Physics ELECTROMAGNETIC WAVES AND ELECTROMAGNETIC SPECTRUM for CBSE-NCERT

### Topic covered

color{blue}{star} ELECTROMAGNETIC WAVES
color{blue}{star} ELECTROMAGNETIC SPECTRUM

### ELECTROMAGNETIC WAVES

color{brown}{bbul{"Sources of electromagnetic waves"}}

color{blue} ✍️Electromagnetic waves neither produce by stationary charges nor charges in uniform motion (steady currents) can be sources of electromagnetic waves.

color{blue} ✍️The former produces only electrostatic fields, while the latter produces magnetic fields that, however, do not vary with time. It is an important result of Maxwell’s theory that accelerated charges radiate electromagnetic waves.

color{blue} ✍️Let's consider a charge oscillating with some frequency. (An oscillating charge is an example of accelerating charge). This produces an oscillating electric field in space, which produces an "oscillating magnetic field," which in turn, is a source of oscillating electric field, and so on.

color{blue} ✍️The oscillating electric and magnetic fields thus regenerate each other, so to speak, as the wave propagates through the space.

color{blue} ✍️The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge.

color{blue} ✍️The energy associated with the propagating wave comes at the expense of the energy of the source – the accelerated charge. From the preceding discussion, it might appear easy to test the prediction that light is an electromagnetic wave.

color{blue} ✍️We might think that all we needed to do was to set up an ac circuit in which the current oscillate at the frequency of visible light, say, yellow light. But, alas, that is not possible.

color{blue} ✍️The frequency of yellow light is about 6 × 10^(14) Hz, while the frequency that we get even with modern electronic circuits is hardly about 10^(11) Hz. This is why the experimental demonstration of electromagnetic wave had to come in the low frequency region (the radio wave region).

### Nature of electromagnetic waves

color{blue} ✍️It can be shown from Maxwell’s equations that electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of propagation.

color{blue} ✍️ It appears reasonable, say from our discussion of the displacement current. Consider Fig. 8.2. The electric field inside the plates of the capacitor is directed perpendicular to the plates.

color{blue} ✍️The magnetic field this gives rise to via the displacement current is along the perimeter of a circle parallel to the capacitor plates.

color{blue} ✍️So B and E are perpendicular in this case. This is a general feature. In Fig. 8.4, we show a typical example of a plane electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate, at a given time t).

color{blue} ✍️The electric field E_x is along the x-axis, and varies sinusoidally with z, at a given time. The magnetic field B_y is along the y-axis, and again varies sinusoidally with z.
The electric and magnetic fields E_x and By are perpendicular to each other, and to the direction z of propagation.

color{blue} ✍️We can write E_x and B_y as follows:

color {blue}{E_x= E_0 sin (kz-omega t)}

.................[8.7(a)]

color {blue}{B_y = B_0 sin (kz-omegat )}

...............[8.7(b)]

color{blue} ✍️Here k is related to the wave length λ of the wave by the usual equation

color {blue}{k = (2pi)/(lambda)}

..............(8.8)

color{blue} ✍️and ω is the angular frequency. k is the magnitude of the wave vector (or propagation vector) k and its direction describes the direction of propagation of the wave.

color{blue} ✍️The speed of propagation of the wave is (ω//k ). Using Eqs. [8.7(a) and (b)] for E_x and B_y and Maxwell’s equations, one finds that

color {blue}{ω = ck,} where, color {blue}{c = 1//sqrt(mu_0ε_0)}

.....................[8.9(a)]

color{blue} ✍️The relation ω = ck is the standard one for waves (see for example, Section 15.4 of class XI Physics textbook).
This relation is often written in terms of frequency, ν (=ω//2π) and wavelength, λ (=2π//k) as

2piV= C((2pi)/(lambda)) or

color {blue}{v lambda=c}

...............[8.9(b)]

color{blue} ✍️It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as

color {blue}{B_0 = (E_0 //C)}

................(8.10)

color{blue} ✍️We here make remarks on some features of electromagnetic waves. They are self-sustaining oscillations of electric and magnetic fields in free space, or vacuum.

color{blue} ✍️They differ from all the other waves we have studied so far, in respect that no material medium is involved in the vibrations of the electric and magnetic fields.

color{blue} ✍️Sound waves in air are longitudinal waves of compression and rarefaction. Transverse waves on the surface of water consist of water moving up and down as the wave spreads horizontally and radially onwards.

color{blue} ✍️Transverse elastic (sound) waves can also propagate in a solid, which is rigid and that resists shear.

color{blue} ✍️Scientists in the nineteenth century were so much used to this mechanical picture that they thought that there must be some medium pervading all space and all matter, which responds to electric and magnetic fields just as any elastic medium does. They called this medium ether.

color{blue} ✍️They were so convinced of the reality of this medium, that there is even a novel called The Poison Belt by Sir Arthur Conan Doyle (the creator of the famous detective Sherlock Holmes) where the solar system is supposed to pass through a poisonous region of ether.

color{blue} ✍️We now accept that no such physical medium is needed. The experiment shows Electric and magnetic fields, oscillating in space and time, can sustain each other in vacuum.

color{blue} ✍️But what if a material medium is actually there, We know that light, an electromagnetic wave, does propagate through glass,
for example.

color{blue} ✍️We have seen earlier that the total electric and magnetic fields inside a medium are described in terms of a permittivity ε and a magnetic permeability μ (these describe the factors by which the total fields differ from the external fields).
These replace ε_0 and μ_0 in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity ε and magnetic permeability μ, the velocity of light becomes,

color {blue}{v= 1/(sqrt(mu in))}

...............(8.11)

color{blue} ✍️Thus, the velocity of light depends on electric and magnetic properties of the medium.

color{blue} ✍️The velocity of electromagnetic waves in free space or vacuum is an important fundamental constant. It has been shown by experiments on electromagnetic waves of different wavelengths that this velocity is the same (independent of wavelength) to within a few metres per second, out of a value of 3×10^8 m//s.

color{blue} ✍️The constancy of the velocity of e m waves in vacuum is so strongly supported by experiments and the actual value is so well known now that this is used to define a standard of length.
Namely, the metre is now defined as the distance travelled by light in vacuum in a time (1//c) seconds = (2.99792458 × 10^8)^(–1) seconds.

color{blue} ✍️This has come about for the following reason. The basic unit of time can be defined very accurately in terms of some atomic frequency, i.e., frequency of light emitted by an atom in a particular process.

color{blue} ✍️The basic unit of length is harder to define as accurately in a direct way. Earlier measurement of c using earlier units of length (metre rods, etc.) converged to a value of about 2.9979246 × 10^8 m//s.

color{blue} ✍️Since c is such a strongly fixed number, unit of length can be defined in terms of c and the unit of time! Hertz not only showed the existence of electromagnetic waves, but also demonstrated that the waves, which had wavelength ten million times that of the light waves, could be diffracted, refracted and polarised.

color{blue} ✍️Thus, he conclusively established the wave nature of the radiation. Further, he produced stationary electromagnetic waves and determined their wavelength by measuring the distance between two successive nodes.

color{blue} ✍️Since the frequency of the wave was known (being equal to the frequency of the oscillator), he obtained the speed of the wave using the formula v = νλ and found that the waves travelled with the same speed as the speed of light.

color{blue} ✍️The fact that electromagnetic waves are polarised can be easily seen in the response of a portable AM radio to a broadcasting station. If an AM radio has a telescopic antenna, it responds to the electric part of the signal. When the antenna is turned horizontal, the signal will be greatly diminished.

color{blue} ✍️Some portable radios have horizontal antenna (usually inside the case of radio), which are sensitive to the magnetic component of the electromagnetic wave. Such a radio must remain horizontal in order to receive the signal. In such cases, response also depends on the orientation of the radio with respect to the station.

color{blue} ✍️Electromagnetic waves carry energy and momentum like other waves. As we know that in a region of free space with electric field E, there is an energy density (ε_0E^2//2). Similarly, as seen in , associated with a magnetic field B is a magnetic energy density (B^2/(2μ0)).

color{blue} ✍️As electromagnetic wave contains both electric and magnetic fields, there is a non-zero energy density associated with it.

color{blue} ✍️Now consider a plane perpendicular to the direction of propagation of the electromagnetic wave (Fig. 8.4). If there are, on this plane, electric charges, they will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave.

color{blue} ✍️The charges thus acquire energy and momentum from the waves. This just illustrates the fact that an electromagnetic wave (like other waves) carries energy and momentum.

color{blue} ✍️Since it carries momentum, an electromagnetic wave also exerts pressure, called radiation pressure. If the total energy transferred to a surface in time t is U, it can be shown that the magnitude of the total momentum delivered to this surface (for complete absorption) is,BB

color {blue}{P = U/C}

..........(8.12)

color{blue} ✍️When the sun shines on your hand, you feel the energy being absorbed from the electromagnetic waves (your hands get warm).

color{blue} ✍️ Electromagnetic waves also transfer momentum to your hand but because c is very large, the amount of momentum transferred is extremely small and you do not feel the pressure.

color{blue} ✍️Radiation pressure of visible light was found to be of the order of 7 × 10^(–6) N//m^2. Thus, on a surface of area 10 cm^2, the force due to radiation is only about 7 × 10^(–9) N.

color{blue} ✍️The great technological importance of electromagnetic waves stems from their capability to carry energy from one place to another.

color{blue} ✍️The radio and TV signals from broadcasting stations carry energy. Light carries energy from the sun to the earth, thus making life possible on the earth.
Q 3178245106

A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 hatj V//m. What is B at this point?
Class 12 Chapter 8 Example 2
Solution:

Using Eq. (8.10), the magnitude of B is

B = E/C
= (6.3V//m)/(3xx10^8m//s) = 2.1xx 10^(8)T

To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction. Since, (+ hatj ) × (+ hatk ) = hati , B is along the z-direction. Thus, B = 2.1 × 10–8 hatkT
Q 3118245109

8.3 The magnetic field in a plane electromagnetic wave is given by By = 2 × 10–7 sin (0.5×10^3x+1.5×10^11t) T.
(a) What is the wavelength and frequency of the wave?
(b) Write an expression for the electric field.
Class 12 Chapter 8 Example 3
Solution:

(a) Comparing the given equation with

B_u = B_0 sin [2pi(x/lambda+t/T)]

We get, lambda= (2pi)/(0.5xx10^3) m = 1.26cm.

and 1/T= V = (1.5 xx 10^11)//2pi= 23.9Ghz

(b) E_0 = B_0c = 2×10^(–7) T × 3 × 10^8 m/s = 6 × 10^1 V//m
The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as
E_z = 60 sin (0.5 × 10^3x + 1.5 × 10^(11) t) V//m
Q 3138345202

Light with an energy flux of 18 W//cm^2 falls on a non reflecting surface at normal incidence. If the surface has an area of 20 cm^2, find the average force exerted on the surface during a 30 minute time span.
Class 12 Chapter 8 Example 4
Solution:

The total energy falling on the surface is

U = (18 W//cm^2) × (20 cm^2) × (30 × 60)

= 6.48 × 10^5 J

Therefore, the total momentum delivered (for complete absorption) is

P = U/C = (6.48xx10^5J)/(3xx10^8m//s) = 2.16 xx 10^(-3) kg m//s

The average force exerted on the surface is

F= p/t = (2.16xx 10^(-3))/(0.18xx10^4) = 1.2 xx 10^(-6) N
How will your result be modified if the surface is a perfect reflector?
Q 3178345206

Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 2.5% and it is a point source.
Class 12 Chapter 8 Example 5
Solution:

The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3 m, the surface area of the surrounding sphere is

A = 4 pir^2 = 4pi(3)^2 = 113^2

The intensity at this distance is

I = ("Power")/("Area") = (100Wxx2.5%)/(113m^2)

= 0.022 W//m2
Half of this intensity is provided by the electric field and half by the magnetic field.

1/2 I = 1/2 (ε_0 E_(rms)^(2) C)

 = 1/2 (0.022W/m^2)

E_(rms) = sqrt((0.022)/(8.85xx10^(12)(3xx10^8)))V/m

The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field, E_0 is

E_0 = sqrt(2 E_(rms)) = sqrt(2 xx 2.9) V//m

Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of TV or FM waves, which is of the order of a few microvolts per metre.

Now, let us calculate the strength of the magnetic field. It is

B_(rms) = (E_(rms))/c = (2.9 Vm^(-1))/(3xx 10^8ms^(-1) = 9.6 xx 10^(-9)T

Again, since the field in the light beam is sinusoidal, the peak magnetic field is B_0 = sqrt2 B_(rms) = 1.4 × 10^(–8) T. Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak.

### ELECTROMAGNETIC SPECTRUM

color{blue} ✍️At the time Maxwell predicted the existence of electromagnetic waves, the only familiar electromagnetic waves were the visible light waves.

color{blue} ✍️The existence of ultraviolet and infrared waves was barely established. By the end of the nineteenth century, X-rays and gamma rays had also been discovered.

color{blue} ✍️We now know that, electromagnetic waves include visible light waves, X-rays, gamma rays, radio waves, microwaves, ultraviolet and infrared waves.

color{blue} ✍️The classification of em waves according to frequency is the electromagnetic spectrum (Fig. 8.5). There is no sharp division between one kind of wave and the next. The classification is based roughly on how the waves are produced and/or detected.

color{blue} ✍️We briefly describe these different types of electromagnetic waves, in order of decreasing wavelengths.

color {brown}{ulbb{"Radio waves"}}
color{blue} ✍️Radio waves are produced by the accelerated motion of charges in conducting wires. They are used in radio and television communication systems.

color{blue} ✍️They are generally in the frequency range from 500 kHz to about 1000 MHz. The AM (amplitude modulated) band is from 530  kHz to 1710 kHz. Higher frequencies upto 54 MHz are used for short wave bands.

color{blue} ✍️TV waves range from 54 MHz to 890 MHz. The FM (frequency modulated) radio band extends from 88 MHz to 108 MHz. Cellular phones use radio waves to transmit voice communication in the ultrahigh frequency (UHF) band. How these waves are transmitted and received is described in Chapter 15.

color {brown}{bbul{"Microwaves"}}
color{blue} ✍️Microwaves (short-wavelength radio waves), with frequencies in the gigahertz (GHz) range, are produced by special vacuum tubes (called klystrons, magnetrons and Gunn diodes).

color{blue} ✍️Due to their short wavelengths, they are suitable for the radar systems used in aircraft navigation. Radar also provides the basis for the speed guns used to time fast balls, tennisserves, and automobiles.

color{blue} ✍️Microwave ovens are an interesting domestic application of these waves. In such ovens, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of the molecules. This raises the temperature of any food containing water.

color {brown}{ulbb{"Infrared waves"}}
color{blue} ✍️Infrared waves are produced by hot bodies and molecules. This band lies adjacent to the low-frequency or long-wave length end of the visible spectrum.

color{blue} ✍️ Infrared waves are sometimes referred to as heat waves. This is because water molecules present in most materials readily absorb infrared waves (many other molecules, for example, CO_2, NH_3, also absorb infrared waves).

color{blue} ✍️After absorption, their thermal motion increases, that is, they heat up and heat their surroundings. Infrared lamps are used in physical therapy. Infrared radiation also plays an important role in maintaining the earth’s warmth or average temperature through the greenhouse effect.

color{blue} ✍️Incoming visible light (which passes relatively easily through the atmosphere) is absorbed by the earth’s surface and reradiated as infrared (longer wavelength) radiations.

color{blue} ✍️This radiation is trapped by greenhouse gases such as carbon dioxide and water vapour. Infrared detectors are used in Earth satellites, both for military purposes and to observe growth of crops.

color{blue} ✍️Electronic devices (for example semiconductor light emitting diodes) also emit infrared and are widely used in the remote switches of household electronic systems such as TV sets, video recorders and hi-fi systems.

color {brown}{bbul{"Visible rays"}}
color{blue} ✍️It is the most familiar form of electromagnetic waves. It is the part of the spectrum that is detected by the human eye. It runs from about 4 × 10^(14) Hz to about 7 × 10^(14) Hz or a wavelength range of about 700 – 400 nm.

color{blue} ✍️Visible light emitted or reflected from objects around us provides us information about the world.
Our eyes are sensitive to this range of wavelengths. Different animals are sensitive to different range of wavelengths. For example, snakes can detect infrared waves, and the ‘visible’ range of many insects extends well into the ultraviolet

color {brown}{ulbb{"Ultraviolet rays"}}
color{blue} ✍️It covers wavelengths ranging from about 4 × 10^(–7) m (400 nm) down to 6 × 10^(–10)m (0.6 nm).
Ultraviolet (UV) radiation is produced by special lamps and very hot bodies. The sun is an important source of ultraviolet light.
But fortunately, most of it is absorbed in the ozone layer in the atmosphere at an altitude of about 40 – 50 km.
UV light in large quantities has harmful effects on humans. Exposure to UV radiation induces the production of more melanin, causing tanning of the skin.

color{blue} ✍️UV radiation is absorbed by ordinary glass. Hence, one cannot get tanned or sunburn through glass windows. Welders wear special glass goggles or face masks with glass windows to protect their eyes from large amount of UV produced by welding arcs. Due to its shorter wavelengths, UV radiations can be focussed into very narrow beams for high precision applications such as LASIK (Laserassisted in situ keratomileusis) eye surgery. UV lamps are used to kill germs in water purifiers.

color{blue} ✍️Ozone layer in the atmosphere plays a protective role, and hence its depletion by chlorofluorocarbons (CFCs) gas (such as freon) is a matter of international concern.

color {brown}{bbul{"X-rays"}}
color{blue} ✍️Beyond the UV region of the electromagnetic spectrum lies the X-ray region. We are familiar with X-rays because of its medical applications. It covers wavelengths from about 10^(–8) m (10 nm) down to 10^(–13) m (10^(–4) nm).
One common way to generate X-rays is to bombard a metal target by high energy electrons. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer.
Because X-rays damage or destroy living tissues and organisms, care must be taken to avoid unnecessary or over exposure.

color {brown}{bbul{"Gamma rays"}}
They lie in the upper frequency range of the electromagnetic spectrum and have wavelengths of from about 10^(–10)m to less than 10^(–14)m.

color{blue} ✍️This high frequency radiation is produced in nuclear reactions and also emitted by radioactive nuclei. They are used in medicine to destroy cancer cells. Table 8.1 summarises different types of electromagnetic waves, their production and detections.
As mentioned earlier, the demarcation between different region is not sharp and there are over laps.