Contents
Physics Topics can help us understand the behavior of the natural world around us.
What is the Speed of an Electromagnetic Wave in Space?
i) Electric field \(\vec{E}\) and magnetic field \(\vec{B}\) are perpendicular to the direction of propagation of the wave. Hence electro-magnetic wave is transverse in nature.
ii) Magnetic field being a varying field, an electric field is induced (following Faraday’s laws of electromagnetic induction) perpendicular to it.
At the same time, varying electric field induces magnetic field (following Maxwell’s law of displacement current).
iii) Let an electromagnetic wave propagate along positive x direction [Fig.]. In that case, \(\vec{E}\) field and \(\vec{B}\) field oscillate parallel to y and z -axes. Expressions for \(\vec{E}\) and \(\vec{B}\) fields at a distance x from the origin 0, are given in scalar form, by
E = E0 sin(ωt – kx) ……. (1)
and B = B0 sin(ωt – kx) …… (2)
where E0, B0, are the amplitudes of electric field and mag-netic field, while ω and k are the angular frequency and the magnitude of propagation vector, respectively.
It is to be noted that none of the two fields exists inde-pendently of the other. A time varying magnetic field induces an electric field. Simultaneously, a time varying electric field induces a magnetic field. Thus, self-sustaining electromagnetic waves travel through space.
Fig. depicts an em wave with its components along the y and z -axes oscillating in the xy -plane and zx -plane and representing equations (1) and (2) respectively. They represent the field vectors in magnitude and direction at different distances from origin at any time t.
iv) Equations (1) and (2) show that the frequencies of the electric and magnetic fields are equal and these two fields are always in same phase.
v) Speed of electromagnetic wave in free space,
c = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\) …… (3)
where µ0 and ε0 are respectively permeability and permittivity of free space.
Here, µ0 = 4π × 10-7 N ᐧ s2 ᐧ C-2
ε0 = 8.854 × 10-2 C2 ᐧ N-1 ᐧ m-2
Substituting the values in equation (3) we get,
c = \(\frac{1}{\sqrt{4 \pi \times 10^{-7} \times 8.854 \times 10^{-12}}}\) = 2,998 × 108
≈ 3.0 × 108m ᐧ s-1
In addition, velocity of an electromagnetic wave can also be obtained from the amplitudes of electric and magnetic fields as,
\(\frac{E_0}{B_0}\) = c …. (4)
From equations (1) and (2) we get,
\(\frac{E}{B}\) = \(\frac{E_0}{B_0}\) = c
vi) If the frequency and the wavelength of an electromagnetic wave in a medium are respectively f and λ, then in that medium the velocity of the wave will be fλ. If the medium changes, the wavelength as well as the velocity will change, but frequency will remain the same, because frequency is a characteristic of the source.
Some Quantities Related to EM Waves
Energy density: We have seen that electric field, E = E0sin(ωt – kx) and magnetic field, B = B0sin(ωt – kx).
Consider a volume element dV of the medium in which an electromagnetic wave is propagating. At a certain moment, the energy carried by the volume is U. As the energy stored is due to both electric and magnetic field, we have
U = UE + UM
Here, UE = stored energy due to electric field in the volume element dV
= \(\frac{1}{2} \epsilon_0 E^2 \cdot d V\)
UM = stored energy due to magnetic field in the volume element dV
= \(\frac{1}{2} \frac{B^2}{\mu_0} \cdot d V\)
So, at any position x of electromagnetic wave at time t, energy density of the electric field,
uE = \(\frac{U_E}{d V}\) = \(\frac{1}{2} \epsilon_0 E^2\) = \(\frac{1}{2} \epsilon_0 E_0^2 \sin ^2(\omega t-k x)\)
The energy stored in unit volume of the space between two plates of a charged air capacitor, uE = \(\frac{1}{2} \epsilon_0 E^2\). We can prove that the total stored energy of the charged capacitor = UE × volume of the space between two plates of capacitor = \(\frac{1}{2} C V_{P d}^2\), where, C = capacitance of the capacitor and Vpd = potential difference between the plates of the capacitor, and the energy density of the magnetic field,
uM = \(\frac{U_M}{d V}\) = \(\frac{1}{2 \mu_0} B^2\) = \(\frac{1}{2 \mu_0} B_0^2 \sin ^2(\omega t-k x)\)
The above equation is used to calculate the energy, stored in an inductor.
Average value of sin2θ or cos2θ in a complete cycle = \(\frac{1}{2}\)
It implies that the energy is equally distributed between electric and magnetic fields during the propagation of electromagnetic waves.
The average energy density of electromagnetic wave,
\(\bar{u}\) = \(\bar{u}_E+\bar{u}_M\) = \(\frac{1}{4} \epsilon_0 E_0^2\) + \(\frac{1}{4} \epsilon_0 E_0^2\) = \(\frac{1}{2} \epsilon_0 E^2\)
For a complete cycle, the average value can be written as u instead of \(\bar{u}\).
Hence energy density,
u = \(\frac{1}{2} \epsilon_0 E^2\) = \(\frac{1}{2 \mu_0} B_0^2\) …. (1)
In general, for an Isotropic medium of permittivily ε, the energy density, u = \(\frac{1}{2} \in E_0^2\), where ε = kε0, k being the dielectric constant of the medium.
Intensity: In an electromagnetic field, if we consider a unit area around a point, perpendicular to the direction of radiation, then the electromagnetic energy incident on that area per second is called intensity of electromagnetic radiation at that point.
Let P is a point in an electromagnetic field. The wave propagates toward the point P with velocity c. A unit area is considered around P, which is perpendicular to the direction of propagation of radiation.
In time dt, the wave travels a distance equal to cdt in the direction of propagation. Imagine a cylinder of length cdt and unit cross sectional area on the path of the wave such that it crosses the cylinder normally. The energy of electromagnetic wave incident at the point P is equal to the energy stored in the volume of the imaginary cylinder (cdt × 1 = cdt) . The intensity of radiation at P is equal to the energy incident per second i.e., the energy stored in the volume, c × 1 = c.
So, the energy contained in that cylinder = cu, where u = energy density.
Therefore, the intensity of electromagnetic wave at the point P, by definition
I = Cu
By using the equation (1), we get,
I = cu = \(\frac{1}{2} c \epsilon_0 E_0^2\) = \(\frac{1}{2 \mu_0} c B_0^2\) …. (2)
Incidentally, unlike in mechanical waves, this energy flow per unit area, per unit time, called energy flux is denoted by S, not I.
Hence, S = \(\frac{d u}{A d t}\) = cu
Average value \(\vec{S}\) is given by \(\vec{S}\) = \(\frac{1}{2 \mu_0} E_0 B_0\) = \(\frac{1}{2} \epsilon_0 c E_0^2\)
\(\frac{1}{2} \epsilon_0 c E_0^2\)
The direction of \(\vec{S}\) is given by the direction of propagation of the wave, as shown in Fig.
If the point P is considered in an isotropic medium of electric permittivity ε and permeability µ instead of free space, then the equation (2) will be
I = vu = \(\frac{1}{2} v \in E_0^2\) = \(\frac{1}{2 \mu} v B_0^2\)
where, v = \(\frac{1}{\sqrt{\epsilon \mu}}\) = velocity of the electromagnetic radiation in that medium.
Unit of the intensity of electromagnetic radiation
= J ᐧ m-2 ᐧ s-1 = W ᐧ m-2
Its dimension = \(\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~L}^2 \mathrm{~T}}\) = MT-3
Radiation Pressure: EM wave has linear momentum as well as energy. This means that radiation can exert force and hence pressure on any surface on which it is incident during its propagation through a medium. The force exerted on unit area of the surface, perpendicular to the direction of propagation is called radiation pressure.
From thermodynamical analysis, we can show that (this analysis is beyond our syllabus),
Radiation pressure, p = \(\frac{1}{3}\)u. where, u = energy density of electromagnetic radiation [see the equation (1)].
The unit of pressure (p) = N ᐧ m-2; the unit of energy density (u) = J ᐧ m-3.
Generally, both the units are same as N ᐧ m = J.
The dimension of both p and u = M L-1T-2.
The force due to the radiation pressure is too small to detect under everyday circumstances. In subjects like astronomy or astrodynamics related works, its importance cannot be Ignored. For example, if the radiation pressure of the sun had been ignored, the space crafts, viking-1 and viking -2 (sent by NASA to the Mars’s orbit for collecting information about Mars) would have missed Mars’s orbit by about 15000 km.
Numerical Examples
Example 1.
If the velocity of em wave in vacuum is 3 × 108 m ᐧ s-1 and the magnetic permeability of vacuum is 4π × 10-7 N ᐧ C-2 ᐧ s2, find the electric permittivity of vacuum.
Solution:
If permittivity and permeability of free space are ε0 and µ0, respectively,
c = \(\frac{1}{\sqrt{\epsilon_0 \mu_0}}\) or, c2 = \(\frac{1}{\epsilon_0 \mu_0}\)
∴ ε0 = \(\frac{1}{\mu_0 c^2}\) = \(\frac{1}{4 \pi \times 10^{-7} \times 9 \times 10^{16}}\) [∵ µ0 = 4π × 10-7N ᐧ C-2 ᐧ s2 and c = 3 × 108 m ᐧ s-1]
= 8.842 × 10-12C2 ᐧ N-1 ᐧ m-2
Example 2.
Ampiltude of electric field of a plane electromagnetic wave is 48 V ᐧ m-1. What is the amplitude of the magnetic field of the wave?
Solution:
In this case, amplitude of electric field,
E0 = 48V ᐧ m-1
As \(\frac{E_0}{B_0}\) = c, amplitude of magnetic field, B0 = \(\frac{E_0}{c}\)
∴ B0 = \(\frac{48}{3 \times 10^8}\) [∵ c = 3 × 108 m ᐧ s-1]
= 16 × 10-8 Wb ᐧ m-2
Example 3.
Electric field of an electromagnetic we is, E = 10-5 sin( 12 × 1015t – 4 × 107x). Find the frequency, velocity, wavelength of the wave. Write the equation of the magnetic field corresponding to this wave. Assume all quantities in SI unit.
Solution:
Electric field of the given electromagnetic wave,
E = 10-5sin(12 × 1015t – 4 × 107x)
Comparing with the expression for electric field,
E = E0sin(ωt – kx) we get,
ω = 12 × 1015 rad ᐧ s-1
∴ 2πf = 12 × 1015
or, f = \(\frac{12 \times 10^{15}}{2 \pi}\) = 1.9 × 1015Hz
Hence, frequency of wave, f = 1.9 × 1015 Hz
Velocity of wave,
v = \(\frac{\omega}{k}\) = \(\frac{12 \times 10^{15}}{4 \times 10^7}\) = 3 × 108 m ᐧ s-1
Wavelength, λ = \(\frac{c}{f}\) = \(\frac{3 \times 10^8}{1.9 \times 10^{15}}\) = 1.58 × 10-7 m
Equation of the corresponding magnetic field,
B = B0sin(ωt – kx) = \(\frac{E_0}{c}\)sin(ωt – kx) [∵ \(\frac{E_0}{B_0}\) = c]
= \(\frac{10^{-5}}{3 \times 10^8}\)sin(12 × 105t – 4 × 107x)
= 3.33 × 10-14 sin(12 × 1015t – 4 × 107x)T
Example 4.
In an electromagnetic field, the amplitude of electric field at a point is 3V ᐧ m-1. Calculate energy density and Intensity of the wave at that point. Given µ0 = 4π × 10-7H ᐧ m-1
Solution:
The amplitude of e’lectric field E0 = 3V ᐧ m-1
B0 = \(\frac{E_0}{c}\) = \(\frac{3}{3 \times 10^8}\) = 10-8Wb ᐧ m-2
So, the energy density,
u = \(\frac{1}{2 \mu_0} B_0^2\) = \(\frac{1}{2 \times 4 \pi \times 10^{-7}}\) × (10-8)2
= 3.98 × 10-11J ᐧ m-3
Intensity, I = cu = (3 × 108) × (3.98 × 10-11)
= 0.012 W ᐧ m-2
Example 5.
A rectangular parallel plate capacitor of dimension 5 cm × 4 cm is charging in such away that the rate of change of electric field between the two plates is 5.65 × 1011V ᐧ m ᐧ s-1. Calculate the displacement current for this capacitor.
Given ε0 = 8.85 × 10-12 F.m-1
Solution:
Area of rectangular plate,
A = (5 × 4) cm2 = 20 cm2 = 0.002 m2
If E be the electric field between two plates, then the electric flux,
ϕE = EA; \(\frac{d \phi_E}{d t}\) = A\(\frac{d E}{d t}\)
Therefore, the displacement current,
Id = ε0\(\frac{d \phi_E}{d t}\) = ε0A\(\frac{d E}{d t}\)
= 0.01 A = 10 mA
Example 6.
Calculate the Intensity and the rms value of electric field of electromagnetic wave at a distance 10 m from a 100 W electric bulb. Given ε0 = 8.85 × 10-12 F ᐧ m-1
Solution:
Area of a spherical surface of radius 10 m
= 4π(10)2 = 400πm2
∴ Intensity, I = \(\frac{100 \mathrm{~W}}{400 \pi \mathrm{m}^2}\) = \(\frac{1}{4 \pi}\) W ᐧ m-2 = 0.08W ᐧ m-2
Here, I = \(\frac{E_0}{\sqrt{2}}\) = \(c \epsilon_0 E_{\mathrm{rms}}^2\) [∵ Erms = \(\frac{E_0}{\sqrt{2}}\)]
∴ Erms = \(\sqrt{\frac{I}{c \epsilon_0}}\) = \(\sqrt{\frac{0.08}{\left(3 \times 10^8\right) \times\left(8.85 \times 10^{-12}\right)}}\)
= 5.5 V ᐧ m-1