Chapter 8 · Electromagnetic Waves

8.2 Electromagnetic Waves and the Electromagnetic Spectrum

Accelerated charges send out electric and magnetic fields that regenerate each other and travel through empty space at $c=1/\sqrt{\mu_0\varepsilon_0}$. The waves are transverse, with $E_0=cB_0$, and they carry energy and momentum. Sorted by wavelength they form the electromagnetic spectrum, from radio waves to gamma rays.

When you look at the Sun you see it as it was about eight minutes ago: light needs $\dfrac{1.5\times10^{11}\ \text{m}}{3.0\times10^{8}\ \text{m/s}}=500$ s to cross the gap. The radio commands sent to a spacecraft orbiting Mars take several minutes to arrive, so the engineers on Earth must plan every manoeuvre in advance. Sunlight, the radio link to the spacecraft, the infrared beam of a TV remote, the microwaves in an oven and the X-rays in a hospital scanner all travel at this same speed through empty space. They are all electromagnetic waves; what tells them apart is only their wavelength, which ranges from kilometres to less than the size of an atomic nucleus.

In Section 8.1 we found that Maxwell's equations allow a changing electric field to make a magnetic field, and a changing magnetic field to make an electric one. In this section we see what kind of charge motion launches such waves, describe the wave itself — its fields, its speed, the relation $E_0=cB_0$ — and work out the energy and momentum it carries. Finally we survey the whole electromagnetic spectrum: how each band is produced and detected, and what it is used for.

Sources of Electromagnetic Waves

A charge at rest has an electric field but no magnetic field. A charge moving with constant velocity — a steady current — has a magnetic field too, but its fields simply travel along with it; they do not break away. To launch a wave, the fields must keep changing, and that requires a charge whose velocity changes: an accelerated charge. Accelerated charges radiate electromagnetic waves.

The simplest source is a charge oscillating to and fro with frequency $\nu$. Its electric field changes with frequency $\nu$, which by the Ampère–Maxwell law makes a magnetic field changing at the same frequency, which by Faraday's law makes a changing electric field further out, and so on. The disturbance spreads outwards as an electromagnetic wave whose frequency equals the frequency of the oscillating charge. In a radio transmitter, an oscillating circuit drives charges up and down a metal rod, the antenna, millions of times a second; in a receiving antenna the arriving electric field pushes the electrons back and forth at the same frequency, giving a tiny alternating current that the receiver amplifies.

Visible light is radiated by charges oscillating far faster, about $10^{15}$ times a second — too fast for any circuit. Such light comes from electrons in atoms jumping between energy levels (Chapter 12) and from the jostling charges in hot bodies. X-rays come from fast electrons that are suddenly stopped in a metal target, and gamma rays from changes inside atomic nuclei.

The first experiments. In 1887 Heinrich Hertz drove sparks across a small gap between two metal spheres connected to an induction coil. Each spark set charges oscillating rapidly, and a separate ring of wire with its own tiny gap, several metres away, sparked in response: electromagnetic waves had crossed the room. Hertz showed that his waves could be reflected, refracted and made to interfere, just like light, and measured their speed. A few years later Jagadish Chandra Bose in Kolkata made waves between about $5$ mm and $25$ mm long, built receivers for them and studied their optics, and Guglielmo Marconi developed radio into a way of sending messages across oceans.

The Nature of Electromagnetic Waves

Far from its source a small piece of an electromagnetic wave looks like a plane wave: on every plane perpendicular to the direction of travel the fields are the same. For a sinusoidal wave travelling along the $+x$ direction, with $\mathbf E$ along $y$, Maxwell's equations allow only one kind of solution: $\mathbf B$ along $z$, oscillating in step with $\mathbf E$, with a fixed ratio of amplitudes and a fixed speed.

📘 Definition 8.2 — Electromagnetic wave
An electromagnetic wave is a pair of oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel, that sustain each other and travel through space carrying energy and momentum. It needs no medium: it travels through a vacuum. Its frequency $\nu$ is fixed by the source; its wavelength is $\lambda=v/\nu$, where $v$ is the speed in the medium ($c$ in vacuum).
📐 Result 8.5 — A plane electromagnetic wave in vacuum

A sinusoidal plane electromagnetic wave of angular frequency $\omega=2\pi\nu$ and wave number $k=2\pi/\lambda$, travelling in the $+x$ direction, has fields

$$E_y=E_0\sin(kx-\omega t),$$ $$B_z=B_0\sin(kx-\omega t).$$

(1) It is transverse: $\mathbf E$ and $\mathbf B$ are perpendicular to each other and to the direction of propagation, which is the direction of $\mathbf E\times\mathbf B$. (2) $\mathbf E$ and $\mathbf B$ oscillate in phase: they reach their maxima and zeros at the same places and times. (3) The speed is

$$c=\frac{\omega}{k}=\nu\lambda=\frac{1}{\sqrt{\mu_0\varepsilon_0}}\approx3.0\times10^8\ \text{m/s},$$

the same for every frequency. (4) At every point and instant the fields are in the ratio

$$\frac{E}{B}=\frac{E_0}{B_0}=c.$$

The ratio $E_0=cB_0$ means that the magnetic field of an electromagnetic wave is numerically tiny: a wave with an electric field of $300$ V/m — strong by everyday standards — has a magnetic field of only $1.0\ \mu$T. Yet, as we shall see, the two fields carry equal shares of the energy.

In a material. Inside a transparent medium of permittivity $\varepsilon=\varepsilon_r\varepsilon_0$ and permeability $\mu=\mu_r\mu_0$, the same reasoning gives a smaller speed.

📐 Result 8.6 — Speed of electromagnetic waves in a medium

In a linear medium of permittivity $\varepsilon$ and permeability $\mu$, $$v=\frac{1}{\sqrt{\mu\varepsilon}}=\frac{c}{\sqrt{\mu_r\varepsilon_r}},\qquad n=\frac{c}{v}=\sqrt{\mu_r\varepsilon_r},$$ where $n$ is the refractive index. For non-magnetic materials ($\mu_r\approx1$), $n\approx\sqrt{\varepsilon_r}$, with $\varepsilon_r$ measured at the frequency of the wave. The frequency does not change when a wave enters a medium; the wavelength becomes $\lambda/n$, and $E_0=vB_0$.

✏️ Example 1 — Reading a wave from its equation
The magnetic field of an electromagnetic wave in vacuum is $B_y=B_0\sin(kx+\omega t)$, with $B_0=2.0\times10^{-7}$ T, $k=5.0\times10^{2}\ \text{m}^{-1}$ and $\omega=1.5\times10^{11}\ \text{s}^{-1}$. (a) In which direction does the wave travel? (b) Find its wavelength and frequency, and check its speed. (c) Write an expression for its electric field. (d) To which part of the spectrum does it belong?
(a) Direction The phase is $kx+\omega t$, so a crest moves to smaller $x$ as $t$ increases: the wave travels in the $-x$ direction.
(b) λ and ν $k=5.0\times10^2\ \text{m}^{-1}$: $\lambda=\dfrac{2\pi}{k}\approx1.3\times10^{-2}$ m $=1.3$ cm. $\omega=1.5\times10^{11}\ \text{s}^{-1}$: $\nu=\dfrac{\omega}{2\pi}\approx2.4\times10^{10}$ Hz. Speed $\dfrac{\omega}{k}=\dfrac{1.5\times10^{11}}{5.0\times10^2}=3.0\times10^8$ m/s ✓.
(c) Electric field $E_0=cB_0=(3.0\times10^8)(2.0\times10^{-7})=60$ V/m. $\mathbf E$ is perpendicular to $\mathbf B$ ($y$) and to the direction of travel ($x$), so it lies along $\pm z$; we need $\mathbf E\times\mathbf B$ along $-x$. Since $\hat{\mathbf k}\times\hat{\mathbf j}=-\hat{\mathbf i}$, $\mathbf E$ is along $+z$ when $\mathbf B$ is along $+y$: $$E_z=E_0\sin(kx+\omega t),$$ with $E_0=60$ V/m and the same $k$ and $\omega$.
(d) Band $\lambda\approx1.3$ cm lies between $1$ mm and $10$ cm: microwaves (used for satellite links).
✏️ Example 2 — A microwave signal in plastic
At microwave frequencies polyethylene, a non-magnetic plastic, has $\varepsilon_r\approx2.25$. A $5.0$ GHz Wi-Fi signal passes from air (take it as vacuum) into a thick polyethylene sheet. Find (a) the speed of the wave in the plastic, (b) its wavelength in air and in the plastic, and (c) its frequency in the plastic.
(a) Speed $\mu_r\approx1$: $n=\sqrt{2.25}=1.5$ and $v=\dfrac{c}{n}=\dfrac{3.0\times10^8}{1.5}=2.0\times10^8$ m/s.
(b) Wavelengths In air $\lambda=\dfrac{c}{\nu}=\dfrac{3.0\times10^8}{5.0\times10^9}=0.060$ m $=6.0$ cm. In the plastic $\lambda'=\dfrac{v}{\nu}=4.0$ cm $=\lambda/n$.
(c) Frequency Unchanged, $5.0$ GHz: the charges in the plastic are driven at the frequency of the arriving wave and re-radiate at that frequency.
Interactive 1 — A Travelling Electromagnetic Wave

A plane microwave travelling along $+x$ in vacuum, drawn in 3-D: the red arrows are $\mathbf E$ (along $y$), the blue arrows are $\mathbf B$ (along $z$), sampled along the line of travel. Both rise and fall together, and $\mathbf E\times\mathbf B$ always points along $+x$. Change the frequency and watch the wavelength shrink as $\lambda=c/\nu$; change $E_0$ and read off $B_0=E_0/c$ and the intensity. The blue arrows are drawn $c$ times longer than to scale, so that they show up at all. The motion is slowed down.

2.560
wavelength $\lambda=c/\nu$—
period $T$—
$B_0=E_0/c$—
intensity $\tfrac12c\varepsilon_0E_0^2$—

Energy, Intensity and Momentum

An electromagnetic wave carries energy: sunlight warms the ground, and a microwave oven heats food. The energy is stored in the fields themselves. In Section 2.4 we found that an electric field has energy density $u_E=\tfrac12\varepsilon_0E^2$; a magnetic field has energy density $u_B=\dfrac{B^2}{2\mu_0}$ (you meet it with the energy of an inductor in Chapter 6). In a wave $B=E/c$ and $c^2=1/(\mu_0\varepsilon_0)$, so

$$u_B=\frac{E^2}{2\mu_0c^2}=\tfrac12\varepsilon_0E^2=u_E.$$

The electric and magnetic fields carry equal shares of the energy, although $B$ looks so small. The total energy density at a point is $u=\varepsilon_0E^2$, and it oscillates because $E$ does. Averaged over a cycle, $\sin^2$ has the value $\tfrac12$, so $\langle u\rangle=\tfrac12\varepsilon_0E_0^2$.

Now imagine a surface of area $A$ facing the wave. In a time $\Delta t$, all the energy in a box of length $c\,\Delta t$ in front of it flows through it: energy $\langle u\rangle\,(Ac\,\Delta t)$. The energy per unit area per unit time is $\langle u\rangle c$.

📘 Definition 8.3 — Intensity of an electromagnetic wave
The intensity $I$ of an electromagnetic wave is the average energy it carries per unit time through unit area held perpendicular to its direction of travel. Its unit is $\text{W/m}^2$.
📐 Result 8.7 — Energy and momentum of an electromagnetic wave

For a sinusoidal electromagnetic wave in vacuum with amplitudes $E_0$ and $B_0=E_0/c$, the average energy density is shared equally between the two fields,

$$\langle u\rangle=\tfrac12\varepsilon_0E_0^2=\frac{B_0^2}{2\mu_0},$$

and the intensity is

$$I=\langle u\rangle c=\tfrac12c\varepsilon_0E_0^2.$$

The wave also carries momentum: when it delivers energy $U$ to a surface that absorbs it completely, it delivers momentum of magnitude

$$p=\frac{U}{c}$$

in its direction of travel. For a point source radiating power $P$ equally in all directions, $I=\dfrac{P}{4\pi r^2}$ at distance $r$, so $E_0\propto1/r$.

Because $I\propto E_0^2$, doubling the field amplitude quadruples the intensity — the same rule as for every wave, where intensity goes as amplitude squared (Class 11 Physics, Chapter 14). Momentum delivery means that light pushes on whatever absorbs it. The push is minute — bright sunlight falling on your palm (about $0.01\ \text{m}^2$) pushes it with roughly $3\times10^{-8}$ N — but it is measurable, and in space, where nothing else acts, it can steer spacecraft and push the tails of comets away from the Sun.

✏️ Example 3 — The fields in sunlight
Above the Earth's atmosphere, sunlight has an intensity of about $1.4\ \text{kW/m}^2$. Treating it as a single sinusoidal wave, find (a) the amplitude of its electric field, (b) the amplitude of its magnetic field, and (c) the average energy density in the sunlight.
(a) E₀ $I=\tfrac12c\varepsilon_0E_0^2\Rightarrow E_0=\sqrt{\dfrac{2I}{c\varepsilon_0}}$ $=\sqrt{\dfrac{2(1.4\times10^3)}{(3.0\times10^8)(8.85\times10^{-12})}}$ $\approx1.0\times10^3$ V/m.
(b) B₀ $B_0=\dfrac{E_0}{c}=\dfrac{1.03\times10^3}{3.0\times10^8}\approx3.4\times10^{-6}$ T $=3.4\ \mu$T — about a fifteenth of the Earth's magnetic field.
(c) Energy density $\langle u\rangle=\dfrac{I}{c}=\dfrac{1.4\times10^3}{3.0\times10^8}\approx4.7\times10^{-6}\ \text{J/m}^3$.
Note Sunlight is a mixture of many frequencies and polarisations, so these are effective (rms-equivalent) amplitudes, not those of a single pure wave.
✏️ Example 4 — Light from a lamp
An LED lamp gives out $2.0$ W of visible light, equally in all directions. Find the intensity and the amplitudes of the electric and magnetic fields $1.5$ m from the lamp.
Intensity The power spreads over a sphere: $$I=\frac{P}{4\pi r^2}=\frac{2.0}{4\pi(1.5)^2}\approx7.1\times10^{-2}\ \text{W/m}^2.$$
Fields $E_0=\sqrt{\dfrac{2I}{c\varepsilon_0}}$ $=\sqrt{\dfrac{2(0.0707)}{2.655\times10^{-3}}}\approx7.3$ V/m, and $B_0=\dfrac{E_0}{c}\approx2.4\times10^{-8}$ T.
Check At $3.0$ m the intensity would be four times smaller and $E_0$ half as large, $3.6$ V/m: $E_0\propto1/r$ for a point source.
✏️ Example 5 — The momentum of a laser beam
A $5.0$ mW laser pointer makes a spot $2.0$ mm in diameter on a black screen, which absorbs all the light. Find (a) the intensity in the spot and the amplitude of the electric field, (b) the energy and momentum delivered to the screen in $10$ s, and (c) the average force on the screen.
(a) Intensity $A=\pi(1.0\times10^{-3})^2=3.14\times10^{-6}\ \text{m}^2$, so $I=\dfrac{5.0\times10^{-3}}{3.14\times10^{-6}}\approx1.6\times10^3\ \text{W/m}^2$ — brighter than sunlight. $E_0=\sqrt{2I/(c\varepsilon_0)}\approx1.1\times10^3$ V/m.
(b) Energy, momentum $U=Pt=(5.0\times10^{-3})(10)=0.050$ J; $p=\dfrac{U}{c}=\dfrac{0.050}{3.0\times10^8}\approx1.7\times10^{-10}$ kg m/s.
(c) Force $F=\dfrac{p}{t}=\dfrac{P}{c}\approx1.7\times10^{-11}$ N — far too small to feel, but not zero.
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — radiation pressure

Not required by the CBSE syllabus. Electromagnetic waves and their characteristics are listed by JEE Main, NEET and JEE Advanced; none names radiation pressure, which follows from the momentum a wave carries.

The force per unit area that a wave exerts on a surface is its radiation pressure. For a wave of intensity $I$ falling perpendicularly on a surface:

$$P_\text{rad}=\frac{I}{c}\quad\text{(complete absorption)},$$ $$P_\text{rad}=\frac{2I}{c}\quad\text{(perfect reflection)}.$$

A reflecting surface gets twice the push, because the wave's momentum is reversed rather than just stopped — like a ball that bounces compared with one that sticks. A perfectly reflecting solar sail $100$ m $\times$ $100$ m facing sunlight of $1.4\ \text{kW/m}^2$ feels $F=\dfrac{2IA}{c}=\dfrac{2(1.4\times10^3)(1.0\times10^4)}{3.0\times10^8}\approx0.093$ N: tiny, but acting for months it can change a spacecraft's speed by kilometres per second. JEE problems also ask for the force on a partly reflecting surface (reflectivity $r$): $P_\text{rad}=(1+r)I/c$.

The Electromagnetic Spectrum

Maxwell's equations put no limit on the frequency of an electromagnetic wave. The waves we know span more than twenty powers of ten in wavelength, and arranged in order they form the electromagnetic spectrum. The traditional bands are named after the way the waves were discovered or are produced; the boundaries between them are approximate, and neighbouring bands overlap (X-rays and gamma rays, for example, are told apart by their origin rather than their wavelength). All of them travel at $c$ in vacuum, obey $c=\nu\lambda$, and are emitted or absorbed in packets of energy $E=h\nu=hc/\lambda$ — the photons of Chapter 11. With $hc=1240$ eV nm, a photon of wavelength $\lambda$ (in nm) has energy $1240/\lambda$ eV.

BandWavelength (approx.)Produced byDetected byMain uses
Radio wavesabove $0.1$ mcharges oscillating in antennas and circuitsreceiving antennas with tuned circuitsAM / FM broadcasting, television, mobile networks, radio astronomy
Microwaves$1$ mm – $0.1$ mmagnetrons, klystrons and special diodesantennas, diode detectorsradar, satellite links, Wi-Fi, microwave ovens
Infrared$700$ nm – $1$ mmwarm and hot bodies; vibrating molecules; IR LEDsthermopiles, bolometers, IR photodiodes, the skinremote controls, night vision, heating therapy, fibre-optic links
Visible light$400$ nm – $700$ nmelectrons in atoms changing levels; hot bodies; LEDsthe eye, photocells, camera sensorsseeing, photography, optical instruments
Ultraviolet$1$ nm – $400$ nmvery hot bodies such as the Sun; gas-discharge lampsphotocells, fluorescent screenssterilising water and surfaces, checking banknotes, eye surgery
X-rays$1$ pm – $1$ nmfast electrons stopped by a metal targetfilm, digital sensors, Geiger countersmedical imaging and CT, airport security, crystal structure
Gamma raysbelow about $1$ pmradioactive nuclei; nuclear reactionsGeiger and scintillation counterscancer therapy, sterilising medical equipment, astronomy

A few features of the spectrum are worth knowing well:

Radio and microwaves pass easily through the atmosphere, clouds and walls, which makes them ideal for communication. Microwaves of $2.45$ GHz are absorbed well by water molecules, which is why a microwave oven heats food; radar uses the echo of a short microwave pulse to find the distance $d=ct/2$ of an aircraft or a rain cloud.

Infrared is the "heat" radiation of warm objects; your body radiates mostly at around $10\ \mu$m. The Earth's surface, warmed by sunlight, radiates infrared that carbon dioxide and water vapour absorb and partly send back down — the greenhouse effect that keeps the planet habitable, and that extra carbon dioxide strengthens.

Visible light is a narrow band — less than one factor of two in wavelength — to which our eyes happen to be sensitive, from red (about $700$ nm) through orange, yellow, green and blue to violet (about $400$ nm). The Sun's output peaks in this band.

Ultraviolet photons carry a few electronvolts or more, enough to break chemical bonds: they cause sunburn and can damage DNA, which is why UV lamps sterilise water. Most of the Sun's ultraviolet is absorbed by the ozone layer, about $15$–$35$ km up.

X-rays and gamma rays have photon energies of thousands to millions of electronvolts. They pass through soft tissue but are absorbed more strongly by bone and metal, which is what makes an X-ray image, and they ionise the atoms they hit — exposure must be kept as low as possible.

✏️ Example 6 — Placing waves in the spectrum
For each of the following, find the missing frequency or wavelength, the photon energy, and the band: (a) a microwave oven, $2.45$ GHz; (b) the infrared radiated by the human body, $10\ \mu$m; (c) a germicidal lamp, $254$ nm; (d) X-rays of wavelength $0.10$ nm. Use $hc=1240$ eV nm and $h=6.63\times10^{-34}$ J s.
(a) Oven $\lambda=\dfrac{c}{\nu}=\dfrac{3.0\times10^8}{2.45\times10^9}\approx0.122$ m $=12.2$ cm (microwaves). $E=h\nu=(6.63\times10^{-34})(2.45\times10^9)$ $\approx1.6\times10^{-24}$ J $\approx1.0\times10^{-5}$ eV.
(b) Body $\nu=\dfrac{3.0\times10^8}{1.0\times10^{-5}}=3.0\times10^{13}$ Hz (infrared). $\lambda=10^4$ nm, so $E=\dfrac{1240}{10^4}\approx0.12$ eV.
(c) Lamp $\nu=\dfrac{3.0\times10^8}{254\times10^{-9}}\approx1.2\times10^{15}$ Hz (ultraviolet). $E=\dfrac{1240}{254}\approx4.9$ eV — enough to break bonds in DNA.
(d) X-rays $\nu=\dfrac{3.0\times10^8}{1.0\times10^{-10}}=3.0\times10^{18}$ Hz. $E=\dfrac{1240}{0.10}=1.24\times10^4$ eV $\approx12$ keV.
Check Interactive 2 gives the same three numbers for each wavelength. Moving up the spectrum by a factor of $10$ in frequency multiplies the photon energy by $10$.
Interactive 2 — Electromagnetic Spectrum Explorer

The whole spectrum on one logarithmic scale: each step along the bar is a factor of $10$ in wavelength, with long waves on the left. Slide the marker (or press Play) and read the wavelength, the frequency $\nu=c/\lambda$ and the photon energy $hc/\lambda$, together with how waves in that band are made, detected and used. The tiny rainbow is all of visible light. Band boundaries are approximate.

−0.90
wavelength $\lambda$—
frequency $\nu$—
photon energy—
band—
⚠️ Common mistakes with electromagnetic waves

1. Thinking $B_0$ is "negligible" because it is small. $B_0=E_0/c$ is numerically tiny, but the magnetic field carries exactly half of the wave's energy.

2. Getting the direction wrong. The wave travels along $\mathbf E\times\mathbf B$, not $\mathbf B\times\mathbf E$. For $\sin(kx-\omega t)$ it travels along $+x$; for $\sin(kx+\omega t)$ along $-x$.

3. Changing the frequency in a medium. Entering glass or water, the speed and wavelength drop by the factor $n$; the frequency stays the same.

4. Forgetting the $\tfrac12$ in the intensity. $I=\tfrac12c\varepsilon_0E_0^2$ uses the amplitude $E_0$; the $\tfrac12$ is the average of $\sin^2$. (With the rms field, $I=c\varepsilon_0E_\text{rms}^2$.)

5. Using the static dielectric constant for light. Water has $\varepsilon_r\approx80$ for steady fields but $n=1.33$ for light: at optical frequencies its molecules cannot follow the field. Use $\varepsilon_r$ at the wave's frequency.

6. Mixing up the order of the bands. In increasing frequency (decreasing wavelength): radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays.

Exercises

I
The wave and its fields. Take $c=3.0\times10^8$ m/s.
  • 1 An FM station broadcasts at $91.1$ MHz. Find the wavelength of its waves.
  • 2 Find the frequency of green light of wavelength $500$ nm.
  • 3 Find the period of the wave in part 2.
  • 4 The electric field amplitude of a wave is $30$ V/m. Find the magnetic field amplitude.
  • 5 The magnetic field amplitude of a wave is $5.0$ nT. Find the electric field amplitude.
  • 6 A wave travels along $+x$ and its electric field at some instant points along $+z$. In which direction does its magnetic field point at that instant?
  • 7 The electric field of a wave is $E_y=50\sin\left[(1.0\times10^{7})x-(3.0\times10^{15})t\right]$ V/m (SI units). Find the wavelength, the frequency and $B_0$, and name the band.
  • 8 Find the speed of electromagnetic waves in a non-magnetic medium with $\varepsilon_r=4.0$, and its refractive index.
1 $\lambda=\dfrac{3.0\times10^8}{9.11\times10^7}\approx3.3$ m.
2 $\nu=\dfrac{3.0\times10^8}{5.00\times10^{-7}}=6.0\times10^{14}$ Hz.
3 $T=\dfrac1\nu\approx1.7\times10^{-15}$ s.
4 $B_0=\dfrac{E_0}{c}=\dfrac{30}{3.0\times10^8}=1.0\times10^{-7}$ T.
5 $E_0=cB_0=(3.0\times10^8)(5.0\times10^{-9})=1.5$ V/m.
6 Need $\mathbf E\times\mathbf B$ along $+x$ with $\mathbf E$ along $+z$: $\hat{\mathbf k}\times(-\hat{\mathbf j})=\hat{\mathbf i}$, so $\mathbf B$ points along $-y$.
7 $\lambda=\dfrac{2\pi}{1.0\times10^7}\approx6.3\times10^{-7}$ m $=628$ nm; $\nu=\dfrac{3.0\times10^{15}}{2\pi}\approx4.8\times10^{14}$ Hz; $B_0=\dfrac{50}{3.0\times10^8}\approx1.7\times10^{-7}$ T (along $z$). Visible (orange-red) light.
8 $v=\dfrac{c}{\sqrt{4.0}}=1.5\times10^8$ m/s; $n=2.0$.
II
Energy, intensity and momentum. $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$.
  • 1 Find the intensity of a wave with $E_0=100$ V/m.
  • 2 Find the average energy density of the wave in part 1.
  • 3 Show that in an electromagnetic wave the electric and magnetic energy densities are equal at every point.
  • 4 A wave has intensity $10\ \text{W/m}^2$. Find $E_0$ and $B_0$.
  • 5 A source radiates $100$ W of electromagnetic waves equally in all directions. Find the intensity and $E_0$ at $5.0$ m.
  • 6 By what factor does $E_0$ change when the intensity of a wave is made four times as great?
  • 7 A solar panel of area $1.5\ \text{m}^2$ absorbs sunlight of intensity $800\ \text{W/m}^2$ for one hour. Find the energy and the momentum it receives.
  • 8 Sound cannot cross a vacuum, but light can. Explain the difference.
1 $I=\tfrac12c\varepsilon_0E_0^2$ $=\tfrac12(3.0\times10^8)(8.85\times10^{-12})(100)^2$ $\approx13\ \text{W/m}^2$.
2 $\langle u\rangle=\tfrac12\varepsilon_0E_0^2=\dfrac{I}{c}\approx4.4\times10^{-8}\ \text{J/m}^3$.
3 $u_B=\dfrac{B^2}{2\mu_0}=\dfrac{(E/c)^2}{2\mu_0}$ $=\dfrac{E^2\mu_0\varepsilon_0}{2\mu_0}$ $=\tfrac12\varepsilon_0E^2=u_E$, using $B=E/c$ and $c^2=1/(\mu_0\varepsilon_0)$.
4 $E_0=\sqrt{\dfrac{2I}{c\varepsilon_0}}=\sqrt{\dfrac{20}{2.655\times10^{-3}}}\approx87$ V/m; $B_0=E_0/c\approx2.9\times10^{-7}$ T.
5 $I=\dfrac{100}{4\pi(5.0)^2}\approx0.32\ \text{W/m}^2$; $E_0=\sqrt{\dfrac{2(0.318)}{2.655\times10^{-3}}}\approx15$ V/m.
6 $E_0\propto\sqrt I$: it doubles.
7 $U=IAt=(800)(1.5)(3600)\approx4.3\times10^6$ J; $p=\dfrac{U}{c}\approx1.4\times10^{-2}$ kg m/s (towards the panel).
8 Sound is a wave of pressure in matter: it needs molecules to push on each other. An electromagnetic wave is made of electric and magnetic fields, which exist in empty space and regenerate each other (Faraday's law and the Ampère–Maxwell law), so it needs no medium.
III
The spectrum. Use $hc=1240$ eV nm.
  • 1 Arrange in order of increasing wavelength: X-rays, microwaves, ultraviolet, radio waves, visible light, gamma rays, infrared.
  • 2 A wave has frequency $5.0\times10^{19}$ Hz. Find its wavelength and name the band.
  • 3 A radar works at a wavelength of $3.0$ cm. Find its frequency and name the band.
  • 4 A photon has energy $2.0$ eV. Find its wavelength and say what colour it is.
  • 5 Which band does a TV remote control use? Why does it not work through a closed door?
  • 6 Why is the ozone layer important to life on Earth?
  • 7 Why do X-ray images show bones clearly?
  • 8 Find the wavelength of the waves in a microwave oven ($2.45$ GHz) and explain why they heat moist food.
1 Gamma rays, X-rays, ultraviolet, visible, infrared, microwaves, radio waves.
2 $\lambda=\dfrac{3.0\times10^8}{5.0\times10^{19}}=6.0\times10^{-12}$ m $=6.0$ pm: X-rays (between $1$ pm and $1$ nm).
3 $\nu=\dfrac{3.0\times10^8}{0.030}=1.0\times10^{10}$ Hz $=10$ GHz: microwaves.
4 $\lambda=\dfrac{1240}{2.0}=620$ nm: visible, orange-red.
5 Infrared (from an LED near $940$ nm). Infrared at this wavelength is absorbed or reflected by the wood of a door; it does not pass through opaque solids, although it reflects off walls well enough to work round a room.
6 Ozone absorbs most of the Sun's ultraviolet, whose photons (several eV) break chemical bonds and damage DNA; without it, sunburn, eye damage and skin cancer would be far more common, and many organisms could not survive on land.
7 X-ray photons pass through soft tissue fairly easily but are absorbed much more strongly by bone, which contains heavier atoms (calcium, phosphorus). Bones therefore cast a "shadow" on the detector.
8 $\lambda=\dfrac{3.0\times10^8}{2.45\times10^9}\approx12$ cm. Water molecules are polar; the oscillating electric field keeps twisting them to and fro, and their collisions turn that motion into heat. Dry materials such as paper and glass absorb little.
IV
Applications and longer problems.
  • 1 The electric field of a wave in vacuum is $E_y=120\sin(2.0x-6.0\times10^{8}t)$ V/m (SI units). Find its wavelength, frequency, band and intensity, and write the expression for its magnetic field.
  • 2 Illustrative values. When Mars is $2.25\times10^{11}$ m from the Earth, how long does a radio command take to reach a rover there?
  • 3 The Sun is $1.5\times10^{11}$ m away. How old is the sunlight you see?
  • 4 A radar pulse returns from an aircraft $2.0\times10^{-4}$ s after it was sent. How far away is the aircraft?
  • 5 A simple receiving antenna works best when its length is half a wavelength. Estimate the best length for receiving a $100$ MHz FM broadcast.
  • 6 Modelling (illustrative values). Treat a mobile phone as a source radiating $1.0$ W equally in all directions. Estimate the intensity and $E_0$ at $2.0$ m.
  • 7 A $1.0$ mW laser beam has a diameter of $1.0$ mm. Find its intensity and the amplitude of its electric field.
1 $\lambda=\dfrac{2\pi}{2.0}\approx3.1$ m, $\nu=\dfrac{6.0\times10^8}{2\pi}\approx9.5\times10^7$ Hz $=95$ MHz (FM radio). $I=\tfrac12c\varepsilon_0E_0^2$ $=\tfrac12(3.0\times10^8)(8.85\times10^{-12})(120)^2$ $\approx19\ \text{W/m}^2$. $B_0=\dfrac{120}{3.0\times10^8}=4.0\times10^{-7}$ T and the wave travels along $+x$ with $\mathbf E$ along $y$, so $\mathbf B$ is along $z$ ($\hat{\mathbf j}\times\hat{\mathbf k}=\hat{\mathbf i}$): $B_z=4.0\times10^{-7}\sin(2.0x-6.0\times10^8t)$ T.
2 $t=\dfrac{2.25\times10^{11}}{3.0\times10^8}=750$ s $=12.5$ min.
3 $t=\dfrac{1.5\times10^{11}}{3.0\times10^8}=500$ s $\approx8.3$ min.
4 The pulse goes there and back: $d=\dfrac{ct}{2}=\dfrac{(3.0\times10^8)(2.0\times10^{-4})}{2}=3.0\times10^4$ m $=30$ km.
5 $\lambda=\dfrac{3.0\times10^8}{1.0\times10^8}=3.0$ m, so the length is about $1.5$ m.
6 $I=\dfrac{1.0}{4\pi(2.0)^2}\approx2.0\times10^{-2}\ \text{W/m}^2$; $E_0=\sqrt{\dfrac{2(0.0199)}{2.655\times10^{-3}}}\approx3.9$ V/m. (Real phones radiate much less than $1$ W on average and not equally in all directions.)
7 $A=\pi(5.0\times10^{-4})^2\approx7.85\times10^{-7}\ \text{m}^2$; $I=\dfrac{1.0\times10^{-3}}{7.85\times10^{-7}}\approx1.3\times10^3\ \text{W/m}^2$; $E_0=\sqrt{\dfrac{2I}{c\varepsilon_0}}\approx9.8\times10^2$ V/m.
🎯 Exam corner — electromagnetic waves and the spectrum BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Name the electromagnetic radiation used (a) to locate aircraft by radar, (b) to kill germs in water purifiers, (c) in the remote control of a television, (d) to study the arrangement of atoms in crystals. Arrange them in decreasing order of frequency.

Names (a) Microwaves; (b) ultraviolet; (c) infrared; (d) X-rays.
Order X-rays $\gt$ ultraviolet $\gt$ infrared $\gt$ microwaves. Exam tip: write the full spectrum order in the margin first — radio, micro, IR, visible, UV, X, gamma — and read off the answer.

E2. Board · 3 marks The electric field of a plane electromagnetic wave in vacuum is $\mathbf E=E_0\sin(kz-\omega t)\,\hat{\mathbf i}$ with $E_0=30$ V/m, $k=1.0\times10^{3}\ \text{m}^{-1}$ and $\omega=3.0\times10^{11}\ \text{s}^{-1}$. (a) Find the wavelength and frequency. (b) Find the amplitude and direction of the magnetic field and write its expression. (c) Name the band.

(a) $\lambda=\dfrac{2\pi}{k}=\dfrac{2\pi}{1.0\times10^3}\approx6.3$ mm; $\nu=\dfrac{\omega}{2\pi}\approx4.8\times10^{10}$ Hz. (Check: $\omega/k=3.0\times10^8$ m/s.)
(b) $B_0=\dfrac{E_0}{c}=\dfrac{30}{3.0\times10^8}=1.0\times10^{-7}$ T. The wave travels along $+z$ and $\mathbf E$ is along $x$; $\hat{\mathbf i}\times\hat{\mathbf j}=\hat{\mathbf k}$, so $\mathbf B$ is along $y$: $$\mathbf B=B_0\sin(kz-\omega t)\,\hat{\mathbf j},$$ with $B_0=1.0\times10^{-7}$ T and the same $k$ and $\omega$.
(c) $6.3$ mm: microwaves. Exam tip: one mark each for $\lambda$/$\nu$, for $B_0$ with its direction, and for the band; always check the direction with $\mathbf E\times\mathbf B$.

E3. JEE Main The amplitude of the electric field of a plane electromagnetic wave in vacuum is $6.0$ V/m. Its intensity is (A) $4.8\times10^{-2}\ \text{W/m}^2$ (B) $9.6\times10^{-2}\ \text{W/m}^2$ (C) $2.4\times10^{-2}\ \text{W/m}^2$ (D) $1.6\times10^{-10}\ \text{W/m}^2$.

Formula $I=\tfrac12c\varepsilon_0E_0^2$ $=\tfrac12(3.0\times10^8)(8.85\times10^{-12})(36)$ $\approx4.8\times10^{-2}\ \text{W/m}^2$.
Answer (A). Exam tip: (B) drops the $\tfrac12$; (D) is the average energy density $\tfrac12\varepsilon_0E_0^2$ in $\text{J/m}^3$ — a different quantity with the wrong unit.

E4. NEET An electromagnetic wave travels in a non-magnetic medium of relative permittivity $4.0$. Its speed in the medium is (A) $3.0\times10^8$ m/s (B) $1.5\times10^8$ m/s (C) $0.75\times10^8$ m/s (D) $6.0\times10^8$ m/s.

Formula $v=\dfrac{c}{\sqrt{\mu_r\varepsilon_r}}=\dfrac{3.0\times10^8}{\sqrt{4.0}}=1.5\times10^8$ m/s.
Answer (B). Exam tip: (C) divides by $\varepsilon_r$ instead of its square root.

E5. JEE Advanced (One or more correct.) For a plane electromagnetic wave travelling in vacuum: (A) the electric and magnetic fields oscillate in phase; (B) the average electric and magnetic energy densities are equal; (C) the wave travels in the direction of $\mathbf B\times\mathbf E$; (D) the ratio of the amplitudes $E_0/B_0$ equals $1/\sqrt{\mu_0\varepsilon_0}$.

Options (A) true: both vary as $\sin(kx-\omega t)$. (B) true: $u_B=B^2/2\mu_0=\tfrac12\varepsilon_0E^2=u_E$ at every point, so the averages are equal too. (C) false: the direction is $\mathbf E\times\mathbf B$; $\mathbf B\times\mathbf E$ points backwards. (D) true: $E_0/B_0=c=1/\sqrt{\mu_0\varepsilon_0}$.
Answer (A), (B), (D). Exam tip: cross-product order is the commonest trap in this topic.

E6. Board · Case study On a clear day, sunlight reaching a rooftop solar panel in Chennai has an intensity of about $1.0\ \text{kW/m}^2$. (i) (2 marks) Treating it as a single sinusoidal wave, find the amplitudes of its electric and magnetic fields. (ii) (1 mark) Which part of the Sun's radiation is largely removed by the ozone layer before it reaches the panel? (iii) (1 mark) How much energy falls on a $2.0\ \text{m}^2$ panel in one minute?

(i) $E_0=\sqrt{\dfrac{2I}{c\varepsilon_0}}$ $=\sqrt{\dfrac{2(1.0\times10^3)}{2.655\times10^{-3}}}$ $\approx8.7\times10^2$ V/m; $B_0=\dfrac{E_0}{c}\approx2.9\times10^{-6}$ T.
(ii) Ultraviolet.
(iii) $U=IAt=(1.0\times10^3)(2.0)(60)=1.2\times10^5$ J. Exam tip: in a case study, keep $c\varepsilon_0=2.655\times10^{-3}$ handy — it appears in every intensity calculation.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How strong are the radio waves near a mobile-phone tower, compared with the exposure limits that regulators set — and why are such waves called "non-ionising"?

Why it matters. Phone towers stand on rooftops in cities from Mumbai to Nairobi to São Paulo, and residents often worry about them. Regulators base their limits on the intensity of the waves: the International Commission on Non-Ionizing Radiation Protection (ICNIRP) publishes reference levels, and some countries, India among them, adopt stricter limits than ICNIRP. Physics lets you estimate the intensity yourself and compare.

What to do.

  1. Model one antenna of a tower as a source radiating $20$ W at $900$ MHz equally in all directions (an illustrative, deliberately simple model — real antennas beam their power into a narrow sector). Find the intensity at $10$ m, $50$ m and $100$ m, and the field amplitude $E_0$ at $10$ m.
  2. Look up the current ICNIRP general-public reference level for $900$ MHz and your own country's limit, and express each of your intensities as a fraction of them.
  3. Find the energy of one $900$ MHz photon in eV. The energy needed to ionise an atom or break a chemical bond is a few eV. Explain why radio waves are "non-ionising" however intense they are, while ultraviolet and X-rays are not.
  4. Real antennas concentrate their power in a beam. Explain qualitatively how that changes your estimate directly in front of the antenna and directly below it.

Tools: Calculator or spreadsheet; the regulators' published limits (search for the ICNIRP guidelines and your national telecom authority).

Hand in: A table of $I$ and $I/I_\text{limit}$ at the three distances, the photon-energy calculation and a half-page explanation for a worried neighbour.

Going further. The heating effect of radio waves on tissue is measured by the specific absorption rate (SAR, in W/kg). Find out what SAR value phones sold in your country must stay below, and why the phone held against your head usually matters more than a distant tower.

Instructor note: $I=P/(4\pi r^2)$: $1.6\times10^{-2}$, $6.4\times10^{-4}$ and $1.6\times10^{-4}\ \text{W/m}^2$ at $10$, $50$, $100$ m; $E_0=\sqrt{2I/(c\varepsilon_0)}\approx3.5$ V/m at $10$ m. The ICNIRP (1998) general-public reference level for $400$–$2000$ MHz is $f/200\ \text{W/m}^2$ ($f$ in MHz), i.e. $4.5\ \text{W/m}^2$ at $900$ MHz; India's limit is one-tenth of that, $0.45\ \text{W/m}^2$ (students should confirm the current values). At $10$ m the model gives about $0.4\%$ of the ICNIRP level and $4\%$ of the Indian limit. Photon energy $h\nu\approx3.7\times10^{-6}$ eV, about a million times too small to ionise, so the only established effect is heating. A beamed antenna gives a higher intensity in front of it and much less directly below.

🧪 Self-check quiz

0/5 correct

Q1. For an electromagnetic wave in vacuum, the ratio $E_0/B_0$ of the field amplitudes is

Q2. Which list is in order of increasing frequency?

Q3. An electromagnetic wave travels along $+x$ and at some point its electric field points along $+y$. Its magnetic field there points along

Q4. If the amplitude of the electric field of a wave is doubled, its intensity

Q5. Which of these does not radiate electromagnetic waves?