DAY TWENTY FIVE

Electromagnetic
Waves
Learning & Revision for the Day
u Electromagnetic Waves and u Maxwell’s Equations u Spectrum of Electromagnetic
their Characteristics u Transverse Nature of Electromagnetic Waves Radiation

Electromagnetic Waves and

their Characteristics
Electromagnetic waves are those waves, in which electric and magnetic fields vary
sinusoidally in space with time. The electric and magnetic fields are mutually
perpendicular to each other and each field is perpendicular to the direction of
propagation of the wave.
●
Maxwell’s theory predicted that electromagnetic waves of all frequencies (and hence
all wavelengths) propagate in vacuum, with a speed given by
1
c= .
µ 0ε 0
where, µ 0 is the magnetic permeability and ε 0 is the electric permittivity of vacuum.
Now, for the vacuum, µ 0 = 4π × 10 −7 TmA −1 and ε 0 = 8.85 × 10 −12 C2 N −1m−2 .
Substituting these values in the above relation, we have PREP
c=
1
~ 3.0 × 10 8 ms −1 MIRROR
[(4π × 10 )(8.85 × 10 −12 )]1/ 2
−7
Your Personal Preparation Indicator
●
All the electromagnetic waves are of the transverse nature whose speed depends u No. of Questions in Exercises (x)—
upon the medium, but their frequency does not depend on the medium. u No. of Questions Attempted (y)—
●
Transverse waves can be polarised. u No. of Correct Questions (z)—
●
Energy is being transported with the electromagnetic waves. (Without referring Explanations)

u Accuracy Level (z / y × 100)—

Conduction Current u Prep Level (z / x × 100)—
It is a current in the electric circuit, which arises due to the flow of electrons in the
connecting wires of the circuit, in a definite closed path. In order to expect good rank in JEE,
your Accuracy Level should be above
85 & Prep Level should be above 75.
DAY TWENTY FIVE ELECTROMAGNETIC WAVES 289

In this, E 0 and B0 are the amplitudes of the fields.

Maxwell’s Displacement Current E 1
Further, c = 0 = = speed of light in vacuum
It is that current which comes into play in the region, B0 ε0 µ 0
whenever the electric field and hence the electric flux is
changing with time.
●
The rate of flow of energy in an electromagnetic wave, is
dφ E described by the vector S called the Poynting vector, which
id = ε 0 is defined by the expression,
dt
1
The generalised form of the Ampere’s law is S= E×B
µ0
 dφ E 
∫ B ⋅ d l = µ 0(ic + id ) = µ 0  i c + ε 0 dt 
 ●
The time average of S over one cycle is known as the wave
intensity. When the average is taken, we obtain an
where, ic is conduction current. expression involving the time average of cos2 (kx − ωt )
1
which equals .
2
Maxwell’s Equations E B E2 c B02
Maxwell in 1862, gave the basic laws of electricity and Thus, I = Sav = 0 0 = 0 = Wm −2
2 µ 0 2 µ 0c 2 µ 0
magnetism in the form of four fundamental equations, which
are known as Maxwell’s equations. ●
The total average energy per unit volume is,
ε E2 B2
●
Gauss’s law for electrostatics This law states that the total u = uE + uB = 0 0 =
electric flux through any closed surface is always equal to 2 2µ0
1 The radiation pressure p exerted on a perfectly absorbing
times the net charged enclosed by that surface. ●

ε0 S
surface, p = .
q c
Mathematically, ∫ E ⋅ dS =
S ε0 ●
If the surface is a perfect reflector and incidence is normal,
●
Gauss’s law for magnetism This law also predicts that the then the momentum transported to the surface in a time t is
isolated magnetic monopole does not exist. 2u 2S
given by, p = and the radiation pressure will be, p = .
i.e. net magnetic flux through any closed surface is always c c
zero. ●
Energy density of electromagnetic wave,
Mathematically, ∫ B ⋅ dS = 0 1 1 B2
S ue = ε 0Eu2B =
2 2 µ0
●
Faraday’s law of electromagnetic induction It states that
u
the induced e.m.f. produced in a circuit is numerically ●
Momentum delivered, p = (absorbing surface)
equal to time rate of change of magnetic flux through it. c
dφ 2u
Mathematically, ∫ E ⋅ dl = − B p= (reflecting surface)
dt c
hc
●
Ampere-Maxwell’s law At an instant in a circuit, the ●
Energy of wave = = hν
conduction current is equal to displacement current. λ
 dφ 
Mathematically, ∫ E ⋅ dl = µ 0  I c + ε 0 E 
 dt  Transverse Nature of
These equations are collectively called Maxwell’s
equations.
Electromagnetic Waves
According to Maxwell, electromagnetic waves consist of time
varying electric and magnetic fields, which are perpendicular
Properties of to each other, as well as direction of wave propagation.
Electromagnetic Waves Y Wave propagation
E B
●
If the electromagnetic wave is travelling along the positive
direction of the X -axis, the electric field is oscillating
parallel to the Y-axis and the magnetic field is oscillating X
parallel to the Z-axis.
E = E 0 sin(ωt − kx) ⇒ B = B0 sin(ωt − kx) Z B E
290 40 DAYS ~ JEE MAIN PHYSICS DAY TWENTY FIVE

detecting the fracture of bones, hidden bullet, needle,

Spectrum of Electromagnetic costly material etc. inside the body, and also used in the
Radiation study of crystal structure.
The array obtained on arranging all the electromagnetic ●
Ultraviolet Rays The major part of the radiations received
waves in an order on the basis of their wavelength is called from sun consists of the ultraviolet radiation. Its other
the electromagnetic spectrum. sources are the electric discharge tube, carbon arc, etc.
The Electromagnetic Spectrum These radiations are mainly used in excitation of
photoelectric effect and to kill the bacteria of many diseases.
Frequency Wavelength ●
Visible Light Visible light is obtained from the glowing
Name Source
Range (Hz) Range (m) bodies, while they are white hot. The light obtained from
Radio 104 to 108 0.1 to 600 Oscillating electric the electric bulbs, sodium lamp, fluorescent tube is the
waves circuits visible light.
Microwaves 109 to 1012 10−3 to 0.3 Oscillating current in ●
Thermal or Infrared Waves A body on being heated, emits
special vacuum tubes out the infrared waves. These radiations have the
Infrared 1011 to 10−6 to Outer electrons in atoms maximum heating effect. The glass absorbs these
5 × 1014 5 × 10−3 and molecules radiations, therefore for the study of these radiations, rock
salt prism is used instead of a glass prism. These waves are
Visible 4 × 1014 to 4 × 10−7 to Outer electrons in atoms mainly used for therapeutic purpose by the doctors because
light 7 × 10 14
8 × 10 −7 of their heating effect.
Ultraviolet 15
10 to 10 17
1.5 × 10−7 to Outer electrons in atoms
●
Microwaves These waves are produced by the spark
−7
discharge or magnetron valve. They are detected by the
3.5 × 10
crystal or semiconductor detector. These waves are used
X-rays 18
10 to 10 20
10−11 to 10−8 Inner electrons in atoms mainly in radar and long distance communication.
and sudden ●
Radio waves They can be obtained by the flow of high
deacceleration of high
frequency alternating current in an electric conductor.
energy free electrons
These waves are detected by the tank circuit in a radio
Gamma 1019 to 1024 10−16 to 10−13 Nuclei of atoms and
receiver or transmitter.
rays sudden deacceleration of
high energy free electrons
Applications of
Various Electromagnetic Radiations Electromagnetic Spectrum
●
Gamma rays The main sources of gamma rays are the ●
Radio waves are used in radar and radio broadcasting.
natural and artificial radioactive substances. These rays ●
Microwaves are used in long distance wireless
affect the photographic plate and mainly used in the communications via satellites.
treatment of cancer disease. ●
Infrared, visible and ultraviolet radiations are used to know
●
X-rays X-rays are produced, when highly energetic the structure of molecules.
cathode rays are stopped by a metal target of high melting ●
Diffraction of X-rays by crystals, gives the details of the
point. They affect the photographic plate and can penetrate structure of crystals.
through the transparent materials. They are mainly used in
DAY TWENTY FIVE ELECTROMAGNETIC WAVES 291

DAY PRACTICE SESSION 1

FOUNDATION QUESTIONS EXERCISE

1 During the porpagation of electromagnetic waves in a (a) 4 × 10−6 T (b) 6 × 10−8 T
medium, ª JEE Main 2014 (c) 9 × 10−9 T (d) 11 × 10−11 T
(a) electric energy density is double of the magnetic energy 9 The magnetic field in a travelling electromagnetic wave
density has a peak value of 20 nT. The peak value of electric field
(b) electric energy density is half of the magnetic energy
strength is ª JEE Main 2013
density
(a) 3 V/m (b) 6 V/m (c) 9 V/m (d) 12 V/m
(c) electric energy density is equal to the magnetic energy
density 10 The rms value of the electric field of the light coming from
(d) Both electric and magnetic energy densities are zero the sun is 720 NC −1. The average total energy density of
2 A perfectly reflecting mirror has an area of 1 cm 2. the electromagnetic wave is
Light energy is allowed to fall on it for 1 h at the rate of (a) 4.58 × 10−6 Jm−3 (b) 6.37 × 10−9 Jm−3
10 W cm −2. The force acting on the mirror is (c) 81.35 × 10−12 Jm−3 (d) 3 .3 × 10−3 Jm−3
. × 10−8 N
(a) 67 (b) 2.3 × 10−4 N 11 A radiation of energy E falls normally on a perfectly
(c) 10−3 N (d) zero reflecting surface. The momentum transferred to the
3 The magnetic field between the plates of radius 12 cm, surface is
separated by a distance of 4 mm of a parallel plate E 2E E
(a) (b) (c) E c (d)
capacitor of capacitance 100 pF along the axis of plates c c c2
having conduction current of 0.15 A, is 12 An electromagnetic wave in vacuum has the electric and
(a) zero (b) 1.5 T (c) 15 T (d) 0.15 T magnetic fields E and B, which are always perpendicular
4 Instantaneous displacement current of 1.0 A in the space to each other. The direction of polarisation is given by X
between the parallel plates of a1µF capacitor, can be and that of wave propagation by k.$ Then, ª AIEEE 2012
established by changing potential difference of (a) X || B and k$ || B×E (b) X || E and k$ || E×B
(a) 10−6 Vs–1 (b) 106 Vs–1 (c) X || B and k$ || E×B (d) X || B and k$ || B×E
(c) 10−8 Vs–1 (d) 108 Vs–1
13 An electromagnetic wave travels in vacuum along
5 A large parallel plate capacitor, whose plates have an z-direction E = (E 1 $i + E 2 $j ) cos (kz − ωt ). Choose the
area of 1 m 2 and are separated from each other by correct option from the following
1 mm, is being charged at a rate of 25 Vs −1. If the
(a) The associated magnetic field is given as
dielectric between the plates has the dielectric constant 1
B = (E1 $i − E $j) cos (kz − ωt )
10, then the displacement current at this instant is C
(a) 25 µA (b) 11µA (c) 2.2 µA (d) 1.1 µA (b) The associated magnetic field is given as
1
6 A parallel plate capacitor with plate area A and B = (E1 $i − E $j) cos (kz − ωt )
separation between the plates d, is charged by a C
(c) The given electromagnetic field is circularly polarised
constant current I. Consider a plane surface of area A /2
(d) The given electromagnetic wave is plane polarised
parallel to the plates and drawn simultaneously between
the plates. The displacement current through this area is 14 Match List I (Electromagnetic wave type) with List II (Its
(a) I (b)
I
(c)
I
(d)
I association/application) and select the correct option
2 4 8 from the choices given below the lists. ª JEE Main 2014
7 Select the correct statement from the following
ª JEE Main (Online) 2013 List I List II
A. Infrared waves 1. To treat muscular strain
(a) Electromagnetic waves cannot travel in vacuum
(b) Electromagnetic waves are longitudinal waves B. Radio waves 2. For broadcasting
(c) Electromagnetic waves are produced by charges C. X-rays 3. To detect fracture of bones
moving with uniform velocity D. Ultraviolet waves 4. Absorbed by the ozone layer of
(d) Electromagnetic waves carry both energy and the atmosphere
momentum as they propagate through space Codes
8 In an apparatus, the electric field was found to oscillate A B C D A B C D
with an amplitude of 18 Vm −1. The magnitude of the (a) 4 3 2 1 (b) 1 2 4 3
oscillating magnetic field will be (c) 3 2 1 4 (d) 1 2 3 4
292 40 DAYS ~ JEE MAIN PHYSICS DAY TWENTY FIVE

15 Arrange the following electromagnetic radiations per Statement II Ultraviolet radiations are absorbed by the
quantum in the order of increasing energy. atmosphere.
A. Blue light B. Yellow light
17 Statement I If the earth did not have atmosphere, its
C. X-ray D. Radio wave average surface temperature would be lower than what is
ª JEE Main 2016 (Offline) now.
(a) D, B, A, C (b) A, B, D, C Statement II Greenhouse effect of the atmosphere would
(c) C, A, B, D (d) B, A, D, C
be absent, if the earth did not have atmosphere.
Direction (Q. Nos. 16-20) Each of these questions contains 18 Statement I Electromagnetic waves exert radiation
two statements : Statement I (Assertion) and Statement II pressure.
(Reason). Each of these questions also has four alternative Statement II Electromagnetic waves carry energy.
choices, only one of which is the correct answer. You have to
select one of the codes (a), (b), (c), (d) given below 19 Statement I Light is a transverse wave, but not an
(a) Statement I is true, Statement II is true; Statement II is electromagnetic wave.
the correct explanation for Statement I Statement II Maxwell showed that speed of
(b) Statement I is true, Statement II is true; Statement II is electromagnetic waves is related to the permeability and
not the correct explanation for Statement I the permittivity of the medium through which it travels.
(c) Statement I is true; Statement II is false 20 Statement I Out of radio waves and microwaves, the
(d) Statement I is false; Statement II is true radio waves undergo more diffraction.
16 Statement I Ultraviolet radiation being higher frequency Statement II Radio waves have greater frequency
waves are dangerous to human being. compared to microwaves. ª JEE Main (Online) 2013

DAY PRACTICE SESSION 2

PROGRESSIVE QUESTIONS EXERCISE

1 You are given a 2 µF parallel plate capacitor. How would 4 The ratio of contributions made by the electric field and
you establish an instantaneous displacement current of magnetic field components to the intensity of an
1 mA in the space between its plates? electromagnetic wave is
(a) By applying a varying potential difference of 500 V/s (a) c:1 (b) c 2 :1 (c) 1:1 (d) c :1
(b) By applying a varying potential difference of 400 V/s
(c) By applying a varying potential difference of 100 V/s 5 An FM radio station, antenna radiates a power of 10 kW
(d) By applying a varying potential difference of 300 V/s at a wavelength of 3 m. Assume the radiated power is
2 A uniform but time varying magnetic field B(t ) exists in a confined to and is uniform over a hemisphere with
circular region of radius a and is B(t) antenna at its centre. E max at a distance of 10 km from
directed into the plane of the paper × × × × × × P antenna is
as shown in the figure. The r
× × × × × ×
(a) 0.62 NC −1 (b) 0.41 NC −1
magnitude of the induced electric × × × × × × ×
(c) 0.31 NC −1 (d) 0.10 NC −1
× ×a × × × ×
field at a point P, a distance r from × × × × × ×
the centre of the circular region ×× × × × × 6 Assume that all the energy from a 1000 W lamp is
×× × × × ×
(a) increases with r × × × × × × radiated uniformly, then the amplitude of electric field of
(b) decreases with r radiation at a distance of 2 m from the lamp is
1
(c) decreases as 2 (a) 245.01 V/m (b) 17 V/m
r (c) 0 (d) 2.96 V/m
(d) zero
7 A red LED emits light at 0.1W uniformly around it. The
3 The magnetic field of a beam emerging from a filter facing amplitude of the electric field of the light at a distance of
a flood light is given by 1m from the diode is ª JEE Main 2015
B = 12 × 10−8 sin (1.20 × 107 z − 3.60 × 1015 t ) T. What is the
(a) 1.73 V/m
average intensity of the beam?
(b) 2.45 V/m
(a) 1.7 W/ m2 (b) 2.3 W/m2 (c) 5.48 V/m
2
(c) 2.7 W/m (d) 3.2 W/m2 (d) 7.75 V/m
DAY TWENTY FIVE ELECTROMAGNETIC WAVES 293

8 In a transverse wave, the distance between a crest and εr1 εr1 εr1 1 εr1 1
(a) =4 (b) =2 (c) = (d) =
neighbouring through at the same instant is 4.0 cm and εr2 εr2 εr2 4 εr2 2
the distance between a crest and trough at the same
10 An electromagnetic wave of frequency ν = 3 .0 MHz
place is 1.0 cm. The next crest appears at the same
passes from vacuum into a dielectric medium with
place after a time interval of 0.4 s. The maximum speed
permittivity ε = 4.0. Then,
of the vibrating particles in the medium is
(a) wavelength is doubled and the frequency remains
ª JEE Main (Online) 2013
unchanged
3π
(a) cm/s (b) 5 π cm/s (b) wavelength is doubled and frequency becomes half
2
π (c) wavelength is halved and frequency remains unchanged
(c) cm/s (d) 2π cm/s
2 (d) wavelength and frequency both remain unchanged
9 An EM wave from air enters a medium. The electric 11 The magnetic field at a point between the plates of a
 z  capacitor at a perpendicular distance R from the axis of
fields are E1 = E 01x$ cos 2πν  − t   in air and
 c  the capacitor plate radius R, having the displacement
E2 = E 02x cos[k ( 2z − ct )] in medium, where the wave
$ current ID is given by
number k and frequency ν refer to their values in air. µ IDr µ 0 ID
(a) (b)
The medium is non-magnetic. 2 πR 2 2 πR
µ 0 ID
(c) (d) zero
If εr1 and εr2 refer to relative permittivities of air and πr 2
medium respectively, which of the following options is
correct? ª JEE Main 2018

ANSWERS
SESSION 1 1 (c) 2 (a) 3 (a) 4 (b) 5 (c) 6 (b) 7 (d) 8 (b) 9 (b) 10 (a)
11 (b) 12 (b) 13 (d) 14 (d) 15 (a) 16 (b) 17 (a) 18 (b) 19 (d) 20 (c)

SESSION 2 1 (a) 2 (b) 3 (a) 4 (c) 5 (d) 6 (a) 7 (b) 8 (b) 9 (c) 10 (c)
11 (b)

Hints and Explanations

SESSION 1 3 As B ∝ r , since the point is on the axis, = 8.85 × 10−8 × 25
1 Both the energy densities are equal, where r = 0, so B = 0. = 2.2 × 10−6 = 2.2 µA
i.e. energy is equally divided between
electric and magnetic field. 4 As, Id = C  V  Q I = dQ 
 
6 Charge on the capacitor plates, at time t is,
t   dt  q = It
2 Let E = energy falling on the surface V I
Electric field between the plates at this
or = d instant,
per second = 10 J t C q It
Momentum of photons, E = =
h h =
1.0
= 106 Vs –1 A ε0 A ε0
p= = 10−6
λ c/f Electric flux through the given area,
5 As, C = ε0KA φ E =   E =
A It
hf E
= = d 2 2 ε0
c c
Therefore, displacement current,
On reflection, change in momentum (8.85 × 10−12 ) × 10 × 1
= dφ
per second = force 10 −3 Id = ε 0 E
2E 2 × 10 dt
= 2p = =
c 3 × 108 = 8.85 × 10−8 F d  It  I
= ε0  =
= 6.7 × 10−8 N d dV dt  2 ε0  2
∴ I = (CV ) = C
dt dt
294 40 DAYS ~ JEE MAIN PHYSICS DAY TWENTY FIVE

7 B E Electromagnetic wave 14 (a) Infrared waves are used to treat 19 In free space or vacuum, the speed of
muscular strain. electromagnetic waves is
Direction of
(b) Radio waves are used for 1
Propagation c = …(i)
broadcasting purposes. µ 0ε0
B E (c) X-rays are used to detect fracture Here, µ 0 = 4 π × 10−7 Ns 2 C – 2 is
As electromagnetic waves contains both of bones.
electric field and magnetic field. It carry (d) Ultraviolet waves are absorbed permeability (constant) of free space.
both energy and momentum according by ozone. ε0 = 8.85418 × 10−12 C2 N –1 m – 2 is the
to de-Broglie wave particle duality of
15 As, we know energy liberated, permittivity of free space. On substituting
radiations.
hc the values in Eq. (i), we have
8 Here, E 0 = 18 Vm –1 E =
λ 1
E0 18 c =
∴ B0 = = i.e. E ∝
1 4 π × 10−7 × 8.85418 × 10−12
c 3 × 108 λ
= 2.99792 × 108 ms –1
= 6 × 10−8 T So, lesser the wavelength, than greater
will be energy liberated by This is same as the speed of light in
9 E = B × c keep value of electric field electromagnetic radiations per vacuum. From this we conclude that
⋅ c | = 20 × 10−9 × 3 × 108
|E| = |B|| quantum. light is an electromagnetic wave.
= 6 V/m As, order of wavelength is given by
20 The frequency of radio waves less than
X-ray, VIBGYOR, Radio waves the frequency of microwaves.
10 Total average energy = ε0 E 2rms (C) (A) (B) (D)
−12
Q Frequency of radio waves = 3 × 108 Hz
= 8.85 × 10 × (720) 2
∴Order of increasing energy of and frequency of microwaves = 1010 Hz
= 4.58 × 10 −6
Jm −3 electromagnetic radiations per
∴ ν radio waves < ν microwaves
quantum.
11 Initial momentum of surface, p i = E /c ⇒ D<B<A<C
SESSION 2
where, c = velocity of light (constant). 16 Ultraviolet radiations are
Since, the surface is perfectly reflecting, 1 Given, capacitance of capacitor, C = 2µF
electromagnetic waves. The wavelength
so the same momentum will be ° Displacement current, Id = 1 mA
of these waves ranges between 4000 A Charge, q = CV
reflected completely. Final momentum, °
p f = E /c [negative value] to 100 A, i.e. of smaller wavelength and Id dt = CdV [Q q = It ]
∴ Change in momentum, higher frequency. They are absorbed by dV
or Id = C
∆ p = p f − pi ozone layer of stratosphere in dt
atmosphere. They cause skin diseases dV
=−
E
−
E
=−
2E 1 × 10 = 2 × 10−6 ×
−3
and they are harmful to eye and may dt
c c c
cause permanent blindness. dV 1 +3
Thus, momentum transferred to the or = × 10 = 500 V/s
surface is 17 Earth is heated by sun’s infrared dt 2
2E radiation.The earth also emits radiation Clearly, by applying a varying potential
∆ p ′ = |∆ p| = most in infrared region. These
c difference of 500 V/s, we would produce
radiations are reflected back by heavy a displacement current of desired value.
12 In electromagnetic wave, the direction gases like CO2 in atmosphere. These
of propagation of wave, electric field back radiation keep the earth’s surface 2 A time varying magnetic field produces
and magnetic field are mutually warm at night. This phenomenon is an electric field. The magnitude of the
perpendicular, i.e. wave propagates called greenhouse effect. When the electric field at a distance r from the
perpendicular to E and B or along E×B. atmosphere were absent, then centre of a circular region of radius a,
While polarisation of wave takes place where a time varying field B exists, is
temperature of the earth falls.
given by
parallel to electric field vector.
18 Electromagnetic waves have linear a2 dB
E =
13 Here, in electromagnetic wave, the momentum as well as energy. From this 2r dt
electric field vector is given as we conclude that, we can exert At r = a,
E = (E1 $i+E2 $j )cos (kz − ωt ) radiation pressure by making a beam of
E =  
a dB
electromagnetic radiation fall on an  2  dt
In electromagnetic wave, the associated object. Let us assume that object is free
magnetic field vector, to move and that the radiation is This is the value of E at the edge of the
E E $i+E2 $j entirely absorbed in the object during circular region. For r > a, E decreases
B= = 1 cos(kz − ωt )
c c time interval ∆t. The object gains an with r.
Also, E and B are perpendicular to each energy ∆U from the radiation. Maxwell 3 Magnetic field, B = B 0 sin ω t .
other and the propagation of showed that the object also gains linear
Given equation,
electromagnetic wave is perpendicular momentum, the magnitude ∆p of the
change in momentum of the object is B = 12 × 10−8
to E as well as B, so the given
related to the energy change ∆U as sin (1.20 × 107 z − 3.60 × 1015t ) T
electromagnetic wave is plane
∆U
polarised. ∆p = (total absorption)
c
DAY TWENTY FIVE ELECTROMAGNETIC WAVES 295

On comparing this equation with 9 Speed of progressive wave is given by,

6 Poynting vector, S = E × H
standard equation, we get ω
v =
B 0 = 12 × 10−8 = EH sin 90° = EH k
1000 W As electric field in air is,
The average intensity of the beam, Energy of lamp =
πr 2 2 πνz
I av =
1 B 20
⋅c E1 = E 01 x$ cos  − 2 πνt 
2 µ0 =
1000
Jm −2 s −1  c 
−8 2 π × 22 2 πν
1 (12 × 10 ) × 3 × 10 8
∴ Speed in air = =c
= × S represents energy flow per unit area  2 πν 
2 4 π × 10−7  
per second, we have  c 
= 1.7 W/m2 1000 1
EH = = 79.61, Also, c = …(i)
π × 22 µ 0εr1 ε0
4 Intensity in terms of electric field,
1 E
U av = ε0E 20 = 377 In medium,
2 H E2 = E 02 x$ cos (2kz − kct )
E
Intensity in terms of magnetic field, EH × = 79.61 × 377 = 300159. kc c
H ∴ Speed in medium = =
1 B 20 2k 2
U av = ⇒ E = 300159 .
2µ0 c 1
Also, = …(ii)
= 173.25V/m 2 µ 0εr2 ε0
Now, taking the intensity in terms of Amplitude of electric field of radiation
electric field, is As medium is non-magnetic medium,
µ medium = µ air
1 E 0 = E 2 = 245.01 V/m
(U av ) electric field = ε0E 20 On dividing Eq. (i) by Eq. (ii), we have
2
1 7 Consider the LED as a point source of εr2 ε 1
= ε0(cB 0 )2 (Q E 0 = cB 0) 2= ⇒ r1 =
light. Let power of the LED is P. εr1 εr2 4
2
Intensity at r from the source,
1
But, c = P 10 In vacuum, ε0 = 1
µ 0ε0 I = …(i)
4 πr 2 In medium, ε = 4
1 1
∴ (U av ) electric field = ε 0 × B 20 As we know that , ∴ Refractive index,
2 µ 0ε0 1
I = ε0E 20c …(ii) ε 4
1 B 20 2 µ = = =2
= ⋅ ε0 1
2 µ0 From Eqs.(i) and (ii), we can write
P 1 Wavelength,
Thus, the energy in electromagnetic = ε0E 20c
4 πr 2 2 λ λ
wave is divided equally between λ′ = =
electric field vector and magnetic field 2P µ 2
or E 20 =
vector. Therefore, the ratio of 4 πε0r 2c and wave velocity,
contributions by the electric field and c
v = =
c Q µ = c 
magnetic field components to the 2 × 01
. × 9 × 109 µ  v 
= 2
intensity of an electromagnetic wave 1 × 3 × 108 Hence, it is clear that wavelength and
is 1:1.
or E 20 =6 velocity will become half, but frequency
ρ 104 10−4 remains unchanged when the wave is
5 As, I = = = ⇒ E0 = 6 = 2.45 V /m
2 πr 2
2 π (10 × 10 ) 3 2
2π passing through any medium.

1 8 Given, λ = 4 cm 11 According to the Ampere-Maxwell’s law,

and I = ε0 E 20 c 4
2 for a closed surface,
∴ λ = 16 cm and T = 0.4 s
⇒ E0 =
2I
As, f λ × T = 2 π ∫ B⋅dl = µ 0I D
ε0c 2π 5 π −1
⇒ f = = s As, B (2πR ) = µ 0 I D
10−4 16 × 0.4 16 µ I
= ⇒ B= 0 D
π × 8.85 × 10−12 × 3 × 108 5π 2 πR
Now, v = f λ= × 16 = 5π cm/s
16
. NC −1
or E 0 = 010