Chapter 34

Electromagnetic Waves

Level I

Recall that anywhere there is an electric charge, there exists an electric field around it throughout space.  If the charge moves, the electric field (or any field-line) around it moves as well.  The motion of the charge causes change in the electric field intensity (E) at any point in space that varies with distance and time.  Also, recall that the motion of a charge creates a magnetic field (B) that is perpendicular to the direction of motion of the charge.  The important aspect is that the two effects (E) and consequently (B) turn out to be perpendicular to each other and they coexist.

Now, if the motion of the charge is of oscillatory nature [with a sinusoidal equation: y = yo sin(ωt) ], an up-and-down motion for example,  the variations of E and B will also be sinusoidal as well with equations of the type

E = Eo sin (ωt)      and      B = Bo sin (ωt)      where      ω = 2πf.

Since such oscillations cause other charges at some distance away ( no matter how far ) to oscillate accordingly, we believe that oscillations of an electric charge creates waves that propagate through space.  The graphical representation of the propagation of such waves called the "electromagnetic waves" through space may be pictured as shown below. This is the way fields variations are sensed by a distant charge as the waves pass by it.

Note that E is much greater than B.  In fact E = cB where c is the speed of light (3.00x108 m/s in vacuum). The scale on the E-axis is therefore much greater than the scale on the B-axis ( an order of 108 times greater).  This means that the magnetic effect is much weaker than the electric effect.  Also, this figure shows one propagation direction only.  

Note that the E and  B that reach a distant charge as a result of E&M waves propagation cause that distant charge to oscillate accordingly.  The transmission of this effect is very fast but not instant.  The speed of propagation of charge oscillations, E&M Waves,  is measured to be 3.00x108 m/s in vacuum. This means 300,000 km/s or 186,000 miles /sec.  You may put 186,000 miles on your car in 8 to10 years.  Electromagnetic waves ( Light being one type of it ) travel that distance in one second!  All radio transmitters and cellular phone systems take advantage of E&M waves.  The trick is to mount sound waves (voice) onto the E&M waves (called modulation) and send them at the speed of light ( E&M waves).  This is the maximum possible speed according to the "Einstein's Theory of Relativity."

 

Wave Speed:

Recall that wave speed ( v ) is related to wavelength ( λ ) pronounced " lambda" and frequency ( f ) by

v = f λ

For electromagnetic E&M waves, letter ( c ) is commonly used for the wave speed.    c = f λ

Since c is a constant for any given medium, if f increases, then  λ  has to decrease in that medium.

Example 1: An AC source is running at a frequency of 60.0Hz.  This causes the current ( moving charges )  in wires connected to this generator to flow back-and-forth at that frequency.  The charge oscillations in such wires produce E&M waves that as we know propagate at the speed of light c = 3.00x108 m/s.  Find the wavelength, λ, for such waves.

Solution:  From c = f λ,   we get    λ  = c /f   =   (3.00x108 / 60.0 ) m   =   5,000,000 m   =  3100 miles

This is a very long wavelength and therefore very weak!  Only 60.0 of such waves pass by a given point in space every second. This means that if there is a charge at one point in space, it oscillates only 60.0 times per second as such waves keep coming to it. The shorter the wavelength, the more energetic the wave is or the more energy it carries.   Shorter wavelengths are associated with higher frequencies ( c = f λ ).   Higher frequencies make a distant charge to oscillate faster; thus, imparting more energy to that charge.

Example 2:  Waves transmitted or received by cell-phones have wavelengths of about 15 cm or 6.0 inches.  Calculate the frequency of such waves and express it in MHz.   Note that  MHz means Mega Hertz or Million Hertz.

Solution:  c = f λ    ;    f =  c / λ  =  (3.00x108 / 0.15 )s -1  =  2.00x109 Hz = 2000 MHz

Example 3:  White light is a mixture of a large number of different electromagnetic waves whose wavelengths range from about 400nm ( violet ) to about 700nm ( red ).  Find the corresponding frequency for each of these two limiting values in the visible range.

Solution: To be solved by students.

 

Speed of E&M Waves by Calculation:

By solving the wave equation ( a differential equation not shown here ), it is possible to show that speed of E&M waves is given by 

Solution: Use the values of  εo  and  μo  (Form previous chapters or the front or back cover of your text) to verify that  c = 3.00x108 m/s.

 

Energy carried by Electromagnetic Waves:

The energy carried by a wave is proportional to the square of its amplitude (A2).  For  E&M waves, it will be (Eo2), or (Bo2), or ( EoBo) where Eo and Bo are the maximum values of the electric and magnetic fields intensities.   Eo= cBo.  This energy transfer is expressed in units of Joules per second per meter squared through space.  Since Joules per second is watt, we may say that the energy carried by a wave is expressed in watts / m2.   It can be shown that the formula for maximum energy density carried by an electromagnetic wave is either of the three following forms:

Io = c εo Eo 2 = c Bo 2 / μo = Eo Bo / μo

For a continuous sinusoidal wave, we may calculate an RMS ( Root Mean Square) value.  Since  (rms) power is  1/2 of the max. power, we may write:        [subscript (o) denotes maximum value]

I rms = (1/2) c εo Eo 2 = c Bo 2 / 2μo = Eo Bo/2μo

Example 4:  A 10-kilowatt AM radio station is tuned on by a radio set at a distance of 5.0 miles = 8.0 km from the transmitter antenna.  Isotropic wave propagation in space means that waves are sent out by the transmitter uniformly in all directions.  This is not really the case with actual antennas, but for simplicity here, we suppose isotropic propagation of wave energy in space.  Assuming isotropic, calculate (a) the wave intensity (I rms) in watts / m2 at the 8.0-km radius, and (b) the magnitude of the electric and magnetic field strength ( Eo and Bo ) at that radius.

Solution: Visualize a huge sphere ( of 8000m radius or 5 miles ) that 10,000 watts of energy is to be distributed over its surface.  How much energy will every m2 of it receive?  This simple division will give us the value of  Irms. 

Irms = 10,000watts  / [ 4(3.14 x 80002 )m2 ] = 1.2 x 10-5 watts / m2.    (or 12 micro-watts per m2)

Since I rms   = (1/2)cεoEo2 , solving for Eo we get : Eo = [2I rms/cεo]1/2  =  9.4 x 10-3 V/m.

(Can you verify that Volt/meter is the same as N/Coul.?)

Since E = c B, we get:  Bo = Eo / c = (9.4x10-3 / 3.00x108)  =  3.1 x 10-11 T.

 

Test Yourself 1:     click here.

1) An electromagnetic wave is a result of the oscillation of (a) an electron  (b) a proton  (c) a neutron   (d) a & b.

2) When a charged particle moves, its electric field (a) moves accordingly  (b) remains unchanged  (c) varies as sensed by other charges elsewhere  (d) a & c.    click here.

3) If a charged particle oscillates, its equation of motion is (a) quadratic in time  (b) sinusoidal in time  (c) neither a nor b.

4) A particle oscillating in the y-direction at a frequency of (f) Hz and an amplitude of (A) meters follows (a) y=Asin(2πft)  (b) y=Asint  (c) y=Asin(ft).

5) If a charged particle oscillates, the ripples generated in its electric field follow (a) E =Eosin (ωt)  where ω = 2πf   (b) E = (1/2)Eot2 + ωt   (c) E = (1/2)Eot2.    click here.

6) Anywhere a charged particle moves, it generates a magnetic effect that is (a) parallel to its direction of motion  (b) perpendicular to its direction of motion  (c) neither a nor b.

7) The magnetic effect of an electromagnetic wave is (a) separable from its electric effect  (b) is not separable from its electric effect  (c) can only be separated at high frequency oscillations of a charged particle.

8) The magnetic effect of an E&M wave is (a) much stronger than its electric effect  (b) much weaker than its electric effect  (c) has the same strength as its electric effect.    click here.

9) The speed of E&M waves in vacuum is (a) the same as the speed of light  (b) 3.00x1010 cm/s  (c) 3.00x105 km/s  (d) 186,000mi/s  (e) a, b, & c.

10) Light is (a) an E&M wave  (b) a mechanical wave and cannot travel in vacuum  (c) is a longitudinal wave  (d) is a transverse wave  (e) a & d.    click here.

11) The formula for wave speed is (a) v = f λ     (b) v = ω λ     (a) v = f ω.

12) Frequency is defined as (a) the number of meters per second  (b) the number of ωs that occur per second  (c)  the number of wavelengths (full cycles) that occur per second.

13) Wavelength is (a) the distance between any two crests on a wave  (b) the distance between a crest to the next one on a wave  (c) the distance between a trough to the next one on a wave  (d) b & c.    click here.

14) The energy carried by a wave is proportional to (a) its amplitude, A  (b) its amplitude squared, A2  (c) neither a nor b.

15) When a charged particle oscillates at one point in space, other charges in space (a) oscillate instantly as a result  (b) will oscillate accordingly at some later time depending on their relative distances  (c) both a & b.    click here.

16) Since E&M waves move at a constant velocity in a medium with fixed properties (v = 300,000km/s in vacuum), they do not accelerate (a = 0), and the equation of motion for them is (a) x = (1/2)a t2 + vi t   (b) x =  vi t   (c) x = Rθ.

17) If a charge starts oscillating now here on the Earth, a charge that is on the Moon, an average distance of 384,000km away, will start oscillating (a)1.2 min. later   (b) 1.28s later  (c) one month later.    click here.

18) If a charge starts oscillating now here on the Earth, a charge that is on the Sun, an average distance of 150,000,000km away, will start oscillating (a)8.3 min. later   (b) 500s later  (c) both a & b.

19) A light year is the distance light travels in 1yr.   If a charge starts oscillating now here on the Earth, a charge that is on the star Alpha Centauri, 4 light-years away, will start oscillating (a) 3x108s. later   (b) 2 light-years later  (c) neither a nor b.

20) From the Earth, it takes a radio signal (An E&M wave) 5.0s to reach a space station an back.  The space station is (a) 1,500,000km away  (b) 3,000,000km away  (c) 750,000km away.    click here.

21) From the Earth, it takes a radio signal (An E&M wave) 5.0min. to reach a space station an back.  The space station is (a) 45,000,000km away  (b) 90,000,000km away  (c) 75,000,000km away.

22) The frequency of E&M waves used foe cellular phones is about 2000MHz.  This frequency is (a) 2x109Hz  (b) 2x106Hz  (c)2x1012Hz.    click here.

23) The wavelength of the E&M waves used for cellular phones is (a) 15cm  (b) 6.0in  (c) 0.15m  (d) a, b, & c.

24) The wavelength of a certain red light (of course, an E&M wave) is 680nm.  Its frequency is (a) 4.4x1014 s-1  (b)4.4x1014/s  (c)  4.4x1014Hz   (d) a, b, & c.

25) If you are solving for the frequency of an E&M wave, and you come up with f = 2.7x10-9/s, for example, (a) you accept the answer and think it must be correct  (b) you doubt the answer thinking that order of 10-9 is extremely small to be the frequency of an E&M wave  (c) you may think that a charge oscillating once every 109 seconds means practically motionless  (d) b & c.    click here.

26) The frequency of a certain violet light (of course, an E&M wave) is 7.3x1014/s.   Its wave length is (a) 4.1x10-9m   (b)41.0nm  (c) 410nm  (d) 160nm click here.

27) The wavelength of a wave is 750m in vacuum and it occurs 400,000 times per second.  The wave (a) has a speed of 3.0x108m/s  (b) has a speed of 3.0x105km/s  (c) is electromagnetic because only E&M waves can travel at that speed in vacuum  (d) a, b, & c.

28) The wavelength of a wave is 1500m in vacuum and it occurs 200,000 times per second.  The wave (a) has a speed of 3.0x108m/s  (b) has a speed of 1.86x105mi/s  (c) is electromagnetic because only E&M waves can travel at that speed in vacuum  (d) a, b, & c.

29) The wavelength of a wave is 3000m in vacuum and it occurs 100,000 times per second.  The wave (a) has a speed of 3.0x108m/s  (b) has a speed of 3.0x105km/s  (c) is electromagnetic because only E&M waves can travel at that speed in vacuum  (d) a, b, & c.    click here.

30) Ultraviolet rays (of course, E&M waves) have frequencies more than that of violet (fv = 7.5x1014/s).   An E&M wave of frequency 9.5x1014/s is of course UV and not visible.  It has a wavelength of (a) 3.2E-7m  (b) 320nm  (c) both a & b.

31) X- rays (of course, E&M waves) have frequencies more than that of ultraviolet (fUV > 7.5x1014/s).  An E&M wave of frequency 6.5x1016/s is of course of X-rays type, and not visible.  It has a wavelength of (a) 4.6E-9m  (b) 4.6nm  (c) both a & b.    click here.

32) Gamma rays (of course, E&M waves) have frequencies more than that of X-Rays (fX > 1016/s).   An E&M wave of frequency 5.0x1021/s is of course of Gamma type, not visible, and very penetrable.  It's wavelength is (a) 6.0x10-14m  (b) 60fm  (c) a & b.  Note: fm means femtometer that is10-15m.  click here.

33) In general, for E&M waves, the speed is constant (3.00x108m/s in vacuum).  An E&M wave of  (a) lower frequency has of course a greater wavelength  (b) higher frequency has of course a smaller wavelength  (c) both a & b   (d) neither a nor b.     click here.

 

Level II

Maxwell's Equations:

The combination of 4 laws or formulas form the "Maxwell's Equations" as follows:

We have so far studied (1) the Gauss's Law for an electric field as

If we apply Gauss's law to any closed surface in a magnetic field, what will the result be?  Think for a moment.  Magnetic field lines emerge from the N-pole of a magnet and enter its S-pole.  Remember that magnetic poles coexist.  This means that any B field-line that enters a closed surface, must exit that closed surface in order to return to back to the magnet.  Investigate this by drawing a bar-magnet with its field-lines and then once put a closed surface somewhere in its field-lines and once put the entire magnet inside that closed surface and see what conclusion you can draw.  By now you may have figured out the answer.

You are right, the net magnetic flux will be 0.  We may write the Gauss's law for a Magnetic field as

Maxwell used the Faraday's law of electromagnetic induction as his law (3) in the following form:

 

and finally, the modified Ampere law called the "Ampere-Maxwell Law" as shown below:

The above 4 equations must be satisfied for the completeness of the solution to any electromagnetic problem.  The details of formulas (3) and (4) are given below:

Faraday's Law of Magnetic Induction (3):

Once more, let's look at Faraday's Law of magnetic induction, simply

 V= -dφB/dt.  Note that φm or φB means magnetic flux. The greater and faster the change in φm, the greater the induced voltage, V.  Recall that when a magnet is moved toward or away from a closed loop of wire such that φm through the loop changes with time, an induced voltage develops in the loop.  Fig. (1) supports this statement.

In Fig. (1) on the right, as the bar magnet moves to the right, the changing B induces an electric field E normal to the B that results in the current I in the wire as sensed by the galvanometer.


In Fig. (2) on the right, a varying current is forced into a solenoid by a varying source that causes a varying magnetic field B. The varying magnetic field B generates an electric field E normal to itself such that

the line integral of E∙dℓ = -d φB/d t  

for any closed loop as shown.  The E field-lines around B in (2) are circular and the line integral of any closed loop around B results in -d φB/d t.   Faraday's formula can be shown in two forms as shown under (1) or (2) on the right.

The equation on the right is the 3rd equation listed under Maxwell's equations.

 

Ampere-Maxwell Law (4):

Displacement Current:

     When a capacitor is being charged, we know that there is a varying current in the wires that connect the battery to it; however, the current from one plate of the capacitor to the other is via electric filed lines or electric flux, φE as shown in Fig. (3) on the right.

Applying Ampere's law to the wire portion can calculate the current in the wire portion for us with no problem (Fig. 3) by integrating or measuring the magnetic field that the current produces around it.  Current I passes through the flat circular surface (Fig. 3) that is chosen for integration, and the closed loop is the circle around it along which the integral of B∙dℓ must equal μoI.

     Now, if we apply Ampere's law to the dielectric section, no provision is made in Ampere's formula for incorporating the fact that the varying electric flux in between the plates generates a measurable magnetic field.  Maxwell pointed out this shortcoming by using a semispherical surface as shown in (Fig. 4) for which the closed path of integration is still the same circle.  As can be seen, there is no actual current I flowing through the semispherical surface.  The wire ends at the left plate.  The current I in Fig. 3 crosses the flat circular surface, but in Fig. 4 it does not cross the semispherical surface.  This means that I = 0 for case 2, and the same integral of B∙dℓ equals 0 that is not true.  We know that current exists in between the plates but in the form of electric flux.

To clarify this contradiction, Maxwell named the current in between the plates as the "Displacement Current, ID."  He modified the Ampere's formula as follows:

The second term in parenthesis is derived as follows:

For a parallel-plates capacitor, Q = CV and since V = Ed, we get:

Q = CEd.    Since  C = εoA/d,  this makes Q = εoAE or since φE = AE, we get: Q = εoφE.   The rest is continued on the right.

 

Q = εoφE. (In between the plates)

As electric charge Q increases on the plates, φE increases and dQ/dt becomes:

dQ/dt = εodφE /dt and consequently:

ID = εodφE /dt.

 

Mathematical Form of Electromagnetic Waves:

By applying Faraday's as well as Ampere-Maxwell's equations, it is not that difficult to arrive at the following form that is the form of a one-dimensional transverse wave equation propagating in the x-direction at speed c in vacuum:

The solution to these equations are E = Eosin(kx - ωt)     and    B = Bosin(kx - ωt) where E and B are related by E = cB.

Of course, ω = 2πf is the angular frequency and k = is called the the "wave number."

Example 5: Show that each of the above solutions satisfy the corresponding wave equation.

Solution: Solution is left for students.

Energy Stored in a Capacitor:

Recall the formula for the derivation of the electric energy stored in a capacitor:

 Ue = (1/2)CV2  (6) where V is the maximum capacitor voltage during charging from 0 to V.

Note that V = Ed where E is the constant electric field in between the plates and d the the gap for a parallel-Plates capacitor.  If there is vacuum in between the plates,

C = εoA/d.  Substituting for C and V in (6) yields:

Ue = (1/2)εoE2Ad  (verify) where Ad is the volume of the space between the platesDividing both sides by Ad, results in: 

Ue/Ad = (1/2)εoE2.    The left side is the energy per unit volume in the capacitor space.  It it called the "electric energy density, ue=Ue/Ad."

Electric energy density:   ue = (1/2)εoE2.    (7)

Energy Stored in an Inductor:

In a somewhat similar way, it can be shown that the magnetic energy stored in an inductor is:

Um = (1/2)LI2   (8) where L = μon2A and I can be solved for from the solenoid formula: B = μonI to yield  I = B/μon.

Substitute for L and I in (8) show that Um = (1/2)[B2/μo]Aℓ.   Here, A is the volume inside the inductor (solenoid) and dividing both sides by A gives us the "magnetic energy density, um" or the magnetic energy per unit volume.  Show that the result is:

Magnetic energy density:   um=(1/2)B2/μo.    (9)

Energy Transport and The Poynting Vector:

Electromagnetic waves like other waves transport energy.  To calculate the energy transmission by E&M waves, we will use both the electric and magnetic energy densities: Equations 7 and 9.

Since E = cB = B/(εoμo)1/2, we may conclude that the instantaneous values of ue and um are equal.  (Verify).  The total energy density is therefore

u = ue + um  = 2ue = 2um

= εoE2 = B2o= (εoμo)1/2EB.  (10)

Now, within every volume Adx in space normal to the direction of wave propagation, the energy contained is dU = uAdx.  The rate of change of this energy with time is, of course, dU/dt.

dU/dt = uAdx/dt   or,   dU/dt = uAc    {where c = dx/dt is the wave speed (here speed of light).}

 

The direction of S is the direction of the energy flow.

Fig. (9)

Since E = Eosin(kx - ωt) and B = Bosin(kx - ωt),

 the product EB = EoBosin2( kx - ωt). 

The average of this product over one period is 1/2. (Verify.)

The average power incident per unit area normal to the direction of propagation is therefore

 

Radiation Pressure:

     It can be shown that E&M waves transfer energy to the surfaces they are incident on.  Crooke's radiometer (on the right) is a device that shows how energy is received from light when incident on the vanes of a small propeller placed inside an air- vacuumed glass  container.  Each vane has one side of it painted black and the other side white. Click on the following link for a demo:  http://www.youtube.com/watch?v=cey-JBeHrww . 

For the apparatus shown, if you send light toward the vanes in the direction you are looking at them right now, the propeller will start spinning.  A person who is looking down from the top of the apparatus, will see it spin counterclockwise.  As it spins, you see the black sides of the vanes on the right and the white sides on the left.  The black sides absorb the incident light energy, but the white sides reflect it.  " The light energy received by the black vanes causes counterclockwise rotation as viewed from the top.  The energy received by the white vanes get almost completely reflected or sent back."  This reasoning is because of the observation we make on the basis of the counterclockwise rotation from a top view.  You may verify this by trying the above link.

It is important to note that the momentum change for the light incident on the white vane is twice that of the black vane. One may erroneously conclude that the rotation will be clockwise; however, the momentum transfer or exchange is not between two mechanical objects like two billiard balls.  It is between the electromagnetic waves (considered to be of zero mass) and a mechanical object like the vanesAlso, each reflected photon from any of the white vanes is not the same photon that was incident on it before collision.  The interpretation of the process requires more detailed analysis.

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The linear momentum (p) that an electromagnetic wave carries is related to the energy it transports by

If the energy of an incident E&M wave is fully absorbed upon perpendicular collision on a nonreflecting surface, the above equation can calculate its imparted momentum to that surface.  On the other hand, if the energy of the incident E&M wave is fully reflected, its imparted momentum is

 

To find the radiation pressure, we may calculate the force that such momentum transfer exerts on a surface.  Since F = Ma, we may write: