K2KARAKORAM
मराठीEnglish
Open the app

State Board · Class 11 · Physics

Lesson 13: Electromagnetic Waves and Communication System

7 concepts · 56 practice questions with answers · Free to read

This lesson is also in: मराठी

Maxwell's equations, displacement current and sources of waves

From daily life

A mobile tower sends EM waves — Maxwell's equations are behind it. Link to earlier classes : the light, sound and radio waves of classes 9-10 now extend to Maxwell's equations and communication formulas.

Simple definition

Displacement current is the current-like term from a changing electric field that also produces B.

Displacement currentE ↑I_d = ε₀ dΦ_E/dt∮E·dA = q/ε₀∮B·dA = 0∮E·dl = −dΦ_B/dt∮B·dl = μ₀(I + I_d)

In the figure : A capacitor and the four equations.

Rule 1: Maxwell's equations — Four : Gauss (electric), Gauss (magnetic), Faraday, Ampère-Maxwell. They give c = 1/√(μ₀ε₀).

  • Easy : c = 1/√(4π×10⁻⁷ × 8.85×10⁻¹²) = 3.00 × 10⁸ m/s.
  • Medium : Changing B ⇒ E (Faraday); changing E ⇒ B (Maxwell).
  • Exam (CET) : c = ? (1) 1/√(μ₀ε₀) (2) √(μ₀ε₀) (3) μ₀/ε₀ (4) ε₀/μ₀ → answer 1)

Rule 2: Displacement current — I_d = ε₀ dΦ_E / dt; between capacitor plates I_d equals the conduction current I.

  • Easy : I = 2 A in the wire ⇒ I_d = 2 A between the plates.
  • Medium : dΦ_E/dt = 10¹¹ V m/s ⇒ I_d = 0.885 A.
  • Exam (CET) : Between plates the current is? (1) displacement (2) conduction (3) neither (4) ohmic → answer 1)

Rule 3: Sources of waves — An accelerated charge produces EM waves; a static or uniformly moving one does not. An oscillating charge radiates the same f.

  • Easy : Charge oscillating at 1 MHz ⇒ wave of 1 MHz.
  • Medium : Steady DC ⇒ no wave.
  • Exam (CET) : EM waves come from? (1) accelerated charge (2) static charge (3) uniform motion (4) static magnet → answer 1)

Derivation : Maxwell's correction

  • Step 1 : Ampère : ∮B·dl = μ₀ I; between plates I = 0 ⇒ contradiction.
  • Step 2 : Between plates E grows : Φ_E = q/ε₀.
  • Step 3 : dΦ_E/dt = (1/ε₀) dq/dt = I/ε₀ ⇒ ε₀ dΦ_E/dt = I.
  • Step 4 : ∮B·dl = μ₀(I + I_d).
Remember

I_d = ε₀ dΦ_E/dt; c = 1/√(μ₀ε₀).

Common mistake / CET trap

Conduction current is 0 between plates; I_d = I.

Shortcut

Memory clue : "Acceleration = wave".

Spot the difference

LawSays
Gauss (E)Charge ⇒ E
Gauss (B)No monopole
FaradayChanging B ⇒ E
Ampère-MaxwellI and changing E ⇒ B

Practice questions

Easy
  1. 1What is displacement current?
  2. 2What produces EM waves?
  3. 3Formula for c?
Medium
  1. 4I = 5 A : I_d between plates?
  2. 5dΦ_E/dt = 2 × 10¹¹ : I_d?
  3. 6Source 50 Hz : wave f?
Challenging
  1. 7Why correct Ampère's law?
  2. 8Does steady current radiate?
Show answers (try them yourself first)
  1. 1A current-like term due to a changing E
  2. 2An accelerated charge
  3. 31/√(μ₀ε₀)
  4. 45 A
  5. 51.77 A
  6. 650 Hz
  7. 7I = 0 between plates
  8. 8No

Characteristics of electromagnetic waves

From daily life

Sunlight crosses empty space — no medium needed.

Simple definition

An EM wave is a travelling oscillation of mutually perpendicular E and B.

EM wavexEBc = E₀/B₀ = fλE ⊥ B ⊥ c

In the figure : E (blue) and B (red) in phase.

Rule 1: Characteristics — Transverse; E ⊥ B ⊥ propagation; E and B in phase; no medium needed; speed in vacuum c = 3 × 10⁸ m/s.

  • Easy : f = 100 MHz ⇒ λ = c/f = 3 m.
  • Medium : λ = 600 nm ⇒ f = 5 × 10¹⁴ Hz.
  • Exam (CET) : EM wave is? (1) transverse (2) longitudinal (3) mechanical (4) needs a medium → answer 1)

Rule 2: Relation of E and B — E₀ = c B₀; instantaneous E/B = c.

  • Easy : E₀ = 300 V/m ⇒ B₀ = 300 ÷ 3×10⁸ = 10⁻⁶ T.
  • Medium : B₀ = 2 × 10⁻⁷ T ⇒ E₀ = 60 V/m.
  • Exam (CET) : E/B = ? (1) c (2) 1/c (3) c² (4) 1 → answer 1)

Rule 3: Energy and intensity — Energy is shared equally by E and B; I = ½ ε₀ c E₀²; energy density u = ε₀E².

  • Easy : E₀ = 300 V/m ⇒ I = ½ × 8.85×10⁻¹² × 3×10⁸ × 300² = 119 W/m².
  • Medium : E₀ doubled ⇒ I four times.
  • Exam (CET) : I ∝ ? (1) E₀² (2) E₀ (3) 1/E₀ (4) E₀³ → answer 1)

Derivation : E₀ = c B₀

  • Step 1 : Faraday : the E loop equals −dΦ_B/dt.
  • Step 2 : Wave E = E₀ sin(kx − ωt), B = B₀ sin(kx − ωt).
  • Step 3 : k E₀ = ω B₀.
  • Step 4 : E₀/B₀ = ω/k = c.
Remember

c = E₀/B₀ = fλ = 1/√(μ₀ε₀); I = ½ε₀cE₀².

Common mistake / CET trap

E and B are in phase (not 90°); all three directions are perpendicular.

Shortcut

Memory clue : "E ⊥ B ⊥ c".

Spot the difference

PropertyEM waveSound
TypeTransverseLongitudinal
MediumNot neededNeeded
Speed in vacuum3 × 10⁸Cannot travel

Practice questions

Easy
  1. 1Speed of EM waves in vacuum?
  2. 2Relation of E, B and c?
  3. 3Is a medium needed?
Medium
  1. 4f = 30 MHz : λ?
  2. 5E₀ = 150 V/m : B₀?
  3. 6λ = 3 cm : f?
Challenging
  1. 7B₀ = 4 × 10⁻⁷ T : I?
  2. 8Ratio of E and B energies?
Show answers (try them yourself first)
  1. 13 × 10⁸ m/s
  2. 2E₀ = cB₀
  3. 3No
  4. 410 m
  5. 55 × 10⁻⁷ T
  6. 610¹⁰ Hz
  7. 719.1 W/m²
  8. 81 : 1

The electromagnetic spectrum and its ranges

From daily life

The rainbow is a tiny part of the full spectrum.

Simple definition

The EM spectrum arranges all EM waves by λ or f.

SpectrumRadioMicroIRVisUVXγλ largef largeE = hf (energy →)

In the figure : f increases left to right.

Rule 1: Order — From longer to shorter λ : radio, microwave, IR, visible, UV, X-rays, γ-rays.

  • Easy : Visible : 400 nm (violet) to 700 nm (red).
  • Medium : f = c/λ : 400 nm ⇒ 7.5 × 10¹⁴ Hz.
  • Exam (CET) : Highest f? (1) γ (2) radio (3) IR (4) visible → answer 1)

Rule 2: Photon energy — E = h f = h c / λ; E (eV) = 1240 ÷ λ (nm); h = 6.63 × 10⁻³⁴ J s.

  • Easy : λ = 620 nm ⇒ E = 2 eV.
  • Medium : f = 10¹⁵ Hz ⇒ E = 6.63 × 10⁻¹⁹ J = 4.14 eV.
  • Exam (CET) : E ∝ ? (1) f (2) λ (3) f² (4) 1/f → answer 1) Trick : shorter λ ⇒ more E.

Rule 3: Ranges (approx.) — Radio > 0.1 m; microwave 1 mm-0.1 m; IR 700 nm-1 mm; visible 400-700 nm; UV 1-400 nm; X 10⁻³-10 nm; γ < 10⁻² nm.

  • Easy : λ = 0.5 mm ⇒ IR.
  • Medium : λ = 0.1 nm ⇒ X-rays.
  • Exam (CET) : λ = 500 nm? (1) visible (2) UV (3) IR (4) X → answer 1)

Derivation : E (eV) = 1240/λ

  • Step 1 : E = hc/λ.
  • Step 2 : hc = 6.63×10⁻³⁴ × 3×10⁸ = 1.99×10⁻²⁵ J m.
  • Step 3 : In eV ÷ 1.6×10⁻¹⁹ ⇒ 1.24×10⁻⁶ eV m.
  • Step 4 : E (eV) = 1240 ÷ λ (nm).
Remember

c = fλ; E = hf = 1240/λ (nm) eV.

Common mistake / CET trap

All travel at the same speed in vacuum; only f and λ differ.

Shortcut

Memory clue : "Rad-Mic-IR-Vis-UV-X-Gam".

Spot the difference

BandλPhoton energy
Radio> 0.1 mVery low
Visible400-700 nm1.8-3.1 eV
X-rays~ 0.1 nm~ 12 keV

Practice questions

Easy
  1. 1Order of the spectrum?
  2. 2λ of visible light?
  3. 3Formula for E?
Medium
  1. 4λ = 400 nm : E (eV)?
  2. 5f = 3 × 10¹⁵ Hz : λ?
  3. 6λ = 3 m : f?
Challenging
  1. 7λ = 0.2 nm : E (eV)?
  2. 8Why do γ-rays have more E?
Show answers (try them yourself first)
  1. 1Radio → γ
  2. 2400-700 nm
  3. 3hf
  4. 43.1 eV
  5. 5100 nm
  6. 610⁸ Hz
  7. 76200 eV
  8. 8Higher f ⇒ larger E = hf

Sources, properties and uses of the bands

From daily life

The wave that heats food in an oven is a bit shorter than radio.

Simple definition

Each band has a source and use set by its f.

Uses of the bandsRadiobroadcastMicroradar, ovenIRremoteVisseeingUVsteriliseXbone imageγcancer therapy

In the figure : Band and use.

Rule 1: Radio and microwaves — Radio : oscillating circuits, broadcasting. Microwave : klystron/magnetron; radar, ovens, satellites.

  • Easy : FM radio ~ 100 MHz; λ = 3 m.
  • Medium : Oven 2.45 GHz ⇒ λ = 12.2 cm.
  • Exam (CET) : Radar uses? (1) microwaves (2) X (3) γ (4) UV → answer 1)

Rule 2: IR, visible, UV — IR : hot bodies, remotes, thermal imaging. Visible : seen by eye. UV : Sun; sterilising; absorbed by ozone.

  • Easy : TV remote = IR.
  • Medium : UV to purify water.
  • Exam (CET) : The ozone layer absorbs? (1) UV (2) IR (3) radio (4) visible → answer 1)

Rule 3: X and γ rays — X : fast electrons hitting a target; bone imaging. γ : from nuclei; cancer therapy, sterilisation.

  • Easy : Bone image = X-rays.
  • Medium : From the nucleus = γ.
  • Exam (CET) : From the nucleus? (1) γ (2) X (3) UV (4) IR → answer 1)

Derivation : Oven wavelength

  • Step 1 : f = 2.45 × 10⁹ Hz.
  • Step 2 : λ = c / f.
  • Step 3 : = 3 × 10⁸ ÷ 2.45 × 10⁹.
  • Step 4 : λ = 12.2 cm.
Remember

Radio/micro = messages; IR = heat; X = diagnosis; γ = therapy.

Common mistake / CET trap

Do not mix X and γ : γ is from the nucleus.

Shortcut

Memory clue : "Rad-Mic-IR-Vis-UV-X-Gam".

Spot the difference

BandSourceUse
RadioOscillating circuitBroadcast
MicroMagnetronRadar, oven
IRHot bodiesRemote
XElectron impactDiagnosis
γNucleusTherapy

Practice questions

Easy
  1. 1Which wave is used in an oven?
  2. 2In a TV remote?
  3. 3Use of X-rays?
Medium
  1. 4f = 90 MHz : λ?
  2. 5Source of γ?
  3. 6One use of UV?
Challenging
  1. 7Why is IR called heat waves?
  2. 8Why is ozone important?
Show answers (try them yourself first)
  1. 1Microwaves
  2. 2IR
  3. 3Bones / diagnosis
  4. 43.33 m
  5. 5Nucleus
  6. 6Sterilisation
  7. 7Hot bodies emit IR
  8. 8It absorbs UV

Propagation of electromagnetic waves

From daily life

A radio catches distant stations, but TV fades far from the tower.

Simple definition

Propagation is how a wave reaches the receiver — decided by f.

Propagationsky wavespace waveground waveionosphere

In the figure : Three modes.

Rule 1: Three modes — Ground wave (< 2 MHz); sky wave (2-30 MHz, reflected by the ionosphere); space wave (> 30 MHz, line of sight).

  • Easy : AM radio at 1 MHz ⇒ ground wave.
  • Medium : Short wave 10 MHz ⇒ sky wave.
  • Exam (CET) : TV at 100 MHz? (1) space (2) ground (3) sky (4) any → answer 1)

Rule 2: Line-of-sight range — d = √(2 R h); two antennas : d = √(2Rh_T) + √(2Rh_R); R = 6.4 × 10⁶ m.

  • Easy : h = 100 m ⇒ d = √(2 × 6.4×10⁶ × 100) = 35.8 km.
  • Medium : h_T = 100, h_R = 25 ⇒ d = 53.7 km.
  • Exam (CET) : h four times ⇒ d? (1) double (2) four times (3) half (4) same → answer 1) Trick : d ∝ √h.

Rule 3: Ionosphere and maximum f — A wave is reflected only if f ≤ f_c = 9 √N_max (N = electrons/m³).

  • Easy : N_max = 10¹² ⇒ f_c = 9 MHz.
  • Medium : f = 20 MHz > f_c ⇒ passes through.
  • Exam (CET) : f > f_c ⇒ ? (1) passes through (2) reflected (3) absorbed (4) scattered → answer 1)

Derivation : d = √(2Rh)

  • Step 1 : Tower height h, Earth radius R; tangent length d.
  • Step 2 : Pythagoras : (R + h)² = R² + d².
  • Step 3 : R² + 2Rh + h² = R² + d² ⇒ d² = 2Rh + h².
  • Step 4 : h ≪ R ⇒ d = √(2Rh).
Remember

d = √(2Rh); f_c = 9√N.

Common mistake / CET trap

d ∝ √h, not h; add the two terms for two antennas.

Shortcut

Memory clue : "Low f = ground, medium = sky, high = straight".

Spot the difference

ModefUse
Ground< 2 MHzAM
Sky2-30 MHzShort wave
Space> 30 MHzTV, FM, satellite

Practice questions

Easy
  1. 1Three modes of propagation?
  2. 2Which mode for TV?
  3. 3Value of R?
Medium
  1. 4h = 64 m : d?
  2. 5N_max = 4 × 10¹¹ : f_c?
  3. 6h_T = h_R = 100 m : d?
Challenging
  1. 7d = 16 km : h?
  2. 8Why use satellites for TV?
Show answers (try them yourself first)
  1. 1Ground, sky, space
  2. 2Space
  3. 36.4 × 10⁶ m
  4. 428.6 km
  5. 55.69 MHz
  6. 671.6 km
  7. 720.0 m
  8. 8It extends the line of sight

Elements and terms of a communication system

From daily life

On a phone call your voice travels through wires or waves.

Simple definition

A communication system carries information from a source to a destination.

Communication systemSourceTransmitterChannelReceivernoiseanalog / digital

In the figure : Four main blocks and noise.

Rule 1: Elements — Source → transmitter → channel → receiver → destination; noise enters the channel.

  • Easy : Phone : mic = transducer; air = channel.
  • Medium : Wires, optical fibre, free space = channel.
  • Exam (CET) : Noise enters in the? (1) channel (2) source (3) receiver only (4) none → answer 1)

Rule 2: Key terms — Transducer, signal, attenuation, amplification, range, bandwidth, repeater.

  • Easy : Signal weakening in a wire = attenuation.
  • Medium : A repeater boosts and resends.
  • Exam (CET) : Signal getting weaker? (1) attenuation (2) amplification (3) modulation (4) noise → answer 1)

Rule 3: Bandwidth — Bandwidth = width of frequencies used. Sound 20 Hz-20 kHz; speech ~ 3 kHz; TV ~ 6 MHz.

  • Easy : Speech 300 Hz-3.1 kHz ⇒ BW = 2.8 kHz.
  • Medium : TV BW = 6 MHz ⇒ more information.
  • Exam (CET) : BW = ? (1) f_high − f_low (2) f_high + f_low (3) f_high × f_low (4) f_high ÷ f_low → answer 1)

Derivation : Antenna length

  • Step 1 : Effective radiation : l ≈ λ/4.
  • Step 2 : λ = c/f.
  • Step 3 : f = 1.5 MHz ⇒ λ = 200 m.
  • Step 4 : l = λ/4 = 50 m (hence modulation).
Remember

Source → Tx → channel → Rx; noise in the channel.

Common mistake / CET trap

Attenuation ≠ noise : one weakens, one adds.

Shortcut

Memory clue : "Src-Tx-Chan-Rx".

Spot the difference

ElementJob
TransmitterPrepares/modulates signal
ChannelCarries it
ReceiverRecovers the signal
RepeaterBoosts the signal

Practice questions

Easy
  1. 1Elements of a communication system?
  2. 2What is noise?
  3. 3What is a transducer?
Medium
  1. 4f = 3 MHz : λ/4?
  2. 5300 Hz-3300 Hz : BW?
  3. 6How to reduce attenuation?
Challenging
  1. 7Why is a direct antenna impractical?
  2. 8Analog versus digital?
Show answers (try them yourself first)
  1. 1Source, transmitter, channel, receiver
  2. 2Unwanted signal
  3. 3Converts one energy form to another
  4. 425 m
  5. 53000 Hz
  6. 6Repeaters / amplifiers
  7. 7Large λ ⇒ very long antenna
  8. 8Continuous versus 0/1

Modulation and demodulation

From daily life

Many radio stations run together — each with its own carrier f.

Simple definition

Modulation varies a carrier's amplitude or f by the message; demodulation reverses it.

AM modulationm = A_m / A_cf_c ± f_m ; BW = 2 f_mP_t = P_c (1 + m²/2)

In the figure : The carrier amplitude follows the message.

Live figure

Modulation and demodulation

In the app you can move this figure with your finger.

Open in the app

Rule 1: Why modulate — Low-f signals are hard to send directly (huge antenna, mixing). We ride them on a high-f carrier.

  • Easy : Speech ~ 3 kHz; carrier 1 MHz.
  • Medium : Many stations on different f.
  • Exam (CET) : Why modulate? (1) smaller antenna, better propagation (2) lower f (3) less power (4) add noise → answer 1)

Rule 2: Amplitude modulation — The carrier amplitude follows the message. m = A_m / A_c (≤ 1); sidebands f_c ± f_m; BW = 2 f_m.

  • Easy : f_c = 1 MHz, f_m = 5 kHz ⇒ 0.995 and 1.005 MHz; BW = 10 kHz.
  • Medium : A_c = 10 V, A_m = 6 V ⇒ m = 0.6.
  • Exam (CET) : m > 1 ⇒ ? (1) over-modulation (2) good (3) FM (4) same → answer 1)

Rule 3: Power, FM and demodulation — P_t = P_c (1 + m²/2). FM : f varies, amplitude constant — less noise. Demodulation = recovering the message (detector).

  • Easy : P_c = 100 W, m = 0.5 ⇒ P_t = 112.5 W.
  • Medium : m = 1 ⇒ P_t = 1.5 P_c.
  • Exam (CET) : In FM what is constant? (1) amplitude (2) f (3) phase (4) nothing → answer 1)

Derivation : AM sidebands

  • Step 1 : Carrier A_c sin ω_c t; message A_m sin ω_m t.
  • Step 2 : c_m = (A_c + A_m sin ω_m t) sin ω_c t.
  • Step 3 : sin A sin B ⇒ cos(ω_c − ω_m)t and cos(ω_c + ω_m)t.
  • Step 4 : Three f : f_c, f_c ± f_m; BW = 2f_m.
Remember

m = A_m/A_c; BW = 2f_m; P_t = P_c(1 + m²/2).

Common mistake / CET trap

m ≤ 1; BW = 2f_m, not f_m.

Shortcut

Memory clue : "Amplitude = AM, Frequency = FM".

Spot the difference

FeatureAMFM
VariesAmplitudeFrequency
NoiseMoreLess
BWSmallLarge

Practice questions

Easy
  1. 1What is modulation?
  2. 2BW of AM?
  3. 3What is demodulation?
Medium
  1. 4A_c = 8, A_m = 4 : m?
  2. 5f_c = 800 kHz, f_m = 4 kHz : sidebands?
  3. 6P_c = 200 W, m = 0.6 : P_t?
Challenging
  1. 7Prove BW = 2f_m.
  2. 8Why is FM less noisy?
Show answers (try them yourself first)
  1. 1Imposing the message on a carrier
  2. 22 f_m
  3. 3Recovering the message
  4. 40.5
  5. 5796 and 804 kHz
  6. 6236 W
  7. 7sin A sin B expansion
  8. 8Amplitude is constant; noise changes amplitude

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Displacement currentI_d = ε₀ dΦ_E/dtA; A
Speedc = 1/√(μ₀ε₀) = fλm/s; L T⁻¹
E, BE₀ = cB₀V/m; T
IntensityI = ½ε₀cE₀²W/m²; M T⁻³
PhotonE = hf = 1240/λ eVJ; M L² T⁻²
Line of sightd = √(2Rh)m; L
f_c9 √N_maxHz; T⁻¹
AMm = A_m/A_c; BW = 2f_m—; Hz
PowerP_t = P_c(1 + m²/2)W; M L² T⁻³

🎯 Most important points for the exam

  • I_d = ε₀ dΦ_E/dt; accelerated charge radiates.
  • c = E₀/B₀ = fλ; E ⊥ B ⊥ c, in phase.
  • Radio → micro → IR → visible → UV → X → γ : f increases.
  • E = hf = 1240/λ (nm) eV.
  • Ground < 2 MHz, sky 2-30 MHz, space > 30 MHz; d = √(2Rh).
  • Source → Tx → channel → Rx; noise in the channel.
  • AM : m ≤ 1, BW = 2f_m, P_t = P_c(1 + m²/2).
Study mantra

Maxwell → c → spectrum → uses → propagation → communication → modulation.