Electromagnetic Waves


DISPLACEMENT CURRENT

(i) Circuit with Capacitor

Let us consider a circuit containing the capacitor as shown in fig. 21.1

Circuit Diagram: A capacitor connected in a circuit with current flowing through connecting wires but not through the space between plates

(ii) Ampere's Circuital Law

According to Ampere's circuital law, the line integral of magnetic field along any closed path is μ₀ times the total current enclosed by the closed path. Mathematically:

Ampere's Circuital Law

In this law it is assumed that conduction current flows through the connecting wires charging the condenser plates but no current flows in the space in between the plates. Actually it is not true.

Illustration 1: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency 2 × 10¹⁰ Hz and amplitude 48 V/m. The amplitude of oscillating magnetic field will be:

Options:

(A) 1.6 × 10⁻⁸ Wb/m²

(B) 16 × 10⁻⁸ Wb/m²

(C) 12 × 10⁻⁷ Wb/m²

(D) 1.2 × 10⁻⁷ Wb/m²

Solution:

(B) Oscillating magnetic field

B = E/c = 48/(3 × 10⁸) = 16 × 10⁻⁸ Wb/m²

(iii) Maxwell's Modification

When the circuit is closed, conduction current flows from the plate P of the capacitor to the other plate Q through the conducting wires. Maxwell suggested that due to time varying electric field between the plates, an electric current, called displacement current (I_D), also flows across the space between the plates of the capacitor.

(iv) Continuous Current Flow

Thus, there is a continuous flow of current in a capacitive circuit also, through the conducting wire there is flow of conduction current I_C and through the space across the plates of capacitor, there is flow of displacement current I_D.

(v) Ampere-Maxwell's Circuital Law

Maxwell pointed out that in Ampere's circuital law, the current I should be treated as total current i.e., the sum of the conduction current I_C and displacement current I_D and modified the law as:

∮ B⃗ · dl⃗ = μ₀(I_C + I_D)

It is called the Ampere-Maxwell's circuital law.

(vi) Displacement Current Definition

The displacement current is defined as:

I_D = ε₀ (dΦ_E/dt)

where Φ_E is the electric flux linked between the plates of the capacitor at any instant.

Therefore Ampere-Maxwell circuital law may be expressed as:

∮ B⃗ · dl⃗ = μ₀[I_C + ε₀(dΦ_E/dt)]

Illustration 2: A parallel plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400V. If the plate area is 60 cm², then the value of displacement current for 10⁻⁶ sec. will be:

Options:

(A) 1.062A

(B) 1.062 × 10⁻²A

(C) 1.062 × 10⁻³A

(D) 1.062 × 10⁻⁴A

Solution:

(D) I_D = ε₀(dΦ_E/dt) = ε₀(d(EA)/dt) = ε₀A(dE/dt) = ε₀A(V/d)/t

I_D = (8.85 × 10⁻¹² × 400 × 60 × 10⁻⁴)/(2 × 10⁻³ × 10⁻⁶) = 1.062 × 10-4A

Illustration 3: In an electric circuit, there is a capacitor of reactance 100Ω connected across the source of 220V. The displacement current will be:

Options:

(A) 2.2A

(B) 0.22A

(C) 4.2A

(D) 2.4A

Solution:

(A) Displacement current and conduction current are equal.

ID = E/Z = 220/100 = 2.2A

(vii) & (viii) Important Properties

  • The conduction current and the displacement current are always equal, i.e., IC = ID
  • Like conduction current, the displacement current is also the source of magnetic field

Also Read: Electrostatic Potential and Capacitance

MAXWELL'S EQUATIONS AND LORENTZ FORCE

The existence of electro-magnetic waves that propagate through the space in the form of varying electric and magnetic fields has been predicted by the four basic laws of electromagnetism which are called Maxwell's equations.

(i) Maxwell's First Equation - Gauss's Law in Electrostatics

It states that the total electric flux through any closed surface is equal to 1/ε₀ times the net charge enclosed by

Mathematically,

∮ E⃗ · ds⃗ = q/ε₀

This equation is called Maxwell's first equation.

(ii) Maxwell's Second Equation - Gauss's Law in Magnetism

It states that the net magnetic flux crossing any closed surface is always zero.

Mathematically,

∮ B⃗ · ds⃗ = 0

This equation is called Maxwell's second equation. A direct consequence of this equation is that the magnetic monopoles do not exist.

(iii) Maxwell's Third Equation - Faraday's Law of Electromagnetic Induction

It states that the induced emf produced in a circuit is numerically equal to the rate of change of magnetic flux through it.

Mathematically,

ε = -dΦ_B/dt

But ε = ∮ E⃗ · dl⃗ = Line integral of electric field.

∮ E⃗ · dl⃗ = -dΦ_B/dt

This equation is called Maxwell's third equation.

The negative sign in this equation indicates that the induced emf produced opposes the rate of change of magnetic flux.

Illustration 4: A point source of electromagnetic radiation has an average power output of 800W. The maximum value of electric field at a distance 3.5m from the source will be:

Options:

(A) 56.7 V/m

(B) 62.6 V/m

(C) 39.3 V/m

(D) 47.5 V/m

Solution:

(B) Intensity of electromagnetic wave given is by

 Intensity of electromagnetic wave given is by

Illustration 5: In the above problem, the maximum value of magnetic field will be:

Options:

(A) 2.09 × 10⁻⁵ T

(B) 2.09 × 10⁻⁶ T

(C) 2.09 × 10⁻⁷ T

(D) 2.09 × 10⁻⁷ T

Solution:

(C) The maximum value of magnetic field is given by

B_m = E_m/c = 62.6/(3 × 10⁸) = 2.09 × 10⁻⁷ T

(iv) Maxwell's Fourth Equation - Maxwell-Ampere Circuital Law

It states that the line integral of magnetic field along a closed path is equal to μ₀ times the total current (i.e., sum of conduction and displacement currents threading the surface bounded by that closed path)

Mathematically,

∮ B⃗ · dl⃗ = μ₀[I_C + ε₀(dΦ_E/dt)]

This equation is called Maxwell's fourth equation.

(v) Lorentz Force

The vector sum of electric force and magnetic force on any charged particle is called the Lorentz force.

F⃗ = q[E⃗ + (v⃗ × B⃗)]

The above five equations give a complete description of all electromagnetic interactions.

Do Check - Current Electricity

PRODUCTION OF ELECTROMAGNETIC WAVES

(i) Maxwell's Theory

According to Maxwell, an accelerated charge sets up a magnetic field in its neighbourhood. The magnetic field, in turn, produces an electric field in that region. Both these fields vary with time and act as sources for each other.

(ii) Oscillating Charge

As oscillating charge is accelerated continuously, it will radiate electromagnetic waves continuously.

(iii) Hertz's Demonstration (1888)

In 1888, Hertz demonstrated the production of electromagnetic apparatus is shown schematically in fig.

Hertz's Apparatus:

Input → Transmitter (+/-) → Receiver

Induction coil connected to spherical electrodes

(iv) Transmitter Operation

An induction coil is connected to two spherical electrodes with a narrow gap between them. It acts as a transmitter. The coil provides short voltage surges to the spheres making one positive and the other negative. A spark is generated between the spheres when the voltage between them reaches the breakdown voltage for air. As the air in the gap is ionised, it conducts more rapidly and the discharge between the spheres becomes oscillatory.

(v) LC Circuit Equivalent

The above experiment arrangement is equivalent to an LC circuit, where the inductance is that of the loop and the capacitance is due to the spherical electrodes.

(vi) Wave Radiation

Electromagnetic waves are radiated at very high frequency (100 MHz) as a result of oscillation of free charges in the loop.

(vii) & (viii) Detection

  • Hertz was able to detect these waves using a single loop of wire with its own spark gap (the receiver)
  • Sparks were induced across the gap of the receiving electrodes when the frequency of the receiver was adjusted to match that of the transmitter

HISTORY OF ELECTROMAGNETIC WAVES

(i) 1865 - Maxwell's Prediction

In year 1865, Maxwell predicted the electromagnetic waves theoretically. According to him, an accelerated charge sets up a magnetic field in its neighborhood.

(ii) 1887 - Hertz's Production

In 1887, Hertz produced and detected electromagnetic waves experimentally at wavelength of about 6m.

(iii) J.C. Bose's Contribution

Seven year later, J.C. Bose became successful in producing electromagnetic waves of wavelength in the range 5 mm to 25 mm.

(iv) 1896 - Marconi's Discovery

In 1896, Marconi discovered that if one of the spark gap terminals is connected to an antenna and the other terminal is earthed, the electromagnetic waves radiated could go upto several kilometers.

(v) Antenna System

The antenna and the earth wires form the two plates of a capacitor which radiates radio frequency waves. These waves could be received at a large distance by making use of an antenna earth system as detector.

(vi) 1899 - First Wireless Communication

Using these arrangements; in 1899 Marconi first established wireless communication across the English channel i.e. across a distance of about 50 km.

PROPERTIES OF ELECTROMAGNETIC WAVES

(i) Wave Equations

The electric and magnetic fields satisfy the following wave equations, which can be obtained from Maxwell's third and fourth equations.

Wave Equations

(ii) Speed of Light

Electromagnetic waves travel through vacuum with the speed of light c, where

c = 1/√(μ₀ε₀) = 3 × 10⁸ m/s

(iii) Transverse Nature

The electric and magnetic fields of an electromagnetic wave are perpendicular to each other and also perpendicular to the direction of wave propagation. Hence, these are transverse waves.

(iv) Relationship between E and B

The instantaneous magnitude of E⃗ and B⃗ in an electromagnetic wave are related by the expression

E/B = c

(v) Energy Flow - Poynting Vector

Electromagnetic waves carry energy. The rate of flow of energy crossing a unit area is described by the Poynting vector S⃗

where

S⃗ = (1/μ₀)(E⃗ × B⃗)

(vi) Radiation Pressure

Electromagnetic waves carry momentum and hence can exert pressure (P) on surfaces, which is called radiation pressure. For a perfectly absorbing surface

P = S/c

and if incident on a perfectly reflecting surface

P = 2S/c

(vii) Sinusoidal Wave Representation

The electric and magnetic fields of a sinusoidal plane electromagnetic wave propagating in the positive x-direction can also be written as

E⃗ = E_m sin(kx - ωt)

B⃗ = B_m sin(kx - ωt)

where ω is the angular frequency of the wave and k is wave number which are given by

ω = 2πf and k = 2π/λ

(viii) Intensity

The intensity of a sinusoidal plane electromagnetic wave is defined as the average value of Poynting vector taken over one cycle.

Intensity

(ix) Fundamental Sources

The fundamental sources of electromagnetic waves are accelerating electric charges. For example radio waves emitted by an antenna arise from the continuous oscillations (and hence acceleration) of charges within the antenna structure.

(x) Superposition Principle

Electromagnetic waves obey the principle of superposition.

(xi) Light Vector

The electric vector of an electromagnetic field is responsible for all optical effects. For this reason electric vector is called a light vector.

Illustration 6: DIn an electromagnetic wave, the amplitude of electric field is 1V/m. The frequency of wave is 5 × 10¹⁴ Hz. The wave is propagating along z-axis. The average energy density of electric field, in Joule/m³, will be:

Options:

(A) 1.1 × 10⁻¹¹

(B) 2.2 × 10⁻¹²

(C) 3.3 × 10⁻¹³

(D) 4.4 × 10⁻¹⁴

Solution:

(B) Average energy density is given by

Average energy density is given by

Illustration 7: To establish an instantaneous displacement current of 2A in the space between two parallel plates of 1μF capacitor, the potential difference across the capacitor plates will have to be changed at the rate of:

Options:

(A) 4 × 10⁴ V/s

(B) 4 × 10⁶ V/s

(C) 2 × 10⁴ V/s

(D) 2 × 10⁶ V/s

Solution: (D)

ELECTROMAGNETIC SPECTRUM

(i) Definition

The orderly distribution of electromagnetic waves (on the basis of wavelength or frequency) in the form of distinct groups having widely differing properties is called the electromagnetic spectrum.

The following table gives a complete and detailed picture of electromagnetic spectrum

(ii) Various parts of electromagnetic spectrum

S. No. Radiation Discoverer How produced Wavelength range Frequency range Energy range Properties Application
1. γ-Rays Henry Becquerel and Madam Curie Due to decay of radioactive nuclei 10⁻¹⁴ m to 10⁻¹⁰ m 3 × 10²² Hz to 3 × 10¹⁸ Hz 10⁷eV – 10⁴ eV (a) High penetrating power
(b) Uncharged
(c) Low ionising power
(a) Gives information on nuclear structure
(b) Medical treatment etc.
2. X-Rays Roentgen Due to collisions of high energy electrons with heavy targets 6 × 10⁻¹⁰ m to 10⁻⁹ m 5 × 10¹⁹ Hz to 3 × 10¹⁷ Hz 2.4 × 10⁵ eV to 1.2 × 10³ eV (a) Low penetrating power
(b) Other properties similar to γ-rays except wavelength
(a) Medical diagnosis and treatment
(b) Study of crystal structure
(c) Industrial radiography
3. Ultraviolet Rays Ritter By ionised gases, sun lamp spark etc. 6 × 10⁻¹⁰ m to 3.8 × 10⁻⁷ m 3 × 10¹⁷ Hz to 5 × 10¹⁹ Hz 2×10³ eV to 3eV (a) All properties similar to visible light
(b) Photo-electric effect
(a) To detect adulteration
(b) Sterilization of water due to its destructive action on bacteria
4. Visible Light Newton Outer orbit electron transitions in atoms, gas discharge tube, incandescent 3.8 × 10⁻⁷ m to 7.8 × 10⁻⁷ m 8 × 10¹⁴ Hz to 4 × 10¹⁴ Hz 3.2 eV to 1.6eV (a) Sensitive to human eye (a) To see objects
(b) To study molecular structure
- (a) Violet - - 3.9×10⁻⁷ m to 4.55×10⁻⁷ m 7.69×10¹⁴ Hz to 6.59×10¹⁴ Hz - - -
- (b) Blue - - 4.55×10⁻⁷ m to 4.92×10⁻⁷ m 6.59×10¹⁴ Hz to 6.10×10¹⁴ Hz - - -
- (c) Green - - 4.92×10⁻⁷ m to 5.77×10⁻⁷ m 6.10×10¹⁴ Hz to 5.20×10¹⁴ Hz - - -
- (d) Yellow - - 5.77×10⁻⁷ m to 5.97×10⁻⁷ m 5.20×10¹⁴ Hz to 5.03×10¹⁴ Hz - - -
- (e) Orange - - 5.97×10⁻⁷ m to 6.22×10⁻⁷ m 5.03×10¹⁴ Hz to 4.82×10¹⁴ Hz - - -
- (f) Red - - 6.22×10⁻⁷ m to 7.80×10⁻⁷ m 4.82×10¹⁴ Hz to 3.84×10¹⁴ Hz - - -
5. Infra-Red waves William Herschel (a) Rearrangement of outer orbitals electrons in atoms and molecules
(b) Change of molecular vibrational and rotational energies
(c) By bodies at high temperature
7.8×10⁻⁷ m to 10⁻³ m 4×10¹⁴ Hz to 3×10¹¹ Hz 1.6 eV to 1.6×10⁻³ eV (a) Thermal effect
(b) All properties similar to those of light except wavelength
(a) Used in industry, medicine and astronomy
(b) Used for fog or have photography
6. Microwaves Hertz Special electronic devices such as klystron tube 10⁻³ m to 0.3m 3×10¹¹ Hz to 10⁹ Hz 10⁻³eV to 10⁻⁵eV (a) Phenomena of reflection, refraction and diffraction (a) Radar and telecommunication
(b) Analysis of fine details of molecular structure
7. Radio waves Marconi Oscillating circuits 0.3m to few kms 10⁹ Hz to few Hz 10⁻³eV to 0 (a) Exhibit wave like properties more than particle like properties (a) Radio communication
(b) Radar, Radio and satellite communication (Microwaves)

Subparts of Radio spectrum

Type Wavelength Frequency Applications
(a) Super High (SHF) 0.01m to 0.1m 3×10¹⁰ Hz to 3×10⁹ Hz Radar and television broadcast
(b) Ultra High (UHF) 0.1m to 1m 3×10⁹ Hz to 3×10⁸ Hz Short distance communication
(c) Very High (VHF) 1m to 10m 3×10⁸ Hz to 3×10⁷ Hz Television communication
High frequency 10m to 100m 3×10⁷ Hz to 3×10⁶ Hz Medium distance communication
Medium frequency 100m to 1000m 3×10⁶ Hz to 3×10⁵ Hz Marine and navigation use
Low Frequency 1000m to 10000m 3×10⁵ Hz to 3×10⁴ Hz Long range communication
Very low frequency 10000m to 30000m 3×10⁴ Hz to 10³ Hz Long distance communication

EARTH'S ATMOSPHERE AND ELECTROMAGNETIC WAVES

(i) Earth's Atmosphere

The gaseous envelop surrounding the earth is called earth's atmosphere.

(ii) Composition

It mainly consists of nitrogen 78% and oxygen 21% alongwith a little portion of argon, carbon-di-oxide, water vapour, hydrocarbons, sulphur compounds and dust particles.

(iii) Density Variation

The density of atmospheric air goes on decreasing gradually as we go up.

(iv) Atmospheric Layers

The earth's atmosphere has no sharp boundary. However, it has been divided into various regions as given below:

  • (a) Troposphere

It extends upto a height of 12 km from earth's surface. The temperature in this region decreases from 298K to 220K and conductivity increases. All climatic changes occur in this region.

  • (b) Stratosphere

It extends from 12 km to 50 km after troposphere. At the upper part of this region, approximately 20km thick, most of ozone of atmosphere is concentrated. This layer is called as ozone layer. This layer absorbs very large portion of ultraviolet radiations coming from sun, therefore its temperature increases from 220K to 280K.

  • (c) Mesosphere

It extends from 50km to 80 km after stratosphere. In this region the temperature decreases from 280K to 180K

  • (d) Ionosphere

It extends from 80 km to 400 km after mesosphere. The temperature of this region rises from 180K to 700K. In this region ultraviolet radiation coming from sun cause ionisation, therefore this part mostly consists of free electrons and positive ions. The concentration of free electrons is found to be very large in a region beyond 110 km from earth's surface which extends vertically for a few kilometers and is called Kennelly Heaviside layer. Beyond this layer the concentration of free electrons decreases considerably until a height of about 250 km. Beyond it there is another layer of electrons, called Appleton layer.

(v) Greenhouse Effect

The atmosphere is transparent to visible radiations, but most infrared (heat) radiations are not allowed to pass through. The energy from the sun heats the earth which then starts emitting radiations like any other hot body. However, since the earth is much colder than sun, its radiations are mainly in the infra red region. These radiations are unable to cross the lower atmosphere and are reflected back. Low lying clouds also reflect back the infra red radiations. As such, the earth's surface warm at night. This phenomenon is called the Green house effect.

(vi) Propagation of Radio waves

  • (a) Low frequency waves - the AM band

Radio waves having wavelengths of 10m or more (frequency less than 30 MHz) are said to constitute the AM band. The lower atmosphere is transparent to these waves, but the ionosphere reflects them back. A signal transmitted from a certain point can be received at another point in two possible ways - directly along the surface of the earth (called ground wave) and after reflection from ionosphere (called sky wave). Waves having frequencies upto about 1500kHz (Wavelength above 200m) are mainly transmitted through ground because low frequency sky waves lose their energy very quickly than the sky waves. Therefore, higher frequencies are mainly transmitted through sky. These two regions of the AM band are called medium wave and short wave bands respectively.

  • (b) High frequency waves - Television transmission

Above a frequency of about 40MHz the ionosphere does not reflect the wave toward the earth. The television signals have frequencies in the range 100-200 MHz. Therefore TV transmission via the sky is not possible - only direct reception via the ground is possible. Therefore, in order to have larger coverage, the transmission has to be done through very tall antennas. The height of transmitting antenna for TV telecast is given by

h = d2/(2Re)

where d is the radius of the area to be covered for TV telecast and Re is the radius of earth.

Illustration 8: A T.V. tower has a height of 100m. How much population is covered by T.V. broadcast, if the average population density around the tower is 1000/km²?

Options:

(A) 39.5 × 105

(B) 19.5 × 106

(C) 29.5 × 107

(D) 9 × 104

Solution:

(A) Radius of the area covered by T.V. telecast

d = √(2hRe)

Total population covered = πd² × population density

= 2πhRe × population density = 2 × 3.14 × 100 × 6.4 × 10⁶ × (1000/10⁶) = 39.503 × 10⁵

Illustration 9: An electromagnetic radiation has an energy 14.4 KeV. To which region of electromagnetic spectrum does it belong?

Options:

(A) Infra red region

(B) Visible region

(C) X-ray region

(D) Y-ray region

Solution:

 An electromagnetic radiation has an energy 14.4 KeV. To which region of electromagnetic spectrum does it belong?

ASSIGNMENT Questions

Question: If E and B are the electric and magnetic field vectors of electromagnetic waves then the direction of propagation of electromagnetic wave is along the direction of:

(A) E

(B) B

(C) E × B

(D) None of these

Solution: (C)

Question: The charge on a parallel plate capacitor is varying as q = q₀ sin 2πnt. The plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is:

(A) q₀/ε₀A

(B) q₀ sin2πnt/ε₀

(C) 2πnq₀ cos2πnt

(D) 2πnq₀ cos2πnt/ε₀

Solution: (C) I_D = dq/dt = d/dt(q₀sin2πnt) = 2πnq₀ cos2πnt

Question: The value of magnetic field between plates of capacitor, at distance of 1m from centre where electric field varies by 10¹⁰ V/m/s will be:

(A) 5.56μT

(B) 5.56T

(C) 5.56mT

(D) 55.6nT

Solution: (D) B = (μ₀ε₀r/2) × (dE/dt) = (1/(2 × 9 × 10¹⁶)) × 1 × 10¹⁰ = 5.56 × 10⁻⁸ T

Question: The electromagnetic waves do not transport:

(A) Energy

(B) Charge

(C) Momentum

(D) Information

Solution: (B)

Question: A capacitor is connected in an electric circuit. When key is pressed, the current in the circuit is:

(A) Zero

(B) Maximum

(C) Any transient value

(D) Depends on capacitor used

Solution: (B)

Question: Displacement current is continuous:

(A) When electric field is changing in the circuit

(B) When magnetic field is changing in the circuit

(C) In both types of fields

(D) Through wires and resistance only

Solution: (A)

Question: Instantaneous displacement current 1A in the space between the parallel plates of 1μF capacitor can be established by changing the potential difference at the rate of:

(A) 0.1 V/s

(B) 1 V/s

(C) 10⁶ V/s

(D) 10⁻⁶ V/s

Solution: (C) I_D = C(dV/dt) or dV/dt = I_D/C = 1/(10⁻⁶) = 10⁶ V/s

Question: The magnetic field between the plates of a capacitor when r > R is given by:

(A) μ₀I_D r/(2πR²)

(B) μ₀I_D/(2πR)

(C) μ₀I_D/(2πr)

(D) None of these

Solution: (C) According to Ampere's law, when r > R, B = μ₀I_D/(2πr)

Question: The magnetic field between the plates of a capacitor is given by B = μ₀I_D r/(2πR²):

(A) r ≥ R

(B) r ≥ R

(C) r < R

(D) r = R

Solution: (C) According to Ampere's law, when r < R, B = μ₀I_D r/(2πR²)

Question: The conduction current is the same as displacement current when the source is:

(A) A.C. only

(B) D.C. only

(C) Both A.C. and D.C.

(D) Neither for A.C. nor for D.C.

Solution: (A)

Question: The wave function (in S.I. units) for an electromagnetic wave is given as: ψ(x, t) = 10³ sin (3 × 10⁶x – 9 × 10¹⁴t) The speed of the wave is:

(A) 9 × 10¹⁴ m/s

(B) 3 × 10⁸ m/s

(C) 3 × 10¹⁶ m/s

(D) 3 × 10⁷ m/s

Solution: (B) c = ω/k = (9 × 10¹⁴)/(3 × 10⁶) = 3 × 10⁸ m/s

Question: In the above problem, wavelength of the wave is:

(A) 666 nm

(B) 666Å

(C) 666μm

(D) 6.66 nm

Solution: (A) ψ(x,t) = 10³ sin[3 × 10⁶(x – 3 × 10⁸ t)]
Comparing it with ψ(x,t) = a sin[2π/λ(x – vt)]
2π/λ = 3 × 10⁶
or λ = (2π × 10⁹)/(3 × 10⁶) = 666 nm

Question: The Maxwell's four equations are written as:
(i) ∮ E⃗ · ds⃗ = q/ε₀
(ii) ∮ B⃗ · ds⃗ = 0
(iii) ∮ E⃗ · dl⃗ = -d/dt ∮ B⃗ · ds⃗
(iv) ∮ B⃗ · ds⃗ = μ₀I + μ₀ε₀ d/dt ∮ E⃗ · ds⃗
The equations which have sources of E⃗ and B⃗

(A) (i), (ii), (iii)

(B) (i), (ii)

(C) (i) and (iii)

(D) (i) and (iv)

Solution: (D)

Question: Out of the above four equations which do not contain source field are:

(A) (i) and (ii)

(B) (ii) only

(C) All of four

(D) (iii) only

Solution: (B)

Question: Out of four Maxwell's equations above, which one shows non-existence of monopoles?

(A) (i) and (iv)

(B) (ii) only

(C) (iii) only

(D) (iv) only

Solution: (B)

Question: Which of the above Maxwell's equations shows that electric field lines do not form closed loops?

(A) (i) only

(B) (ii) only

(C) (iii) only

(D) (iv) only

Solution: (A)

Question: In an electromagnetic wave the average energy density is associated with:

(A) Electric field only

(B) Magnetic field only

(C) Equally with electric and magnetic fields

(D) Average energy density is zero

Solution: (C)

Question: In an electromagnetic wave the average energy density associated with magnetic field will be:

(A) (1/2)LI²

(B) B²/(2μ₀)

(C) (1/2μ₀)B²

(D) (1/2)qB²

Solution: (B)

Question: In the above problem, the energy density associated with the electric field will be:

(A) (1/2)CV²

(B) (1/2)q²/C

(C) (1/2)εE²

(D) (1/2)ε₀E²

Solution: (D)

Question: If there were no atmosphere, the average temperature on earth surface would be:

(A) Lower

(B) Higher

(C) Same

(D) 0°C

Solution: (A) The green house effect would not have been possible without atmosphere. Hence temperature would be lower.

Question: In which part of earth's atmosphere is the ozone layer present?

(A) Troposphere

(B) Stratosphere

(C) Ionosphere

(D) Mesosphere

Solution: (B)

Question: Kennelly's Heaviside layer lies between:

(A) 50Km to 80 Km

(B) 80Km to 400 Km

(C) Beyond 110 Km

(D) Beyond 250 Km

Solution: (C)

Question: The ozone layer in earth's atmosphere is crucial for human survival because it:

(A) Has ions

(B) Reflects radio signals

(C) Reflects ultraviolet rays

(D) Reflects infra red rays

Solution: (C)

Question: The frequency from 3 × 10⁹ Hz to 3 × 10¹⁰ Hz:

(A) High frequency band

(B) Super high frequency band

(C) Ultra high frequency band

(D) Very high frequency band

Solution: (B)

Question: The frequency from 3 to 30 MHz is known as:

(A) Audio band

(B) Medium frequency band

(C) Very high frequency band

(D) High frequency band

Solution: (D)

Question: The AM range of radio waves have frequency:

(A) Less than 30 MHz

(B) More than 30 MHz

(C) Less than 20000 Hz

(D) More than 20000 Hz

Solution: (A)

Question: The displacement current flows in the dielectric of a capacitor

(A) Becomes zero

(B) Has assumed a constant value

(C) Is increasing with time

(D) Is decreasing with time

Solution: (C)

Question: Select wrong statement from the following Electromagnetic waves:

(A) Are transverse

(B) Travel with same speed in all media

(C) Travel with the speed of light

(D) Are produced by acceleration charge

Solution: (B)

Question: The waves related to tele-communication are:

(A) Infra red

(B) Visible light

(C) Microwaves

(D) Ultraviolet rays

Solution: (C)

Question: Electromagnetic waves do not transport:

(A) Energy

(B) Charge

(C) Momentum

(D) Information

Solution: (B)

Question: The nature of electromagnetic wave is:

(A) Longitudinal

(B) Longitudinal stationary

(C) Transverse

(D) Transverse stationary

Solution: (C)

Question: Greenhouse effect keeps the earth surface:

(A) Cold at night

(B) Dusty and cold

(C) Warm at night

(D) Moist

Solution: (C)

Question: A parallel plate capacitor consists of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source. The charging current is constant and is equal to 0.15 A. The rate of change of potential difference between the plates will be:

(A) 8.173 × 10⁷ V/s

(B) 7.817 × 10⁸ V/s

(C) 1.873 × 10⁹ V/s

(D) 3.781 × 10¹⁰ V/s

Solution: (C)

dV/dt = I_d/(ε₀A) = (0.15 × 5 × 10⁻³)/(8.85 × 10⁻¹² × 3.14 × 0.0144) = 1.873 × 10⁹ V/s

Question: In the above problem, the displacement current is:

(A) 15A

(B) 1.5A

(C) 0.15A

(D) 0.015A

Solution: (C) I_D = I_C = 0.15A

Question: The wave emitted by any atom or molecule must have some finite total length which is known as the coherence length. For sodium light, this length is 2.4cm. The number of oscillations in this length will be:

(A) 4.068 × 10⁵

(B) 4.068 × 10⁶

(C) 4.068 × 10⁷

(D) 4.068 × 10⁸

Solution: (B) No. of oscillations in coherence length

= l/λ = 0.024/(5.9 × 10⁻⁷) = 4.068 × 10⁶ Hz

Question: In the above problem, the coherence time will be:

(A) 8 × 10⁻⁸s

(B) 8 × 10⁻⁹s

(C) 8 × 10⁻¹⁰s

(D) 8 × 10⁻¹¹s

Solution: (D) The coherence time t = l/c = 0.024/(3 × 10⁸) = 8 × 10⁻¹¹s

Answer: (D)

Question: A parallel plate capacitor made of circular plates each of radius R = 6cm has capacitance C = 100pF. The capacitor is connected to a 230V A.C. supply with an angular frequency of 300 rad/s. The r.m.s. value of conduction current will be:

(A) 5.7μA

(B) 6.3μA

(C) 9.6μA

(D) 6.9μA

Solution: (D) I_RMS = E_RMS/X_C = ωCE_RMS
= 300 × 10⁻¹⁰ × 230 = 6.9μA

38Question: In the above problem, the displacement current will be:

(A) 6.9μA

(B) 9.6μA

(C) 6.3μA

(D) 5.7μA

Solution: (A) I_D = I_C = 6.9μA

Question: In Q. 37, the value of B at a point 3 cm from the axis between the plates will be:

(A) 1.63 × 10⁻⁸T

(B) 1.63 × 10⁻⁹T

(C) 1.63 × 10⁻¹⁰T

(D) 1.63 × 10⁻¹¹T

Solution: (D)

B₀ = (μ₀I_D(peak)r)/(2πR²) = (4π × 10⁻⁷ × √2 × 6.9 × 10⁻⁶ × 2)/(2π × 36 × 10⁻⁴) = 1.63 × 10⁻¹¹T

Answer: (D)

Question: A plane electromagnetic wave of frequency 40 MHz travels in free space in the X-direction. At some point and at some instant, the electric field E⃗ has its maximum value of 750 N/C in Y-direction. The wavelength of the wave is:

(A) 3.5 m

(B) 5.5 m

(C) 7.5 m

(D) 9.5 m

Solution: (C) λ = C/f = (3 × 10⁸)/(4 × 10⁷) = 7.5m

41Question: In the above problem, the period of the wave will be:

(A) 2.5 μs

(B) 0.25 μs

(C) 0.025 μs

(D) None of these

Solution: (C) T = 1/f = 1/(4 × 10⁷) = 0.025μs

Question: In Q. 40, the magnitude and direction of magnetic field will be:

(A) 2.5 μT in X-direction

(B) 2.5 μT in Y-direction

(C) 2.5 μT Z-direction

(D) None of these

Solution: (C) B_m = E_m/C = 750/(3 × 10⁸) = 2.5μT Z-direction

Question: In Q. 40, the angular frequency of e.m.f. wave will be: (in rad/s)

(A) 8 × 10⁷

(B) 4 × 10⁶

(C) 4 × 10⁵

(D) 8 × 10⁴

Solution: (A) ω = 2πf = 2π × 4 × 10⁷ = 8 × 10⁷ rad/s.

Question: In Q. 40, the propagation constant of the wave will be:

(A) 8.38m⁻¹

(B) 0.838m⁻¹

(C) 4.19m⁻¹

(D) 0.419m⁻¹

Solution: (B) K = 2π/λ = (2 × 3.14)/7.5 = 0.838 m⁻¹

45Question: The sun delivers 10³ W/m² of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions 8m × 20m, will be:

(A) 6.4 × 10³W

(B) 3.4 × 10⁴W

(C) 1.6 × 10⁵W

(D) None of these

Solution: (C) Power P = I × A = 10³ × 8 × 20 = 1.6 × 10⁵W

Question: In the above problem, the radiation force on the roof will be:

(A) 3.33 × 10⁻⁵N

(B) 5.33 × 10⁻⁴N

(C) 7.33 × 10⁻³N

(D) None of these

Solution: (B)

F = PA/c = SA/c = (1.6 × 10⁵)/(3 × 10⁸) = 5.33 × 10⁻⁴N

Question: In Q. 45, the solar energy incident on the roof in 1 hour will be:

(A) 5.76 × 10⁸J

(B) 5.76 × 10⁷J

(C) 5.76 × 10⁶J

(D) 5.76 × 10⁵J

Solution: (A) E = power × time = 1.6 × 10⁵ × 3600 = 5.76 × 10⁸J

Question: The sun radiates electromagnetic energy at the rate of 3.9 × 10²⁶W. Its radius is 6.96 × 10⁸m. The intensity of sun light at the solar surface will be:

(A) 1.4 × 10⁴

(B) 2.8 × 10⁵

(C) 4.2 × 10⁶

(D) 5.6 × 10⁷

Solution: (D)

I_surface = P/A = (3.9 × 10²⁶)/(4π × (6.96 × 10⁸)²) = 5.6 × 10⁷ W/m²

Question: In the above problem, if the distance from the sun to the earth is 1.5 × 10¹¹ m, then the intensity of sunlight on earth's surface will be-(in W/m²)

(A) 1.38 × 10³

(B) 2.76 × 10⁴

(C) 5.52 × 10⁵

(D) None of these

Solution: (A)

I_earth = P/(4πr²) = (3.9 × 10²⁶)/(4π × 2.25 × 10²²) = 1.38 × 10³ W/m²

Question: A laser beam can be focussed on an area equal to the square of its wavelength. A He-Ne laser radiates energy at the rate of 1mW and its wavelength is 632.8 nm. The intensity of focussed beam will be:

(A) 1.5 × 10¹³ W/m²

(B) 2.5 × 10⁹ W/m²

(C) 3.5 × 10¹⁷ W/m²

(D) None of these

Solution: (B) Area through which the energy of beam passes
= (6.328 × 10⁻⁷)² = 4 × 10⁻¹³ m²

I = P/A = 10⁻³/(4 × 10⁻¹³) = 2.5 × 10⁹ W/m²

Question: A flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave:
E_x = 36sin (1.20 × 10⁷z – 3.6 × 10¹⁵t) V/m
The average intensity of the beam will be:

(A) 0.86 W/m²

(B) 1.72 W/m²

(C) 3.44 W/m²

(D) 6.88 W/m²

Solution: (B)

I_av = (ε₀cE₀²)/2 = (8.85 × 10⁻¹² × 3 × 10⁸ × 36²)/2 = 1.72 W/m²

Question: An electric field of 300 V/m is confined to a circular area 10 cm in diameter. If the field is increasing at the rate of 20 V/m-s, the magnitude of magnetic field at a point 15cm from the centre of the circle will be:

(A) 1.85 × 10⁻¹⁵ T

(B) 1.85 × 10⁻¹⁶ T

(C) 1.85 × 10⁻¹⁷ T

(D) 1.85 × 10⁻¹⁸ T

Solution: (D)

B = (μ₀ε₀/4) × (dE/dt) × (πd²/r) = (4π × 10⁻⁷ × 8.85 × 10⁻¹² × π × 0.01 × 20)/(4 × 0.15) = 1.85 × 10⁻¹⁸T

Question: A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 10m from the lamp will be:

(A) 1.34 V/m

(B) 2.68 V/m

(C) 5.36 V/m

(D) 9.37 V/m

Solution: (A)

S_av = P/(4πR²) = (1/2)ε₀cE₀²
E₀ = √(2P/(4πR²ε₀c)) = √(2π × 3 × 8.85 × 10⁻¹² × 3 × 10⁸) = 1.34 V/m

Question: A plane electromagnetic wave of wave intensity 6W/m² strikes a small mirror of area 40 cm², held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be:

(A) 6.4 × 10⁻⁷ kg-m/s

(B) 4.8 × 10⁻⁸ kg-m/s

(C) 3.2 × 10⁻⁹ kg-m/s

(D) 1.6 × 10⁻¹⁰ kg-m/s

Solution: (D) In one second

P = 2S_av A/c = 2U/c = (2 × 6 × 40 × 10⁻⁴)/(3 × 10⁸) = 1.6 × 10⁻¹⁰ Kg-m/s

Question: In the above problem, the radiation force on the mirror will be:

(A) 6.4 × 10⁻⁷ N

(B) 4.8 × 10⁻⁸ N

(C) 3.2 × 10⁻⁹ N

(D) 1.6 × 10⁻¹⁰ N

Solution: (D) Momentum per sec is force
F = 1.6 × 10⁻¹⁰ Newton

Question: In the above problem, the wavelength of the wave will be:

(A) 1.5m

(B) 66.6m

(C) 1.5cm

(D) 66.6cm

Solution: (C) Wavelength of electromagnetic wave

λ = c/v = (3 × 10⁸)/(2 × 10¹⁰) = 1.5 × 10⁻² = 1.5 cm Hence correct answer will be (C)

Question: In Q. 5, the energy density at a distance 3.5m from the source will be_ (in joule/m³)

(A) 1.73 × 10⁻⁵

(B) 1.73 × 10⁻⁶

(C) 1.73 × 10⁻⁷

(D) 1.73 × 10⁻⁸

Solution: (D) Energy density at 3.5m is given by

u = (1/2)ε₀E_m² = (1/2) × 8.85 × 10⁻¹² × (62.6)² = 1.73 × 10⁻⁸ Hence the correct answer will be(D)

Question: A 100 pF capacitor is connected to a 230V, 50 Hz A.C. source. The r.m.s. value of conduction current will be:

(A) 7.2 × 10⁻⁶A

(B) 3.6 × 10⁻⁵A

(C) 1.8 × 10⁻⁴A

(D) 0.9 × 10⁻³A

Solution: (A) The r.m.s. value of conduction current

I = V_RMS/X_C = ωCV_RMS = 2πfCV_RMS
or I = 2 × 3.14 × 50 × 100 × 10⁻¹² × 230 = 7.2 × 10⁻⁶ A Hence the correct answer will be(A)

Question: What should be the height of transmitting antenna if the T.V. telecast is to cover a radius of 128 km?

(A) 1560m

(B) 1280m

(C) 1050m

(D) 979m

Solution: (B) Height of transmitting antenna

h = d²/(2R_e) = (128 × 10³)²/(2 × 6.4 × 10⁶) = 1280m Hence the correct answer will be(B)

Question: The area to be covered for T.V. telecast is doubled, then the height of transmitting antenna (T.V. tower) will have to be:

(A) Doubled

(B) Halved

(C) Quadrupled

(D) Kept unchanged

Solution: (A) The area of transmission surrounding the T.V. tower

A = πd² = π(2hR_e)
A ∝ h Hence the correct answer will be (A)