Unit 3 Electromagnetic Waves in Vacuum

UNIT 3
ELECTROMAGNETIC
WAVES IN VACUUM
Structure
3.1 Introduction Electromagnetic Plane Wave
Expected Learning Outcomes Travelling in the k̂ Direction
3.2 Electromagnetic Waves in Vacuum Electromagnetic Wave which is
Electromagnetic Plane Waves Not Plane
Electromagnetic Plane Polarised 3.3 Summary
Plane Waves 3.4 Terminal Questions
Complex Notation 3.5 Solutions and Answers

3.1 INTRODUCTION
A momentous consequence of Maxwell’s equations (obtained in Unit 1) is that
electric and magnetic fields satisfy the three-dimensional wave equation. The
waves generated by time-varying electric and magnetic fields are called
electromagnetic waves. Once the e.m. waves are generated by moving
charges, they can travel away all by themselves, even long after all charges
stop moving. These waves do not require any medium for their propagation
and their speed depends on constants that emerge purely from electrostatic
and magnetostatic experiments. This speed turns out to be equal to the speed
of light, indicating that light is an electromagnetic wave.
The demonstration that light is an electromagnetic wave brought about a
highly satisfactory unification of electrodynamics and optics. Visible light has
frequencies in a very narrow range. Understanding of electromagnetic waves
opened a window for detecting and creating electromagnetic waves in a wide
range of wavelengths ranging from 10 12 m for gamma rays to 10 5 m for radio
waves.
A proper understanding of electromagnetic waves and its propagation in free
space (vacuum) and in different media has played a crucial role in creation of
our present knowledge about various aspects of the Universe from the
smallest to the largest scales. It is electromagnetic waves of various
frequencies that bring us information from distant stars as well as about
structure of atoms. A grand inference is that the laws governing Nature are the
same everywhere in this vast Universe. The microwave background radiation 79
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
and its detailed distribution provide strong experimental constraints on
cosmological models.
Much of our modern technological life depends on our ability to detect and
generate electromagnetic waves of different frequencies. Some examples of
commonplace gadgets and conveniences that depend on electromagnetic
waves are: X-rays, radios, televisions, microwave ovens, remote-control
devices, satellite communication, internet, mobile phones etc. It will not be an
exaggeration to say that the difference between the lives of a person today, as
compared to a person two century ago is largely due to our understanding of
electromagnetic waves.
In this Unit, you will study the propagation of plane electromagnetic waves in
vacuum. In general, the solutions of the three-dimensional wave equation can
be quite complex. However, if we assume that the electric and magnetic fields
depend on one coordinate only, we get plane wave solutions, which satisfy the
one-dimensional wave equation. In Sec. 3.2, you will learn to obtain one-
dimensional wave equations and discover the characteristics of its plane wave
solutions. The solutions in this case are simple, with simple physical
interpretation. Any solution of the three-dimensional wave equation can be
expressed as a superposition of plane waves in different directions. You will
learn about one of the important consequences of Maxwell’s equations. In a
plane electromagnetic wave, the electric field and the magnetic field, at any
point and at any time, are perpendicular to the direction of propagation of the
wave and to each other. However, mutual orthogonality of the electric field,
magnetic field and the direction of propagation need not hold if the
electromagnetic wave is not plane. You will also learn to compute energy
density, Poynting vector and momentum density of electromagnetic fields in
electromagnetic waves and their interpretation for progressive waves.
In the next unit, you will deduce the electromagnetic wave equation in different
media before discussing electromagnetic wave propagation in such media.
Expected Learning Outcomes
After studying this unit, you should be able to:
™ use Maxwell’s equations to show that electric and magnetic fields satisfy
the three-dimensional wave equation;
™ compute the phase velocity of the electromagnetic wave;
™ show that for a plane wave, at any point and at any time, the electric field
vector and the magnetic field vector are perpendicular to the direction of
propagation of the wave and to each other;
™ show that the ratio of the magnitude of the electric field to the magnitude
of the electric field is equal to the speed of light;
™ compute the energy density, Poynting vector and momentum density for a
plane electromagnetic wave; and
™ understand that if the wave is not plane, the electric field and the magnetic
field need not be perpendicular to the direction of propagation and to
80 each other.
Unit 3 Electromagnetic Waves in Vacuum

3.2 ELECTROMAGNETIC WAVES IN VACUUM

One of the great successes of Maxwell’s equations was that they predicted the
existence of electromagnetic waves in 1864 long before these were generated or
detected in experiments. Almost 25 years later, it was Heinrich Hertz who first
generated and detected these waves experimentally in 1887. Maxwell also
predicted that all electromagnetic waves would travel at a speed that was very
close to the speed of visible light in the air. And Maxwell correctly asserted that
visible light was an electromagnetic wave. Now we know that radio waves,
infrared, visible, ultraviolet, X-rays and gamma rays are all electromagnetic
waves differing only in frequency.

To show how Maxwell’s equations led to the prediction of electromagnetic waves,

we use these equations to derive an equation which is just the wave equation
and understand
& & its physical meaning. In vacuum (free space), where
U 0, j 0 , the Maxwell’s equations reduce to
& &
’.E 0 (3.1)
& &
’.B 0 (3.2)
&
& & wB
’uE  (3.3)
wt
&
& & wE
’ u B P0H0 (3.4)
wt
If we take curl of Eq. (3.3) and use Eq.(3.4), we can write,
& &
& & & w & & w§ wE · w 2E
’ u (’ u E )  (’ u B)  ¨¨ P0H0 ¸ P0H0 (3.5)
wt wt © wt ¸¹ wt 2

Using Eq. (3.1), we can write,

& & & & & & & &
’ u (’ u E ) ’(’.E )  ’2E ’2E (3.6)

Eqs. (3.5) and (3.6) imply

&
& w 2E
’2E P0 H 0 (3.7)
wt 2 Recall the vector
Example 3.1 identity &
& &
’ u (’ u A)
& & & & &
& w 2B ’( ’. A )  ’ 2 A
Show that ’2B P0 H 0 .
wt 2
&
& & wE & & /
Solution : We know that ’ u B P0H0 if j 0.
wt
& &
& & & w & & w wB w 2B
Therefore, ’ u (’ u B ) P 0 H 0 (’ u E ) P 0 H 0 (  ) P0H0
wt wt wt wt 2
& & & & & & & & & &
But ’ u (’ u B ) ’(’.B )  ’2B ’2B (’.B 0)
&
& w 2B
Hence, 2
’ B P0 H0
wt 2

Expressed in Cartesian coordinates, Eq. (3.7) is 81

Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
& & & &
w 2 E w 2E w 2 E w 2E
  P0H0 (3.8)
wx 2 wy 2 wz 2 wt 2
&
If E do not depend on y and z coordinates, Eq. (3.8) gives,
& &
w 2E w 2E
P0 H 0 (3.9)
wx 2 wt 2
& &
w 2B w 2B
Similarly, P0 H0 (3.10)
wx 2 wt 2
Now, recall from your UG physics course that the one-dimensional wave
equation is given by
w 2\ 1 w 2\
(3.11)
wx 2 v 2 wt 2
This equation describes a wave travelling with speed v. The most general
solution of Eq. (3.11) is
\ f ( x  vt )  g ( x  vt ) (3.12)

The first term on the RHS of Eq. (3.12) represents a wave travelling in the
direction of positive x whereas the second term represents a wave travelling in
the direction of negative x. In a wave what travels is a constant phase x – vt.
The phase will have the same value at (x, t) and ( x  dx, t  dt ) if
dx
x  vt x  dx  v (t  dt ) , i.e., if v , showing that v is the phase velocity of
dt
the wave.
If we compare Eqs. (3.9) and (3.10) with Eq. (3.11), we find that Eqs. (3.9) and
(3.10) are wave equations for time-varying & electric
& and magnetic fields which
propagate like waves in space. Thus, if E , B do not depend on y and z, they
satisfy the one-dimensional wave equation. We write the most general solution
of Eqs. (3.9) and (3.10) on the basis of Eq. (3.12) as
& & &
E f ( x  vt )  g ( x  vt ) (3.13)
& & &
B F ( x  vt )  G( x  vt ) (3.14)

In this way, we find that Maxwell’s equations suggest the possibility of self-
sustaining travelling electromagnetic fields, which travel in space like waves.
This is how the concept of an ‘electromagnetic wave’ was born. Now we think
of electromagnetic waves as structures consisting of electric and magnetic
fields that travel freely through vacuum.
Further, comparing Eqs. (3.9) and (3.10) again with Eq. (3.11), we get the
speed of the electromagnetic waves as
1
v
P0H0
&
As E has the same value at all points of a plane at x constant, the solution
&
[Eq. (3.13)] is called a plane wave solution and the same holds true for B . The
& &
term ‘plane’ is meant to indicate that field vectors E and B at each point in
space lie in a plane. Also the planes at any two different points are parallel to
82 each other.
Unit 3 Electromagnetic Waves in Vacuum
&
In general, E satisfies Eq. (3.7), which being a generalisation of the one-
dimensional wave equation, is called the three-dimensional wave equation.
Example 3.2
& & & 1
Show that E f ( x  vt )  g ( x  vt ) , where v , is a solution of
P0 H0
& &
w 2E w 2E
P0 H 0 .
wx 2 wt 2
& & &
Solution : E f ( x  vt )  g ( x  vt ) /

Then, & &

wE & & w 2E & &
f ' ( x  vt )  g ' ( x  vt ) Ÿ f ' ' ( x  vt )  g ' ' ( x  vt )
wx wx 2
& &
wE & & w 2E & &
vf ' ( x  vt )  vg ' ( x  vt ) Ÿ v 2f ' ' ( x  vt )  v 2g ' ' ( x  vt )
wt wt 2
& & &
w 2E 1 w 2E w 2E
? P0H0
wx 2 v 2 wt 2 wt 2
& & & 1
Hence, E f ( x  vt )  g ( x  vt ) , where v , is a solution of
P0 H0
& &
w 2E w 2E
P0 H 0 .
wx 2 wt 2

Example 3.3
1
Show that the value of c (the speed of light).
P0 H0

Solution : We know from electrostatics that /

(Coulomb)2
H0 8.85 u 10 12 . It Is for this value of H 0 , that the mutual
Newton (Meter )2
force on two charges, each having magnitude 1 Coulomb, separated by a
1
distance of 1 Meter in free space, will be Newton . The constant H 0 is
4SH0
called the permittivity of free space.
Newton
We know from magnetostatics that P0 4S u 10 7. It Is for this
( Ampere )2
value of P 0 , that the mutual force per Meter on two parallel infinite wires, each
carrying current of magnitude 1 Ampere, separated by a distance of 1 Meter in
P
free space, will be 0 Newton / Meter . The constant P 0 is called the
2S
permeability of free space. Therefore,
 21  21
1 § Newton · § Coulomb2 ·
¨ 4S u 10 7 ¸ ¨¨ 8.85 u 10 12 ¸
P0H0 © Ampere 2 ¹ © Newton(Meter )2 ¸¹

Meter ª Ampere Coulomb º

3 u 108
Second «¬ Second »¼

= c (the speed of light)

83
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
3.2.1 Electromagnetic Plane Waves
& &
For a plane electromagnetic wave travelling in the x-direction, E and B do not
depend on y and z coordinates. Therefore,

& & wE x wEy wEz wE x

Eq. (3.1) ’.E 0Ÿ   0Ÿ 0 (3.15)
wx wy wz wx

& & wBx wBy wBz wBx

Eq. (3.2) ’.B 0Ÿ   0Ÿ 0 (3.16)
wx wy wz wx
&
& & wB wBx § wE wE y ·
Eq. (3.3) ’uE  Ÿ ¨ z  ¸ 0 (3.17)
wt wt © wy wz ¹
&
& & wE wE x 1 § wBz wBy ·
Eq. (3.4) ’uB P0H0 Ÿ ¨  ¸ 0 (3.18)
wt wt P0H0 © wy wz ¹
& &
So, for a plane electromagnetic wave travelling in the x-direction, E , B, and
therefore, E x , B x , do not depend on y and z. Also, Eqs. (3.15) to (3.18) show
that E x , Bx also do not depend on x and t. This implies that E x and B x are
constant. As constants can always be added to any solution of Maxwell’s
equations, we take these constant to be zero, so that E x Bx 0. Thus, we
find that the components of the E and B fields of the electromagnetic wave in
the direction of propagation are zero. This implies that the direction of
propagation of an electromagnetic wave is perpendicular to the directions of
the E and B fields that constitute it. In other words, for a plane wave,
E and B are perpendicular to the direction of propagation. Hence,
electromagnetic plane waves are transverse. Also, the directions of the
electric field component and magnetic field component of the electromagnetic
&
wave
& are perpendicular to each other. Fig. 3.1 depicts the orientation of E and
B fields in a plane electromagnetic wave travelling along the x-direction.

y
&
E

&
B
t 0 x
z
y
&
E

&
B x
z
t t1

Fig. 3.1: Travelling plane electromagnetic wave.

84
Unit 3 Electromagnetic Waves in Vacuum
Note that Fig. 3.1 shows the travelling electromagnetic wave at two different
times t and t 1 ( t  ' t ). As time passes, the entire pattern slides to the right
because ( x  c t ) has the same value at ( x  'x ) and t  ' t as it had at x
and t, provided that 'x c ' t . This is an example of a plane electromagnetic
wave. In other words, we have a plane electromagnetic wave travelling with a
constant speed c in the x-direction.

After having studied the characteristics of plane electromagnetic waves, let us

now find out how to describe plane polarised plane electromagnetic waves.

3.2.2 Electromagnetic Plane Polarised Plane Waves

You may recall that the polarised waves are those that have vibrations in one
plane. Plane-polarised light is composed of waves where the direction of time-
varying electric (and magnetic) field is in the same plane for each wave. For a
plane electromagnetic wave travelling in the positive x-direction, from
Eq. (3.13), we get
& & & x & x
E ( x, t ) f ( x  ct ) f §¨ 0  c §¨ t  ·¸ ·¸ E §¨ 0, t  ·¸ (3.19)
© © c ¹¹ © c¹

where,
& c is the speed of electromagnetic waves in vacuum. The electric field
E at time t, on a plane characterized by x, will be the same as on the plane
x 0 at an earlier time. How much earlier? The answer is: The time taken by
the wave to travel distance x, which is ( x c ).

As the electric field is perpendicular to the direction of propagation, we can

write
&
E (0, t ) E y (t )yˆ  E z (t )zˆ (3.20)

Then, from Eq. (3.19), we can write

& § x· § x·
E ( x, t ) E y ¨ t  ¸ yˆ  Ez ¨ t  ¸zˆ (3.21)
© c¹ © c¹

In general, the direction of the electric field can change in the y-z plane with
time and position. Such an electromagnetic wave is called unpolarised.
However, if the ratio E y (t ) : Ez (t ) is constant, the direction of the electric field
will remain fixed; it will not depend on time and position. In this case, the
electromagnetic plane wave is called plane polarised.

For a plane polarised electromagnetic plane wave travelling in the x-direction,

it is possible to choose the coordinate system in such a way that the direction
of the electric field is along the ŷ direction, so that
&
E (0, t ) E y t yˆ (3.22)
& x
and therefore E ( x, t ) E y §¨ t  ·¸ yˆ (3.23)
© c¹

Now, let us consider a case of great importance, that of monochromatic

sinusoidal waves propagating in x-direction with electric field in the y-direction:
Ey t E0 sin(Zt  G) (3.24) 85
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
where, E 0 is magnitude of the electric field and Z is the frequency of the
wave and G is phase. Note that Eq. (3.24) describes a simple harmonic
oscillation of the electric field at each point on the plane x 0. Substituting
Eq. (3.24) in Eq. (3.22), we get,
&
E (0, t ) E y t yˆ E0 sin( Zt  G)yˆ (3.25)

And Eq. (3.23) takes the form

& x x
E ( x, t ) E y §¨ t  ·¸ yˆ E 0 sin§¨ Z §¨ t  ·¸  G ·¸ yˆ
© c¹ © © c¹ ¹
E0 sin Zt  kx  G yˆ (3.26)

Z
where, k (3.27)
c

The electric field in Eq. (3.26) completely determines the corresponding

magnetic field of the plane wave. To write the magnetic field, recall that the
y-component of the Faraday’s law [Eq. (3.3)] is
wBy wE wE
§¨ x  z ·¸ 0 (3.28)
wt © wz wx ¹

So, we can write, By ( x,t ) h( x ) (3.29)

Further, Eq. (3.4) gives

& & wE z § wB y wB x ·
(’ u B ) z P0H0 Ÿ ¨¨  ¸¸ 0 Ÿ h' ( x ) 0 (3.30)
wt © wx wy ¹

Hence, By ( x,t ) h( x ) 0 (3.31)

You may ask: Why do we take h( x ) 0 ? Eq. (3.28) implies only that h(x ) is
constant. This constant can be taken to be zero as constant fields are of no
interest in discussions about electromagnetic waves. Constant fields can
always be added to electric and magnetic fields without affecting the Maxwell’s
equations.

Similarly, applying the z-component of the Faraday’s law [Eq. (3.3)], we get,

wBz § wE y wE x · w
¨  ¸ E0 sin(Zt  kx  G)
wt © wx wy ¹ wx

kE0 cos( Zt  kx  G) (3.32)

Therefore,
k
Bz ( x, t ) E0 sin(Zt  kx  G)  h( x ) (3.33)
Z

And, the y-component of generalised Ampere’s law [Eq. (3.4)] gives

& & wE y
(’ u B ) y P0H0
wt
§ wB x  wB z · w
Ÿ ¨ ¸ P0H0 E 0 sin( Zt  kx  G )
86 © wz wx ¹ wt
Unit 3 Electromagnetic Waves in Vacuum
w §k ·
Ÿ  ¨ E0 sin(Zt  kx  G)  h( x ) ¸ P 0 H 0 ZE0 cos(Zt  kx  G)
wx © Z ¹

k2
Ÿ E 0 cos(Zt  kx  G )  h c( x ) P 0 H 0 ZE 0 cos(Zt  kx  G )
Z
(3.34)

Z 1
Since c (3.35)
k P0H0
Eq. (3.34) gives h c( x ) 0 , implying that h(x) is constant. This constant can be
taken to be zero because constant fields are of no interest in discussions
about wave motion. So, Eq. (3.33) reduces to
k
Bz ( x, t ) E0 sin(Zt  kx  G) (3.36)
Z
Thus, the magnetic field of the one-dimensional plane electromagnetic wave
propagating along x-direction having electric field polarised along the y-
direction is
& k 1
B( x, t ) E0 sin(Zt  kx  G) zˆ E0 sin(Zt  kx  G) zˆ While you study
Z c physics, you should
develop a habit of
B 0 sin( Zt  kx  G ) zˆ (3.37)
checking the
E0 dimensions of all
where B0 (3.38) the terms in
c
equations. This
Comparing Eqs. (3.26) and (3.37), we see that the electric field and magnetic helps in detecting
field in a plane polarised electromagnetic plane wave are perpendicular to errors. It also helps
each other, they are in phase and in reinforcing
& & definitions and
E ( x, t ) c B( x, t ) (3.39) concepts. This is
particularly useful in
To concretise your understanding of what you have studied above, go though electrodynamics in
the example below. which there are
many systems of
Example 3.4 units. Recall that
the dimensions of
the electric and
For
& the plane polarised electromagnetic plane wave described by
& magnetic fields can
E( x, t ) E0 sin Zt  kx  G yˆ , B( x, t ) B0 sin( Zt  kx  G ) zˆ, compute a) the be obtained using
electromagnetic energy density; b) the Poynting vector; c) intensity (average the Lorentz force
electromagnetic energy crossing unit area per unit time); d) electromagnetic law. Thus the
momentum density; e) electromagnetic pressure. dimension of the
electric field is that
Solution : a) The energy density for an electromagnetic field is given by / of Force/Charge
[refer to Eq. (2.31)]: and that of the
magnetic field is
H0 2 1 2 that of
u E  B Force/(charge u
2 2P 0
velocity).
E2
As B 2 P0H0E 2 , we get u H0E 2 H0E02 sin2 Zt  kx  G .
c2

b) The Poynting vector is given by [refer to Eq. (2.32)]: 87

Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
& 1 & & 1
P (E u B ) ^E0 sin Zt  kx  G yˆ` u ^B0 sin(Zt  kx  G) zˆ`
P0 P0

1
E0 2 sin2 Zt  kx  G xˆ cH 0 E0 2 sin2 Zt  kx  G xˆ cu xˆ
P 0c

Recall that Poynting vector gives the electromagnetic energy flowing

through unit area per unit time. In time 't electromagnetic energy
contained in the volume (c't )A passes through an area A. This energy is
equal to u (c't )A . Hence the energy flowing through unit area per unit time
is equal to cu .

c) The average electromagnetic energy crossing unit area per unit time for a
wave, called the intensity of the wave, is given by
& 1
I P cH0 E02 sin2 Zt  kx  G cH0 E02
2
d) The electromagnetic momentum density is given by [refer to Eq. (2.54)]:
& &
dpem u
P0H0P P0H0cu xˆ xˆ
dV c
[Recall that in relativistic mechanics E 2 p2c 2  m20c 4 ; for a massless
particle E = pc].

e) Electromagnetic Pressure is the force per unit area exerted by the

electromagnetic field on a surface. Force is equal to the momentum
imparted in unit time. In time 't electromagnetic momentum in the volume
(c't )A passes through an area A. This momentum is equal to
dpem
(c't )A . Hence the momentum flowing through unit area per unit
dV
dpem 1 I
time is equal to c u H0E02 sin2 Zt  kx  G H 0 E0 2 .
dV 2 c

Example 3.5

The intensity of sunlight falling on earth is 1300 W/m 2. What pressure does
sunlight exert when it falls on (a) perfect absorber (b) perfect reflector?

Solution : a) Sunlight is an electromagnetic wave./Let a Cartesian

coordinate system be so chosen that the x-axis is along the propagation
direction of the wave. The momentum flowing through unit area per unit time
I
is equal to xˆ where, I is the electromagnetic intensity. When sunlight falls
c
on a perfect absorber of area A, having normal along the x-direction, the
& I
momentum transferred to the absorber in time 't is given by 'p A 't xˆ .
c
Pressure exerted on the absorber is equal to force per unit area. Force is
& 'p& I
given by F A xˆ . Therefore, pressure ( S xx component of the stress
't c
tensor) is equal to

88
Unit 3 Electromagnetic Waves in Vacuum
F I 1300 Wm2
4.33 u 10  6 (Nms 1)(m 2 )(ms1)1
A c 3 u 10 8 ms1

4.33 u 10 6 Nm 2 .

b) When sunlight falls on a perfect reflector, in time 't , the momentum of the
& I & I
wave changes from 'p A 't xˆ to 'p  A 't xˆ . Therefore, the
c c
momentum imparted to the perfect reflector in time 't is given by
& I
'p 2 A 't xˆ . Therefore, the pressure exerted on the perfect reflector is
c
equal to 8.67 u10  6 Nm 2 .

So, we find that when we require that the solution of wave equation representing
electromagnetic wave should satisfy Maxwell’s equations, we get information that
if the electric field has only y-component, the magnetic field will have only
z-component. In other words, electromagnetic waves are transverse in nature.
Also, the magnitudes of the electric field and magnetic field associated with
electromagnetic wave are related as given by Eq. (3.39).
Now, let us discuss the representation of electric and magnetic fields in complex
notation and see how such a representation makes the analysis of
electromagnetic waves simpler.

3.2.3 Complex Notation

You have learnt complex numbers and their properties in the first semester
course entitled Mathematical Methods in Physics (MPH-001). You know that
use of complex number representation of physical quantities makes the
analysis of physical phenomena lot easier. In the instant case of
electrodynamics, it is very useful to use Euler’s formula e i T cos T  i sin T to
express electric and magnetic fields as real or imaginary parts of complex
functions.
Thus, we can write the electric and magnetic fields as
& & &
E E 0 sin Zt  kx  G ^
Im E 0 e i Zt  kx G ` (3.38)
& & &
B B0 sin Zt  kx  G ^
Im B0 e i Zt  kx G ` (3.39)

Note that we have not shown the real or imaginary part explicitly. However, we
will keep this fact in mind while interpreting the results. As you can see, the
advantage of using complex notation arises due to the fact that mathematical
manipulation of exponential functions is easier than sine and cosine functions.
To appreciate this fact, go through the following example.

Example 3.6
& & & & & &
If E E 0 e i Zt  kx G , (a) compute ’.E , (b) show that ’.E 0 implies E x 0.
&
Solution: a) Let E0 E0 x xˆ  E0 y yˆ  E0z zˆ . Then,

Ex E 0 x e i Zt  kx  G , E y E 0 y e i Zt  kx  G , E z E 0 z e i Zt  kx  G 89
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
& & w w
’.E E 0 x e i Zt  kx  G  E 0 y e i Zt  kx  G
wx wy
w
 E0zei Zt  kx G ikE0 x ei Zt kx G
wz
& &
b) From the result of (a), it follows that ’.E 0 implies E0 x 0 and
therefore E x 0 .

Before proceeding further, solve an SAQ.

SAQ 1
& & & &
If E E0e i Zt  kx G , and B B0 e i Zt  kx G , show that
& & &
(a) ’ u E ik ( xˆ u E0 )ei Zt kx G
&
& & wB & k & 1 &
(b) ’ u E  implies B0 ( xˆ u E0 ) ( xˆ u E0 )
wt Z c

Till now, we have described the propagation of a single electromagnetic plane

wave by considering a coordinate frame in such a manner that the x-axis is
along the direction of propagation. Such a choice makes the mathematical
manipulations simpler. However, in more complex situations, for example
superposition of two or more waves travelling in different directions, it may be
necessary to describe waves that are not travelling along a coordinate axis.
How do we do that? You will learn it now.

3.2.4 Electromagnetic Plane Wave Travelling in the k̂

Direction

Let us consider an electromagnetic plane wave travelling in the k̂ direction

(not to be confused with z-direction). To describe such a plane wave, the only
change required is that kx , in the above discussion of plane wave & &travelling in
the x-direction [Eqs. (3.38) and (3.39)] needs to be replaced by k .r . To justify
&
this, we need to show&that if r is the position
&& vector of a point on a plane
perpendicular to the k direction, then k .r has the same value for all the points
on the plane.
&
The vector k is called the propagation vector of the wave. The magnitude
&& of
&
k is the wave number which denotes the decrease in phase Z t  k .r per
unit distance in the direction of propagation. In the following
& example, we
discuss the characteristics of the propagation vector k and its relation with the
&
position vector r .

Example 3.7
a) Show that If a surface is described by \( x, y, z) constant, then
&
’\( x, y, z) is normal to the surface at the point ( x, y, z) .
& &
b) Show that if ’\( x, y , z ) k (a constant vector) for all points on a surface,
then the surface is described by \( x, y, z) k .r = constant. Explain why
90 this surface is a plane.
Unit 3 Electromagnetic Waves in Vacuum
&&
c) Show that at each point on the plane k .r kd , where d is the distance of
this plane from the origin O.
&&
d) Show that the rate of change of k .r per unit distance in the direction of
wave propagation is equal to k.

Solution: a) Let the position vector of a point P on the surface

&
\( x, y, z) constant be described by r xxˆ  yyˆ  zzˆ and of a neighbouring
point Q on the surface by
& &
r  'r ( x  'x )xˆ  ( y  'y )yˆ  (z  'z)zˆ

so that
&
'r 'x xˆ  'y yˆ  'z zˆ

Then,
'\ \(Q )  \(P ) \( x  'x, y  'y , z  'z )  \( x, y , z )

w\ w\ w\
'x  'y  'z
wx wy wz
Therefore, \( x, y, z) constant, implies
w\ w\ w\ & &
'x  'y  'z 0 Ÿ ’\.'r 0.
wx wy wz
& &
In the limit 'r o 0 , it lies in the tangent plane of the surface
\( x, y, z) constant.

Hence ’ψ(( x, y , z ) is normal to the surface at the point ( x, y, z) .

& & & w\ w\ w\
b) If ’\( x, y, z) k k x xˆ  ky yˆ  kz zˆ , ’\ xˆ  yˆ  zˆ implies
wx wy wz
w\ w\ w\
kx, ky  kz .
wx wy wz
w\
Now k x Ÿ \( x, y, z) k x x  f ( y, z) .
wx
w\ wf
Then, ky Ÿ ky Ÿ f ( y, z) ky y  g(z)
wy wy
so that \( x, y, z) k x x  f ( y, z) k x x  k y y  g ( z)

w\
Then, kz Ÿ g (z) kz z  constant
wz
&&
so that \( x, y , z ) k x x  k y y  k z z  const k .r  const.

Hence the surface \( x, y, z) constant can be described by

&&
k .r constant.
& && &
[Verify that ’(k .r ) k ]. This surface is a plane because the normal at
each point on the surface is the same.
&&
c) Let P0 be a point on the plane k .r = constant such that the distance of P0
from the origin O is equal to d, where d is the distance
& & of the plane from
the origin. As OP0 is perpendicular to the plane k .r = constant, it must lie 91
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
&
along the k vector. Therefore, the position vector of the point P0 is given
& && &&
by r0 d kˆ . Hence at P0 , k .r (k kˆ ).(d kˆ ) kd . But k .r is constant on the
&&
plane, therefore for each point on the plane k .r kd , where d is the
distance of this plane from the origin. [For a geometrical interpretation of
this result, notice that if P is a point on the plane, then the projection of OP
on k̂ is OP0 . ]
&
d) Consider two planes perpendicular to the propagation vector k , so that
the distance of one plane from the origin O is d and of the other is d  'd .
&& &&
Then on the first plane k .r kd and on the second k .r k (d  'd ) so
&& &&
that the change '(k .r ) k 'd . Hence, the rate of change of k .r per unit
&&
'(k .r ) '(kd )
distance in the direction of wave propagation k .
'd 'd
&&
Alternatively, the rate of change of k .r per unit distance in the direction of
& && &
wave propagation is given by ’(k .r ).kˆ k .kˆ k.

To concretise
& & your understanding of propagation vector and its relation with
the E and B fields, solve the following SAQ.

SAQ 2
& & && & & & &
a) If E E0ei Zt k .r G , (i) Compute ’.E . (ii) Show that ’.E 0 implies
& &
k .E 0.
& & && & & & &
b) If E E0ei Zt k .r G , B B0 e i Zt k .r  G , show that (i)
&
& & & & & & wB & k & 1 ˆ &
’uE  ik u E , and (ii) ’ u E  implies B (kˆ u E ) (k u E ).
wt Z c

Till now, we have confined our discussion of the electromagnetic waves and
its propagation by considering it as a plane wave. For a plane electromagnetic
wave, the electric field, the magnetic field and the direction of propagation are
perpendicular to each other. Before we end our discussion on the propagation
of electromagnetic waves in vacuum, a note of caution is & appropriate.
& & You
should not over-generalize the mutual orthogonality of E , B and k (the
direction of propagation) and believe that this property holds for all
electromagnetic waves. Let us briefly discuss other situations where this may
not be true.

3.2.5 Electromagnetic Wave which is Not Plane

If an electromagnetic wave travels in the x-direction and the electric fields and
magnetic fields depend on the y and z coordinates, then it is not necessary
that the x-component of the fields is zero. Therefore, the electromagnetic
wave in this case is not a transverse wave; the electric field and the magnetic
field need not be perpendicular to the direction of propagation. For such
situations, Maxwell’s equations tells us that each of the transverse
components of the electric and magnetic fields can be expressed in terms of
92 partial derivatives of the longitudinal components of the electric and magnetic
Unit 3 Electromagnetic Waves in Vacuum
fields. Thus, if the longitudinal components of the electric and magnetic fields
are given, the transverse components can be computed. The longitudinal
components, like all components of the electric and magnetic fields, have to
satisfy the three-dimensional wave equation. Functions of y and z, which
satisfy the three-dimensional wave equation for a wave propagating in the
x-direction can be easily determined. From these, it is possible to find some
functions of y and z for which the electric and magnetic field are not
perpendicular to each other. You will learn the details of such situations in
TQ 4.
Let us now summarise what you have learnt in this unit.

3.3 SUMMARY
„ Maxwell’s equations imply that, in regions where the charge density
and current density are zero, the electric and magnetic fields satisfy the
wave equation with phase velocity equal to

1
3 u108 ms1 c (the speed of light)
P0 H 0

„ As the value of H 0 emerges from electrostatic experiments and of

P 0 from magnetostatic experiments, the computed value of phase
velocity equalling the speed of light indicated that light is an
electromagnetic wave.

„ For a plane electromagnetic wave, as a consequence of Maxwell’s

equation, the electric field and magnetic fields are perpendicular to the
direction of propagation and to each other.

„ If a monochromatic plane electromagnetic wave of angular frequency

&
Z is travelling in k̂ direction with wave number k, then k k kˆ is called
& k & 1 ˆ &
the propagation vector and B (kˆ u E ) (k u E ) .
Z c

„ For an electromagnetic wave which is not plane, it is not necessary

that the electric field and magnetic field are perpendicular to the
direction of propagation or to each other.

3.4 TERMINAL QUESTIONS

1. An electromagnetic plane wave is travelling in the x̂ direction. The electric
field at each point on the plane x 0 is given by
&
E ( 0, t ) E0 cos(Zt ) yˆ  E0 sin( Zt ) zˆ .
& & & &
(a) Determine E ( x , t ). (b) Determine B( x, t ). (c) Evaluate E( x, t ).B( x, t ).

2. An electromagnetic x̂
& & plane wave is travelling in the direction in a region
where U 0, j 0 . The electric field at each point on the plane x 0 is
&
given by E(0, t ) f (t ) yˆ  g(t ) zˆ .
& & & &
(a) Determine E ( x, t ). (b) Determine B( x, t ). (c) Evaluate E ( x, t ). B( x, t ). 93
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
& & &
E0 sin(k z z ) ei (Zt k x x ) yˆ , B( x, y, z, t )
3. Given E ( x, y , z, t ) B0 ( y, z) ei Zt k x x
& &
in a region where U 0, j 0

a) What condition must k x and k z satisfy so that

&
E ( x, y , z, t ) E0 sin(k z z ) ei (Zt k x x ) yˆ satisfies the wave equation for an
electromagnetic wave?
& & &
b) Determine B0 ( y , z ) so that E, B satisfy Maxwell’s equations.
& &
c) Evaluate E.B.
& &
d) Interpret E, B in terms of real functions.
& &
4. The most general expression for the E and B fields of an electromagnetic
wave, which is not necessarily plane, travelling in the x-direction is given
by
& & & &
E( x, y , z, t ) E0 ( y , z )ei (Zt k x x ) , B( x, y , z, t ) B0 ( y , z ) ei (Zt k x x )
&
where E0 ( y , z ) E0 x ( y , z ) xˆ  E0 y ( y , z ) yˆ  E0z ( y , z ) zˆ
&
B0 ( y , z ) B0 x ( y , z ) xˆ  B0 y ( y , z ) yˆ  B0z ( y , z ) zˆ

& § w2 w 2 Z2 ·& & &

a) Show that E0 satisfies ¨¨ 2  2  2  k x 2 ¸¸E0 0 and B0 satisfies
© wy wz c ¹
§ w2 w2 Z2 ·& &
¨¨    k x 2 ¸¸B0 0.
© wy 2 wz 2 c 2 ¹
b) Using
& Maxwell’s
& equations obtain relations between the components of
E0 and B0 .
c) Manipulate the equations obtained in part (b) to express E0y , E 0 z ,
B0y and B0 z in terms of E0 x and B0 x .
& &
d) Use the results obtained in part (a) to evaluate E.B in terms of
E0 x and B0 x .
e) Try different functions E0 x ( y , z ) and B0 x ( y , z ) which satisfy the
& &
equations in part (a) to see if you can get E.B z 0.

5. An electromagnetic
& & wave is travelling in the z-direction in a region where
U 0, j 0 . The wave is cylindrically symmetric about the z-axis. In the
&
plane z 0 , E ( r , T,0, t ) E 0 ( r )e iZt rˆ. Determine the electric and
magnetic field for this electromagnetic wave.
& &
6. In cylindrical coordinates, the most general expression for the E and B
fields of an electromagnetic wave, which is not necessarily plane, travelling
in the z-direction is given by
& & & &
E(r , T, z, t ) E0 (r , T) ei (Zt kz) , B(r , T, z, t ) B0 (r , T) ei (Zt kz)
&
where E0 (r , T) E0r (r , T) rˆ  E0T (r , T) Tˆ  E0z (r , T) zˆ
&
B0 (r , T) B0r (r , T) rˆ  B0T (r , T) Tˆ  B0z (r , T) zˆ

a) Using
& Maxwell’s
& equations obtain relations between the components of
E0 and B0 in cylindrical coordinates.
94
Unit 3 Electromagnetic Waves in Vacuum
b) Manipulate the equations obtained in part (a) to express E 0r , E 0T ,
B0 r and B0 T in terms of E 0 z and B0 z .
7. Imagine an infinite sheet of charge in the y-z plane. At time t 0 , it starts
moving with a constant velocity. This moving sheet of charge will produce
a constant magnetic field in positive z-direction for positive x and in
negative z-direction for negative x . Consider only the region of positive x .
According to the Ampere’s law without Maxwell’s modification, a uniform
magnetic field pointing in the z-direction will emerge instantaneously at
t 0 , which is not possible. Due to Maxwell’s modification of Ampere’s
law, the field will propagate in the x-direction at speed c so that the
magnetic field is uniform in the region 0  x  ct and 0 for x ! ct . As a
consequence of Maxwell’s equations, this magnetic field will be
accompanied by an electric field pointing in the y-direction, which is
uniform in the region 0  x  ct and 0 for x ! ct .
Confining attention to the region x ! 0 , show that
&
E( x, y, z, t ) E0 ^1  4( x  ct )` yˆ
&
B( x, y, z, t ) B0 ^1  4( x  ct )` zˆ
(where 4(x ) is the step function defined by 4( x ) 0 for x  0 and 1 for
x ! 0) satisfy Maxwell’s equations if E0 cB0 .
&
[Notice that the definition of the step function gives E ( x, y , z, t ) E 0 yˆ for
&
x  ct and 0 for x ! ct . Similarly, B( x, y , z, t ) B0 zˆ for x  ct and 0 for
x ! ct ].

3.5 SOLUTIONS AND ANSWERS

Self-Assessment Questions
& & § wE z wE y · § wE wE · § wE wE y ·
1. a) ’uE ¨  ¸ xˆ  ¨ x  z ¸yˆ  ¨ z  ¸zˆ
© wy wz ¹ © wz wx ¹ © wy wz ¹

­w w ½
® E0ze i Zt kx G  E0y e i Zt kx G ¾xˆ
¯ wy wz ¿

w w
­
® E0 x e i Zt  kx  G  E0 z e i Zt  kx  G ½
¾yˆ
¯ wz wx ¿

­w w ½
® E0 y e i Zt  kx G  E0 x e i Zt  kx G ¾zˆ
¯ wx wy ¿

ikE0 z e i Zt  kx  G yˆ  ikE0 y e i Zt  kx  G zˆ
&
xˆ u E0 xˆ u (E0 x xˆ  E0 y yˆ  E0z zˆ ) E0 y zˆ  E0z yˆ
& & &
Hence ’ u E ik ( xˆ u E0 )ei Zt kx G .
&
wB w & i Zt  kx G &
b)   B0 e iZB0 e i Zt  kx G
wt wt 95
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
&
& & wB
Therefore ’ u E  implies
wt
& &
 ik ( xˆ u E0 )ei Zt  kx G iZB0 ei Zt  kx G .
& k & 1 &
Hence B0 ( xˆ u E0 ) ( xˆ u E0 ).
Z c
&
2. a) (i) Let E0 E0 x xˆ  E0 y yˆ  E0 z zˆ . Then,

Ex E0x ei Zt k x x k y y k z z G , Ey E0y ei Zt k x x k y y k z z G ,

Ez E0z e i Zt  k x x  k y y  k z z  G
& & w w
’.E E0 x e i Zt  k x x  k y y  k z z  G  E0 y e i Zt  k x x  k y y  k z z  G
wx wy

w
 E0zei Zt  k x x ky y  kz z  G
wz

i (k x E0x  k y E0y  kzE0z )ei Zt kx x ky y kzz  G

& & & &
ik .E0e i Zt  k x x  k y y  k z z  G ik .E.
& & & &
(ii) From the result of (i) it follows that ’.E 0 implies k .E 0.
&
So, we find that the concept of propagation vector k conforms to the
fact that, for a plane electromagnetic wave, electric field is
perpendicular to the direction of propagation.

b) (i) We know that

& & § wE z wE y · § wE wE · § wE wE y ·
’uE ¨  ¸ xˆ  ¨ x  z ¸yˆ  ¨ z  ¸zˆ
© wy wz ¹ © wz wx ¹ © wy wz ¹

­w w ½
® E0zei Zt k x x k y y k z z  G  E0y ei Zt k x x k y y k z z  G ¾xˆ
¯ wy wz ¿
w w
­
® E0 x e i Zt  k x x  k y y  k z z  G  E0z e i Zt  k x x  k y y  kz z  G ½
¾yˆ
¯ wz wx ¿

­w w ½
® E0y ei Zt k x x  ky y kz z  G  E0 x e i Zt k x x ky y  kz z  G ¾zˆ
¯ wx wy ¿
 i ^ ky E0z  kzE0y xˆ  kzE0x  k xE0z yˆ

 k x E0y  k y E0x zˆ`ei Zt k x x k y y k z z G

& & & &
i (k u E0 )e i Zt  k x x  k y y  kz z  G ik u E
&
wB w & i Zt  k&.r&  G & && &
(ii)  
B0 e iZB0 e i Zt  k .r  G iZB
wt wt
&
& & wB & & &
Therefore ’ u E  implies  ik u E iZB .
wt
& k ˆ & 1 ˆ &
Hence B (k u E ) (k u E0 ).
96 Z c
Unit 3 Electromagnetic Waves in Vacuum
These results indicate that the propagation vector conforms to the
other fact that, for a plane electromagnetic wave, magnetic field is
perpendicular to the direction of propagation as well as to the electric
field. Another result in this example which is noteworthy is that the grad
& & & &
operator ’, when acting on E 0 e i Zt k .r  G , can be replaced by
&
 ik . We shall be using this result in this course very often without
going through the detailed mathematical manipulations every time.

Terminal Questions
1. a) We know that for an electromagnetic plane wave travelling in the
x̂ direction
& &§ x·
E ( x, t ) E ¨ 0, t  ¸
© c ¹
&
As E(0, t ) E0 cos(Zt ) yˆ  E0 sin(Zt ) zˆ , we get,

& & x x x
E ( x, t ) E §¨ 0, t  ·¸
E 0 cos Z §¨ t  ·¸ yˆ  E 0 sin Z §¨ t  ·¸ zˆ
© c¹ © c¹ © c¹
&
& & wB
b) From Faraday’s law ’ u E  , we get,
wt

wB y wE wE z · wE z w ­E sin Z § t  x · ½
§¨ x  ¸ ® 0 ¨ ¸¾
wt © wz wx ¹ wx wx ¯ © c ¹¿

Z x
 E 0 cos Z §¨ t  ·¸
c © c ¹

1 x
Therefore, B y ( x, t )  E 0 sin Z §¨ t  ·¸  f ( x ) . Hence,
c © c¹

wB y Z x
E 0 cos Z §¨ t  ·¸  f c( x ) .
wx c 2 © c ¹
& &
& & wE 1 wE
From generalised Ampere’s law ’ u B P0H0 , we get,
wt c 2 wt
§ wBy wBx · 1 wEz
¨  ¸ . Therefore,
© wx wy ¹ c 2 wt

wB y 1 wE z 1 w ­ x Z x
®E 0 sin Z §¨ t  ·¸ ½
¾ E 0 cos Z §¨ t  ·¸
wx c 2 wt c 2 wt ¯ © c ¹¿ c2 © c¹

Hence, f c( x ) 0 . This implies that f ( x ) 0 [ Note that the constant of

integration is taken to be zero as addition of constant fields to solutions
of Maxwell’s equations give fields which also satisfy Maxwell’s
equations, but are of no interest in discussions of wave solution].

1 x
Therefore, By ( x, t )  E0 sin Z§¨ t  ·¸.
c © c ¹ 97
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
The z-component of the magnetic field can be obtained similarly.
&
& & wB
From Faraday’s law ’ u E  , we get,
wt
wB z § wE y wE x · wE y w ­ § x ·½
 ¨¨  ¸¸   ®E 0 cos Z ¨ t  ¸ ¾
wt © wx wz ¹ wx wx ¯ © c ¹¿

Z x
 E 0 sin Z §¨ t  ·¸
c © c¹
1 x
Therefore, B z ( x, t ) cos Z §¨ t  ·¸  f ( x ) . Hence,
c © c ¹
wB z Z x
E 0 sin Z §¨ t  ·¸  f c( x ) .
wx c 2 © c ¹
& &
& & wE 1 wE
From generalised Ampere’s law ’ u B P0H0 , we get,
wt c 2 wt
§ wBx  wBz · 1 wE y
¨ ¸ . Therefore,
© wz wx ¹ c 2 wt

wB z 1 wE y 1 w ­ § x ·½ Z x
  ®E 0 cos Z ¨ t  ¸ ¾ E 0 sin Z §¨ t  ·¸
wx c 2 wt c wt ¯
2 © c ¹¿ c 2 © c¹

Hence, f c( x ) 0 . This implies that f ( x ) 0 [ Again, the constant of

integration is taken to be zero as addition of constant fields to solutions
of Maxwell’s equations give fields which also satisfy Maxwell’s
equations, but are of no interest in discussions of wave propagation].
1 x
Therefore, B z ( x , t ) E 0 cos Z §¨ t  ·¸ .
c © c¹
& 1 x 1 x
Hence, B( x, t )  E 0 sin Z §¨ t  ·¸ yˆ  E 0 cos Z §¨ t  ·¸ zˆ.
c © c¹ c © c¹
& & ­E cos Z § t  x · yˆ  E sin Z § t  x · zˆ ½
c) E ( x, t ). B ( x, t ) ® 0 ¨ ¸ 0 ¨ ¸ ¾
¯ © c¹ © c¹ ¿
1 § x· 1 § x· ½
.­
®  E 0 sin Z ¨ t  ¸ yˆ  E 0 cos Z ¨ t  ¸ zˆ ¾
¯ c © c¹ c © c¹ ¿
1 2 x x
 E 0 cos Z §¨ t  ·¸ sin Z §¨ t  ·¸
c © c¹ © c¹
1 2 x x
 E 0 cos Z §¨ t  ·¸ sin Z §¨ t  ·¸
c © c¹ © c¹
0

[You should pause for a moment and visualise the electric and
magnetic fields described in this problem. Notice that, how at any point,
the electric field and the magnetic field vectors vary with time. Also
visualize how they vary with position at a fixed time. As at any point,
the electric and magnetic field vectors rotate with angular frequency Z ,
this wave is called a circularly polarized wave].
98
Unit 3 Electromagnetic Waves in Vacuum
2. a) We know that for an electromagnetic plane wave travelling in the
x̂ direction
& &§ x·
E ( x, t ) E ¨ 0, t  ¸
© c¹
&
As E(0, t ) f (t ) yˆ  g (t ) zˆ , we get,
& &§ x· § x· § x·
E ( x, t ) E ¨ 0, t  ¸ f ¨ t  ¸ yˆ  f ¨ t  ¸zˆ
© c¹ © c¹ © c¹
&
& & wB
b) From Faraday’s law ’ u E  , we get,
wt
wB y wE wE z · wE z w ­ § x ·½ 1 § x
§¨ x  ¸ ®g ¨ t  ¸ ¾  g c¨ t  ·¸
wt © wz wx ¹ wx wx ¯ © c ¹¿ c © c¹
1 § x
Therefore, B y ( x, t )  g ¨ t  ·¸  h ( x ) . Hence,
c © c¹
wB y 1 x
g c§¨ t  ·¸  h c( x ) .
wx c 2 © c¹
& &
& & wE 1 wE
From generalised Ampere’s law ’ u B P0H0 , we get,
wt c 2 wt
§ wBy wBx · 1 wEz
¨  ¸ . Therefore,
© wx wy ¹ c 2 wt
wB y 1 wE z 1 w ­ § x ·½ 1 x
®g ¨ t  ¸ ¾ g c§¨ t  ·¸
wx c 2 wt c 2 wt ¯ © c ¹¿ c 2 © c ¹
Hence, h c( x ) 0 . This implies that h( x ) 0 (the constant of
integration is taken to be zero as addition of constant fields to solutions
of Maxwell’s equations give fields which also satisfy Maxwell’s
equations, but are of no interest in discussions of wave solution).
1 x
Therefore, By ( x, t )  g §¨ t  ·¸.
c © c¹
The z-component of the magnetic field can be obtained similarly.
&
& & wB
From Faraday’s law ’ u E  , we get,
wt
wB z § wE y wE x · wE y w ­ § x ·½ 1 § x
 ¨¨  ¸¸   ®f ¨ t  ¸ ¾ f c¨ t  ·¸
wt © wx wz ¹ w x wx ¯ © c ¹¿ c © c ¹
1 § x
Therefore, Bz ( x, t ) f ¨ t  ·¸  h( x ) . Hence,
c © c¹
wB z 1 § x
 f c¨ t  ·¸  h c( x ) .
wx c2 © c¹
& &
& & wE 1 wE
From generalised Ampere’s law ’ u B P0H0 , we get,
wt c 2 wt
§ wBx  wBz · 1 wE y
¨ ¸ . Therefore,
© wz wx ¹ c 2 wt
wB z 1 wE y 1 w ­ § x 1 § x
  ®f ¨ t  ·¸ ½
¾  f c¨ t  ·¸
wx c 2 wt c 2 wt ¯ © c ¹¿ c2 © c¹
99
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
Hence, h c( x ) 0 . This implies that h( x ) 0 (the constant of
integration is taken to be zero as addition of constant fields to solutions
of Maxwell’s equations give fields which also satisfy Maxwell’s
equations, but are of no interest in discussions of wave solution).
1 § x
Therefore, Bz ( x, t ) f ¨ t  ·¸ .
c © c¹
& 1 x 1 x
Hence, B( x, t )  g §¨ t  ·¸yˆ  f §¨ t  ·¸zˆ.
c © c¹ c © c¹
& & ­f § t  x · yˆ  g § t  x ·zˆ ½
c) E ( x, t ). B( x, t ) ® ¨ ¸ ¨ ¸ ¾
¯ © c¹ © c¹ ¿
1 § x· 1 § x· ½
.­
® g ¨ t  ¸ yˆ  f ¨ t  ¸zˆ ¾
¯ c © c ¹ c © c ¹ ¿
1 § x x 1 x x
 f ¨ t  ·¸g §¨ t  ·¸  g §¨ t  ·¸f §¨ t  ·¸
c © c¹ © c¹ c © c¹ © c¹
0
& & &
& w 2E w 2 E w 2E § w2 w2 w2 ·
3. a) ’ 2E   ¨¨   ¸¸
wx 2 wy 2 wz 2 © wx 2 wy 2 wz 2 ¹
^E0 sin(kz z) ei (Zt kx x )yˆ`
(k x 2  kz 2 )^E0 sin(kz z) ei (Zt kx x )yˆ `
&
w 2E w2
^E0 sin(kz z) ei (Zt k x x )yˆ` Z2 ^E0 sin(kz z) ei (Zt k x x )yˆ`
wt 2 wt 2

&
E0 sin(k z z ) ei (Zt k x x ) yˆ satisfies the
Therefore, E ( x, y , z, t )
&
& 1 w 2E Z2
electromagnetic wave equation ’ E 2 if k x 2  k z 2 .
c 2 wt 2 c2
&
& & wB
b) Using the Faraday’s law ’ u E  , we get,
wt
§ wE z wE y · wBx
¨  ¸ 
© wy wz ¹ wt
w
Ÿ  ^E0 sin(kz z) ei (Zt kx x )`  w ^B0 x (y, z) ei (Zt kx x )`
wz wt
Ÿ  k z E0 cos (k z z ) e i ( Zt  k x x ) iZB0 x ( y , z )e i ( Zt  k x x )
ik
Ÿ  z E0 cos (k z z )
B0 x ( y , z )
Z
§ wE x  wE z ·  wBy
¨ ¸
© wz wx ¹ wt
w
Ÿ 0 
wt
^
B0 y ( y, z) ei (Zt  k x x ) `
Ÿ 0 iZB0 y e i (Zt  k x x )

Ÿ B0 y 0

§ wE y wE x · wBz
¨  ¸ 
100 © wx wy ¹ wt
Unit 3 Electromagnetic Waves in Vacuum
w
Ÿ ^E0 sin(kz z) ei (Zt kx x )`  w ^B0z (y, z) ei (Zt kx x )`
wx wt
Ÿ  ik x E0 sin(k z z) ei (Zt  k x x ) iZB0z ( y, z)e i (Zt k x x )
kx
Ÿ B0 z ( y, z) E0 sin (k z z)
Z
Therefore,
&
B0 ( y , z ) B0 x ( y , z ) xˆ  B0 y ( y , z ) yˆ  B0z ( y , z ) zˆ

§  ik z E cos (k z ) · xˆ  § k x E sin (k z ) ·zˆ

¨ 0 z ¸ ¨ 0 z ¸
© Z ¹ © Z ¹
Hence,
& &
B( x, y, z, t ) B0 ( y, z) e i Zt  k x x
§  ik z E cos (k z ) e i Zt  k x x · xˆ
¨ 0 z ¸
© Z ¹
k
 §¨ x E0 sin (k z z )e i Zt  k x x ·¸zˆ
© Z ¹
& &
We need to verify that E, B satisfy Maxwell’s equations.
& & wE x wEy wEz w
’.E   E0 sin(k z z) ei (Zt k x x ) 0
wx wy wz wy
& & wBx wBy wBz w § ik z
’.B   ¨ E0 cos (k z z ) e i Zt k x x ·¸
wx wy wz wx © Z ¹
w § kx
 ¨ E0 sin (k z z )e i Zt  k x x ·¸
wz © Z ¹
k k k k
 x z E0 cos (k z z) ei Zt k x x  x z E0 cos (kz z) ei Zt k x x
Z Z
0
&
& & wB & & & &
’uE  was used to determine B from E , so E, B satisfy this
wt
equation.
& & § wB wBy · 1 wE x
’uB x ¨ z  ¸ 0
© wy w z ¹ c 2 wt
& & wB wB w § ik z
’ u B y §¨ x  z ·¸ ¨ E0 cos (k z z ) e i Zt  k x x ·¸
© w z wx ¹ wz © Z ¹
w § kx
 ¨ E0 sin (k z z )e i Zt  k x x ·¸
wx © Z ¹

ik z 2 ik 2
E0 sin (k z z ) e i Zt  k x x  x E0 sin (k z z ) e i Zt  k x x
Z Z
iZ Z2
E0 sin (k z z ) e i Zt  k x x ( k x 2  k z 2 )
c2 c2
1 w 1 wE y wE y
E0 sin (k z z ) e i Zt  k x x P0H0
c 2 wt c 2 wt wt
& & § wBy wBx · 1 wEz
’uB z ¨  ¸ 0
© wx wy ¹ c 2 wt 101
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
[This question shows that if an electromagnetic wave is not plane, then
it need not be transverse. If you try solving this question without using
complex notation, you will appreciate how much easier it is to solve
problems using the complex notation].
& &
c) E.B E x Bx  E y By  E z Bz 0
& & &
[In this example E and B are perpendicular to each&other, &but B is not
perpendicular to the direction of propagation. Are E and B always
perpendicular to each other even for electromagnetic waves that are
not plane? To answer this question, either prove this statement or find
an example where it does not hold.]
d) Re^E0 sin(kz z) ei (Zt k x x )yˆ ` E0 sin(kz z) cos(Zt  k x x )yˆ

Re­§ ik z E cos (k z ) e i Zt  k x x · xˆ  § k x E sin (k z )e i Zt  k x x ·zˆ ½

®¨  0 z ¸ ¨ 0 z ¸ ¾
¯© Z ¹ © Z ¹ ¿

Re­§ ik z E cos (k z ) cos Zt  k x  i sin Zt  k x · xˆ

®¨  0 z x x ¸
¯© Z ¹
k
 §¨ x E0 sin (k z z )e i Zt  k x x ·¸zˆ ½¾
© Z ¹ ¿

Re­§ ik z E cos (k z ) cos Zt  k x  k z E cos (k z ) sin Zt  k x · xˆ

®¨  0 z x 0 z x ¸
¯© Z Z ¹
k
 §¨ x E0 sin (k z z )e i Zt  k x x ·¸zˆ ½¾
© Z ¹ ¿
kz k
E0 cos (k z z) sin Zt  k x x xˆ  x E0 sin (k z z) cos Zt  k x x zˆ
Z Z
&
You can verify yourself that E E0 sin(kz z) cos(Zt  k x x ) yˆ and
& kz k
B E0 cos (k z z) sin Zt  k x x xˆ  x E0 sin(k z z) cos Zt  k x x zˆ
Z Z
satisfy Maxwell’s equations.
Similarly,
Im^E0 sin (kz z ) ei (Zt k x x )yˆ ` E0 sin(kz z) sin(Zt  k x x )yˆ

Im ­§ ik z E cos (k z ) e i Zt  k x x · xˆ  § k x E sin (k z )e i Zt  k x x ·zˆ ½

®¨  0 z ¸ ¨ 0 z ¸ ¾
¯© Z ¹ © Z ¹ ¿

Im ­§ ik z E cos (k z ) cos Zt  k x  i sin Zt  k x · xˆ

®¨  0 z x x ¸
¯© Z ¹
k
 §¨ x E0 sin (k z z )e i Zt  k x x ·¸zˆ ½¾
© Z ¹ ¿

Im ­§ ik z E cos (k z ) cos Zt  k x  k z E cos (k z ) sin Zt  k x · xˆ

®¨  0 z x 0 z x ¸
¯© Z Z ¹
k
 §¨ x E0 sin (k z z )e i Zt  k x x ·¸zˆ ½¾
© Z ¹ ¿
k k
 z E0 cos (k z z) cos Zt  k x x xˆ  x E0 sin(k z z) sin Zt  k x x zˆ
102 Z Z
Unit 3 Electromagnetic Waves in Vacuum
&
Therefore, E E0 sin(kz z) sin(Zt  k x x ) yˆ and
& k k
B  z E0 cos(kz z) cos Zt  k x x xˆ  x E0 sin(kz z) sin Zt  k x x zˆ
Z Z
satisfy Maxwell’s equations. [You should work out the details and verify
this.]
&
& & w 2E
4. a) As E satisfies the electromagnetic wave equation ’ 2 E P 0 H 0 ,
wt 2
for the given electromagnetic fields, we get,
§ w2 w2 w2 · & 1 w2 &
¨¨   ¸ E ( y , z )e i ( Zt  k x x ) E0 ( y , z )e i ( Zt  k x x )
wx 2 wy 2 wz 2¸ 0 c 2 wt 2
© ¹

§ w2 w 2 · & i ( Zt  k x x ) Z2 & i ( Zt  k x x )
Ÿ ¨¨  k x 2   ¸ E0e  E0e
© wy 2 wz 2 ¸¹ c2

§ w2 w2 Z2 ·& &
Ÿ ¨¨    k x 2 ¸¸E0 0
© wy 2 wz 2 c 2 ¹

§ w2 w2 Z2 ·& &
Similarly, ¨¨ 2  2  2  k x 2 ¸¸B0 0.
© wy wz c ¹
& & wE x wE y wEz
b) ’.E   0
wx wy wz
w w
Ÿ E0 x ( y , z ) e i ( Zt  k x x )  E0 y ( y , z ) e i ( Zt  k x x )
wx wy

w
 E0 x ( y, z) e i (Zt  k x x ) 0
wz
w
Ÿ  ik x E0 x ( y , z ) e i ( Zt  k x x )  E0 y ( y , z ) e i ( Zt  k x x )
wy

w
 E0 x ( y, z) e i (Zt  k x x ) 0
wz
wE0y wE0 x
Ÿ  ik x E0 x   0
wy wz
& & wB0y wB0 x
Similarly, ’.B 0 implies  ik x B0 x   0.
wy wz
& & wBx wEz wE y wBx
’uE x  Ÿ  
wt wy wz wt
w w
Ÿ E0 z ( y , z ) e i ( Zt  k x x )  E0 y ( y , z ) e i ( Zt  k x x )
wy wz

w
 B0 x ( y , z ) e i (Zt  k x x )
wt
wE0z i (Zt k x x ) wE0y i (Zt k x x )
Ÿ e  e iZB0 x ei (Zt k x x )
wy wz

wE0z wE0y
Ÿ  iZB0 x
wy wz 103
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
& & wBy wE x wE z wBy
’uE y  Ÿ  
wt wz wx wt
w w
Ÿ E0 x ( y, z) ei (Zt k x x )  E0z ( y, z) ei (Zt k x x )
wz wx
w
 B0y ( y, z) e i (Zt  k x x )
wt
wE0 x i (Zt k x x )
Ÿ e  ik x E0zei (Zt k x x ) iZB0y ei (Zt k x x )
wz
wE0 x
Ÿ  ik x E0z iZB0y
wz
& & wB wE y wE x wBz
’uE z  z Ÿ  
wt wx wy wt
w w
Ÿ E0 y ( y , z ) e i ( Zt  k x x )  E0 x ( y , z ) e i ( Zt  k x x )
wx wy
w
 B0z ( y , z ) e i (Zt  k x x )
wt
wE0 x i ( Zt  k x x )
Ÿ  ik x E0 y e i ( Zt  k x x )  e iZB0z e i ( Zt  k x x )
wy
wE0 x
Ÿ  ik x E0 y  iZB0 z
wy
& &
& & wE x 1 wE x § wB wBy · 1 wE x
’uB x P0H0 Ÿ ¨ z  ¸
wt c wt
2
© wy wz ¹ c 2 wt
w w
Ÿ B0 z ( y , z ) e i ( Zt  k x x )  B0 y ( y , z ) e i ( Zt  k x x )
wy wz
1 w
E0 x ( y , z ) e i ( Zt  k x x )
c 2 wt
wB0z i (Zt k x x ) wB0y i (Zt k x x ) iZE0 x i (Zt k x x )
Ÿ e  e e
wy wz c2
wB0z wB0y iZE0 x
Ÿ 
wy wz c2
& &
& & wE y 1 wE y wB wB 1 wE y
’uB y P0 H0 Ÿ §¨ x  z ·¸
wt c wt
2 © wz wx ¹ c 2 wt
w w
Ÿ B0 x ( y, z) ei (Zt k x x )  B0z ( y, z) ei (Zt k x x )
wz wx
1 w
E0 y ( y , z ) e i ( Zt  k x x )
c 2 wt
wB0 x i ( Zt  k x x ) iZE0 y i ( Zt  k x )
Ÿ e  ik x B0 z e i ( Zt  k x x ) e x
wz c2
wB0 x iZE0 y
Ÿ  ik x B0 z
wz c2
& &
& & wE z 1 wE z § wBy wBx · 1 wE z
’uB z P0H0 Ÿ ¨  ¸
wt c wt
2
© wx wy ¹ c 2 wt
104
Unit 3 Electromagnetic Waves in Vacuum
w w
Ÿ B0 y ( y , z ) e i ( Zt  k x x )  B0 x ( y , z ) e i ( Zt  k x x )
wx wy
1 w
E0z ( y , z ) e i ( Zt  k x x )
c 2 wt
wB0 x i ( Zt  k x x ) iZE0z i ( Zt  k x x )
Ÿ  ik x B0 y e i ( Zt  k x x )  e e
wy c2
wB0 x iZE0 z
Ÿ  ik x B0 y 
wy c2
c) To obtain an expression for E0y in terms of E0 x and B0 x , from
amongst the equations obtained in part (b), we look for equations that
contain E0y , but not as derivatives of E0y . These equations are:
wE0 x
 ik x E0 y  iZB0 z
wy
wB0 x iZE0 y
and  ik x B0z
wz c2
In these two equations, if we eliminate B0 z , we will be left with terms
containing E0y , E0 x and B0 x . Thus, we will be able to express E0y
in terms of E0 x and B0 x . To eliminate B0 z , we multiply the first
equation by k x and the second by Z , to get
wE0 x
 ik x 2E0 y  k x iZk x B0 z
wy

wB0 x iZ2E0 y
and Z  ik x ZB0z
wz c2
Hence,
wE0 x wB0 x iZ2E0 y
 ik x 2E0 y  k x Z 
wy wz c2
§ Z2 · wE0 x wB
Ÿ i ¨¨  k x 2 ¸¸E0 y kx  Z 0x
2 wy wz
©c ¹
i § wE0 x wB ·
E0 y  ¨ kx  Z 0x ¸
Z2 © wy wz ¹
 kx2
c 2

Following the above approach, you can show that

i § wE0 x wB ·
E0 z  ¨ kx  Z 0x ¸
Z2 2© wz wy ¹
 kx
c2
i § wB0 x Z wE0 x ·
B0 y  ¨ kx  ¸
Z2 2© wy c 2 wz ¹
 kx
c2
i § wB0 x Z wE0 x ·
B0 z  ¨kx  ¸
Z2 2© wz c 2 wy ¹
 kx
c2
& &
[Given E0 x ( y , z ) and B0 x ( y , z ), using these equations, E and B can
105
be determined. Notice that in deriving these equations only
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
& &
& & wB & & 1 wE
’uE  and ’ u B were used. Therefore, E0 x ( y , z ) and
wt c 2 wt & &
B0 x ( y , z ) must be so chosen, that E and B obtained from them satisfy
& & & &
’.E 0 and ’.B 0 if E0 x ( y , z ) and B0 x ( y , z ) satisfy the equations in
part (a).
& & & &
d) E.B E0ei (Zt k x x ) . B0ei (Zt k x x )
& &
E 0 .B 0 E 0 x B0 x  E 0 y B0 y  E 0 z B0 z
2
§ Z2 · § wE 0 x wB ·
E 0 x B 0 x  ¨¨  k x 2 ¸¸ ¨kx  Z 0x ¸
©c 2 ¹ © wy wz ¹
§ wB0 x Z wE0 x ·
¨ kx  ¸
© wy c 2 wz ¹
2
§ Z2 · § wE0 x wB ·§ wB0 x Z wE0 x ·
 ¨¨  k x 2 ¸¸ ¨kx  Z 0 x ¸¨ k x  ¸
©c 2
¹ © wz wy ¹© wz c 2 wy ¹

2
§ Z2 · § 2 wE 0 x wB 0 x Z 2 wE 0 x wB 0 x ·
E 0 x B 0 x  ¨¨  k x 2 ¸¸ ¨¨ k x  ¸
©c 2 ¹ © wy wy c 2 wz wz ¸¹
2
§ Z2 · § 2 wE0 x wB0 x Z2 wE0 x wB0 x ·
 ¨¨  k x 2 ¸¸ ¨¨ k x  ¸
©c 2
¹ © wz wz c 2 wy wy ¸¹
2
§ Z2 · ­§ 2 Z 2 · wE 0 x wB 0 x
E 0 x B 0 x  ¨¨  k x 2 ¸¸ ®¨¨ k x  2 ¸¸
©c 2 ¹ ¯© c ¹ wy wy
§ Z2 · wE0 x wB0 x ½
 ¨¨ k x 2  ¸ ¾
© c 2 ¸¹ wz wz ¿
1
§ Z2 · § wE 0 x wB 0 x wE 0 x wB 0 x ·
E 0 x B 0 x  ¨¨  k x 2 ¸¸ ¨  ¸
© c2 ¹ © wy wy wz wz ¹
e) The result of part (d) can suggest choices of E0 x ( y , z ) and
& &
B0 x ( y , z ) which would lead to E.B z 0 . It is clear that both
E0 x ( y , z ) and B0 x ( y , z ) should be non-zero. If the second term in the
& &
result of part (d) is zero, but the first is not, then although E and B will
not be perpendicular to each other, their projections on the y-z plane
will be perpendicular to each other. However, if the second term is also
non-zero, then these projections will also be not perpendicular to each
other. Of course, E0 x ( y , z ) and B0 x ( y , z ) have to satisfy the equations
in part (a). A simple choice is:
E0x ( y, z) E0 sin ky y , B0x ( y, z) B0 sin ky y
Z2
where k y 2  kx2
c 2

For this choice,

i § wE0 x wB · i w
E0 y  ¨ kx  Z 0x ¸  kx E0 sin k y y
Z2 2© wy wz ¹ ky 2 wy
 kx
c2

106
Unit 3 Electromagnetic Waves in Vacuum

i § wB0 x Z wE0 x · i § Z w ·
B0z  ¨ kx  ¸  ¨ E0 sin k y y ¸
Z2 2© wz c 2 wy ¹ ky 2 © c 2 wy ¹
 kx
c2
i Z
 E0 cos k y y
ky c2

Hence,
& & ­ ik x
E( x, y, z, t ) E0 ( y , z )ei (Zt k x x ) ®E0 sin k y y xˆ  E0 cos k y y yˆ
¯ ky

iZ ½
 B0 cos k y y zˆ ¾e i ( Zt  k x x )
ky ¿
& & ­ ik x
B ( x, y , z, t ) B0 ( y , z ) e i ( Zt  k x x ) ®B0 sin k y y xˆ  B0 cos k y y yˆ
¯ ky

iZ ½
 B0 cos k y y zˆ ¾e i ( Zt  k x x )
ky c 2
¿
& &
[You should verify that E( x, y, z, t ) and B( x, y , z, t ) satisfy Maxwell’s
equations. You may also work out to express these as real fields by
taking the real part or imaginary part of the equations and again verify
that these fields satisfy the Maxwell’s equations.]
& &
E0 .B0 E0 x B0 x  E0 y B0 y  E0z B0z

§ ik ·
E0 sin k y y B0 sin k y y  ¨¨  x E0 cos k y y ¸¸
© ky ¹
§ ik x · § iZ ·§ i ·
¨ B cos k y ¸ ¨ ¸¨ ¸
¨ k 0 y ¸ ¨ k B0 cos k y y ¸¨ k E 0 cos k y y ¸
© y ¹ © y ¹© y ¹

kx2 Z2
E0B0 sin2 k y y  E0B0 cos2 k y y  E0B0 cos 2 k y y
2 2 2
ky ky c

§ Z2 2 ·
E0B0 sin2 k y y  E0B0 cos2 k y y ¨¨ c 2  k x k y 2 ¸¸
© ¹
E0B0
& & & & & &
? E.B E0ei (Zt k x x ) . B0ei (Zt k x x ) E0 .B0 e2i (Zt k x x ) z 0

[Check that you get the same result by using the result of part (d). It is
a good habit to check results in different ways. That helps in detecting
errors.]
5. The electric field is given by
z
& & z iZ§¨ t  ·¸
E ( r , T, z, t ) E §¨ r , T,0, t  ·¸ E 0 ( r )e © c ¹ rˆ .
© c¹

107
Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
& &
The electromagnetic fields E(r , T, z, t ) and B(r , T, z, t ) have to satisfy
Maxwell’s equations. In cylindrical coordinates,
& & 1 w 1 wE T wE z
’.E ( rE r )   .
r wr r wT wz
& & 1d
Therefore, ’.E 0 implies ^rE0 (r )` 0 . Hence, r E0 (r ) D , where
r dr
D
D is a constant. Thus, E0 (r ) . Therefore,
r
z z
& iZ§¨ t  ·¸ § ·
D iZ¨© t  c ¸¹
E ( r , T, z, t ) E 0 ( r )e © c¹ rˆ e rˆ
r
In cylindrical coordinates,
& & § 1 wE z wE T · § wE wE z ·
’ u E ¨¨  ¸¸ rˆ  ¨ r  ¸ Tˆ
© r wT wz ¹ © wz wr ¹
§ · z
1§ w wE r · iZ D iZ¨© t  c ¸¹ ˆ
 ¨ rET  ¸z 
ˆ e T
r © wr wT ¹ c r
& & § z·
& & wB wB iZ D iZ¨© t  c ¸¹
Therefore, ’ u E  implies e T̂.
wt wt c r
§ z·
& 1 D iZ¨© t  c ¸¹ ˆ &
Hence, B(r , T, z, t ) e T  f (r , T, z ) . As the second term in the
& c r
expression for B( r , T, z, t ) does not depend on time, it is of no interest in
discussions of wave propagation and can be taken to be zero.
& &
[You should work out the details and verify that E(r , T, z, t ) and B(r , T, z, t )
&
& & 1 wE
satisfy the remaining Maxwell’s equation ’ u B ]. Thus, the
c 2 wt
§ z·
& D iZ¨© t  c ¸¹
required electric and magnetic fields are E (r , T, z, t ) e rˆ and
r
§ z·
& 1 D iZ¨© t  c ¸¹ ˆ
B(r , T, z, t ) e T.
c r
& &
6. a) In cylindrical coordinates, ’.E 0 implies
& & 1 w 1 wE T wE z
’.E ( rE r )  
r wr r wT wz
1 w 1 w
rE0r (r , T)ei (Zt kz)  E0T (r , T)ei (Zt kz)
r wr r wT
w
 E0z (r , T)e i (Zt  kz)
wz
­ 1 w rE (r , T)  1 w E (r , T)  ik E (r , T)½e i ( Zt  kz)
® 0r 0T 0z ¾
¯ r wr r wT ¿
1 w 1 wE0T
Ÿ rE0r   ik E0z 0
r wr r wT
& &
Similarly, ’.B 0 implies
1 w 1 w
rB0r (r , T)  B0T (r , T)  ik B0z (r , T) 0.
r wr r wT
108 In cylindrical coordinates,
Unit 3 Electromagnetic Waves in Vacuum
& & § 1 wE z wE T · § wE r  wE z · Tˆ  1 § w rE  wE r · zˆ
’uE ¨  ¸ rˆ  ¨ ¸ ¨ T ¸
© r wT wz ¹ © wz wr ¹ r © wr wT ¹
Therefore,
& & wBr 1 wE z wE T wBr
(’ u E )r  Ÿ  
wt r wT wz wt
1 w w
Ÿ E0z (r , T)ei (Zt  kz)  E0T (r , T)e i (Zt  kz)
r wT wz
w
 r B0r (r , T)e i (Zt  kz)
wt

Ÿ ­ 1 w E (r , T)  ikE (r , T)½e i ( Zt  kz) iZB0r (r , T)e i (Zt  kz)

® 0z 0T ¾
¯ r wT ¿
1 wE0z
Ÿ  ikE0T iZB0r
r wT
Similarly,
& & wBT wE
(’ u E )T  Ÿ  ikE0r  0z iZB0T
wt wr
& & wBz 1 w wE
(’ u E )z  Ÿ (r E0T )  0r iZB0z
wt r wr wT
& & 1 wEr 1 wB0z iZ
(’ u B )r Ÿ  ikB0T E0r
2
c wt r wT c2
& & 1 wET wB iZ
(’ u B )T Ÿ ikB0r  0z E0T
c wt
2 wr c2
& & 1 wEz 1 w wB iZ
(’ u B)z Ÿ (r B0T )  0r E0z
c 2 wt r wr wT c2
b) To express E 0r in terms of E 0 z and B0 z , consider the equations
obtained in part (a) that have E 0 r , but not a derivative of E 0r .
These equations are:
wE0z
 ikE0r  iZB0T
wr
and
1 wB0z iZ
 ikB0T E0 r
r wT c2
If B0T is eliminated from these two equations, we would get a
relationship connecting E 0r with E 0 z and B0 z . This can be done
by multiplying the first equation by k and the second equation by
Z to get,
wE0z
 ik 2E0r  k ikZB0T
wr
and
Z wB0z iZ2
 ikZB0T E 0r
r wT c2
Therefore,

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Block 1 Maxwell’s Equations, Electromagnetic Waves and Dynamics of Charged Particles
wE0z Z wB0z iZ2
 ik 2E0r  k  E 0r
wr r wT c2
Hence,
i § k wE 0 z  Z wB0 z ·
E 0r ¨ ¸
Z2 2© wr r wT ¹
k
c2
Following this method, you can easily obtain,

E 0T
i § k wE 0 z  Z wB0 z ·
¨ ¸
Z2 2 © r wT wT ¹
k
c2
i § k wB0 z  Z wE 0 z ·
B0r ¨ ¸
Z2 2© wr c 2 r wT ¹
k
c2

B0T
i § k wB0 z  Z wE 0 z ·
¨ ¸
Z2 2 © r wT c 2 wr ¹
k
c2
If functions E 0 z and B0 z are chosen so that they satisfy the wave
equation in cylindrical coordinates, the electric and magnetic fields can
be determined using the above four& equations
& satisfying Maxwell’s
equations in a region where U 0, j 0.
& &
7. To show that the given E and B satisfy Maxwell’s equations, it will be
necessary to compute the derivatives of the step function. Recall that
d
4( x ) G( x ) where G(x ) is the Dirac delta function defined by G( x ) 0
dx
f
d
if x z 0 and ³ G( x ) dx 1 . Clearly
dx
4( x ) 0 if x z 0 and
f
f
d d
³ dx
4( x ) dx 4(f)  4( f) 1  0 1 . Therefore,
dx
4( x ) G( x ) .
f
Using this result, you &can follow
& the steps used in above solutions and
verify that the given E and B satisfy Maxwell’s equations:
& & wE x wE y wE z w
’.E   E0 ^1  4( x  ct )` 0
wx wy wz wy
& & wBx wBy wBz w
’.B   B0 ^1  4( x  ct )` 0
wx wy wz wz
& & wE z wE y w wBx
’uE x  E0 ^1  4( x  ct )` 0 
wy wz wz wt

110