frequencies 3 × 109 – 3 × 108 1 × 10–1 – 1
Very high Radio frequencies 3 × 109 – 3 × 107 1 – 10
Radio frequencies 3 × 107 – 3 × 104 10 – 104

7. VISIBLE SPECTRUM
Colour Wavelength (m)
Violet 4 × 10–7 – 4.5 × 10–7
Blue 4.5 × 10–7 – 5 × 10–7
Green 5.5 × 10–7 – 5.7 × 10–7
Yellow 5.7 × 107 – 5.9 × 10–7
Orange 5.9 × 10–7 – 6.2 × 10–7
Red 6.2 × 10–7 – 7.5 × 10–7
9. PROPOGATION OF EM WAVE IN ATMOSPHERE
Electromagnetic wave of frequency ranging from
a few KHz to a few MHz are called radio waves
the various frequency ranges used in radiowave
of microwave communication system are
(i) Medium Freq. Band (MF) 300 to 3000 KHz
(ii) High Freq. Band (HF) 3 to 30 MHz
(iii) Very High Freq. Band (VHF) 30 to 300 MHz
(iv) Ultra High Freq. Band (UHF) 300 to 3000 MHz
Radiowaves emitted from a transmilter antenna
can reach to the receiver antenna by any one of
the following modes of propogation
(a) Ground wave propogation
(b) Sky wave propogation
(c) Space wave progation
(a) Ground wave propogation :
The radiowaves which progress along the surface
of the earth are known as ground wave or surface
wave ground wave can have frequency upto 1500
KHz. Not suitable for long distance
communication
(b) Sky wave propogation :
The propogation of radio waves from one place to
another place via reflection from ionosphere is
called sky wave propogation.
Sky waves are suitable for progation of signals
having freq. band from 1500 KHz to 40 MHz.
Suitable for long distance communication up to
4000 km.
(c) Space wave communication :
The space wave communication can travell
through atmosphere from transmitter antenna to
receiver antena either directly or by reflection from
earth’s troposphere
space ware propogation is utilized in very high
frequency band (30 MHz to 300 MHz), Ultra high
frequency band etc.
Space wave communication is utilized in TV and
radar communication
* Relation between height of antenna and
distance up to which signal can be recieved
directly
Let PQ be a TV transmitter antenna of hight h
located on earth surface. The signal transmitted
by this antenna can be recieved within a circle of
radius QS or TQ (As hight of antenna << Radius
of earth)
Antenna
Earth
R
T
h
R
R
S
P
Q
O
From right angletriangle OTQ We have
(OQ)2 = (OT)2 + (QT)2 ..... (1)
Here,
QT = QS = d, PQ = h
 OQ = OP + PQ = (R + h)
Substituting in equation (1)
(R + h)2 = R2 + d2
Or d2 = h2 + 2Rh
d2 2RH (R >> h)
d = Rh
2

8. The area covered for TV transmission
= d2 or  (2hR)
Population covered = Population density × Area
covered.
Electromagnetic spectrum
Ex.9 The height of transmitting antenna for
transmitting a TV telecast so that it covers a
radius of 128 kM is (Radius of earth,
R = 6.4 × 106 m)
(A) 128 m (B) 1028 m
(C) 1280 m (D) 1208 m
Sol. (C)
The height of transmitting anterna is given by
h =
R
2
d2
= 6
2
3
10
4
.
6
2
)
10
128
(



= 1280 m
Hence correct ans is C
Ex.10 Tv transmission is possible by-
(A) Ground wave (B) Sky wave
(C) space wave (D) None
Sol. (C)
Ex.11 A TV tower has a hight of 100 m. The average
population density arround the tower is 1000
person/ Km2 and radius of earth is 6.4 × 106.
the population covered is-
(A) 4 × 106 (B) 6 × 104
(C) 8 × 106 (D) 16 × 104
Sol. (A)
The distance ‘d’ upto which TV transmission
can be viewed is given by
d = hR
2
Area is which transmission is viewed
A = d2 =  × 2hR = 2hR
A = 2 × 3.14 × 100 × 6.4 × 106
 Population = (P.D) × Area = 4 × 106 person
Hence correct ans is (A)

9. SOLVED EXAMPLES
Ex.1 Elecromagnetic waves travel in a medium with
a speed of 2 × 108 m/s. The relative
permeability of the medium is 1. What is the
relative permittivity of the medium-
Sol. The speed of electromagnetic waves and in a
medium is given by
 =
)
µ
(
1

Where µ and are absolute permeability and
absolute permittivity of the medium.
We know that, µ = µ0µr and = 0r. Hence
=
 
r
0
r
0 .
µ
µ
1


=
 
0
0
µ
1

×
 
r
r
µ
1

or v =
 
r
r
µ
c

or r =
)
µ
(
c
r
2
2

 r =
1
)
10
2
(
)
10
3
(
2
8
2
8



= 2.25
Ex.2 Fig. shows a capacitor made of two circular
plates each of radius 12 cm and separated
by 5.0 mm. The capacitor is being charged
by an external source (not shown in the
figure). The charging current is constant and
equal to 0.15 A.
(a) Calculate the capacitance and the rate of
change of potential difference between
the plates
(b) Obtain the displacement current across
the plates.
(c) Is Kirchhoff’s first rule valid at each plate
of the capacitor ?
Given 0 = 8.85 × 10–12 C2 N–1 m–2
Sol. Here, R = 12 cm = 0.12 m,
d = 5.0 mm = 5 × 10–3 m.,
I = 0.15 A,
0 = 8.85 × 10–12 C2N–1m–2
Area, A = R2 = 3.14 × (0.12)2m2 ]
(a) We know that capacity of a parallel plate
capacitor is given by
C =
d
A
0

= 3
2
12
10
5
)
12
.
0
(
14
.
3
10
85
.
8






= 80.1 × 10–12 F = 80.1 pF
Now q = CV
or
dt
dq
= C ×
dt
dV
or I = C
dt
dV
(I =
dt
dq
)
or
dt
dV
=
C
I
= 12
10
1
.
80
15
.
0


= 1.87 × 109 Vs–1
(b) Displacement current is equal to
conduction current i.e. 0.15 A.
(c) Yes, Kirchhoff’s first rule is valid at each
plate of the capacitor provided. We take the
current to be the sum of the conduction and
displacement currents.
Ex.3 A magnetic field in a plane electromagnetic
wave is given by
By = 2 × 10–7 sin(0.5 × 103x + 1.5 × 1011t)
tesla.
(a) What is wavelength and frequency of the
wave ?
(b) Write down an expression for the electric
field. (x is in metre and t in second.)
Sol. Given
By = 2 × 10–7 sin (0.5 × 103x + 1.5 × 1011)
(a) Comparing it with a standard equation for
a progresive wave travelling along the negative
direction of x-axis is
y = r sin


2
(x + vt)
= r sin 









 vt
2
x
2
= r sin 









vt
2
x
2
We have,

x
2
= 0.5 × 103 x
or  = 3
10
5
.
0
2


= 12.56 × 10–3x
= 12.56 mm
2 = 1.5 × 1011

10.  =


2
10
5
.
1 11
= 23.9 × 109 Hz = 23.9 Hz
(b) E0 = cB0 = (3 × 108) ×(2 × 10–7)
= 60 Vm–1
 Ez = E0 sin


2
(x + vt)
= E0sin 




 



T
t
2
x
2
or Ez = 60 sin(0.5 × 103 x + 1.5 × 1011 t) V/m.
Ex.4 The electric field of a plane electromagnetic
wave in vacuum is represented by
Ex = 0, Ey = 0.5 cos[2 × 108 (t – x/c)]
and Ez = 0
(a) What is the direction of propagation of
electromagnetic wave ?
(b) Determine the wavelength of the wave.
(c) Compute the component of associated
magnetic field
Sol. (a) The given equation
Ey = 0.5 cos[2 × 108 (t – x/c)] ..... (1)
indicates that the electromagnetic waves are
propagating along the positive direction of
X-axis.
(b) The standard equation of electromagnetic
wave is given by
Ey = E0 cos(t – x/c) ..... (2)
Comparing the given eq. (1) with the standard
eq. (2), we get
 = 2 × 108
or 2 = 2 × 108
  = 108 per second
Now,  =

c
= 8
8
10
10
3
= 3 m
(c) Here, the wave is propagating along
X-axis and electric field is along Y-axis and
hence the magnetic field must be along
Z-axis. So the components of associated
magnetic field are :
Bx = 0, By = 0
and Bz = B0 cos 














c
x
t
10
2 8
or Bz =
c
E0
cos 














c
x
t
10
2 8
or Bz = 8
10
3
5
.
0

cos 














c
x
t
10
2 8
Ex.5 A light beam travelling in the X-direction is
described by the electric field
Ey = (300 V/m) sin(t – x/c). An electron is
constrained to move along the Y-direction with
a speed of 2.0 × 107 m/s. Find the maximum
electric force and the maximum magnetic
force on the electron.
Sol. The maximum value of magnetic field (B0) is
given by
B0 =
c
E0
= 8
10
3
300

= 10–6 tesla
The magnetic field will be along Z-axis
The maximum magnetic force on the electron
is given by
Fe = qE0 = (1.6 × 10–19) × 300
= 4.8×10–17 N
The maximum magnetic force on teh electron
is
Fb = |q (v × B)| = q  B0
= (1.6 × 10–19) × (2.0 × 107) × (10–6)
= 3.2 × 10–18 N
Ex.6 The TV transmission tower at a particular
staion has a height of 160 m.
(a) What is its coverage rang ?
(b) How much population is covered by
transmission, if the average population
density around the tower is 1200/km2.
(c) By how much the height of tower be
increased to double its coverage range ?
Sol. (a) We know that
d = )
hR
2
(
= )
10
4
.
6
(
160
2 6



= 45255 m
(b) Population covered = d2 × population
density
= 3.14 × (45.255)2 × 1200
= 77.29 lakh
(c) In this case, d’ = )
R
h
2
(  = 2d
 )
R
h
2
(  = 2 × )
R
h
2
( 
h = 4h = 4 × 160 = 640 m
Increase in height of tower = h– h
= 640 – 160 = 480 m.