Electromagnetic Waves: Class 12 Physics · Chapter 8

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 28 Sep 2026 · Updated 28 Sep 2026

95 min
17 practice questions
27 MCQs

What you will learn in Electromagnetic Waves

Explain why Ampère's circuital law fails for a charging capacitor and how Maxwell's displacement current removes the contradiction
Calculate displacement currents and the magnetic fields they produce, including the field between the plates of a charging capacitor
State Maxwell's four equations in integral form and describe what each one says about the sources of electric and magnetic fields
Explain how accelerated charges produce electromagnetic waves and describe the experiments of Hertz, J. C. Bose and Marconi
Describe a plane electromagnetic wave, with perpendicular electric and magnetic fields in phase, and use E0 = cB0, c = 1/√(μ0ε0) and v = 1/√(με)
Write and interpret wave equations for E and B using k = 2π/λ, ω = 2πν and c = νλ, and find the energy the fields carry
Name the regions of the electromagnetic spectrum in order, with rough wavelength ranges, sources, detectors, uses and hazards
Relate photon energy to frequency and explain why different regions of the spectrum interact with matter so differently

Original companion text aligned to NCERT Physics Class 12, Chapter 8, pp. 201–214. Not affiliated with or endorsed by NCERT.

8.1 Introduction

Pick up a phone. Signals from a mobile tower and a Wi-Fi router are reaching it, its screen is sending light to your eyes, and the sunlight through the window is warming your hand. The X-rays in a hospital scanner and the gamma rays used to treat cancer belong to the same family. All of these are electromagnetic waves (वैद्युतचुंबकीय तरंगें): travelling patterns of electric and magnetic field that differ only in wavelength. This chapter explains what such a wave is, how Maxwell predicted it on paper before anyone had made one, and how the whole family is arranged into a spectrum.

The road to these waves begins with a small puzzle about a capacitor. Apply Ampère's circuital law to the wire that is charging it, and the answer depends on which surface you happen to draw, which a law of nature must never do (Section 8.2). Maxwell's cure was to let an electric field that changes with time act as a source of magnetic field, just as a current does. He called the new term the displacement current. Fig. 8.1 shows why this single addition matters so much: it closes a loop of cause and effect between the two fields.

3/4Maxwell's new link: an electric field that changes with time creates a magnetic field.

Fig. 8.1Charges and currents create fields; changing fields create each other. Maxwell added the last link, and with it the fields can travel on their own.NCERT p. 201

Two circles stand side by side: an orange one for the electric field E on the left and a blue one for the magnetic field B on the right. Step 1: a box labelled charges on the far left points an orange arrow into E, and a box labelled currents on the far right points a blue arrow into B; nothing joins the two circles. Step 2: an orange arc below the circles, labelled changing B and Faraday, 1831, runs from B to E. Step 3: a blue arc above them, labelled changing E and Maxwell, 1862, runs from E to B. Step 4: the charges and currents boxes fade, the two arcs thicken and flow round in a loop, and a wavy arrow along the bottom, labelled speed c, shows the pair of fields travelling away as a wave.

With the loop closed, the two fields can keep each other going far from any charge or current. A changing $𝐄$ sustains a changing $𝐁$ , and that changing $𝐁$ sustains a changing $𝐄$ in turn. Maxwell's equations, the four laws in Table 8.1, predict that such a disturbance travels at $1 / \sqrt{μ_{0} ε_{0}}$ . Both constants come from bench experiments with charges and currents, yet the result is $3.00 \times 10^{8} m / s$ . In 1849 Hippolyte Fizeau had timed a beam of light over a $17 k m$ round trip near Paris and found about $3.15 \times 10^{8} m / s$ . The match convinced Maxwell that light is an electromagnetic wave, and optics became part of electromagnetism.

Two decades passed before anyone tested the prediction. In 1887 Heinrich Hertz made and detected electromagnetic waves in his laboratory, and radio, radar, television and mobile phones grew out of his apparatus. We begin with the displacement current and then look at how the waves are launched and what they look like. The chapter ends with a tour of the spectrum, from gamma rays shorter than a trillionth of a metre to radio waves many kilometres long.

Warm-up

1 of 2

Which link between electric and magnetic fields did Maxwell add to what Oersted, Ampère and Faraday had already found?

An electric current produces a magnetic fieldA changing magnetic field produces an electric fieldA changing electric field produces a magnetic fieldA steady electric field produces a steady magnetic field

Show answer

Answer (C) A changing electric field produces a magnetic field

Currents making magnetic fields (Oersted, Ampère) and changing magnetic fields making electric fields (Faraday) were known before him. Maxwell's new idea was the reverse of Faraday's: a changing electric field makes a magnetic field. A steady electric field makes no magnetic field at all.

8.2 Displacement current

Ampère's circuital law, from Chapter 4, relates the magnetic field around a closed loop to the current that passes through the loop:

Ampère's circuital law

\oint 𝐁 \cdot d 𝐥 = μ_{0} i (t)

(8.1)

where

$i (t)$: the current through any surface whose boundary is the loop

The words any surface matter. A loop is the rim of countless surfaces: a flat disc, a shallow dome, a deep bag. The law claims that every one of them is crossed by the same current. For a steady current in a wire this is true, because charge cannot pile up anywhere; whatever flows into a closed region flows out again. Maxwell asked what happens when charge does pile up, as it does on the plates of a capacitor while it is being charged.

Ampère's law asks for two choices: a closed loop, and a surface that has the loop as its edge. Make them for the capacitor in Fig. 8.2, which is charged by a current $i (t)$ flowing up a straight wire. For the loop, take a horizontal circle of radius $r$ centred on the wire, a little below the lower plate. Turning the apparatus about the wire leaves it unchanged, so the field must wrap around this circle with a single strength $B$ . The left side of (8.1) then becomes $2 π r B$ . For the surface, the obvious choice is the flat disc $S_{1}$ that fills the circle. The wire pierces it, so

B (2 π r) = μ_{0} i (t)

(8.2)

Now keep the same loop but take the balloon-shaped surface $S_{2}$ , which rises from the loop, passes around the lower plate and closes in the gap between the plates. No wire crosses $S_{2}$ anywhere, so the current through it is zero, and Ampère's law would give $B = 0$ on the very same loop. One loop, one field, two answers: as it stands, the law contradicts itself whenever charge is building up somewhere.

3/3What crosses $S_{2}$ is a growing electric field. Its flux changes at a rate for which $ε_{0} d Φ_{E} / d t = i$ .

Fig. 8.2One loop, two surfaces. The wire crosses the flat surface but not the balloon, yet both must give the same magnetic field around the loop.NCERT Fig. 8.1 · p. 203

A vertical wire carries a current i upward into the lower plate of a parallel-plate capacitor drawn in perspective; the lower plate carries plus signs and the upper plate minus signs. A blue circular loop of radius r surrounds the wire below the lower plate, with arrows showing the magnetic field B circulating around it. A panel on the right states what Ampere’s law gives. Step 1: the loop is spanned by a flat green disc S1 that the wire pierces, and the panel reads: through S1, current i, so B times 2 pi r equals mu nought i. Step 2: the disc is replaced by a purple balloon-shaped surface S2 with the same rim, which rises around the lower plate and closes in the gap between the plates; no wire crosses it, and the panel reads: through S2, no current, B times 2 pi r equals zero, with a question mark. Step 3: orange electric field lines appear in the gap, pointing up from the positive to the negative plate and crossing S2, and the panel reads: through S2, changing electric flux, epsilon nought times d Phi E by d t equals i.

Something must cross $S_{2}$ that plays the part of the current. The only thing in the gap is the electric field between the plates, and while the capacitor charges, that field is growing. Let us work out how fast its flux through $S_{2}$ grows.

DerivationThe current hidden in the gap

Let each plate have area $A$ , and let the charges on the plates be $\pm Q$ at time $t$ . Between the plates the field is uniform, with magnitude $E = Q / (ε_{0} A)$ , pointing from the positive plate to the negative one. Outside the plates it is negligible, if we ignore the small fringe field near the edges. Field lines cross only the part of $S_{2}$ that lies in the gap, at right angles to it, so the electric flux through $S_{2}$ is

Φ_{E} = E A = \frac{Q}{ε_{0} A} A = \frac{Q}{ε_{0}}

(8.3)

The charge on the plates grows at exactly the rate at which the wire delivers it, $d Q / d t = i$ . Differentiating (8.3) with respect to time gives $d Φ_{E} / d t = (1 / ε_{0}) d Q / d t = i / ε_{0}$ , that is

ε_{0} \frac{d Φ_{E}}{d t} = i

(8.4)

The quantity $ε_{0} d Φ_{E} / d t$ through $S_{2}$ has the units of current, and it is exactly equal to the conduction current through $S_{1}$ . If we count it as a current in Ampère's law, the contradiction disappears: both surfaces carry the same total current $i$ .

Definition

Displacement current

The quantity $i_{d} = ε_{0} d Φ_{E} / d t$ , where $Φ_{E}$ is the electric flux through a surface. No charge flows, yet it produces a magnetic field exactly as a conduction current of the same size would.

ExampleWhile a capacitor charges at $0.50 A$ , a displacement current of $0.50 A$ crosses the gap between its plates.

The familiar current, carried by moving charges, is now called the conduction current $i_{c}$ . Maxwell's generalisation is that both kinds of current produce a magnetic field, and that the current in Ampère's law is their sum:

i = i_{c} + i_{d} = i_{c} + ε_{0} \frac{d Φ_{E}}{d t}

(8.5)

LawAmpère–Maxwell law

The line integral of $𝐁$ around any closed loop equals $μ_{0}$ times the total current (conduction plus displacement) through any surface bounded by that loop.

Ampère–Maxwell law

\oint 𝐁 \cdot d 𝐥 = μ_{0} i_{c} + μ_{0} ε_{0} \frac{d Φ_{E}}{d t}

(8.6)

Apply (8.6) once more to the loop of Fig. 8.2. The flat surface $S_{1}$ lies outside the plates, where there is no electric field, so only conduction current crosses it. The balloon $S_{2}$ is crossed by no wire, so only displacement current crosses it. Both equal $i$ , and both now give $B (2 π r) = μ_{0} i$ . The field no longer depends on which surface we happen to choose, which is what a law of nature must guarantee. This form of the law is called the Ampère–Maxwell law.

Compare

Two kinds of current

Conduction current $i_{c}$

Charges actually move
Found in wires, electrolytes and ionised gases
Can be perfectly steady
Heats the conductor it flows in ( $I^{2} R$ )
Produces a magnetic field around it

Displacement current $i_{d}$

No charge moves; the electric flux through a surface changes
Found wherever an electric field changes: capacitor gaps, insulators, empty space
Zero whenever the electric field is steady
Produces no heating in empty space
Produces a magnetic field in exactly the same way

In real materials, which are never perfect conductors or perfect insulators, the two kinds of current usually flow side by side. But the second column hides the key to this chapter. Picture the space between the Sun and the Earth, with no wires and almost no charges in it. If the electric field there changes, the Ampère–Maxwell law says that a magnetic field must circle the change, exactly as it would circle a wire carrying a current $ε_{0} d Φ_{E} / d t$ . Magnetic fields can therefore arise in regions with no conduction current at all, which is what allows electromagnetic waves to exist.

Is the displacement current real? Its magnetic field is. Between circular plates that are being charged, a magnetic field circles the axis of the capacitor, as in Fig. 8.3. At any distance from the axis larger than the plate radius, the field in the plane of the gap has the same value as at that distance from the wire. Careful measurements confirm this. When the charging stops, the electric field stops changing and the magnetic field in the gap disappears; when the capacitor discharges, it returns with its direction reversed.

1/3While the capacitor charges, $𝐄$ grows and a magnetic field circles the gap, anticlockwise when seen from above.

Fig. 8.3A magnetic field circles the gap only while the electric field between the plates is changing, and its direction depends on whether the field is growing or shrinking.NCERT Fig. 8.2 · p. 204

Left: a side view of a parallel-plate capacitor with the negative plate on top and the positive plate below, orange electric field arrows pointing up across the gap, and a bar gauge showing the strength of the field. Right: the gap seen from above, with the electric field shown as orange dots (pointing out of the page) and blue circles for the magnetic field around the axis. Step 1: current flows up the wires, the gauge reads rising, and the blue circles carry arrows running anticlockwise. Step 2: no current flows, the gauge is full and reads steady, and the blue circles have vanished. Step 3: current flows down the wires, the gauge reads falling, and the blue circles return with arrows running clockwise.

DerivationMagnetic field between circular plates

Let the plates be discs of radius $R$ , charged by a current $i$ . Take a circular loop of radius $r$ in the gap, centred on the axis and parallel to the plates. By symmetry $B$ is tangent to the loop and has the same size all round it. The flat disc spanning the loop is crossed by displacement current only.

If $r \geq R$ , the disc catches all of the electric flux between the plates, so the enclosed displacement current is the whole of $i$ and

B = \frac{μ_{0} i}{2 π r} (r \geq R)

If $r < R$ , the field between the plates is uniform, so the disc catches only the fraction $π r^{2} / π R^{2}$ of the flux, and of the displacement current:

B (2 π r) = μ_{0} i \frac{r^{2}}{R^{2}} ⟹ B = \frac{μ_{0} i r}{2 π R^{2}} (r < R)

The field grows in proportion to $r$ from zero on the axis to its largest value, $μ_{0} i / 2 π R$ , at the rim of the plates. Beyond the rim it falls off as $1 / r$ , exactly like the field around the wire (Fig. 8.4).

3/3The field is zero on the axis, largest at the rim of the plates, and falls off outside.

Fig. 8.4Between the plates,

B

rises linearly up to the rim and then falls off as

1 / r

, just like the field around the wire.

A graph of magnetic field B (up) against distance r from the axis (across), with a dashed vertical line at the plate radius R. On the right, a small top view of the plate shows which part of the displacement current a loop encloses. Step 1: only the outer part of the curve is drawn, falling off as mu nought i over 2 pi r beyond R; the inset shows a loop larger than the plate enclosing the whole shaded plate, so the enclosed displacement current equals i. Step 2: the straight line from the origin up to R is drawn, labelled mu nought i r over 2 pi R squared; the inset shows a small loop enclosing only a shaded inner disc, the fraction r squared over R squared. Step 3: both parts are drawn together with a dot at the peak at r equals R, labelled B max equals mu nought i over 2 pi R.

Example

The field inside a charging capacitor

A capacitor with circular plates of radius $5.0 c m$ is charged by a steady current of $1.2 A$ . Find the magnetic field in the gap (a) $2.0 c m$ and (b) $10 c m$ from the axis. (c) Where between the plates is the field strongest, and how strong is it there? Take $μ_{0} = 4 π \times 10^{- 7} T m / A$ .

Try it yourself first. Then reveal the solution one step at a time.

InteractiveField around a charging capacitor

An interactive capacitor with circular plates of radius 5 cm, charged through a vertical wire. You choose whether a circular loop sits around the wire or in the gap between the plates, and whether a flat disc or a bag through the gap spans it. Sliders set the loop radius from 0.5 to 12 cm and the current from minus 2 to plus 2 amperes. Readouts show the conduction current, the displacement current and the total current through the chosen surface, and the magnetic field on the loop with its direction. A graph plots the field against loop radius in the gap and around the wire, with a dot for the current setting.

Fig. 8.5Move a loop around the wire or into the gap, change its radius and the current, and compare the conduction and displacement currents it encloses.

Look back at Fig. 8.1. Before Maxwell, only one of the arrows between the two fields existed: a changing magnetic field makes an electric field (Faraday's law). The Ampère–Maxwell law supplies the arrow in the other direction, so each field can now be a source of the other.The symmetry is not perfect. Electric field lines start and end on charges, but no isolated magnetic pole (a monopole) has ever been found, so magnetic field lines always close on themselves. Together with Gauss's laws for the two fields, these two laws form the complete set in Table 8.1.

Table 8.1Maxwell's equations in vacuum, in integral form

Law	Equation	What it says
Gauss's law for $𝐄$	$\oint 𝐄 \cdot d 𝐀 = Q / ε_{0}$	Electric field lines begin and end on charges; the outward flux measures the charge enclosed.
Gauss's law for $𝐁$	$\oint 𝐁 \cdot d 𝐀 = 0$	There are no magnetic charges; magnetic field lines form closed loops.
Faraday's law	$\oint 𝐄 \cdot d 𝐥 = - \frac{d Φ_{B}}{d t}$	A changing magnetic flux produces an electric field that circulates around it.
Ampère–Maxwell law	$\oint 𝐁 \cdot d 𝐥 = μ_{0} i_{c} + μ_{0} ε_{0} \frac{d Φ_{E}}{d t}$	Currents and changing electric flux produce a magnetic field that circulates around them.

Add the Lorentz force, $𝐅 = q (𝐄 + 𝐯 \times 𝐁)$ , which tells charges how to respond to the fields, and you have the whole of classical electromagnetism in five lines. The next section shows what the two middle equations do when no charges or currents are present at all.

Quick check

1 of 4

A capacitor is being charged. Which statement about the region between its plates is correct?

A conduction current flows across the gapNo current of any kind crosses the gap, so there is no magnetic field thereA displacement current flows, but it is always smaller than the current in the leadsA displacement current equal to the current in the leads crosses the gap

Show answer

Answer (D) A displacement current equal to the current in the leads crosses the gap

No charge crosses the gap, so there is no conduction current there. But the electric flux in the gap changes at exactly the rate needed for

ε_{0} d Φ_{E} / d t = i

, so the displacement current equals the current in the leads and produces a magnetic field in the gap.

Maxwell's equations

Match each of Maxwell's equations with what it tells us.

Tap an item on the left, then its match on the right.

Show answer

Answer

Gauss's law for E : Charges are the sources of electric field lines
Gauss's law for B : There are no magnetic charges
Faraday's law : A changing magnetic flux produces an electric field
Ampère–Maxwell law : Currents and changing electric flux produce a magnetic field

Each law gives one kind of source: charges for E, nothing for B (no monopoles), changing B for circulating E, and currents plus changing E for circulating B.

8.3 Electromagnetic waves

A changing electric field makes a magnetic field, and a changing magnetic field makes an electric field. Put the two facts together and a remarkable possibility opens up: a disturbance in which each field keeps creating the other, moving on through space with nothing to carry it. This section asks what produces such a disturbance and what it looks like.

8.3.1 Sources of electromagnetic waves

What does a transmitter mast actually do? Inside its antenna (ऐन्टेना), an oscillator drives electrons up and down the metal many millions of times a second. As they surge, the electric field near the antenna keeps changing. A changing electric field is a displacement current, so a changing magnetic field curls around it. By Faraday's law, that changing magnetic field drives a changing electric field a little further out, which in turn drives a magnetic field further out still. Each field keeps the other going, and the pattern spreads outwards at the speed of light (Fig. 8.6). Two rods fed at the middle like this form an oscillating electric dipole, the simplest source of electromagnetic waves.

4/4Every cycle adds one wavelength, so $λ = c T = c / ν$ and the wave has the frequency $ν$ of the source.

Fig. 8.6An oscillating charge sends out one more wavelength of field pattern with every cycle, so the wave has the frequency of its source.

On the left, a vertical antenna made of two metal rods is fed at its middle by an oscillator of frequency nu. A dashed horizontal line runs to the right from the antenna. Step 1: the oscillator is off, the antenna is idle and there is no field pattern. Step 2: double-headed red arrows beside the rods show charge surging up and down; one wavelength of orange electric field arrows, alternately pointing up and down, stretches from the antenna to a dashed wave front that moves away at speed c. Step 3: the front has moved twice as far and two wavelengths of the pattern are present. Step 4: three wavelengths are present, and a bracket under the second one is labelled lambda.

Table 8.2Which moving charges radiate?

How the charge moves	Fields it produces	Does it radiate?
At rest	a static electric field	No
At constant velocity (a steady current)	electric and steady magnetic fields that travel along with it	No
Accelerating: speeding up, slowing down, turning or oscillating	changing fields that break away and spread outwards	Yes

The rule in Table 8.2 comes from Maxwell's equations: only accelerated charges radiate. Acceleration includes a change of direction, so even a charge circling at constant speed radiates.

Whatever energy the wave carries away has to be supplied by the source. That is why a transmitter keeps drawing power from its supply even though no wire joins it to the radios that pick up its signal. The figure also shows why the wave's frequency equals the source's: each up-and-down cycle of the electrons launches exactly one more wavelength.

Every wave in Fig. 8.6 has the frequency of the charges that launch it. To make green light ( $530 n m$ ) with an antenna, its electrons would have to swing back and forth $5.7 \times 10^{14}$ times a second, completing each swing in under two femtoseconds. No circuit comes close: today's fastest transistors reach about $10^{12} H z$ , and the spark circuits of the 1880s a few hundred megahertz. Light is made by oscillating charges, but inside atoms, not in wires we can build. A test of Maxwell's claim therefore had to run the other way round. Make much slower waves with a circuit, then show that they travel at the speed of light and behave like light in every other way.

Heinrich Hertz did exactly that in 1887 (Fig. 8.7). An induction coil charged two metal rods, separated by a small air gap, until a spark jumped across. While the spark lasted, charge surged back and forth between the rods at a very high frequency, and the rods radiated. Metres away, a simple loop of wire with its own tiny gap showed faint sparks: the waves had crossed the room. By reflecting them from a metal sheet, Hertz set up standing waves, measured the spacing of their nodes to find the wavelength, and combined it with the frequency of his oscillator. The speed came out close to that of light. He also showed that his waves could be reflected, refracted and polarised, just like light.

3/4Metres away, the waves drive a current round the ring and a tiny spark jumps its gap.

Fig. 8.7Hertz's test of Maxwell's prediction: sparks make waves, the waves make a spark across the room, and standing waves reveal their wavelength.NCERT p. 206

On the left, a transmitter: two metal rods in a line, each ending in a large metal ball, with a small gap between their inner ends; an induction coil below is wired across the gap. On the right, a receiver: a ring of wire with a tiny gap at the top. Step 1: the coil charges the rods, the left ball positive and the right ball negative, and nothing else happens. Step 2: a spark jumps across the transmitter gap and dashed orange wave fronts spread out to the right, not yet reaching the ring. Step 3: the wave fronts reach the ring and a tiny spark appears in its gap. Step 4: a metal sheet stands on the far right; the wave fronts are replaced by a row of humps showing the strength of the standing wave between the transmitter and the sheet, zero at the sheet and at points spaced half a wavelength apart, with the ring placed above a hump and a bracket marking lambda over 2 between two zeros.

Compare

After Hertz: two directions

Shorter waves: J. C. Bose, Kolkata

Generated waves 5–25 mm long (about 12–60 GHz), in today's microwave range
Showed on the bench that they reflect, refract and polarise like light
Invented the horn antenna and a galena crystal detector

Longer range: G. Marconi, Italy and England

Chose longer waves and tall, earthed antennas to reach far
Crossed the Atlantic in 1901, from Cornwall to Newfoundland
Built radio into a service for ships, and shared the 1909 Nobel Prize in Physics

Try at home

A radio that hears sparks

You’ll need

an AM (medium-wave) radio
a 9 V battery
a short piece of wire or a steel key
a spark-type kitchen gas lighter (optional)

Predict first

Each time you brush the wire across the battery terminals near the radio, what will you hear?

Nothing: a battery gives a steady currentA steady hum for as long as the wire touchesA click or crackle each time the circuit is made or broken

Tune an AM (medium-wave) radio to a quiet spot between stations and turn the volume up.
Hold a 9 V battery about $20 c m$ from the radio. Brush a short piece of wire across its two terminals for a moment, again and again.
Repeat from further away, and then try clicking a kitchen gas lighter (the spark type) near the radio.

8.3.2 Nature of electromagnetic waves

Far from its source, a small patch of a wave looks like a plane wave: the fields are the same everywhere on any plane perpendicular to the direction of travel. You can see why the fields in it must be perpendicular to each other. Faraday's law makes an electric field circulate around a changing magnetic field, and the Ampère–Maxwell law makes a magnetic field circulate around a changing electric field. A field that curls around another lies at right angles to it. Maxwell's equations also show that neither field has a component along the direction of travel. So $𝐄$ , $𝐁$ and the direction of travel are mutually perpendicular, and an electromagnetic wave is a transverse wave.

Take a wave travelling along the $+ x$ axis with its electric field along $y$ . Fig. 8.8 shows a snapshot. The magnetic field then points along $z$ , and at every point the two fields rise and fall together:

E_{y} = E_{0} \sin (k x - ω t)

(8.7(a))

B_{z} = B_{0} \sin (k x - ω t)

(8.7(b))

2/4 $𝐁$ points along $\pm z$ (out of or into the page). It is largest where $𝐄$ is largest and zero where $𝐄$ is zero: the fields are in phase.

Fig. 8.8The electric field, the magnetic field and the direction of travel point three mutually perpendicular ways, and the two fields rise and fall together.NCERT Fig. 8.3 · p. 206

A horizontal line runs across the figure, with an arrow at the top right showing that the wave travels to the right. Orange arrows standing on the line show the electric field at equally spaced points: they point up and down with lengths that follow a sine curve, and one crest arrow is drawn thicker in red. Step 1: only the electric field arrows are shown. Step 2: a second row below shows the magnetic field at the same points as blue circles, a dot in a circle for out of the page where E points up and a cross in a circle for into the page where E points down, largest where the E arrows are longest. Step 3: a quarter period later the whole pattern has shifted a quarter wavelength to the right; a dashed line marks where the red crest was and a bracket labelled lambda over 4 shows how far it moved. Step 4: the rows fade almost completely and a large set of three arrows appears: E pointing up, B pointing out of the page (drawn slanting towards the viewer and labelled out of page) and the direction of travel pointing right.

Here $E_{0}$ and $B_{0}$ are the amplitudes, $ω = 2 π ν$ is the angular frequency, and $k$ is the wave number, also called the propagation constant. The vector of size $k$ that points along the direction of travel is the wave vector $𝐤$ . The direction of travel is always the direction of $𝐄 \times 𝐁$ ; here $\hat{𝐣} \times \hat{𝐤} = \hat{𝐢}$ , along $+ x$ as required.

DerivationWavelength, speed and frequency

At any fixed instant, the pattern (8.7) repeats whenever $k x$ increases by $2 π$ . That distance is the wavelength, so

k = \frac{2 π}{λ}

(8.8)

Now follow a single crest. A crest is a point of fixed phase, $k x - ω t = constant$ . Differentiating with respect to time gives $k d x / d t - ω = 0$ , so the crest moves at $d x / d t = ω / k$ . When (8.7) is substituted into Maxwell's equations, they are satisfied only if this speed has one particular value, fixed by $μ_{0}$ and $ε_{0}$ alone:

ω = c k, c = \frac{1}{\sqrt{μ_{0} ε_{0}}}

(8.9(a))

Writing $ω = 2 π ν$ and $k = 2 π / λ$ turns $ω = c k$ into $2 π ν = 2 π c / λ$ , the familiar wave relation

c = ν λ

(8.9(b))

With $μ_{0} = 4 π \times 10^{- 7} T m / A$ and $ε_{0} = 8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ , (8.9a) gives $c = 2.998 \times 10^{8} m / s$ .

3/3At a fixed point, $E_{y}$ oscillates in time with period $T = 1 / ν$ .

Fig. 8.9A snapshot shows the wavelength; watching one point shows the period. The pattern moves one wavelength in one period, so

c = λ / T = ν λ

Top: a graph of E y against x at one instant, a sine curve of amplitude E nought, with a bracket marking one wavelength lambda between neighbouring crests. Step 1: only this orange curve for t equals 0 is shown. Step 2: a dashed red curve for a quarter period later is added; it is the same shape shifted to the right, and an arrow between neighbouring crests is labelled lambda over 4. Step 3: a second graph appears below, showing E y against time t at the single point x equals 0: a green sine curve whose neighbouring crests are one period T apart, marked by a bracket.

Maxwell's equations also link the two amplitudes:

B_{0} = \frac{E_{0}}{c}

(8.10)

Because the fields are in phase, $E = c B$ holds at every point and every instant, not just at the peaks. The factor $c$ makes the magnetic field look tiny in SI units: a strong radio signal with $E_{0} = 3.0 V / m$ has $B_{0} = 1.0 \times 10^{- 8} T$ , about one five-thousandth of the Earth's field. Yet, as we shall see, the two fields carry equal shares of the wave's energy. Use Fig. 8.10 to watch a wave travel, change its frequency and the medium, and read both fields at one point.

InteractiveA travelling electromagnetic wave

An animated three-dimensional view of a plane electromagnetic wave travelling along x, with orange electric field arrows along y and blue magnetic field arrows along z, which you can turn by dragging or by choosing a 3-D, side or top view. Controls pause the wave, step it forward, reverse its direction of travel, show or hide each field, set the frequency from 100 to 500 megahertz and the electric field amplitude from 1 to 60 volts per metre, and put the wave in vacuum or in materials with relative permittivity 2.25 or 4. Readouts give the wavelength, speed, period, magnetic amplitude, wave number and angular frequency, and a probe shows both fields at one point with a scrolling graph of their values in time.

Fig. 8.10A plane wave in slow motion. The probe shows

E_{y}

and

B_{z}

at one point; the graph below it traces them in time.

Example 8.1NCERT Example 8.1 · p. 208

The magnetic partner of a known electric field

A radio wave of frequency $150 M H z$ travels due north through air (take its speed as $c$ ). At a certain point and instant its electric field is $4.8 V / m$ , pointing vertically upward. Take $x$ towards the east, $y$ towards the north and $z$ vertically up. Find (a) the wavelength and (b) the magnetic field at that point and instant, in magnitude and direction.

Try it yourself first. Then reveal the solution one step at a time.

Example 8.2NCERT Example 8.2 · p. 208

Reading a wave equation

The electric field of a plane electromagnetic wave in vacuum is $E_{y} = (45 V / m) \sin [(1.5 \times 10^{2} m^{- 1}) x - (4.5 \times 10^{10} s^{- 1}) t]$ . (a) Find its wavelength and frequency, and name the region of the spectrum. (b) In which direction does it travel? (c) Write an expression for its magnetic field.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

An electromagnetic wave in vacuum has an electric field amplitude of

21 V / m

. What is its magnetic field amplitude, in nanotesla?

Show answer

Answer 70 nT

B_{0} = E_{0} / c = 21 / (3.0 \times 10^{8}) = 7.0 \times 10^{- 8} T = 70 n T

Sound needs air and a wave on a string needs the string, but an electromagnetic wave needs nothing. Its fields are self-sustaining: each is kept going by the changes in the other. That is why light from the Sun, and from galaxies billions of light years away, reaches us across nearly empty space.

Light slows down in glass, to about $2.0 \times 10^{8} m / s$ . The reason is that the charges in the atoms of the glass are pushed back and forth by the wave's electric field, and their own small fields add to it. In Maxwell's equations this is taken care of by replacing $ε_{0}$ with the permittivity $ε = ε_{r} ε_{0}$ of the material, and $μ_{0}$ with its permeability $μ = μ_{r} μ_{0}$ . The wave speed then becomes

Speed in a material

v = \frac{1}{\sqrt{μ ε}} = \frac{c}{\sqrt{μ_{r} ε_{r}}}

(8.11)

Most transparent materials are non-magnetic ( $μ_{r} \approx 1$ ) and have $ε_{r} > 1$ , so waves travel more slowly in them than in vacuum. $ε_{r}$ depends on frequency. For water it is about 80 for slowly varying fields but only about 1.8 at the frequencies of light, which is why water's refractive index is 1.33 rather than $\sqrt{80}$ . The frequency cannot change at a boundary, because every crest that arrives must pass on, so the wavelength shrinks in the same proportion as the speed (Fig. 8.11). The ratio $c / v$ is the refractive index of the material, the starting point of Chapter 9.

2/3In a non-magnetic material with $ε_{r} = 2.25$ , $v = c / \sqrt{2.25} = c / 1.5$ , so the wavelength shrinks to $λ_{0} / 1.5$ .

Fig. 8.11Crossing into a material does not change the frequency, so a slower speed means a shorter wavelength.

An orange sine wave travels from left to right along a dashed line. A bracket under the left part marks one wavelength in vacuum, lambda nought. Step 1: the whole width is vacuum and the wave has the same wavelength everywhere. Step 2: the right half becomes a pale blue slab of a material with relative permittivity 2.25; inside it the wave is squeezed to two-thirds of its vacuum wavelength, marked by a second bracket, and the speed is labelled c over 1.5. Step 3: the slab is replaced by a purple material with relative permittivity 4; the wavelength inside is now half the vacuum value and the speed is c over 2. In both material steps the wave joins smoothly at the boundary and a label notes that the frequency is unchanged.

Example

Signals in a cable

The space inside a coaxial cable carrying a $500 M H z$ television signal is filled with polythene, for which $ε_{r} = 2.25$ and $μ_{r} = 1$ . Find (a) the speed of the signal in the cable, (b) its wavelength in air and in the cable, and (c) the delay introduced by $30 m$ of cable.

Try it yourself first. Then reveal the solution one step at a time.

Experiments with radio waves, light, X-rays and gamma rays find no difference at all between their speeds in vacuum. Physicists now trust $c$ so completely that since 1983 it has not been measured but defined as exactly $299 792 458 m / s$ . The metre is now the distance light covers in vacuum in $1 / 299 792 458$ of a second.

Electromagnetic waves carry energy. Sunlight warms the ground and powers photosynthesis, and a broadcast signal delivers a tiny amount of power to every antenna it reaches. From Chapters 2 and 6, an electric field stores energy $\frac{1}{2} ε_{0} E^{2}$ per unit volume and a magnetic field stores $B^{2} / 2 μ_{0}$ . How is a wave's energy shared between its two fields?

DerivationEqual shares of energy

At any point in the wave $B = E / c$ , and $c^{2} = 1 / μ_{0} ε_{0}$ . The magnetic energy density is therefore

u_{B} = \frac{B^{2}}{2 μ_{0}} = \frac{E^{2}}{2 μ_{0} c^{2}} = \frac{μ_{0} ε_{0} E^{2}}{2 μ_{0}} = \frac{1}{2} ε_{0} E^{2} = u_{E}

The two fields hold equal amounts of energy at every point and every instant, so the total energy density is $u = ε_{0} E^{2}$ . Because $E^{2} = E_{0}^{2} \sin^{2} (k x - ω t)$ averages to $\frac{1}{2} E_{0}^{2}$ over a cycle, the average energy density of the wave is

Average energy density

\overset{ˉ}{u} = {\overset{ˉ}{u}}_{E} + {\overset{ˉ}{u}}_{B} = \frac{1}{4} ε_{0} E_{0}^{2} + \frac{1}{4} ε_{0} E_{0}^{2} = \frac{1}{2} ε_{0} E_{0}^{2}

Example

Energy near a transmitter

Near a broadcasting tower, the electric field of a radio wave has an amplitude of $6.0 V / m$ . Find (a) the magnetic field amplitude, (b) the average energy densities of its electric and magnetic fields, and (c) the total average energy density. ( $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ , $μ_{0} = 4 π \times 10^{- 7} T m / A$ )

Try it yourself first. Then reveal the solution one step at a time.

Try at home

Measure the speed of light in your kitchen

You’ll need

a microwave oven (with an adult's permission)
a large flat bar of chocolate, or grated cheese on a plate
a ruler

Predict first

The waves bounce between the metal walls of the oven and set up standing waves. Where will the chocolate melt first?

In spots about half a wavelength apartEvenly over the whole barOnly at the centre, where the turntable's axle was

Take the turntable out of a microwave oven and put an upturned plate in its place, so that the food cannot rotate.
Lay a large flat bar of chocolate (or a plate evenly covered with grated cheese) on it.
Heat on a low setting for 15–25 s, until a few soft or melted spots appear, then take it out.
Measure the distance $d$ between neighbouring melted spots, read the frequency $ν$ on the oven's rating label (usually $2450 M H z$ ), and calculate $c = 2 d ν$ .

Table 8.3Your measurement

	Value
Spot spacing $d$ (m)
Frequency $ν$ (Hz)
$c = 2 d ν$ (m/s)

Your entries are saved automatically.

Quick check

1 of 4

A non-magnetic plastic has

ε_{r} = 4.0

. At what speed do electromagnetic waves travel through it, in metres per second?

m/s

Show answer

Answer 150000000 m/s

v = c / \sqrt{ε_{r}} = 3.0 \times 10^{8} / 2 = 1.5 \times 10^{8} m / s

8.4 Electromagnetic spectrum

Seven names are in everyday use for electromagnetic waves: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. They were discovered one at a time over about a century, often by people who had no idea that they were all looking at the same thing. They are one phenomenon, differing only in frequency or, equivalently, wavelength. Arranged in order, they form the electromagnetic spectrum (वैद्युतचुंबकीय स्पेक्ट्रम).

Timeline

Filling in the spectrum

1800
Herschel: infrared
A thermometer placed beyond the red end of a prism's spectrum warmed up more than one in the visible colours.
1801
Ritter: ultraviolet
Silver chloride darkened fastest just beyond the violet end, where nothing could be seen.
1865
Maxwell's prediction
Electromagnetic waves should exist and travel at the speed of light; light itself is one of them.
1887
Hertz: radio waves
Waves from a spark gap were detected across a laboratory, and their speed was measured.
1895
Röntgen: X-rays
Rays from a discharge tube passed through card and flesh and showed the bones of a hand on a photographic plate.
1895
Bose: millimetre waves
At a public demonstration in Kolkata, his short-wavelength waves rang a bell and fired a pinch of gunpowder in another room.
1900
Villard: gamma rays
Radium gave out a very penetrating radiation that a magnet could not deflect.
1901
Marconi: across the Atlantic
A radio signal sent from Cornwall in England was received in Newfoundland, across the ocean.

Fig. 8.12 lays the spectrum out on a logarithmic scale, where each step is a factor of ten in wavelength; visible light is a sliver less than a factor of two wide. The names label regions, not sharp boundaries, and neighbouring regions overlap. A photon of wavelength $0.01 n m$ is called an X-ray if it came from an X-ray tube and a gamma ray if it came from a nucleus, yet the wave itself is the same. Below, we move through the regions from the longest wavelengths to the shortest. Fig. 8.13 lets you explore them for yourself.

2/4Frequency increases from left to right as wavelength decreases, since $ν = c / λ$ .

Fig. 8.12Each step along the scale is a factor of ten. The regions merge into one another; visible light is a very thin slice.NCERT Fig. 8.4 · p. 209

A long horizontal bar is divided into coloured regions, from radio waves on the left through microwaves, infrared, a thin yellow visible slice, ultraviolet, X-rays and gamma rays on the right; X-rays and gamma rays overlap. Below the bar a wavelength scale is marked 1 km, 1 m, 1 mm, 1 micrometre, 1 nm and 1 pm, each step a factor of ten. Step 1: the bar with the region names. Step 2: a frequency scale appears above the bar, from 10 to the 6 hertz on the left to 10 to the 21 hertz on the right. Step 3: a row of objects below the scale shows sizes comparable to the wavelengths above them: a house, a coin, a living cell, an atom and an atomic nucleus. Step 4: the objects are replaced by an enlarged strip of the visible slice, running from red at 700 nm on the left through orange, yellow, green and blue to violet at 400 nm on the right.

InteractiveSpectrum explorer

An interactive spectrum bar running on a logarithmic scale from 10 kilometres on the left to 10 to the minus 14 metres on the right, with frequency marks above and wavelength marks below. Dragging a marker along the bar, moving a slider, tapping an example such as an FM station, a microwave oven, a TV remote, a germicidal lamp or cobalt-60, or typing a value in hertz, metres or electronvolts, shows the region's name, the wavelength, frequency and photon energy, an object of about the same size, and how such waves are made, detected and used. In the visible region a swatch shows the colour.

Fig. 8.13Drag along the bar, tap an example or type a frequency, wavelength or photon energy to see where it falls.

Table 8.4The electromagnetic spectrum at a glance (boundaries are approximate)

Region	Wavelength	Photon energy	An example source	An example use
Radio	more than $0.1 m$	below $1.2 \times 10^{- 5} e V$	charges oscillating in an antenna	broadcasting, mobile phones
Microwave	$1 m m$ to $0.1 m$	$1.2 \times 10^{- 5}$ to $1.2 \times 10^{- 3} e V$	a magnetron	radar, ovens, Wi-Fi
Infrared	$700 n m$ to $1 m m$	$1.2 \times 10^{- 3}$ to $1.8 e V$	a warm body, an infrared LED	remote controls, thermal cameras
Visible	$400 n m$ to $700 n m$	$1.8$ to $3.1 e V$	the Sun, a flame, an LED	seeing, photography, lighting
Ultraviolet	about $1 n m$ to $400 n m$	$3.1 e V$ to about $1 k e V$	the Sun, a welding arc, a mercury lamp	disinfecting water, LASIK
X-rays	about $10^{- 4} n m$ to $10 n m$	hundreds of eV to MeV	fast electrons hitting a metal target	medical imaging, airport scanners
Gamma rays	below about $0.1 n m$	above about $10 k e V$	radioactive nuclei such as cobalt-60	cancer therapy, sterilisation

8.4.1 Radio waves

Radio waves have wavelengths from about $10 c m$ to many kilometres, that is, frequencies below about $3 G H z$ . They are made by charges accelerated back and forth in antennas and circuits, exactly as in Section 8.3.1. So many services share the radio region that frequencies are allotted by international agreement; Table 8.5 shows some of them.

Table 8.5Who uses which radio frequencies (approximate)

Service	Frequencies	Wavelengths
Medium-wave AM (including many All India Radio stations)	$0.53$ to $1.6 M H z$	about $570 m$ to $190 m$
Short-wave broadcasting, heard across continents	$3$ to $30 M H z$	$100 m$ to $10 m$
FM radio	$88$ to $108 M H z$	$3.4 m$ to $2.8 m$
Television (VHF and UHF channels)	about $50$ to $900 M H z$	$6 m$ to $0.33 m$
Satellite navigation (GPS; India's NavIC)	about $1.2$ to $2.5 G H z$	$25 c m$ to $12 c m$
Mobile phones	about $0.7$ to $3.6 G H z$ , running into microwaves	$43 c m$ to $8 c m$

An antenna radiates best when its length is comparable to the wavelength, typically a half or a quarter of it. That is why an FM antenna on a car is under a metre long, while a medium-wave transmitter needs a mast tens of metres high. The same rule of thumb runs through the whole spectrum. The wavelength emitted tends to be comparable to the size of whatever emits it, from antennas down to atomic nuclei, as the object sizes in Fig. 8.12 suggest. Atoms are the notable exception: about $0.1 n m$ across, they emit visible light thousands of times longer.

Example

Tuning in

(a) Find the wavelength of an FM station broadcasting at $98.3 M H z$ , and the length of a quarter-wave antenna for it. (b) Repeat for a medium-wave station at $819 k H z$ , and comment on the result.

Try it yourself first. Then reveal the solution one step at a time.

8.4.2 Microwaves

Your kitchen and your pocket both use microwaves. Microwaves (सूक्ष्म तरंगें) sit at the short-wavelength end of the radio region: roughly $1 m m$ to $10 c m$ , or $3$ to $300 G H z$ . The boundary is loose, and the name is often stretched to waves a few tens of centimetres long, such as those in a microwave oven. Wi-Fi routers work at $2.4$ and $5 G H z$ , satellite television dishes receive signals near $11$ – $12 G H z$ , and newer mobile networks use $3.3$ – $3.6 G H z$ . Making them needs special sources: vacuum tubes called magnetrons, found in every microwave oven, and klystrons, used in large radars and satellite transmitters, as well as solid-state devices such as Gunn diodes.

A short wavelength brings a big advantage: a dish only a metre or two across can focus microwaves into a narrow beam. That makes them ideal for radar, which sends out a short pulse and times its echo. Air-traffic controllers track aircraft this way, and the Doppler weather radars of the India Meteorological Department follow cyclones as they approach the coast.

Example

Timing an echo

A weather radar operating at $5.6 G H z$ sends out a pulse and receives the echo from a storm cloud $0.80 m s$ later. (a) How far away is the cloud? (b) What is the wavelength of the radar waves?

Try it yourself first. Then reveal the solution one step at a time.

A microwave oven uses waves of $2.45 G H z$ , with a wavelength of about $12 c m$ . A water molecule is an electric dipole: slightly negative at its oxygen end and slightly positive at its hydrogen end. The oscillating electric field of the wave twists these dipoles back and forth billions of times a second (Fig. 8.14). As they turn, the molecules jostle their neighbours, and the energy taken from the wave ends up as random thermal motion: the food heats up. You may read that the oven's frequency is chosen to match a natural frequency of water molecules. The fuller picture is that water absorbs strongly over a broad range of microwave frequencies; $2.45 G H z$ is a band set aside for such use, and it penetrates a few centimetres into food. Dry glass and ceramic dishes absorb far less and stay cooler, while metals reflect microwaves.

1/3Water molecules are electric dipoles: the field turns each one so that its positive (hydrogen) end points along $𝐄$ .

Fig. 8.14The oscillating electric field keeps twisting the polar water molecules; their jostling turns the wave energy into heat.NCERT p. 209

Seven water molecules, each a large red oxygen atom with two small green hydrogen atoms, float in a panel; a key at the bottom shows the oxygen end is negative and the hydrogen end positive. An orange arrow on the left shows the electric field of the microwaves. Step 1: the field points up and every molecule has turned so that its hydrogen end points up. Step 2: the field has reversed to point down and every molecule has turned upside down to follow it. Step 3: the field arrow becomes double-headed, flipping billions of times a second, the molecules point in all directions with small motion marks beside them, and a warm glow fills the panel.

Quick check

1 of 3

An FM station broadcasts at

93.5 M H z

. What is the wavelength of its waves, in metres?

Show answer

Answer 3.21 m

λ = c / ν = (3.0 \times 10^{8}) / (9.35 \times 10^{7}) = 3.21 m

8.4.3 Infrared waves

Stand near a hot tawa or an iron without touching it, and your face feels warm. What reaches you is infrared (अवरक्त) radiation: waves just beyond the red end of the visible spectrum, from about $700 n m$ to $1 m m$ . Every warm object gives it out, and the hotter the object, the more it emits and the shorter its wavelengths. Your own body glows most strongly near $10 μ m$ , which is why a thermal camera can find people in the dark.

Infrared is everywhere in technology. Remote controls use infrared light-emitting diodes near $940 n m$ . Optical fibres carry internet traffic at $1.3$ and $1.55 μ m$ , where glass is most transparent. Night-vision and thermal cameras detect it, physiotherapists warm stiff muscles with infrared lamps, and earth-observation satellites use it to map crops, forests and water.

Why does infrared, rather than light, feel warm? Its frequencies are close to the natural frequencies at which molecules vibrate and rotate. Water, carbon dioxide and many other molecules absorb it strongly, and the energy goes into shaking whole molecules rather than only their electrons; that is, it becomes heat. Since most materials contain some water, most materials warm up under infrared, which is why it is often called heat radiation.

The same absorption keeps our planet habitable. Without its greenhouse gases, the Earth's surface would average about −18 °C; in fact it averages about +15 °C. The difference is the greenhouse effect (Fig. 8.15). Air lets most of the incoming sunlight through to warm the ground, but the warm ground radiates at infrared wavelengths, which water vapour and carbon dioxide absorb and partly send back down.Adding carbon dioxide to the air strengthens the greenhouse effect, which is the main cause of present-day global warming.

3/3 $C O_{2}$ and $H_{2} O$ molecules absorb some of that infrared and re-emit it in all directions, some back down: the greenhouse effect.

Fig. 8.15The atmosphere lets most sunlight in but absorbs part of the infrared the warm ground sends back out.NCERT p. 210

The Sun is at the top left, a pale blue band of atmosphere crosses the middle and the green ground runs along the bottom. Step 1: three orange wavy arrows with short wavelengths, labelled visible, pass from the Sun straight through the atmosphere to the ground. Step 2: the ground takes on a warm tint and three red wavy arrows with much longer wavelengths, labelled infrared, rise from it through the atmosphere. Step 3: one red arrow still escapes to space, but the other two end at a carbon dioxide molecule and a water molecule in the atmosphere, which send new red arrows back down to the ground and one upward.

Try at home

See your remote control's light

You’ll need

a TV or air-conditioner remote control
a phone with a camera

Predict first

You look straight at the remote's bulb and press a button. What do you see?

Nothing at allA flashing red lightA steady white glow

Point a TV or air-conditioner remote at your eye (from a little distance) and press a button. Watch the small bulb at its front.
Now look at the same bulb through your phone's camera screen while you press the button.
Try the front (selfie) camera and the rear camera. Does one show the flashes more brightly?

8.4.4 Visible rays

The only part of the spectrum we sense directly is a narrow band: red light near $700 n m$ ( $4.3 \times 10^{14} H z$ ) at one end, violet near $400 n m$ ( $7.5 \times 10^{14} H z$ ) at the other. Sunlight, flames, hot filaments and LEDs all produce it as electrons in atoms, molecules and solids give up energy, dropping to lower energy levels.

Why do our eyes use this band and not another? Sunlight that reaches the ground through the atmosphere is richest in visible wavelengths, and human eyes are most sensitive near $555 n m$ , in the yellow-green, close to the middle of that output. Eyes evolved to make the most of the light that is most plentiful. Other animals see a different slice. Reindeer see ultraviolet, which makes lichen and wolves stand out against snow, and vampire bats sense the infrared warmth of blood vessels close to the skin of their prey.

8.4.5 Ultraviolet rays

Beyond the violet end lies ultraviolet (पराबैंगनी) radiation, which shades into X-rays at wavelengths of about $1 n m$ . By far the largest source we meet is the Sun, but welding arcs, mercury-vapour lamps and other very hot sources also emit it. Its photons carry enough energy to break chemical bonds (see photon energies), so its effects depend strongly on wavelength. For living things it is usually split into three bands, compared in Table 8.6.

Table 8.6Three bands of ultraviolet

Band	Wavelength	Photon energy	Sunlight's share of it	Effects and uses
UV-A	$315$ – $400 n m$	$3.1$ – $3.9 e V$	mostly reaches the ground; partly passes window glass	tanning, ageing of the skin; makes banknote security marks glow
UV-B	$280$ – $315 n m$	$3.9$ – $4.4 e V$	mostly absorbed by ozone; stopped by window glass	sunburn and skin-cancer risk; helps the skin make vitamin D
UV-C	$100$ – $280 n m$	$4.4$ – $12 e V$	completely absorbed by ozone and air	germicidal lamps near $254 n m$ disinfect water and air

Ultraviolet is useful as well as harmful. Pulses from an ultraviolet laser reshape the cornea in LASIK eye surgery, removing tissue a fraction of a micrometre at a time. The short wavelength allows a very fine focus, and the energetic photons break bonds without heating the tissue around them. Against the harm there are several shields. Skin exposed to ultraviolet makes extra melanin, a dark pigment that absorbs the radiation before it reaches deeper cells; a tan is the visible sign. Window glass absorbs UV-B, so a closed window prevents sunburn, and a welder facing an arc that pours out ultraviolet looks through a visor of special dark glass.

The biggest shield of all is the ozone layer in the stratosphere, a few tens of kilometres up, which keeps UV-C and most UV-B away from us. Chlorofluorocarbons (CFCs) from old refrigerators and aerosol sprays were found to be destroying it. The world agreed to phase them out in the Montreal Protocol of 1987, which India joined in 1992. The layer is now slowly recovering.

Quick check

1 of 3

What is the main reason infrared is called heat radiation?

It is given out only by objects hotter than 100 °CIt travels more slowly than light and so delivers more energyIt sets whole molecules vibrating, raising their internal energyIt is absorbed only by metals

Show answer

Answer (C) It sets whole molecules vibrating, raising their internal energy

Infrared frequencies match molecular vibrations, so absorbed infrared becomes thermal energy. Every warm object, including you at about 37 °C, gives out infrared, and it travels at the same speed as light.

8.4.6 X-rays

Wilhelm Röntgen discovered X-rays in 1895 while passing electric discharges through evacuated glass tubes. On today's spectrum they occupy the range between roughly $10 n m$ and $10^{- 13} m$ ( $10^{- 4} n m$ ), overlapping ultraviolet on one side and gamma rays on the other. A modern X-ray tube (Fig. 8.16) boils electrons off a hot filament, accelerates them through a potential difference of tens of kilovolts, and stops them suddenly in a metal target such as tungsten. The sharply decelerated electrons radiate X-rays, and electrons knocked out of the inner shells of the target atoms add X-rays of particular wavelengths.

4/4The electrons are stopped in the target. The sudden deceleration, and electron jumps into inner shells of the target atoms, produce X-rays.

Fig. 8.16In an X-ray tube, fast electrons are stopped suddenly in a metal target, and part of their energy comes out as X-rays.NCERT p. 211

A sealed glass tube lies across the figure. On the left is a coiled filament, the negative electrode; on the right a metal block with a slanted target face, the positive electrode. Wires from both run up to a high-voltage supply drawn as a battery symbol at the top. Step 1: the filament is cold and dark and nothing happens. Step 2: the filament glows red and a small cloud of blue electrons gathers beside it. Step 3: the electrons stream across the tube towards the target, shown by a row of dots and an arrow. Step 4: purple wavy arrows with very short wavelengths leave the target face and pass out through the bottom of the tube, labelled X-rays.

X-rays pass easily through soft tissue but are absorbed strongly by bone, whose calcium atoms are heavier than the hydrogen, carbon and oxygen atoms of flesh. That contrast gives the familiar shadow picture of a skeleton, and CT scanners combine many such pictures into three-dimensional images. X-rays also screen luggage at airports, reveal how atoms are arranged in crystals, and in larger doses destroy cancer cells. The same energy damages healthy cells, so exposure is kept as low as possible: technicians stand behind shielding, and patients wear lead aprons over the parts not being examined.

8.4.7 Gamma rays

Atomic nuclei, like atoms, have energy levels, but they are roughly a million times further apart. When a nucleus drops to a lower level, during radioactive decay or a nuclear reaction, it gives out a photon of gamma rays, the most energetic region of Fig. 8.12. Their wavelengths run from about $10^{- 10} m$ to below $10^{- 14} m$ , the size of a nucleus itself. Gamma rays also reach us from violent events in space, such as exploding stars. Their penetrating power makes them useful. In hospitals, beams from cobalt-60 sources are aimed at tumours to destroy cancer cells, and gamma radiation sterilises syringes, gloves and other supplies inside their sealed packets.

Compare

X-rays and gamma rays

X-rays

Come from electrons: rapid deceleration, or jumps between inner atomic shells
Photon energies typically about $0.1$ to a few hundred keV
Made to order in X-ray tubes

Gamma rays

Come from atomic nuclei: radioactive decay and nuclear reactions
Photon energies typically tens of keV to many MeV
Given out by radioactive sources such as cobalt-60

Where the two ranges overlap, an X-ray and a gamma ray of the same wavelength are exactly the same kind of wave; the different names describe only where they came from.

Why do waves that differ only in frequency behave so differently? Chapter 11 will show that radiation is emitted and absorbed in packets called photons, each carrying an energy $E = h ν$ , where $h = 6.63 \times 10^{- 34} J s$ is Planck's constant. Measured in electronvolts ( $1 e V = 1.6 \times 10^{- 19} J$ ), a handy form is

Photon energy

E = h ν = \frac{h c}{λ} \approx \frac{1240 e V n m}{λ}

An FM radio photon carries less than a millionth of an electronvolt, far too little to affect any single molecule. Radio waves and microwaves act on matter through the combined push of their fields on many charges. Photons of visible light, about $2$ to $3 e V$ , match the energies of the outer electrons in atoms and molecules, which is why light can drive vision and photosynthesis. Ultraviolet photons have enough energy to break chemical bonds. X-ray photons, thousands of electronvolts, come from the inner electrons of heavy atoms, and gamma-ray photons, millions of electronvolts, come from rearrangements inside nuclei. The energy of the photons reflects the energy scale of the system that emits them.

Example

Photon energies across the spectrum

Find the energy of one photon, in electronvolts, for (a) FM radio at $100 M H z$ , (b) a microwave oven at $2.45 G H z$ , (c) green light of wavelength $530 n m$ , (d) a germicidal lamp at $254 n m$ , (e) X-rays of wavelength $0.10 n m$ and (f) gamma rays of wavelength $1.0 \times 10^{- 12} m$ . Take $h = 6.63 \times 10^{- 34} J s$ , $1 e V = 1.6 \times 10^{- 19} J$ and $h c \approx 1240 e V n m$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Put these in order of increasing photon energy.

Visible light
Ultraviolet
X-rays
Microwaves
Gamma rays
Infrared
Radio waves

Show answer

Answer

Radio waves
Microwaves
Infrared
Visible light
Ultraviolet
X-rays
Gamma rays

Photon energy

E = h ν

rises with frequency, so the order is the order of increasing frequency, which is the order of decreasing wavelength.

Flash cards · 9

Key relations

1 / 9

Recap

Electromagnetic Waves: key points and summary

Ampère's law gives contradictory answers for a charging capacitor. Maxwell removed the contradiction by adding the displacement current $i_{d} = ε_{0} d Φ_{E} / d t$ (Section 8.2).
Ampère–Maxwell law: $\oint 𝐁 \cdot d 𝐥 = μ_{0} i_{c} + μ_{0} ε_{0} d Φ_{E} / d t$ . Displacement current produces a magnetic field exactly as conduction current does, and exists wherever an electric field changes, even in vacuum (Law).
Between circular plates of radius $R$ carrying charging current $i$ , $B = μ_{0} i r / 2 π R^{2}$ inside the rim and $μ_{0} i / 2 π r$ outside it (Derivation).
Gauss's laws for $𝐄$ and $𝐁$ , Faraday's law and the Ampère–Maxwell law, with the Lorentz force, describe all of classical electromagnetism (Table 8.1).
Charges at rest or in uniform motion do not radiate; accelerated charges do. An oscillating charge sends out waves of its own frequency, and the energy comes from the source (Section 8.3.1).
Hertz produced and detected electromagnetic waves in 1887 and measured their speed; J. C. Bose made millimetre waves and Marconi sent signals across the Atlantic (Fig. 8.7).
Electromagnetic waves are transverse: $𝐄$ , $𝐁$ and the direction of travel are mutually perpendicular, with $𝐄 \times 𝐁$ along the travel. For a plane wave, $E_{y} = E_{0} \sin (k x - ω t)$ and $B_{z} = B_{0} \sin (k x - ω t)$ (Section 8.3.2).
$k = 2 π / λ$ , $ω = 2 π ν$ and $c = ω / k = ν λ = 1 / \sqrt{μ_{0} ε_{0}} \approx 3.0 \times 10^{8} m / s$ ; the fields are in phase with $B_{0} = E_{0} / c$ (Derivation).
In a material the speed is $v = 1 / \sqrt{μ ε}$ ; the frequency is unchanged and the wavelength shrinks in proportion ((8.11)).
The electric and magnetic fields carry equal shares of the energy; the average energy density is $\frac{1}{2} ε_{0} E_{0}^{2}$ (Derivation).
In order of decreasing wavelength the spectrum runs through radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays, with no sharp boundaries (Section 8.4).
Photon energy $E = h ν$ rises from around a millionth of an electronvolt for radio waves to millions of electronvolts for gamma rays, and it decides how each region interacts with matter (photon energies).
Uses and hazards differ across the spectrum: ozone absorbs most solar ultraviolet, infrared drives the greenhouse effect, and X-rays and gamma rays image and treat the body but must be used sparingly (Section 8.4.5).

Practice · 17 questions

Practice questions with answers (NCERT “Exercises”)

Questions 1–10 practise the same skills as the NCERT exercises with new situations and numbers; questions 11–17 are extra practice. Take $c = 3.0 \times 10^{8} m / s$ , $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ , $μ_{0} = 4 π \times 10^{- 7} T m / A$ and $h c \approx 1240 e V n m$ .

1
NumericalMediumNCERT Exercises 8.1
In a school laboratory, two square aluminium sheets, each $20 c m \times 20 c m$ , are held $4.0 m m$ apart by thin plastic spacers to make an air capacitor. A constant-current supply charges it at $60 m A$ .
1. What is its capacitance, in picofarads?
  pF
  Show answer
  Answer 88.5 pF
  $C = ε_{0} A / d = (8.85 \times 10^{- 12}) (0.040) / (4.0 \times 10^{- 3}) = 8.85 \times 10^{- 11} F \approx 89 p F$ .
2. At what rate does the potential difference between the plates rise, in volts per second?
  V/s
  Show answer
  Answer 678000000 V/s
  $d V / d t = i / C = 0.060 / (8.85 \times 10^{- 11}) = 6.8 \times 10^{8} V / s$ .
3. What is the displacement current between the plates?
  Show solution
  $Φ_{E} = E A = (V / d) A$ , so $i_{d} = ε_{0} d Φ_{E} / d t = (ε_{0} A / d) d V / d t = C d V / d t = 60 m A$ , the same as the charging current.
4. Charge flows into one plate, but none crosses the gap. Does Kirchhoff's junction rule hold at a plate?
  Show solution
  For conduction current alone, no: $60 m A$ flows into the plate and none flows out, so charge piles up there. If the displacement current leaving the plate across the gap is counted as well, the total current in equals the total current out at every instant, and the junction rule holds.
2
NumericalHardNCERT Exercises 8.2
A school physics club wants to detect the magnetic field of a displacement current. They make a capacitor of $150 p F$ from two circular brass discs of radius $7.5 c m$ . An audio signal generator drives it at $50 V$ (rms) and an angular frequency of $1.0 \times 10^{4} r a d / s$ (about $1.6 k H z$ ). A tiny search coil will sit between the plates, $5.0 c m$ from their common axis.
1. What rms current will flow in the leads, in microamperes?
  µA
  Show answer
  Answer 75 µA
  $I_{rms} = ω C V_{rms} = (1.0 \times 10^{4}) (1.5 \times 10^{- 10}) (50) = 7.5 \times 10^{- 5} A = 75 μ A$ .
2. How will this current compare with the displacement current between the plates?
  They are equal at every instantThey are equal only when the voltage is zeroThe displacement current is always smaller
  Show answer
  Answer (A) They are equal at every instant
  $i_{d} = ε_{0} d Φ_{E} / d t = C d V / d t$ , which is exactly the charging current at every instant; both are sinusoidal with the same amplitude and phase.
3. What amplitude of magnetic field must the search coil be able to detect, in tesla?
  T
  Show answer
  Answer 1.89e-10 T
  The peak current is $I_{0} = \sqrt{2} \times 75 μ A = 106 μ A$ . Inside the rim, $B_{0} = \frac{μ_{0}}{2 π} \frac{I_{0} r}{R^{2}} = (2.0 \times 10^{- 7}) \frac{(1.06 \times 10^{- 4}) (0.050)}{(0.075)^{2}} = 1.9 \times 10^{- 10} T$ . That is about 250 000 times weaker than the Earth's field, which is why such measurements are difficult.
Quick answerEasyNCERT Exercises 8.3
A lightning stroke a few kilometres away gives out a crackle of radio waves (about $300 m$ long), a flash of visible light (about $500 n m$ ) and, as instruments carried on aircraft have shown, brief bursts of gamma rays (about $1 p m$ ). What do the crackle, the flash and the gamma burst have in common as they cross the air towards you?
Their frequencyTheir speedTheir energy per photonThe size of antenna needed to receive them
Show answer
Answer (B) Their speed
All electromagnetic waves travel at $c = 3.0 \times 10^{8} m / s$ in vacuum, and very nearly that in air, which is why an AM radio crackles at the very instant you see the flash. Their frequencies, photon energies and the detectors that suit them differ enormously.
4
Short answerEasyNCERT Exercises 8.4
A $75 M H z$ radio signal from a transmitter due east of you travels west towards you through air. Take $x$ towards the east, $y$ towards the north and $z$ vertically up.
1. Which directions can its electric and magnetic fields take? If the transmitting antenna is vertical, so that $𝐄$ is vertical, which way does $𝐁$ point at an instant when $𝐄$ points up?
  Show solution
  Both fields are perpendicular to the east–west line of travel, so they lie in the vertical north–south plane, at right angles to each other and in phase. The wave travels along $- x$ , so $𝐄 \times 𝐁$ must point along $- x$ . With $𝐄$ along $+ z$ , this needs $𝐁$ along $+ y$ , since $\hat{𝐤} \times \hat{𝐣} = - \hat{𝐢}$ : $𝐁$ points north when $𝐄$ points up, and south when $𝐄$ points down.
2. What is its wavelength, in metres?
  m
  Show answer
  Answer 4 m
  $λ = c / ν = (3.0 \times 10^{8}) / (7.5 \times 10^{7}) = 4.0 m$ .
5
NumericalEasyNCERT Exercises 8.5
Air-traffic controllers talk to pilots on VHF radio channels between $118 M H z$ and $137 M H z$ . What is the corresponding band of wavelengths?
1. Longest wavelength, in metres:
  m
  Show answer
  Answer 2.54 m
  $λ = c / ν = (3.0 \times 10^{8}) / (1.18 \times 10^{8}) = 2.54 m$ .
2. Shortest wavelength, in metres:
  m
  Show answer
  Answer 2.19 m
  $λ = (3.0 \times 10^{8}) / (1.37 \times 10^{8}) = 2.19 m$ . The band runs from about $2.19 m$ to $2.54 m$ , just shorter than the FM band.
6
Short answerEasyNCERT Exercises 8.6
Charges in the antenna of a Wi-Fi router oscillate $5.0 \times 10^{9}$ times each second.
1. At what frequency does the router radiate?
  Show solution
  The same as that of the oscillating charges: $5.0 \times 10^{9} H z = 5.0 G H z$ .
2. What is their wavelength, in centimetres?
  cm
  Show answer
  Answer 6 cm
  $λ = c / ν = (3.0 \times 10^{8}) / (5.0 \times 10^{9}) = 0.060 m = 6.0 c m$ , in the microwave region.
NumericalEasyNCERT Exercises 8.7
A small search-coil probe near an FM transmitter finds that the magnetic field of the wave oscillates with an amplitude of $42 n T$ . What amplitude would an electric-field probe at the same spot record, in volts per metre?
V/m
Show answer
Answer 12.6 V/m
$E_{0} = c B_{0} = (3.0 \times 10^{8}) (4.2 \times 10^{- 8}) = 12.6 V / m$ .
8
NumericalMediumNCERT Exercises 8.8
Some distance from a short-wave transmitter, its $20 M H z$ signal can be treated as a plane wave travelling through air along $+ y$ . Its electric field points along $z$ and has an amplitude of $75 V / m$ . Find:
1. $λ$ , in metres:
  m
  Show answer
  Answer 15 m
  $λ = c / ν = (3.0 \times 10^{8}) / (2.0 \times 10^{7}) = 15 m$ .
2. $k$ , in radians per metre:
  rad/m
  Show answer
  Answer 0.419 rad/m
  $k = 2 π / λ = 2 π / 15 = 0.419 r a d / m$ (check: $ω / c$ gives the same).
3. $ω$ , in radians per second:
  rad/s
  Show answer
  Answer 125700000 rad/s
  $ω = 2 π ν = 2 π (2.0 \times 10^{7}) = 1.26 \times 10^{8} r a d / s$ .
4. $B_{0}$ , in tesla:
  T
  Show answer
  Answer 2.5e-7 T
  $B_{0} = 75 / (3.0 \times 10^{8}) = 2.5 \times 10^{- 7} T$ .
5. Write expressions for $𝐄$ and $𝐁$ .
  Show solution
  $E_{z} = (75 V / m) \sin [(0.419 r a d / m) y - (1.26 \times 10^{8} r a d / s) t]$ .
  $𝐁$ is perpendicular to $z$ and to $y$ , so it lies along $x$ , in phase with $𝐄$ . Since $\hat{𝐤} \times \hat{𝐢} = \hat{𝐣}$ (the direction of travel), a positive $E_{z}$ goes with a positive $B_{x}$ : $B_{x} = (2.5 \times 10^{- 7} T) \sin [(0.419 r a d / m) y - (1.26 \times 10^{8} r a d / s) t]$ .

Long answerMediumNCERT Exercises 8.9

Using $E = h ν = h c / λ$ , find the photon energy in electronvolts for: radio waves of wavelength $30 m$ , microwaves of $1.0 c m$ , infrared of $10 μ m$ , red light of $650 n m$ , X-rays of $0.050 n m$ and gamma rays of $2.0 p m$ . How do these energy scales relate to the sources of each kind of radiation?

Show solution

Table 8.7

Radiation	Wavelength (nm)	Photon energy	Typical source
Radio	$3.0 \times 10^{10}$	$4.1 \times 10^{- 8} e V$	currents in circuits and antennas
Microwave	$1.0 \times 10^{7}$	$1.2 \times 10^{- 4} e V$	rotating molecules, electronic oscillators
Infrared	$1.0 \times 10^{4}$	$0.12 e V$	vibrating molecules in warm bodies
Red light	$650$	$1.9 e V$	outer electrons of atoms changing levels
X-rays	$0.050$	$2.5 \times 10^{4} e V = 25 k e V$	inner electrons of heavy atoms, fast electrons stopping
Gamma rays	$2.0 \times 10^{- 3}$	$6.2 \times 10^{5} e V = 0.62 M e V$	atomic nuclei

Each energy matches the spacing of energy levels in the system that emits it. Radio waves come from the collective motion of vast numbers of electrons in a circuit, where each photon's energy is negligible. Microwave and infrared photons match the rotational and vibrational energies of molecules, and visible photons (a few eV) the levels of outer electrons. X-ray photons (keV) match the tightly bound inner electrons of heavy atoms, and gamma-ray photons (MeV) the energy levels inside nuclei.

10
NumericalMediumNCERT Exercises 8.10
Close to the horn antenna of a $12 G H z$ microwave link between two telephone towers, the electric field of the beam has an amplitude of $36 V / m$ . Treat the beam as a plane wave travelling through air at $c$ .
1. How long are the waves, in centimetres?
  cm
  Show answer
  Answer 2.5 cm
  $λ = c / ν = (3.0 \times 10^{8}) / (1.2 \times 10^{10}) = 0.025 m = 2.5 c m$ .
2. How strong is the magnetic field at its peak, in tesla?
  T
  Show answer
  Answer 1.2e-7 T
  $B_{0} = E_{0} / c = 36 / (3.0 \times 10^{8}) = 1.2 \times 10^{- 7} T$ .
3. Compare the energy stored per cubic metre in the electric field, averaged over a cycle, with that stored in the magnetic field.
  Show solution
  ${\overset{ˉ}{u}}_{E} = \frac{1}{2} ε_{0} \overline{E^{2}} = \frac{1}{4} ε_{0} E_{0}^{2} = \frac{1}{4} (8.85 \times 10^{- 12}) (36)^{2} = 2.9 \times 10^{- 9} J / m^{3}$ .
  ${\overset{ˉ}{u}}_{B} = \overline{B^{2}} / 2 μ_{0} = B_{0}^{2} / 4 μ_{0} = (1.2 \times 10^{- 7})^{2} / (4 \times 4 π \times 10^{- 7}) = 2.9 \times 10^{- 9} J / m^{3}$ .
  In general, $B_{0}^{2} / 4 μ_{0} = E_{0}^{2} / (4 μ_{0} c^{2}) = \frac{1}{4} ε_{0} E_{0}^{2}$ because $c^{2} = 1 / μ_{0} ε_{0}$ , so the two are always equal.
Quick answerEasy
A capacitor is connected to an ac supply. Between its plates there is
no current of any kinda conduction current carried by electrons jumping the gapa displacement current only at the moment the supply is switched ona displacement current that equals the current in the leads at every instant
Show answer
Answer (D) a displacement current that equals the current in the leads at every instant
The electric field between the plates changes all the time, so $ε_{0} d Φ_{E} / d t$ is never zero for long and equals the current in the leads at every instant. No electrons cross the gap.
12
Quick answerEasy
Match each region of the spectrum with a typical source.
Tap an item on the left, then its match on the right.
Show answer
Answer
Radio waves : charges oscillating in an antenna
Microwaves : a magnetron
Infrared : a warm body
Visible light : electrons changing levels in atoms
X-rays : fast electrons stopped in a metal target
Gamma rays : radioactive nuclei
Moving down the list, the sources get smaller and their energy levels further apart, from circuits to molecules, atoms and nuclei.
Quick answerMedium
At a certain radio frequency a non-magnetic liquid has $ε_{r} = 2.56$ . How fast do radio waves of that frequency travel through it, in metres per second?
m/s
Show answer
Answer 187500000 m/s
$v = c / \sqrt{ε_{r}} = (3.0 \times 10^{8}) / 1.6 = 1.9 \times 10^{8} m / s$ .
Quick answerEasy
An electron moving round a circle at constant speed does not radiate, because its speed never changes.
TrueFalse
Show answer
Answer False
Its velocity keeps changing direction, so it is accelerating (towards the centre) and does radiate. This radiation is a nuisance in circular particle accelerators and a useful source of X-rays in synchrotrons.
Quick answerMedium
At a certain point, the electric field of a wave in vacuum points along $+ y$ while its magnetic field points along $- x$ . In which direction is the wave travelling?
$+ z$ $- z$ $+ x$ $- y$
Show answer
Answer (A) $+ z$
The wave travels along $𝐄 \times 𝐁$ : $\hat{𝐣} \times (- \hat{𝐢}) = - (\hat{𝐣} \times \hat{𝐢}) = - (- \hat{𝐤}) = + \hat{𝐤}$ . It cannot travel along $x$ or $y$ , since the fields are perpendicular to the direction of travel.
Quick answerEasy
An airport radar receives the echo of its pulse from an aircraft $0.24 m s$ after sending it. How far away is the aircraft, in kilometres?
km
Show answer
Answer 36 km
The pulse goes out and back, so $d = c t / 2 = (3.0 \times 10^{8}) (2.4 \times 10^{- 4}) / 2 = 3.6 \times 10^{4} m = 36 k m$ .
Long answerHard
A classmate says that the displacement current is only a mathematical trick, because no charge actually moves across the gap of a capacitor. Give two pieces of evidence that it has real physical effects.
Show solution
First, it produces a real magnetic field. A magnetic field circles the gap of a charging capacitor, and its value at a given distance beyond the rim of the plates equals the field at that distance from the wire, exactly as the Ampère–Maxwell law requires. Second, without it Maxwell's equations would not predict electromagnetic waves. With it, they predict waves travelling at $1 / \sqrt{μ_{0} ε_{0}} = 3.0 \times 10^{8} m / s$ ; Hertz produced and detected such waves, measured their speed, and found the speed of light. Every radio, phone and microwave oven relies on the displacement current in empty space.

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercises 8.1	Practice Q1
NCERT Exercises 8.2	Practice Q2
NCERT Exercises 8.3	Practice Q3
NCERT Exercises 8.4	Practice Q4
NCERT Exercises 8.5	Practice Q5
NCERT Exercises 8.6	Practice Q6
NCERT Exercises 8.7	Practice Q7
NCERT Exercises 8.8	Practice Q8
NCERT Exercises 8.9	Practice Q9
NCERT Exercises 8.10	Practice Q10

Quick check: MCQs on Electromagnetic Waves

Warm-up2 questions
8.2 Displacement current4 questions
Maxwell's equations1 question
8.3.1 Sources of electromagnetic waves3 questions
8.3.2 Nature of electromagnetic waves3 questions
8.3.2 Nature of electromagnetic waves4 questions
8.4.2 Microwaves3 questions
8.4.5 Ultraviolet rays3 questions
8.4.7 Gamma rays4 questions

Glossary · 15

Key terms (English–Hindi)

Ampère–Maxwell lawऐम्पियर-मैक्सवेल नियमempiyar-maxwell niyam: $\oint 𝐁 \cdot d 𝐥 = μ_{0} i_{c} + μ_{0} ε_{0} d Φ_{E} / d t$ : currents and changing electric flux produce magnetic fields.First seen in the chapter →
antennaऐन्टेनाantenna: A conductor in which oscillating charges send out, or are driven by, electromagnetic waves.First seen in the chapter →

conduction currentचालन धाराchalan dhara: Current carried by moving charges, such as electrons in a wire.First seen in the chapter →

displacement currentविस्थापन धाराvisthapan dhara: The term $i_{d} = ε_{0} d Φ_{E} / d t$ in the Ampère–Maxwell law: a changing electric flux that produces a magnetic field like a current, although no charge flows.First seen in the chapter →

electromagnetic spectrumवैद्युतचुंबकीय स्पेक्ट्रमvaidyut-chumbakiya spectrum: The whole range of electromagnetic waves arranged by frequency or wavelength, from radio waves to gamma rays.First seen in the chapter →
electromagnetic waveवैद्युतचुंबकीय तरंगvaidyut-chumbakiya tarang: A self-sustaining pattern of oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel, moving at $c$ in vacuum.First seen in the chapter →

gamma raysगामा किरणेंgama kiranen: The shortest electromagnetic waves, emitted by atomic nuclei.First seen in the chapter →

infraredअवरक्तavrakt: Waves from about $700 n m$ to $1 m m$ , given out by warm objects; they set molecules vibrating and so heat what absorbs them.First seen in the chapter →

Maxwell's equationsमैक्सवेल समीकरणmaxwell samikaran: The four laws (Gauss's laws for E and B, Faraday's law and the Ampère–Maxwell law) that describe how electric and magnetic fields arise.First seen in the chapter →
microwavesसूक्ष्म तरंगेंsukshma tarangen: Short radio waves, roughly $1 m m$ to $10 c m$ long (the boundary is loose), used in radar, ovens and Wi-Fi.First seen in the chapter →

photonफ़ोटॉनphoton: A packet of electromagnetic energy $h ν$ .First seen in the chapter →

radio wavesरेडियो तरंगेंradio tarangen: Electromagnetic waves longer than about $10 c m$ , made by charges oscillating in antennas and circuits.First seen in the chapter →

ultravioletपराबैंगनीparabainganee: Waves from about $400 n m$ down to $1 n m$ , energetic enough to break chemical bonds.First seen in the chapter →

wave numberतरंग संख्याtarang sankhya: $k = 2 π / λ$ , the change of phase per unit distance along a wave.First seen in the chapter →

X-raysएक्स-किरणेंeks-kiranen: Waves from about $10 n m$ down to $10^{- 13} m$ , produced when fast electrons are stopped in a metal target.First seen in the chapter →

8.1 Introduction

8.2 Displacement current

8.3 Electromagnetic waves

8.3.1 Sources of electromagnetic waves

8.3.2 Nature of electromagnetic waves

8.4 Electromagnetic spectrum

8.4.1 Radio waves

8.4.2 Microwaves

8.4.3 Infrared waves

8.4.4 Visible rays

8.4.5 Ultraviolet rays

8.4.6 X-rays

8.4.7 Gamma rays

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Quick check: MCQs on Electromagnetic Waves

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