8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

1.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere? 43 Reprint 2025-26 Physics 1.21 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 mC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere? 1.22 An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density. 1.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates? Reprint 2025-26 Chapter Two ELECTROSTATIC POTENTIAL AND CAPACITANCE 2.12.12.12.12.1 IIINTRODUCTIONIINTRODUCTIONNTRODUCTIONNTRODUCTIONNTRODUCTION In Chapters 5 and 7 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, that work gets stored as potential energy of the body. When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces. Coulomb force between two (stationary) charges is also a conservative force. This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field. Consider an electrostatic field EEEEE due to some charge configuration. First, for simplicity, consider the field E due to a charge Q placed at the origin. Now, imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. With reference Reprint 2025-26 Physics to Fig. 2.1, this will happen if Q and q are both positive or both negative. For definiteness, let us take Q, q > 0. Two remarks may be made here. First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q fromFIGURE 2.1 A test charge q (> 0) is moved from the point R to the R to P, we apply an external force Fext just enough to point P against the repulsive counter the repulsive electric force FE (i.e, Fext= –FE). force on it by the charge Q (> 0) This means there is no net force on or acceleration of placed at the origin. the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. In this situation, work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. If the external force is removed on reaching P, the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved. Thus, work done by external forces in moving a charge q from R to P is WRP = – = (2.1) This work done is against electrostatic repulsive force and gets stored as potential energy. At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P. Thus, potential energy difference ∆U = U P − U R = W RP (2.2) (Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative, i.e., –WRP .) Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge q from one point to another for electric field of any arbitrary charge configuration. Two important comments may be made at this stage: (i) The right side of Eq. (2.2) depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path. The path-independence of work done by an electrostatic field can be proved using the 46 Coulomb’s law. We omit this proof here. Reprint 2025-26 Electrostatic Potential and Capacitance (ii) Equation (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Clearly, potential energy so defined is undetermined to within an additive constant.What this means is that the actual value of potential energy is not physically significant; it is only the difference of potential energy that is significant. We can always add an arbitrary constant a to potential energy at every point, since this will not change the potential energy difference: (U P + α) − (U R + α) = U P − U R Put it differently, there is a freedom in choosing the point where potential energy is zero. A convenient choice is to have electrostatic potential energy zero at infinity. With this choice, if we take the point R at infinity, we get from Eq. (2.2) Count Alessandro Volta (1745 – 1827) Italian W ∞ P = U P − U ∞ = U P (2.3) physicist, professor at Since the point P is arbitrary, Eq. (2.3) provides us with a Pavia. Volta established that the animal electri- COUNTdefinition of potential energy of a charge q at any point. city observed by LuigiPotential energy of charge q at a point (in the presence of field Galvani, 1737–1798, indue to any charge configuration) is the work done by the experiments with frog external force (equal and opposite to the electric force) in muscle tissue placed in bringing the charge q from infinity to that point. contact with dissimilar metals, was not due to 2.2 ELECTROSTATIC POTENTIAL any exceptional property of animal tissues but ALESSANDROConsider any general static charge configuration. We define was also generated potential energy of a test charge q in terms of the work done whenever any wet body on the charge q. This work is obviously proportional to q, since was sandwiched between the force at any point is qE, where E is the electric field at that dissimilar metals. This VOLTA point due to the given charge configuration. It is, therefore, led him to develop the convenient to divide the work by the amount of charge q, so first voltaic pile, orthat the resulting quantity is independent of q. In other words, battery, consisting of a (1745 work done per unit test charge is characteristic of the electric large stack of moist disks of cardboard (electro-field associated with the charge configuration. This leads to lyte) sandwiched the idea of electrostatic potential V due to a given charge between disks of metal –1827) configuration. From Eq. (2.1), we get: (electrodes). Work done by external force in bringing a unit positive charge from point R to P  U P − U R  = VP – VR = (2.4)  q  where VP and VR are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies: Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point. 47 Reprint 2025-26 Physics In other words, the electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point. The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. To obtain the work done per unit test charge, we should take an infinitesimal test charge FIGURE 2.2 Work done on a test charge q dq, obtain the work done dW in bringing it from by the electrostatic field due to any given infinity to the point and determine the ratio charge configuration is independent dW/dq. Also, the external force at every point of the of the path, and depends only on path is to be equal and opposite to the electrostatic its initial and final positions. force on the test charge at that point. 2.3 POTENTIAL DUE TO A POINT CHARGE Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point P. At some intermediate point P¢ on the path, the electrostatic force on a unit positive charge is FIGURE 2.3 Work done in bringing a unit positive test charge from infinity to the Q × 1 rˆ ′ (2.5) point P, against the repulsive force of 2 4 πε0r ' charge Q (Q > 0), is the potential at P due to the charge Q. where ˆ′r is the unit vector along OP¢. Work done against this force from r¢ to r¢ + Dr¢ is Q ∆W = − 2 ∆′r (2.6) 4 πε0r ' The negative sign appears because for Dr¢ < 0, DW is positive. Total work done (W) by the external force is obtained by integrating Eq. (2.6) from r¢ = ¥ to r¢ = r, r Q Q r Q = dr ′ = ε 0r ′ 2 4 πε0r ′ ∞ 4 πε0r (2.7) W = − ∫4∞ π This, by definition is the potential at P due to the charge Q Q V (r ) = (2.8) 48 4 πε0r Reprint 2025-26 Electrostatic Potential and Capacitance Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are FIGURE 2.4 Variation of potential V with r [in units of in the same direction.] Finally, we (Q/4pe0) m-1] (blue curve) and field with r [in units of (Q/4pe0) m-2] (black curve) for a point charge Q.note that Eq. (2.8) is consistent with the choice that potential at infinity be zero. Figure (2.4) shows how the electrostatic potential ( 1/r) and the electrostatic field (1/r 2 ) varies with r. Example 2.1 (a) Calculate the potential at a point P due to a charge of 4 × 10–7C located 9 cm away. (b) Hence obtain the work done in bringing a charge of 2 × 10–9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought? Solution (a) = 4 × 104 V (b) W = qV = 2 × 10–9C × 4 × 104V = 8 × 10–5 J No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along EXAMPLE r and another perpendicular to r. The work done corresponding to the later will be zero. 2.1

8.3 ELECTROMAGNETIC WAVES 8.3.1 Sources of electromagnetic waves How are electromagnetic waves produced? Neither stationary charges nor charges in uniform motion (steady currents) can be sources of electromagnetic waves. 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. The proof of this basic result is beyond the scope of this book, but we can accept it on the basis of rough, qualitative reasoning. 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. The oscillating electric and magnetic fields thus regenerate each other, so to speak, as the wave propagates through the space. The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. 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. 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. The frequency of yellow light is about 6 × 1014 Hz, while the frequency that we get even with modern electronic circuits is hardly about 1011 Hz. This is why the experimental demonstration of electromagnetic 205 Reprint 2025-26 Physics wave had to come in the low frequency region (the radio wave region), as in the Hertz’s experiment (1887). Hertz’s successful experimental test of Maxwell’s theory created a sensation and sparked off other important works in this field. Two important achievements in this connection deserve mention. Seven years after Hertz, Jagdish Chandra Bose, working at Calcutta (now Kolkata), succeeded in producing and observing electromagnetic waves of much shorter 8.1 wavelength (25 mm to 5 mm). His experiment, like that of Hertz’s, was confined to the laboratory. At around the same time, Guglielmo Marconi in Italy followed Hertz’s work and succeeded in transmitting EXAMPLE electromagnetic waves over distances of many kilometres. Heinrich Rudolf Hertz Marconi’s experiment marks the beginning of the field of (1857 – 1894) German communication using electromagnetic waves. physicist who was the first to broadcast and 8.3.2 Nature of electromagnetic wavesHEINRICH receive radio waves. He It can be shown from Maxwell’s equations that electric produced electro- and magnetic fields in an electromagnetic wave are magnetic waves, sent them through space, and perpendicular to each other, and to the direction of measured their wave- propagation. It appears reasonable, say from ourRUDOLF length and speed. He discussion of the displacement current. Consider showed that the nature Fig. 8.2. The electric field inside the plates of the capacitor of their vibration, is directed perpendicular to the plates. The magnetic reflection and refraction field this gives rise to via the displacement current is was the same as that ofHERTZ along the perimeter of a circle parallel to the capacitor light and heat waves, plates. So B and E are perpendicular in this case. This establishing their identity for the first time. is a general feature. He also pioneered In Fig. 8.3, we show a typical example of a plane research on discharge of electromagnetic wave propagating along the z direction electricity through gases, (the fields are shown as a function of the z coordinate, at and discovered the(1857–1894) a given time t). The electric field Ex is along the x-axis, photoelectric effect. and varies sinusoidally with z, at a given time. The magnetic field By is along the y-axis, and again varies sinusoidally with z. The electric and magnetic fields Ex and By are perpendicular to each other, and to the direction z of propagation. We can write Ex and By as follows: Ex= E0 sin (kz–wt) [8.7(a)] By= B0 sin (kz–wt) [8.7(b)] Here k is related to the wave length FIGURE 8.3 A linearly polarised electromagnetic wave, l of the wave by the usual propagating in the z-direction with the oscillating electric field E equation along the x-direction and the oscillating magnetic field B along the y-direction. 2 π k = (8.8) 206 λ Reprint 2025-26 Electromagnetic Waves 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. The speed of propagation of the wave is (ω/k). Using Eqs. [8.7(a) and (b)] for Ex and By and Maxwell’s equations, one finds that ω = ck, where, c = 1/ µ0ε0 [8.9(a)] The relation ω = ck is the standard one for waves (see for example, Section 14.4 of class XI Physics textbook). This relation is often written in terms of frequency, ν (=ω/2π) and wavelength, λ (=2π/k) as  2π  2 πν = c  λ  or νλ = c [8.9(b)] 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 B0 = (E0/c) (8.10) 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. 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. But what if a material medium is actually there? We know that light, an electromagnetic wave, does propagate through glass, for example. 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 ε0 and µ0 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, 1 v = µε (8.11) Thus, the velocity of light depends on electric and magnetic properties of the medium. We shall see in the next chapter that the refractive index of one medium with respect to the other is equal to the ratio of velocities of light in the two media. 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 3×108 m/s. The constancy of the velocity of em 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. The great technological importance of electromagnetic waves stems from their capability to carry energy from one place to another. 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. 207 Reprint 2025-26 Physics Example 8.1 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 ˆj V/m. What is B at this point? Solution Using Eq. (8.10), the magnitude of B is E B = c 6.3 V/m –8 = 8 = 2.1 × 10 T 3 × 10 m/s 8.1 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, (+ ˆj ) × (+ ˆk ) = ˆi , B is along the z-direction. EXAMPLE Thus, B = 2.1 × 10–8 ˆk T Example 8.2 The magnetic field in a plane electromagnetic wave is given by By = (2 × 10–7) T sin (0.5×103x+1.5×1011t). (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field. Solution (a) Comparing the given equation with   x t   By=B0 Sin 2p +  λ T  spectrum  2π We get, λ = 3 m = 1.26 cm, 0.5 × 10 1 11 and = ν= 1.5 × 10 /2 π = 23.9 GHz T ( ) 8.2 (b) E0 = B0c = 2×10–7 T × 3 × 108 m/s = 6 × 101 V/mElectromagnetic http://www.fnal.gov/pub/inquiring/more/light http://imagine.gsfc.nasa.gov/docs/science/ 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 EXAMPLE Ez = 60 sin (0.5 × 103x + 1.5 × 1011 t) V/m 8.4 ELECTROMAGNETIC SPECTRUM At the time Maxwell predicted the existence of electromagnetic waves, the only familiar electromagnetic waves were the visible light waves. 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. We now know that, electromagnetic waves include visible light waves, X-rays, gamma rays, radio waves, microwaves, ultraviolet and infrared waves. The classification of em waves according to frequency is the electromagnetic spectrum (Fig. 8.4). 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. We briefly describe these different types of electromagnetic waves, in 208 order of decreasing wavelengths. Reprint 2025-26 Electromagnetic Waves FIGURE 8.4 The electromagnetic spectrum, with common names for various part of it. The various regions do not have sharply defined boundaries. 8.4.1 Radio waves Radio waves are produced by the accelerated motion of charges in conducting wires. They are used in radio and television communication systems. 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. 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. 8.4.2 Microwaves Microwaves (short-wavelength radio waves), with frequencies in the gigahertz (GHz) range, are produced by special vacuum tubes (called klystrons, magnetrons and Gunn diodes). 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, tennis- serves, and automobiles. 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 209the molecules. This raises the temperature of any food containing water. Reprint 2025-26 Physics 8.4.3 Infrared waves 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. 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, CO2, NH3, also absorb infrared waves). 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. Incoming visible light (which passes relatively easily through the atmosphere) is absorbed by the earth’s surface and re- radiated as infrared (longer wavelength) radiations. 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. 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. 8.4.4 Visible rays 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 × 1014 Hz to about 7 × 1014 Hz or a wavelength range of about 700 – 400 nm. 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 utraviolet. 8.4.5 Ultraviolet rays It covers wavelengths ranging from about 4 × 10–7 m (400 nm) down to 6 × 10–10m (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. 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 (Laser- assisted in situ keratomileusis) eye surgery. UV lamps are used to kill germs in water purifiers. Ozone layer in the atmosphere plays a protective role, and hence its depletion by chlorofluorocarbons (CFCs) gas (such as freon) is a matter 210 of international concern. Reprint 2025-26 Electromagnetic Waves 8.4.6 X-rays 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. 8.4.7 Gamma rays They lie in the upper frequency range of the electromagnetic spectrum and have wavelengths of from about 10–10m to less than 10–14m. 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 regions is not sharp and there are overlaps. TABLE 8.1 DIFFERENT TYPES OF ELECTROMAGNETIC WAVES Type Wavelength range Production Detection Radio > 0.1 m Rapid acceleration and Receiver’s aerials decelerations of electrons in aerials Microwave 0.1m to 1 mm Klystron valve or Point contact diodes magnetron valve Infra-red 1mm to 700 nm Vibration of atoms Thermopiles and molecules Bolometer, Infrared photographic film Light 700 nm to 400 nm Electrons in atoms emit The eye light when they move from Photocells one energy level to a Photographic film lower energy level Ultraviolet 400 nm to 1nm Inner shell electrons in Photocells atoms moving from one Photographic film energy level to a lower level X-rays 1nm to 10–3 nm X-ray tubes or inner shell Photographic film electrons Geiger tubes Ionisation chamber Gamma rays <10–3 nm Radioactive decay of the -do- nucleus 211 Reprint 2025-26 Physics SUMMARY 1. Maxwell found an inconsistency in the Ampere’s law and suggested the existence of an additional current, called displacement current, to remove this inconsistency. This displacement current is due to time-varying electric field and is given by dΦΕ di = ε0 dt and acts as a source of magnetic field in exactly the same way as conduction current. 2. An accelerating charge produces electromagnetic waves. An electric charge oscillating harmonically with frequency n, produces electromagnetic waves of the same frequency n. An electric dipole is a basic source of electromagnetic waves. 3. Electromagnetic waves with wavelength of the order of a few metres were first produced and detected in the laboratory by Hertz in 1887. He thus verified a basic prediction of Maxwell’s equations. 4. Electric and magnetic fields oscillate sinusoidally in space and time in an electromagnetic wave. The oscillating electric and magnetic fields, E and B are perpendicular to each other, and to the direction of propagation of the electromagnetic wave. For a wave of frequency n, wavelength l, propagating along z-direction, we have E = Ex (t) = E0 sin (kz – w t )   z     z t   = E0 sin 2 π  λ − νt  = E 0 sin 2 π  λ − T   B = By(t) = B0 sin (kz – w t)   z     z t   = B 0 sin 2 π  λ − νt  = B 0 sin  2 π  λ − T   They are related by E0/B0 = c. 5. The speed c of electromagnetic wave in vacuum is related to m0 and e0 (the free space permeability and permittivity constants) as follows: c = 1/ µ0 ε0 . The value of c equals the speed of light obtained from optical measurements. Light is an electromagnetic wave; c is, therefore, also the speed of light. Electromagnetic waves other than light also have the same velocity c in free space. The speed of light, or of electromagnetic waves in a material medium is given by v = 1/ µε where m is the permeability of the medium and e its permittivity. 6. The spectrum of electromagnetic waves stretches, in principle, over an infinite range of wavelengths. Different regions are known by different names; g-rays, X-rays, ultraviolet rays, visible rays, infrared rays, microwaves and radio waves in order of increasing wavelength from 10–2 Å or 10–12 m to 106 m. They interact with matter via their electric and magnetic fields which set in oscillation charges present in all matter. The detailed interaction and so the mechanism of absorption, scattering, etc., depend on the wavelength of the electromagnetic wave, and the nature of the atoms and molecules 212 in the medium. Reprint 2025-26 Electromagnetic Waves POINTS TO PONDER 1. The basic difference between various types of electromagnetic waves lies in their wavelengths or frequencies since all of them travel through vacuum with the same speed. Consequently, the waves differ considerably in their mode of interaction with matter. 2. Accelerated charged particles radiate electromagnetic waves. The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates. Thus, gamma radiation, having wavelength of 10–14 m to 10–15 m, typically originate from an atomic nucleus. X-rays are emitted from heavy atoms. Radio waves are produced by accelerating electrons in a circuit. A transmitting antenna can most efficiently radiate waves having a wavelength of about the same size as the antenna. Visible radiation emitted by atoms is, however, much longer in wavelength than atomic size. 3. Infrared waves, with frequencies lower than those of visible light, vibrate not only the electrons, but entire atoms or molecules of a substance. This vibration increases the internal energy and consequently, the temperature of the substance. This is why infrared waves are often called heat waves. 4. The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun. It is because humans have evolved with visions most sensitive to the strongest wavelengths from the sun. EXERCISES