Physics · Ch 13 — Electromagnetic Waves and Communication System

Characteristics of EM waves

13.2.2

Characteristics of EM waves

EM waves share a common set of defining characteristics, all of which follow from Maxwell's equations. First, the electric field $\vec{E}$ and the magnetic field $\vec{B}$ are always perpendicular to each other and also perpendicular to the direction the wave travels -- so EM waves are transverse waves, never longitudinal. Second, the vector cross product $\vec{E}\times\vec{B}$ gives the direction in which the EM wave travels, and it also represents the energy carried by the wave (its intensity, in the form of the Poynting vector). Third, the $\vec{E}$ and $\vec{B}$ fields both vary sinusoidally with position and time, and they do so in phase -- reaching their maximum and minimum values simultaneously, at the same point in space. Fourth, EM waves are always produced by accelerated electric charges, as described in section 13.2.1. Fifth, unlike mechanical waves such as sound, EM waves need no material medium to travel through: they propagate through free space (vacuum) just as readily as through solids, liquids and gases.

Sixth, in free space every EM wave -- regardless of its frequency -- travels at the same fixed speed $c$ , the speed of light, given by

c=\dfrac{1}{\sqrt{\mu_0\epsilon_0}}\approx 3\times10^{8}\ \text{m/s}

where $\mu_0$ (= $4\pi\times10^{-7}$ T.m/A) is the permeability of free space and $\epsilon_0$ (= $8.85\times10^{-12}$ C $^2$ /N.m $^2$ ) is the permittivity of free space -- so the speed of light turns out to be nothing more than a combination of two purely electric and magnetic constants of empty space. Seventh, when an EM wave travels through a material medium instead of vacuum, its speed $v_m$ is reduced according to $v_m=\dfrac{1}{\sqrt{\mu\epsilon}}$ , where $\mu$ and $\epsilon$ are the permeability and permittivity of that particular medium (both are generally larger than $\mu_0$ and $\epsilon_0$ , so $v_m<c$ ). Eighth, EM waves obey the principle of superposition, exactly like other waves: when two or more EM waves overlap, their fields add vectorially.

Ninth, the ratio of the amplitudes of the electric and magnetic fields is constant at every point of the wave and equals the wave's speed:

|\vec{E}|=c\,|\vec{B}| \quad\Longrightarrow\quad \dfrac{|\vec{E}_0|}{|\vec{B}_0|}=c \qquad \text{--- (13.1)}

where $E_0$ and $B_0$ are the amplitudes of $\vec{E}$ and $\vec{B}$ respectively. Tenth, because the electric field vector $\vec{E}_0$ is, in practice, the component of the wave responsible for essentially all optical effects (it is what interacts most strongly with matter and with our eyes), it is conventionally called the light vector, even though the magnetic field is present with an amplitude fixed by the same ratio. Eleventh, the intensity of a wave is proportional to the square of its amplitude, so the electric and magnetic contributions to intensity are

I_E\propto E_0^{2}, \qquad I_B\propto B_0^{2} \qquad \text{--- (13.2)}

Twelfth, and following directly from equations (13.1) and (13.2), the energy carried by an EM wave is shared equally between its electric and magnetic fields, so $I_E=I_B$ at every point -- neither field carries more of the wave's energy than the other. …

Misc Ex.13.1Velocity of EM waves in vacuum

Worked out. Asks for the numerical value of the velocity of an EM wave in free space. The method substitutes the standard values of the permittivity of free space (epsilon_0 = 8.85x10^-12 C^2/N.m^2) and the permeability of free space (mu_0 = 4pi10^-7 T.m/A) into the formula c = 1/sqrt(mu_0 epsilon_0), evaluating the product mu_0*epsilon_0 first and then taking the reciprocal square root to obtain c approximately equal to 3.00x10^8 m/s -- the familiar speed of light in vacuum, confirming that this speed is a direct consequence of two …

Misc Ex.13.2Magnetic field amplitude and direction from a given electric field

Worked out. An EM wave of frequency 28 MHz travels along the x-direction in free space, with its electric field amplitude E = 9.6 V/m directed along the y-axis. The problem asks for the amplitude and direction of the associated magnetic field B. The method first finds the magnitude of B from the fixed amplitude ratio B = E/c (using c = 3x10^8 m/s), then finds its direction by requiring that E cross B point along the known propagation direction (+x-axis), using the vector identity (+j) x (+k) = i to conclude B must point along the +z-axi …

Misc Ex.13.3Ratio of intensities of two light beams of different amplitude

Worked out. A beam of red light has an amplitude 2.5 times that of a second beam of the same colour. The problem asks for the ratio of the intensities of the two waves. The method uses the intensity-amplitude relation Intensity is proportional to (Amplitude)^2, writing I_2 proportional to a^2 and I_1 proportional to (2.5a)^2, and dividing to get I_1/I_2 = (2.5)^2 = 6.25, so the brighter beam is 6.25 times as intense as the dimmer one even though its amplitude is only 2.5 times …

Misc Ex.13.4Magnetic field vector at a point and instant, given the electric field vector

Worked out. An EM wave of frequency 50 MHz travels in vacuum along the positive x-axis. At a particular point x and instant t, the electric field is given as E = 9.6*j V/m (i.e. 9.6 V/m along +y). The problem asks for the magnitude and direction of the magnetic field B at that same point and instant. The method computes the magnitude from B = E/c = 9.6/(3x10^8) = 3.2x10^-8 T, then fixes the direction using the fact that the wave travels along +x while E points along +y, so B must point along +z (k) for E x B to point along +x, giving the final answer B = 3.2x10^ …

Misc Ex.13.5Wavelength and electric-field expression from a given magnetic-field oscillation

Worked out. For an EM wave propagating along the x-direction, the magnetic field oscillates along the z-direction at a frequency of 3x10^10 Hz with amplitude 10^-9 T. The problem has two parts: (a) find the wavelength of the wave, and (b) write the expression representing the corresponding electric field. The method finds wavelength from lambda = c/frequency = (3x10^8)/(3x10^10) = 10^-2 m, then finds the electric-field amplitude from E_0 = c*B_0 = (3x10^8)(10^-9) = 0.3 V/m, and finally -- since B oscillates along z while the wave travels along x -- concludes E must oscillate along y (so that E x B points along x), writing the full sinusoidal expression E_y = E_0 sin(kx - omega t) with the numeric k and omega subst …

Misc Ex.13.6Peak electric force on a moving charge from an EM wave's magnetic field expression

Worked out. The magnetic field of an EM wave travelling along the x-axis is given as B = k4x10^-4sin(omega t - kx) tesla. The problem asks for the peak value of the electric force acting on a particle of charge 5 micro-coulomb travelling with velocity 5x10^5 m/s along the y-axis. The method reads off the magnetic field amplitude B_0 = 4x10^-4 T directly from the given expression, computes the corresponding electric field amplitude E_0 = cB_0 = (3x10^8)(4x10^-4) = 12x10^4 N/C, and then computes the maximum electric force on the charge as F = qE_0 = (5x10^-6)(12x10^4) = 0.6 N, treating only the wave's electric-field component as responsible for the peak electric force (the given velocity of the charge is not needed for this particular part of the calculation, since it is the electric …

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