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 \( 6*10^{14} \), 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), 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 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 electromagnetic waves over distances of many kilometres. Marconi’s experiment marks the beginning of the field of communication using electromagnetic waves.
8.3.2 Nature of electromagnetic waves
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. 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. The magnetic field this gives rise to via the displacement current is along the perimeter of a circle parallel to the capacitor plates. 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). 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 \( B_{y} \) are perpendicular to each other, and to the direction z of propagation. We can write \( E_{x} \) and \( B_{y} \) as follows:
\[ E_{x} = E_{0} sin(kz-\omega t) \] \[ B_{y} = B_{0} sin(kz- \omega t) \]
Here k is related to the wave length \( \lambda \) of the wave by the usual equation
\[ k = \frac{2\pi}{\lambda} \]
and \( \omega \) 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 \( \omega / k \). Using Eqs. [ 8.7(a) and (b) ] for \( E_{x} \) and \( B_{y} \) and Maxwell’s equations, one finds that
\( \omega = ck \), where, \( c = \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}} \)
The relation \( \omega = ck \) 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, \( v (= \omega /2\pi) \) and wavelength, \( \lambda (=2\pi/k) \)
\[ 2\pi v = c \left( \frac{2\pi}{\lambda} \right) \] \( or, v\lambda = c \)
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
\[ B_{0} = (E_{0}/c) \]
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. Sound waves in air are longitudinal waves of compression and rarefaction. Transverse elastic (sound) waves can also propagate in a solid, which is rigid and that resists shear. 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. 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! We now accept that no such physical medium is needed. The famous experiment of Michelson and Morley in 1887 demolished conclusively the hypothesis of ether. Electric and magnetic fields, oscillating in space and time, can sustain each other in vacuum.
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 \( \varepsilon \) and a magnetic permeability \( \mu \) (these describe the factors by which the total fields differ from the external fields). These replace \( \varepsilon_{0} \) and \( \mu_{0} \) in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity \( \varepsilon \) and magnetic permeability \( \mu \) , the velocity of light becomes,
\[ v = \frac{1}{\sqrt{\mu \varepsilon}} \]
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 \( 3x10^8 m/s \). 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. Namely, the metre is now defined as the distance travelled by light in vacuum in a time \( (1/c) seconds = (2.99792458 × 10^8 )^{–1} seconds \) . 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. 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 \( 2.9979246 × 10^8 m/s \). 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. 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. 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 \( \nu = v \lambda \) and found that the waves travelled with the same speed as the speed of light.
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. 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.
Do electromagnetic waves carry energy and momentum like other waves? Yes, they do. We have seen in chapter 2 that in a region of free space with electric field E, there is an energy density \( (\varepsilon_{0}E^2/2) \). Similarly, as seen in Chapter 6, associated with a magnetic field B is a magnetic energy density \( B^2/2\mu_{0} \). As electromagnetic wave contains both electric and magnetic fields, there is a non-zero energy density associated with it. 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. 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. 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,
\[ p = \frac{U}{c} \]
When the sun shines on your hand, you feel the energy being absorbed from the electromagnetic waves (your hands get warm). 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. In 1903, the American scientists Nicols and Hull succeeded in measuring radiation pressure of visible light and verified Eq. (8.12) . It 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 \( 7 × 10^{–9} N\) .
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.
Example 8.2
A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, \( E = 5.3 \hat{j}~\)V/m . What is B at this point?
Using Eq. (8.10), the magnitude of B is \[ B = \frac{E}{c} \] \[ =\frac{6.3 \mathrm{~V} / \mathrm{m}}{3 \times 10^{8} \mathrm{~m} / \mathrm{s}}=2.1 \times 10^{-8} \mathrm{~T} \]
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, (+ ˆ j ) × (+ ˆ k ) = ˆ i , B is along the z-direction. Thus,
\[ \mathbf{B}=2.1 \times 10^{-8} \mathbf{k} \mathrm{T} \]Example 8.3
The magnetic field in a plane electromagnetic wave is given by \(B_{y}=\left(2 \times 10^{-7}\right) \mathrm{T} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right)\)
(a) What is the wavelength and frequency of the wave?
(b) Write an expression for the electric field.
(a) Comparing the given equation with \[ B_{y}=B_{0} \sin \left[2 \pi\left(\frac{x}{\lambda}+\frac{t}{T}\right)\right] \] We get, \[ \lambda=\frac{2 \pi}{0.5 \times 10^{3}} \mathrm{~m}=1.26 \mathrm{~cm} \] and \[ \frac{1}{T}=v=\left(1.5 \times 10^{11}\right) / 2 \pi=23.9 \mathrm{GHz} \]
(b) \[ E_{0}=B_{0} c=2 \times 10^{-7} \mathrm{~T} \times 3 \times 10^{8} \mathrm{~m} / \mathrm{s}=6 \times 10^{1} \mathrm{~V} / \mathrm{m} \] 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
Example 8.4
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 \( 20cm^{2} \), find the average force exerted on the surface during a 30 minute time span.
The total energy falling on the surface is \[ \begin{aligned} U &=\left(18 \mathrm{~W} / \mathrm{cm}^{2}\right) \times\left(20 \mathrm{~cm}^{2}\right) \times(30 \times 60 \mathrm{~s}) \\ &=6.48 \times 10^{5} \mathrm{~J} \end{aligned} \] Therefore, the total momentum delivered (for complete absorption) is \[ p=\frac{U}{c}=\frac{6.48 \times 10^{5} \mathrm{~J}}{3 \times 10^{8} \mathrm{~m} / \mathrm{s}}=2.16 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \] The average force exerted on the surface is \[ F=\frac{p}{t}=\frac{2.16 \times 10^{-3}}{0.18 \times 10^{4}}=1.2 \times 10^{-6} \mathrm{~N} \] How will your result be modified if the surface is a perfect reflector?
Example 8.5
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.
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 \[ A=4 \pi r^{2}=4 \pi(3)^{2}=113 \mathrm{~m}^{2} \] The intensity I at this distance is \[ \begin{aligned} I=\frac{\text { Power }}{\text { Area }}=\frac{100 \mathrm{~W} \times 2.5 \%}{113 \mathrm{~m}^{2}} \\ =0.022 \mathrm{~W} / \mathrm{m}^{2} \end{aligned} \] Half of this intensity is provided by the electric field and half by the magnetic field. \[ \begin{aligned} \frac{1}{2} I=\frac{1}{2}\left(\varepsilon_{0} E_{r m s}^{2} c\right) & \\ &=\frac{1}{2}\left(0.022 \mathrm{~W} / \mathrm{m}^{2}\right) \\ E_{m s}=& \sqrt{\frac{0.022}{\left(8.85 \times 10^{-12}\right)\left(3 \times 10^{8}\right)}} \mathrm{V} / \mathrm{m} \\ =& 2.9 \mathrm{~V} / \mathrm{m} \end{aligned} \] 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 \[ \begin{array}{c} E_{0}=\sqrt{2} E_{\mathrm{rms}}=\sqrt{2} \times 2.9 \mathrm{~V} / \mathrm{m} \\ =4.07 \mathrm{~V} / \mathrm{m} \end{array} \] 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 \[ B_{m s}=\frac{E_{r m s}}{c}=\frac{2.9 \quad \mathrm{~V} \mathrm{~m}^{-1}}{3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}}=9.6 \times 10^{-9} \mathrm{~T} \] Again, since the field in the light beam is sinusoidal, the peak magnetic field is \(B_{0}=\sqrt{2} B_{r m s}=1.4 \times 10^{-8} \mathrm{~T}\). 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.