Class 12 · Physics · Chapter 8
Electromagnetic Waves
How Maxwell repaired Ampere's law with the displacement current, why that repair predicts self-sustaining electromagnetic waves moving at the speed of light, what those waves look like (E and B perpendicular to each other and to the direction of travel, E0/B0 = c), and the seven bands of the electromagnetic spectrum with how each is produced and used. A short chapter that appears every year as dependable 1- to 3-mark questions, so it is one of the cheapest places in the paper to score full marks.
Before you startWhat the board asks from this chapter
Unit V (Electromagnetic Waves, Chapter 8) is grouped with Unit VI (Optics, Chapters 9 and 10) for a combined 18 marks in the CBSE Class 12 Physics paper. This chapter alone typically contributes 2 to 4 marks: a 1-mark MCQ or assertion–reason on the spectrum or on field directions, and a 2- or 3-mark question on displacement current or on the properties of a plane electromagnetic wave. Occasionally it forms one part of a 5-mark question or a case-based question.
| Question type | Marks | Usually asked |
|---|---|---|
| MCQ | 1 | Direction of E, B and propagation for a given wave; E0/B0 = c; wavelength from frequency; which band has the shortest wavelength or largest penetrating power; which band a given frequency or wavelength belongs to; the expression for displacement current. |
| Assertion–Reason | 1 | Displacement current equals conduction current during charging; EM waves are transverse; infrared as heat waves; speed in vacuum independent of wavelength; EM waves need no medium; microwaves for radar. |
| Very short answer | 2 | Name the band for a given wavelength and give one use; how a particular band is produced; why microwaves suit radar; the physical quantity common to all EM waves in vacuum; conduction and displacement current in a fully charged capacitor; write E or B for a wave given the other. |
| Short answer | 3 | Show Ampere's law is inconsistent for a charging capacitor and obtain id = ε0 dΦE/dt; identify three bands from data and give a use each; a numerical on λ, ν, B0 and the expression of the companion field; show that the electric and magnetic energy densities of the wave are equal; the greenhouse effect and the ozone layer. |
| Long answer | 5 | Rare on its own. When it appears: the full displacement-current argument plus a magnetic-field numerical inside the plates, or production of EM waves by an oscillating charge with a labelled sketch plus a numerical on the wave equation. |
| Case-based | 4 | A passage on Maxwell's correction to Ampere's law, on the Hertz experiment, or on the electromagnetic spectrum (microwave ovens, ozone, greenhouse effect) followed by four one-mark sub-questions. |
Hot topics, in order
- Displacement current: the inconsistency in Ampere's law and the derivation of id = ε0 dΦE/dt
- Ampere–Maxwell law and Maxwell's four equations
- Directions of E, B and propagation; transverse nature; sketch of a plane wave along z
- E0/B0 = c, c = 1/√(μ0ε0), writing B given E and vice versa
- Electromagnetic spectrum: wavelength order, production and one use of each band
- Numericals: λ = c/ν, k = 2π/λ, ω = 2πν, B0 = E0/c
- Equality of the average electric and magnetic energy densities in an EM wave (NCERT Exercise 8.10)
- Greenhouse effect, ozone layer, microwave ovens, welders' goggles
The whole chapter, in orderNotes
01 8.1 Introduction
By the end of Chapter 6 you knew two facts. Oersted showed that an electric current produces a magnetic field. Faraday showed that a changing magnetic field produces an electric field (an induced emf). Maxwell asked the obvious next question: does a changing electric field produce a magnetic field? He found that Ampere's circuital law, as it stood, gave contradictory answers for a capacitor that is being charged. To remove the contradiction he added a new term, the displacement current, and with it the law became consistent and symmetric with Faraday's law.
Once that term was in place, the four equations of electricity and magnetism (now called Maxwell's equations) had a remarkable consequence. A changing electric field produces a changing magnetic field, which produces a changing electric field, and so on. The disturbance sustains itself and travels through empty space as a wave. Maxwell calculated its speed from two laboratory constants, c = 1/√(μ0ε0), and got 3 × 108 m s−1, the measured speed of light. His conclusion in 1864 was that light itself is an electromagnetic wave.
Hertz confirmed the prediction in 1887 by producing and detecting radio-frequency electromagnetic waves in the laboratory. Jagadis Chandra Bose worked at much shorter wavelengths (about 5 mm to 25 mm), and Marconi used the waves for long-distance communication, which is the beginning of the communication age. The chapter therefore has three parts: the displacement current, the nature of the wave, and the spectrum of such waves from radio to gamma rays.
02 8.2 Displacement Current: the inconsistency in Ampere's law and its repair PYQ 2016 · 2014
Ampere's circuital law says ∮B·dl = μ0 i, where i is the current crossing any surface bounded by the closed loop. Now take a parallel-plate capacitor being charged by a current i(t) in the connecting wire. Choose a point P just outside the wire and a circular loop of radius r through P, centred on the wire.
Surface 1: the flat disc bounded by the loop. The wire pierces it, so ∮B·dl = μ0 i and, by symmetry, B(2πr) = μ0 i, giving B = μ0 i/2πr.
Surface 2: a pot-shaped surface with the same loop as its rim, but bulging out so that its bottom lies between the plates. No wire passes through it, so no conduction current crosses it, and the law gives B = 0 at the same point P. Both cannot be right. The law is missing a term that is non-zero between the plates.
What is present between the plates is a changing electric field. For plate area A and charge Q, the field is E = Q/(ε0A), uniform and perpendicular to the plates. The electric flux through the bottom of surface 2 is ΦE = EA = Q/ε0. Differentiating, dΦE/dt = (1/ε0) dQ/dt, so
ε0 dΦE/dt = dQ/dt = i.
Maxwell's insight: the quantity id = ε0 dΦE/dt behaves exactly like a current. He named it the displacement current and added it to the conduction current ic. The generalised (Ampere–Maxwell) law is
∮B·dl = μ0(ic + id) = μ0 ic + μ0ε0 dΦE/dt.
Check it on the two surfaces. Through surface 1 there is conduction current i and no changing flux; through surface 2 there is no conduction current but a displacement current equal to i. Both give B = μ0 i/2πr at P. The total current ic + id is the same through every surface bounded by the loop, and it is continuous around the whole circuit: conduction current in the wire, displacement current in the gap.
The physical statement examiners want in words: a changing electric field produces a magnetic field, just as a changing magnetic field produces an electric field. This is the symmetry that makes electromagnetic waves possible.
03 8.2 Displacement Current (contd.): consequences, the field inside the plates, and Maxwell's equations PYQ 2016 · 2013 · 2012
Four consequences follow, and each has been an exam question.
1. Continuity of current. Kirchhoff's junction rule holds at each plate of the capacitor provided you count the displacement current: conduction current arrives at the plate, an equal displacement current leaves through the gap. This is why an ammeter in series with a capacitor and a dc battery shows a momentary deflection while the capacitor charges, and reads zero once it is fully charged (E between the plates is then constant, so id = 0 and ic = 0).
2. A capacitor conducts ac. With an ac source the charge on the plates keeps changing, so the electric field and hence the displacement current never settle to zero. The circuit carries a continuous current although no charge crosses the gap.
3. Displacement current produces a magnetic field exactly as conduction current does. For circular plates of radius R, at a point between the plates a distance r from the axis (r ≤ R), the displacement current enclosed by a circle of radius r is id(r2/R2) because the field is uniform. So B(2πr) = μ0 id r2/R2, giving B = μ0 id r/(2πR2). Outside the plates (r ≥ R) the full current is enclosed and B = μ0 id/2πr.
NCERT Example 8.1 uses this. Plates of radius 1 m, C = 1 nF, charged through R = 1 MΩ from a 2 V battery; find B halfway to the rim at t = 10−3 s. Time constant τ = RC = 10−3 s, so i = (V/R)e−t/τ = 2 × 10−6 × e−1 = 0.74 μA. Then B = μ0 i r/(2πR2) = 2 × 10−7 × 0.74 × 10−6 × 0.5/1 = 0.74 × 10−13 T.
4. Maxwell's equations. With the new term, the laws of electricity and magnetism in free space are: Gauss's law for electricity ∮E·dA = Q/ε0; Gauss's law for magnetism ∮B·dA = 0 (no magnetic monopoles); Faraday's law ∮E·dl = −dΦB/dt; Ampere–Maxwell law ∮B·dl = μ0 ic + μ0ε0 dΦE/dt. Together with the Lorentz force F = q(E + v × B), they describe every electromagnetic phenomenon in this course. The two circulation laws are now symmetric: a time-varying B gives E, a time-varying E gives B.
04 8.3 Electromagnetic Waves and 8.3.1 Sources of Electromagnetic Waves PYQ 2016 · 2012
Maxwell's equations predict that a time-varying electric field generates a time-varying magnetic field and vice versa. Neither field can then stay confined; the pair propagates outward as an electromagnetic wave. The wave needs no material medium, which is why light reaches us from the Sun across empty space.
Which charges radiate? A stationary charge produces only an electrostatic field. A charge in uniform motion is a steady current and produces a magnetic field that does not change with time. Neither radiates. Only an accelerated charge produces a changing electric field, which produces a changing magnetic field, which in turn produces a changing electric field, and so on. So accelerated charges are the sources of electromagnetic waves.
The important case is a charge oscillating with frequency ν. The electric field it produces oscillates at ν, so does the magnetic field, and the wave that travels out has the same frequency ν. Its wavelength in vacuum is λ = c/ν. The energy carried by the wave is supplied by the source, that is, by whatever does work to keep the charge accelerating. This is why a radio transmitter needs power.
Frequency sets what you can build. Hertz (1887) used a spark-gap oscillator to produce waves of a few metres wavelength and detected them with a loop with a small gap in which sparks appeared. Even modern electronic circuits reach only about 1011 Hz, the radio and microwave range. Visible light, at about 1014 Hz (yellow light is close to 6 × 1014 Hz), is far beyond any circuit; it comes from electrons within atoms and molecules. X-rays come from inner-shell electron transitions and gamma rays from the nucleus, so the source of a wave is a good guide to its band.
05 8.3.2 Nature of Electromagnetic Waves: transverse fields, phase and speed PYQ 2017 · 2012 · 2011
Take a plane wave travelling along the z-axis. The electric field is along x and the magnetic field along y:
Ex = E0 sin(kz − ωt), By = B0 sin(kz − ωt),
where k = 2π/λ is the propagation constant (wave number) and ω = 2πν is the angular frequency. The wave moves along +z with speed ω/k. Three facts about this picture carry marks.
Directions. E, B and the direction of propagation are mutually perpendicular. The wave travels along E × B. So if E is along x and B along y, propagation is along +z; if E is along y and B along z, propagation is along +x. Because the fields oscillate perpendicular to the direction of travel, electromagnetic waves are transverse. This is why light can be polarised, which Chapter 10 uses.
Phase. E and B oscillate in phase: they reach their maxima together and pass through zero together. The wave is not like a see-saw where one field is large when the other is small.
Speed and amplitude relation. Maxwell's equations give the speed in vacuum as c = 1/√(μ0ε0) = 3 × 108 m s−1, the same for every frequency. They also fix the ratio of the amplitudes: B0 = E0/c, i.e. E0/B0 = c. Given one field you can write the other immediately: same phase, same k and ω, amplitude scaled by c, direction fixed by E × B.
In a material medium of permittivity ε and permeability μ the speed becomes v = 1/√(με), which is less than c. The frequency does not change on entering a medium (it is set by the source), so the wavelength decreases to λ/n, where n = c/v is the refractive index.
What is waving in vacuum? Nothing material: the Michelson–Morley experiment found no aether, and electromagnetic waves travel through empty space with no medium at all.
06 8.3.2 Nature of Electromagnetic Waves (contd.): energy density of the wave
An electromagnetic wave carries energy because the fields themselves store energy. In a region with electric field E the electric energy density is uE = ½ε0E2, and in a region with magnetic field B the magnetic energy density is uB = B2/2μ0. In an electromagnetic wave these two are equal at every instant, not only on average.
To see this, use the amplitude relation for the wave, E = cB, together with c2 = 1/(μ0ε0):
uB = B2/2μ0 = (E/c)2/2μ0 = E2/(2μ0c2) = E2μ0ε0/(2μ0) = ½ε0E2 = uE.
So the energy carried by the wave is shared equally between its electric and magnetic parts at every point and at every instant, and therefore on average as well. This is exactly the result NCERT Exercise 8.10(c) asks you to prove for a given numerical E0 and ν, so practise substituting numbers into this same chain: find B0 = E0/c first, then show ½ε0E02 equals B02/2μ0.
07 8.4 Electromagnetic Spectrum: the map PYQ 2017 · 2013 · 2010
All electromagnetic waves are the same phenomenon; they differ only in frequency (or wavelength, since νλ = c in vacuum). The whole range, from wavelengths of kilometres down to less than 10−14 m, is the electromagnetic spectrum. Historical names label its parts, and there is no sharp boundary between one part and the next; neighbouring bands overlap. What distinguishes them in practice is how they are produced and how they are detected.
Memorise the bands in order of decreasing wavelength (increasing frequency and photon energy) with NCERT's approximate wavelength ranges:
- Radio waves: greater than 0.1 m
- Microwaves: 0.1 m to 1 mm
- Infrared: 1 mm to 700 nm
- Visible light: 700 nm to 400 nm (frequency 4 × 1014 Hz to 7 × 1014 Hz)
- Ultraviolet: 400 nm to 1 nm
- X-rays: 1 nm to 10−3 nm
- Gamma rays: less than 10−3 nm
Two quantities let you place any wave. Wavelength and frequency are linked by λ = c/ν, so a frequency of 3 × 1013 Hz means λ = 10−5 m = 10 μm, which is infrared. Photon energy is E = hν = hc/λ; the handy form is E (in eV) = 1240/λ (in nm). A 1 mm microwave photon has about 1.2 × 10−3 eV, a 620 nm red photon 2 eV, a 1 nm X-ray photon 1.24 keV and a 10−3 nm gamma photon 1.24 MeV. These energy scales match the sources: circuit oscillations for radio, molecular rotations and vibrations for microwave and infrared, outer-electron transitions in atoms for visible and ultraviolet, inner-shell transitions for X-rays, and nuclear transitions for gamma rays. Whatever the band, the speed in vacuum is the same c.
08 8.4.1 Radio Waves and 8.4.2 Microwaves PYQ 2016 · 2014 · 2011
Radio waves (wavelength above 0.1 m) are produced by the accelerated motion of charges in conducting wires, that is, by oscillating currents in an antenna fed by an LC oscillator or a transmitter circuit. They are used in radio and television communication. NCERT gives the bands: AM radio 530 kHz to 1710 kHz, TV 54 MHz to 890 MHz, FM radio 88 MHz to 108 MHz, and cellular phones in the UHF (ultra high frequency) band. Detection is by an antenna and a tuned receiver circuit. The frequency of the emitted wave equals the frequency of the oscillating current, so a 100 MHz FM station radiates at λ = c/ν = 3 m.
Microwaves (wavelength 0.1 m to 1 mm, frequency in the gigahertz range) are short-wavelength radio waves. They are produced by special vacuum tubes, the klystron and the magnetron, and by Gunn diodes. Their three standard uses:
- Radar and aircraft navigation. Because the wavelength is short, microwaves can be sent as a narrow, directed beam with little spreading, and they reflect well from objects of aircraft size. The time delay of the echo gives the distance.
- Microwave ovens. Water molecules in food have a natural (resonant) frequency of rotation in the microwave range. The oven's frequency is chosen to match it, so water molecules absorb energy strongly, and the food heats from within through its water content while the dry container stays cooler. Ask: why match the resonant frequency? Because energy transfer is most efficient at resonance.
- Speed guns, satellite links and mobile communication.
09 8.4.3 Infrared Waves and 8.4.4 Visible Rays PYQ 2015 · 2014 · 2009
Infrared waves (1 mm to 700 nm) lie just beyond the red end of the visible spectrum. They are produced by hot bodies and by vibrating molecules; every object above absolute zero emits them. They are called heat waves because water molecules present in most materials readily absorb infrared radiation (so do carbon dioxide, ammonia and other molecules); after absorption the molecules vibrate more energetically, their thermal motion increases, and the material warms up along with its surroundings.
Uses that NCERT lists and the board asks: infrared lamps in physical therapy (treating muscular strain); remote controls of TV sets and other household electronics, where a semiconductor light-emitting diode emits an infrared signal; infrared detectors on Earth satellites for military use and for observing the growth of crops; infrared photography and night-vision equipment.
Infrared also maintains the Earth's average temperature through the greenhouse effect. Sunlight (mostly visible) passes through the atmosphere and warms the ground. The warm ground re-radiates energy as infrared. Greenhouse gases in the atmosphere, chiefly water vapour and carbon dioxide, absorb this infrared and radiate part of it back to the surface, so the Earth stays warmer than it would otherwise be. If the Earth had no atmosphere, the infrared would escape directly to space and the average surface temperature would be lower. This 'higher or lower' question has appeared in a 3-mark set.
Visible light (700 nm to 400 nm, frequency 4 × 1014 Hz to 7 × 1014 Hz) is the part our eyes detect; the retina is a detector tuned to this band. It is emitted by atoms and molecules when their outer electrons drop from a higher to a lower energy level (the Sun, lamps, LEDs, lasers). Red lies at the long end near 700 nm and violet at the short end near 400 nm.
10 8.4.5 Ultraviolet Rays, 8.4.6 X-rays and 8.4.7 Gamma Rays PYQ 2017 · 2015 · 2013 · 2010
Ultraviolet (UV) radiation (NCERT: 4 × 10−7 m down to 6 × 10−10 m, i.e. 400 nm to about 0.6 nm) is produced by special lamps and by very hot bodies. The Sun is an important source, and so is a welding arc, which is why welders wear special goggles or face masks with glass windows that absorb UV. Most of the Sun's UV is absorbed by the ozone layer at the top of the stratosphere; that thin layer is crucial for human survival because UV in large quantities harms living tissue, causing skin cancer and cataracts. In moderate doses UV tans the skin and helps the body make vitamin D. Uses: LASIK eye surgery (UV lasers can be focused into very narrow beams), and UV lamps to kill germs in water purifiers.
X-rays (about 10−8 m, 10 nm, down to 10−13 m, 10−4 nm) are produced when a metal target is bombarded by high-energy electrons in an X-ray tube: the sudden deceleration of the electrons and inner-shell transitions in the target atoms produce the radiation. They penetrate soft tissue but are stopped by bone, so they are used as a diagnostic tool in medicine, and also to treat certain forms of cancer. Because they damage living tissue, exposure must be kept low. They are also used to study crystal structure (X-ray diffraction) and in security scanning.
Gamma rays (about 10−10 m to below 10−14 m) have the shortest wavelength and highest frequency, hence the largest penetrating power. They are produced in nuclear reactions and are emitted by radioactive nuclei during decay. In medicine they are used to destroy cancer cells (radiotherapy) and for imaging with radioactive tracers. They cannot be produced by any electronic circuit; the source is always nuclear.
A pattern for the last three bands: the source moves inward in the atom as the frequency rises, from outer electrons (UV) to inner-shell electrons (X-rays) to the nucleus (gamma).
VocabularyKey terms
Keep this on one pageFormula sheet
| Name | Expression | Notes |
|---|---|---|
| Displacement current | id = ε0 dΦE/dt | Between capacitor plates ΦE = Q/ε0, so id = dQ/dt = ic; zero when E is steady |
| Ampere–Maxwell law | ∮B·dl = μ0 ic + μ0ε0 dΦE/dt | Total current ic + id is the same through any surface bounded by the loop |
| Field inside a charging capacitor | B = μ0 id r/(2πR2) for r ≤ R | Circular plates of radius R; outside, B = μ0 id/2πr |
| Plane wave along z | Ex = E0 sin(kz − ωt), By = B0 sin(kz − ωt) | k = 2π/λ, ω = 2πν; kz + ωt means travel along −z |
| Speed in vacuum | c = 1/√(μ0ε0) = ω/k = νλ = 3 × 108 m s−1 | Same for all wavelengths in vacuum |
| Amplitude ratio | E0/B0 = c | E in V m−1 (= N C−1), B in tesla; fields in phase, direction of travel = E × B |
| Speed in a medium | v = 1/√(με) = c/√(μrεr) = c/n | Frequency unchanged; wavelength becomes λ/n |
| Energy density | uE = ½ε0E2, uB = B2/2μ0 | Equal at every instant, since E = cB and c2 = 1/(μ0ε0) — NCERT Exercise 8.10(c) |
| Photon energy | E = hν = hc/λ; E (eV) ≈ 1240/λ (nm) | Ranks the spectrum: radio ~10−6 eV, visible ~2 eV, X-rays ~keV, gamma ~MeV |
Practise with a pencilDiagrams that earn marks
Charging capacitor with two surfaces bounded by the same loop
Draw a battery, a wire and a parallel-plate capacitor with current i flowing in the wire. Mark a point P outside the wire and a circular Amperian loop through P around the wire. Show surface 1 as the flat disc cut by the wire, and surface 2 as a pot-shaped surface with the same rim that passes between the plates without touching the wire. Label the electric field E between the plates and write B = μ0i/2πr for surface 1 and 'no conduction current' for surface 2. This figure alone earns a mark in the displacement-current derivation.
Magnetic field lines of the displacement current between the plates
Draw the two circular plates face-on with the electric field E pointing out of (or into) the page and marked as increasing. Draw concentric circular magnetic field lines around the axis in the plane of the plates, with arrows in the same sense as those around the wire that feeds the capacitor. Label R (plate radius) and r (distance from the axis) and write B = μ0idr/2πR2 for r ≤ R.
Plane electromagnetic wave propagating along the z-axis
Draw three mutually perpendicular axes x, y, z. Along z draw a sinusoidal E wave in the x–z plane (arrows parallel to x) and a sinusoidal B wave in the y–z plane (arrows parallel to y), both starting at the same point and in phase: crests of E and B occur at the same z. Mark one wavelength λ, label E0 and B0, and draw the propagation arrow along +z, with the note E × B points along propagation. Write Ex = E0 sin(kz − ωt) and By = B0 sin(kz − ωt) beside it.
The electromagnetic spectrum
Draw a horizontal bar with wavelength increasing to the right (or frequency increasing to the left). From left to right write gamma rays, X-rays, ultraviolet, visible, infrared, microwaves, radio waves, with the visible band expanded to show violet (400 nm) to red (700 nm). Under each band write NCERT's wavelength limits (10−3 nm, 1 nm, 400 nm, 700 nm, 1 mm, 0.1 m) and show the bands overlapping slightly at the boundaries. Add one source and one use beneath each band.
TextbookNCERT questions the board keeps asking
8.1 A capacitor made of two circular plates each of radius 12 cm, separated by 5.0 cm, is being charged by an external source. The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain.3 m
(a) A = πr2 = π × (0.12)2 = 4.52 × 10−2 m2. C = ε0A/d = 8.85 × 10−12 × 4.52 × 10−2/0.05 = 8.0 × 10−12 F = 8.0 pF.
Since Q = CV, i = C dV/dt, so dV/dt = i/C = 0.15/(8.0 × 10−12) = 1.87 × 1010 V s−1.
(b) The displacement current equals the conduction current charging the plates: id = ε0 dΦE/dt = dQ/dt = 0.15 A.
(c) Yes, provided the displacement current is included. At each plate the conduction current entering through the wire equals the displacement current leaving through the gap, so the total current into the plate equals the total current out of it.
8.2 A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with an angular frequency of 300 rad s−1. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.3 m
(a) XC = 1/ωC = 1/(300 × 100 × 10−12) = 3.33 × 107 Ω. Irms = Vrms/XC = 230/(3.33 × 107) = 6.9 × 10−6 A = 6.9 μA.
(b) Yes. The displacement current between the plates is always equal to the conduction current in the wires, so id = 6.9 μA (rms).
(c) Peak current I0 = √2 × 6.9 μA = 9.76 μA. Inside the plates, B = μ0 r I0/(2πR2) = (2 × 10−7 × 0.03 × 9.76 × 10−6)/(0.06)2 = 5.86 × 10−14/3.6 × 10−3 = 1.63 × 10−11 T.
8.3 What physical quantity is the same for X-rays of wavelength 10−10 m, red light of wavelength 6800 Å and radio waves of wavelength 500 m?1 m
Their speed in vacuum, c = 3 × 108 m s−1. All electromagnetic waves, whatever their wavelength, travel at the same speed in free space.
8.4 A plane electromagnetic wave travels in vacuum along the z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?2 m
E and B are perpendicular to each other and both lie in the x–y plane, perpendicular to the z-direction of propagation (the wave is transverse). The direction of E × B is along +z.
λ = c/ν = 3 × 108/(30 × 106) = 10 m.
8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?2 m
λ1 = c/ν1 = 3 × 108/(7.5 × 106) = 40 m; λ2 = 3 × 108/(12 × 106) = 25 m. The wavelength band is 25 m to 40 m.
8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?1 m
109 Hz. An oscillating charge radiates electromagnetic waves of the same frequency as its own oscillation.
8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?1 m
E0 = cB0 = 3 × 108 × 510 × 10−9 = 153 V m−1 (= 153 N C−1).
8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N C−1 and that its frequency is ν = 50.0 MHz. (a) Determine B0, ω, k and λ. (b) Find expressions for E and B.3 m
(a) B0 = E0/c = 120/(3 × 108) = 4.0 × 10−7 T = 400 nT.
ω = 2πν = 2π × 50 × 106 = 3.14 × 108 rad s−1.
k = ω/c = 3.14 × 108/(3 × 108) = 1.05 rad m−1.
λ = c/ν = 3 × 108/(50 × 106) = 6.0 m.
(b) Taking propagation along +x, E along y and B along z:
E = (120 N C−1) sin[(1.05 rad m−1)x − (3.14 × 108 rad s−1)t] ĵ
B = (400 nT) sin[(1.05 rad m−1)x − (3.14 × 108 rad s−1)t] k̂.
8.9 Use the formula E = hν for the energy of a photon to obtain the photon energy in eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies related to the sources of electromagnetic radiation?3 m
Use E (eV) = hc/λ = 1240/λ (nm). Taking a typical wavelength in each band:
Radio (λ = 1 m = 109 nm): 1.24 × 10−6 eV
Microwave (1 mm): 1.24 × 10−3 eV
Infrared (1 μm): 1.24 eV
Visible (500 nm): 2.5 eV
Ultraviolet (100 nm): 12.4 eV
X-rays (0.1 nm): 1.24 × 104 eV = 12.4 keV
Gamma rays (10−3 nm): 1.24 × 106 eV = 1.24 MeV.
The photon energy matches the energy scale of the source. Radio photons (μeV) come from electrons oscillating in circuits; visible and ultraviolet photons (eV) from transitions of outer electrons in atoms and molecules; X-ray photons (keV) from inner-shell electron transitions and decelerating electrons; gamma photons (MeV) from transitions inside the nucleus. A source can only radiate photons whose energy is comparable to its own energy-level spacing.
8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m−1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. (c = 3 × 108 m s−1.)3 m
(a) λ = c/ν = 3 × 108/(2.0 × 1010) = 1.5 × 10−2 m = 1.5 cm.
(b) B0 = E0/c = 48/(3 × 108) = 1.6 × 10−7 T.
(c) uE = ½ε0E2 and uB = B2/2μ0. Since E = cB and c2 = 1/(μ0ε0): uE = ½ε0c2B2 = ½ε0B2/(μ0ε0) = B2/2μ0 = uB. This holds at every instant, so it also holds for the averages.
Example 8.2 A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 ĵ V m−1. What is B at this point?2 m
B = E/c = 6.3/(3 × 108) = 2.1 × 10−8 T. The direction of propagation is E × B, so with propagation along +x and E along +y, B must be along +z (ĵ × k̂ = î). Hence B = 2.1 × 10−8 k̂ T.
Previous years and frequently askedBoard questions with model answers
MCQ The displacement current through a surface is given by- μ0 dΦE/dt
- ε0 dΦE/dt
- ε0 dΦB/dt
- μ0ε0 dΦE/dt
Frequently asked1 m
MCQ A plane electromagnetic wave travels in vacuum along the +z direction. Its electric field oscillates along the +x direction. The magnetic field oscillates along- the +y direction
- the −z direction
- the +x direction
- the +z direction
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MCQ Which of the following has the shortest wavelength?- Microwaves
- Ultraviolet rays
- X-rays
- Infrared rays
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MCQ The ratio of the amplitudes of the electric and magnetic fields, E0/B0, of an electromagnetic wave in vacuum is equal to- 1/c
- c
- c2
- √(μ0ε0)
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MCQ The electric field of a plane electromagnetic wave in vacuum has amplitude 6 V m−1. The amplitude of its magnetic field is- 2 × 10−8 T
- 1.8 × 109 T
- 2 × 108 T
- 6 × 10−8 T
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MCQ Which part of the electromagnetic spectrum has the largest penetrating power?- Ultraviolet rays
- X-rays
- Gamma rays
- Microwaves
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MCQ In a plane electromagnetic wave the electric and magnetic fields are- in phase and perpendicular to each other
- in phase and parallel to each other
- out of phase by π/2 and perpendicular to each other
- out of phase by π and parallel to each other
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Assertion–Reason Assertion (A): While a capacitor is being charged, the displacement current between its plates equals the conduction current in the connecting wire. Reason (R): The displacement current arises from the changing electric flux between the plates, ε0 dΦE/dt, which equals dQ/dt.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
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Assertion–Reason Assertion (A): Electromagnetic waves are transverse in nature. Reason (R): In an electromagnetic wave the electric and magnetic fields oscillate perpendicular to the direction of propagation.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
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Assertion–Reason Assertion (A): The speed of electromagnetic waves in vacuum is the same for all wavelengths. Reason (R): Electromagnetic waves require a material medium for their propagation.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
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Assertion–Reason Assertion (A): Infrared waves are often called heat waves. Reason (R): Water molecules present in most materials readily absorb infrared radiation, and the absorbed energy increases their thermal motion.- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Frequently asked1 m
Very short answer How is the speed of electromagnetic waves in vacuum determined by the electric and magnetic fields? Write the relation between the amplitudes of the two fields.CBSE 20172 m
The speed of an electromagnetic wave in vacuum is fixed by the electric and magnetic constants of free space: c = 1/√(μ0ε0) = 3 × 108 m s−1.
It also equals the ratio of the amplitudes of the two fields: c = E0/B0, so B0 = E0/c.
Very short answer Welders wear special goggles to protect their eyes from electromagnetic radiations. Name the radiations and write the range of their frequency.CBSE 20132 m
Ultraviolet radiation, produced in large amounts by the welding arc.
Frequency range: about 8 × 1014 Hz to 3 × 1017 Hz (wavelength 400 nm down to 1 nm). The goggles have glass that absorbs UV, which would otherwise damage the retina and cornea.
Very short answer Why are microwaves considered suitable for radar systems used in aircraft navigation? How are microwaves produced?CBSE 20162 m
Microwaves have short wavelengths (0.1 m to 1 mm), so they can be transmitted as a narrow beam in a definite direction with little spreading, and they are reflected sharply by an aircraft-sized target. The echo time gives the distance and the beam direction gives the bearing.
They are produced by special vacuum tubes, the klystron and the magnetron, and by Gunn diodes.
Very short answer To which part of the electromagnetic spectrum does a wave of frequency 3 × 1013 Hz belong? Give one use of this radiation.CBSE 20142 m
λ = c/ν = 3 × 108/(3 × 1013) = 10−5 m = 10 μm, which lies between 1 mm and 700 nm. The wave is infrared.
Use: infrared lamps in physical therapy, or infrared LEDs in TV remote controls.
Very short answer Identify the electromagnetic waves whose wavelengths lie in the range (a) 10−12 m < λ < 10−8 m, (b) 10−3 m < λ < 10−1 m. Write one use of each.CBSE 20172 m
(a) 10−3 nm to 10 nm: X-rays. Use: diagnostic imaging of bones in medicine (also treatment of some cancers).
(b) 1 mm to 0.1 m: microwaves. Use: radar for aircraft navigation (also microwave ovens).
Short answer Write Maxwell's generalisation of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is id = ε0 dΦE/dt.CBSE 20163 m
Ampere–Maxwell law: ∮B·dl = μ0(ic + id) = μ0 ic + μ0ε0 dΦE/dt, where ic is the conduction current and id the displacement current through the surface bounded by the loop.
Consider a parallel-plate capacitor of plate area A being charged by current i, so the charge on the plates is Q(t). The field between the plates is E = Q/(ε0A), perpendicular to the plates. The electric flux through a surface of area A lying between the plates is
ΦE = EA = Q/ε0.
Differentiating, dΦE/dt = (1/ε0) dQ/dt, so ε0 dΦE/dt = dQ/dt = i.
Thus the changing electric field between the plates is equivalent to a current id = ε0 dΦE/dt, equal in magnitude to the conduction current in the wire. This displacement current completes the circuit through the gap and, added to Ampere's law, removes the contradiction between the flat surface and the pot-shaped surface bounded by the same loop.
Short answer Name the types of electromagnetic radiation which (i) are used in destroying cancer cells, (ii) cause tanning of the skin, and (iii) maintain the Earth's warmth. Write briefly a method of producing any one of these waves.CBSE 20153 m
(i) Gamma rays (high-energy radiation used in radiotherapy).
(ii) Ultraviolet rays.
(iii) Infrared rays, through the greenhouse effect.
Production of infrared: every hot body emits infrared radiation; molecules vibrating or rotating in a heated object radiate at infrared frequencies. (Alternatively: gamma rays are emitted by radioactive nuclei during decay; ultraviolet comes from very hot bodies, the Sun, welding arcs and special UV lamps.)
Short answer (a) Name the electromagnetic waves which are suitable for radar systems used in aircraft navigation. Write the range of frequency of these waves. (b) If the Earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? Explain. (c) Show that in a plane electromagnetic wave in vacuum, the electric energy density equals the magnetic energy density at every instant.Frequently asked3 m
(a) Microwaves, frequency roughly 109 Hz to 1012 Hz (wavelength 0.1 m to 1 mm).
(b) Lower. The ground, warmed by sunlight, radiates infrared. Greenhouse gases in the atmosphere (water vapour, carbon dioxide) absorb this infrared and radiate part of it back to the surface, keeping it warm. Without an atmosphere the infrared would escape directly to space and the average surface temperature would fall.
(c) uE = ½ε0E2 and uB = B2/2μ0. Since E = cB, B = E/c, so uB = (E/c)2/2μ0 = E2/(2μ0c2). Using c2 = 1/(μ0ε0), 1/c2 = μ0ε0, so uB = E2μ0ε0/(2μ0) = ½ε0E2 = uE. So the two energy densities are equal at every instant.
Short answer The magnetic field of a plane electromagnetic wave in vacuum is By = 2 × 10−7 sin(0.5 × 103 x + 1.5 × 1011 t) T. (a) Find the wavelength and frequency of the wave. (b) In which direction does the wave travel? (c) Write the expression for the electric field.Frequently asked3 m
(a) Comparing with B0 sin(kx + ωt): k = 0.5 × 103 rad m−1, ω = 1.5 × 1011 rad s−1.
λ = 2π/k = 2π/500 = 1.26 × 10−2 m = 1.26 cm.
ν = ω/2π = 1.5 × 1011/6.28 = 2.39 × 1010 Hz.
(Check: ω/k = 1.5 × 1011/500 = 3 × 108 m s−1 = c.)
(b) The phase is kx + ωt, so the wave travels along the −x direction.
(c) E0 = cB0 = 3 × 108 × 2 × 10−7 = 60 V m−1. E × B must point along −x with B along +y: k̂ × ĵ = −î, so E is along +z.
Ez = 60 sin(0.5 × 103 x + 1.5 × 1011 t) V m−1.
Long answer (a) A parallel-plate capacitor is being charged by a battery. Show that Ampere's circuital law, as originally stated, gives inconsistent results for the magnetic field at a point near the connecting wire, and explain how Maxwell removed the inconsistency. Write the generalised law. (b) The circular plates of the capacitor have radius 10 cm and the charging current is 0.5 A. Find the magnetic field at a point between the plates 5 cm from the axis.Frequently asked5 m
(a) Ampere's law: ∮B·dl = μ0 i, i being the current through any surface bounded by the loop. Take a circular loop of radius r around the wire through a point P near the wire. For the flat disc bounded by the loop, the wire passes through, so B(2πr) = μ0i and B = μ0i/2πr. For a pot-shaped surface with the same rim whose bottom lies between the plates, no conduction current crosses it, so the law gives B = 0 at the same point P. Two surfaces with the same boundary give different B: the law is inconsistent.
Maxwell noticed that between the plates the electric field is changing. With plate area A, E = Q/(ε0A), so the flux through the bottom of the pot is ΦE = Q/ε0 and ε0 dΦE/dt = dQ/dt = i. He added this displacement current id = ε0 dΦE/dt to the conduction current:
∮B·dl = μ0(ic + ε0 dΦE/dt).
Now the flat surface has ic = i and id = 0, the pot has ic = 0 and id = i, and both give B = μ0i/2πr. The total current is continuous around the circuit and a changing electric field produces a magnetic field.
(b) Inside the plates the displacement current is spread uniformly over the plate area, so a circle of radius r encloses id r2/R2. Then B(2πr) = μ0 id r2/R2, giving B = μ0 id r/(2πR2).
With id = 0.5 A, r = 0.05 m, R = 0.10 m: B = (2 × 10−7 × 0.5 × 0.05)/(0.10)2 = 5 × 10−9/10−2 = 5 × 10−7 T.
Long answer (a) How are electromagnetic waves produced by an oscillating charge? Draw a sketch of a linearly polarised plane electromagnetic wave propagating along the z-direction, indicating the directions of the oscillating electric and magnetic fields. (b) Write the relation between the amplitudes of the two fields and the speed of the wave. (c) A plane electromagnetic wave of frequency 50 MHz has an electric field amplitude of 60 V m−1. Find its wavelength, propagation constant and magnetic field amplitude.Frequently asked5 m
(a) A charge at rest produces only an electric field and a charge in uniform motion produces a steady magnetic field; neither radiates. An oscillating (hence accelerated) charge produces an electric field that changes with time. A changing electric field produces a changing magnetic field (Ampere–Maxwell law), and the changing magnetic field in turn produces a changing electric field (Faraday's law). The two fields regenerate each other and move away from the charge as an electromagnetic wave whose frequency equals that of the oscillating charge. The energy radiated is supplied by the agency that keeps the charge oscillating.
Sketch: axes x, y, z; along z draw E oscillating sinusoidally along x and B oscillating along y, in phase, with the propagation arrow along +z; label E0, B0 and λ; write Ex = E0 sin(kz − ωt), By = B0 sin(kz − ωt). E, B and the direction of propagation are mutually perpendicular; the wave is transverse.
(b) E0/B0 = c, where c = 1/√(μ0ε0) = 3 × 108 m s−1.
(c) λ = c/ν = 3 × 108/(50 × 106) = 6 m.
k = 2π/λ = 2π/6 = 1.05 rad m−1.
B0 = E0/c = 60/(3 × 108) = 2 × 10−7 T.
Long answer (a) Name the seven regions of the electromagnetic spectrum in increasing order of frequency. For any three of them state how they are produced and give one use. (b) Explain why the ozone layer is crucial for life on Earth and how infrared radiation maintains the Earth's warmth. (c) Calculate the energy in eV of a photon of X-rays of wavelength 1 nm. (h = 6.63 × 10−34 J s, c = 3 × 108 m s−1, 1 eV = 1.6 × 10−19 J.)Frequently asked5 m
(a) Radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays.
Radio waves: produced by accelerated charges (oscillating currents) in antennas; used in radio and TV broadcasting.
Microwaves: produced by klystrons, magnetrons and Gunn diodes; used in radar and microwave ovens.
X-rays: produced by bombarding a metal target with high-energy electrons; used for medical diagnosis (imaging bones) and for treating some cancers.
(Any three, with one production method and one use each.)
(b) The ozone layer at the top of the stratosphere absorbs most of the Sun's ultraviolet radiation. UV in large amounts damages living cells, causing skin cancer and eye damage, so without the layer the surface would be unsafe for life.
Sunlight warms the ground, which re-radiates the energy as infrared. Water vapour and carbon dioxide in the atmosphere absorb this infrared and radiate part of it back to the surface (greenhouse effect), so the Earth's average temperature stays higher than it would without an atmosphere.
(c) E = hc/λ = 6.63 × 10−34 × 3 × 108/(1 × 10−9) = 1.989 × 10−16 J.
In eV: 1.989 × 10−16/(1.6 × 10−19) = 1.24 × 103 eV = 1.24 keV.
Case-based Ampere's circuital law states that the line integral of the magnetic field around a closed loop equals μ0 times the current through any surface bounded by the loop. Maxwell found that when a capacitor is being charged, two different surfaces bounded by the same loop give different answers: one is crossed by the charging current, the other, passing between the plates, is crossed by no current at all. He resolved this by noting that a changing electric field between the plates is equivalent to a current, which he called the displacement current, id = ε0 dΦE/dt. With this term the law becomes ∮B·dl = μ0(ic + id), and the total current is continuous through the whole circuit.
(i) During charging, how does the displacement current between the plates compare with the conduction current in the wire?
(ii) A capacitor is connected to a dc battery and is fully charged. What is the displacement current between its plates?
(iii) An ideal capacitor is connected across an ac source. Explain in one line why the current in the circuit does not stop.
(iv) The electric field between plates of area 1 × 10−2 m2 increases uniformly at 1 × 1012 V m−1 s−1. Find the displacement current.Frequently asked4 m
(i) They are equal in magnitude: id = ε0 dΦE/dt = dQ/dt = ic.
(ii) Zero. The electric field between the plates is constant, so the electric flux does not change.
(iii) The charge on the plates, and hence the electric field between them, keeps oscillating with the source, so a displacement current flows continuously across the gap and the conduction current in the wires never stops.
(iv) id = ε0 dΦE/dt = ε0A dE/dt = 8.85 × 10−12 × 10−2 × 1012 = 8.85 × 10−2 A ≈ 0.09 A.
Case-based All electromagnetic waves travel at the same speed in vacuum but differ in wavelength, in the way they are produced and in the way they are detected. Microwaves, with wavelengths from 0.1 m to 1 mm, are produced by klystrons and magnetrons. In a microwave oven the frequency is chosen to match the natural frequency of rotation of water molecules, so water in food absorbs energy strongly and the food heats from within. Infrared waves, emitted by every warm body, are absorbed by water vapour and carbon dioxide in the atmosphere, which is why the atmosphere keeps the Earth warmer than it would otherwise be. Ultraviolet radiation from the Sun is largely absorbed by the ozone layer.
(i) Why is it necessary to match the oven frequency to the resonant frequency of water molecules?
(ii) A microwave oven operates at 2.45 GHz. Find the wavelength of the microwaves.
(iii) Name the effect by which the atmosphere keeps the Earth warm and the gases responsible.
(iv) Arrange microwaves, infrared and ultraviolet in increasing order of frequency.Frequently asked4 m
(i) Energy transfer from the wave to the molecules is most efficient at resonance; at the matching frequency water molecules absorb the microwave energy strongly and their thermal motion rises, heating the food.
(ii) λ = c/ν = 3 × 108/(2.45 × 109) = 0.122 m ≈ 12 cm.
(iii) The greenhouse effect; water vapour and carbon dioxide (also methane and other greenhouse gases).
(iv) Microwaves < infrared < ultraviolet.
Test yourselfPractice quiz
Marks lost every yearCommon mistakes
- Writing the displacement current as μ0 dΦE/dt or as ε0 dΦB/dt. It is ε0 dΦE/dt; μ0 appears only when the current is put into the Ampere–Maxwell law. The examiner gives zero for the wrong constant.
- In the derivation, skipping the two-surface argument and jumping straight to id = ic. The first mark of the 3-mark question is for showing that Ampere's law gives two different values of B at the same point.
- Saying that a moving charge radiates. Only an accelerated charge radiates; a charge in uniform motion produces a steady magnetic field and no wave.
- Getting the direction of the missing field wrong. Propagation is along E × B. For a wave written with kz + ωt the propagation is along −z, and students who ignore the sign lose the mark for direction.
- Mixing up E0 = cB0 with B0 = cE0. The electric field number is always the larger one in SI units; 300 V m−1 goes with 10−6 T, not 9 × 1010 T.
- Claiming that one band of the spectrum travels faster than another, or that microwaves are used in radar because they are fast. All EM waves travel at c in vacuum; microwaves suit radar because of their short wavelength.
- Confusing X-rays and gamma rays in the production question. X-rays: bombarding a metal target with fast electrons. Gamma rays: radioactive nuclei and nuclear reactions.
- Answering 'higher' for the Earth without an atmosphere. Without greenhouse gases the ground's infrared escapes directly, and the average surface temperature would be lower.
How to prepareStrategy for this chapter
This chapter is small and predictable, so aim for full marks with two or three hours of preparation. Begin with the displacement-current derivation and write it out by hand twice, drawing the capacitor with the two surfaces each time. The four-step structure (two surfaces disagree; E = Q/ε0A and ΦE = Q/ε0; ε0 dΦE/dt = dQ/dt; generalised law) is worth 3 marks and has appeared in several papers.
Next, be able to write the four Maxwell equations, the pair Ex = E0 sin(kz − ωt), By = B0 sin(kz − ωt), and the three relations c = ω/k, c = E0/B0 and c = 1/√(μ0ε0). Practise the standard numerical: given one field expression, find λ, ν, the direction of travel, and the other field. Do NCERT 8.8, 8.10 and Example 8.2 until you can finish each in three minutes.
For the spectrum, memorise one table: band, NCERT wavelength range, how it is produced, one detector or use. Say the order aloud both ways. Practise the placement problems (frequency or wavelength given, name the band) and the three-part questions on ozone, greenhouse effect, welders' goggles and microwave ovens.
Finally, be able to show that the electric and magnetic energy densities of the wave are equal at every instant: substitute E = cB into uE = ½ε0E2, then use c2 = 1/(μ0ε0) to reach uB = B2/2μ0, as NCERT Exercise 8.10(c) asks. In the exam, budget about five minutes for the 1-mark items from this chapter and no more than eight for a 3-mark derivation.