Chapter 34

Electromagnetic Waves

Level I

Anywhere there is an electric charge, there exists an electric field around it throughout the entire space.  If the charge moves, its electric field (its effect of attraction or repulsion on other charges) moves or changes as well.  The motion of the charge causes change in its distance from any other charge.  Consequently, the electric field intensity E of the charge at any point in space changes 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 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 any other charge at some distance away (no matter how far) to oscillate accordingly at a later time, we believe that the repeated oscillation of an electric charge creates waves that propagate throughout the entire space.   Such waves are called the "electromagnetic waves."   The graphical representation of such 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.

Figure 1

Note that E = cB where c is the speed of light (3.00x108 m/s in vacuum).    This figure shows one out of almost countless propagation directions.  See the foot note regarding Figure 1 at the end of this chapter.

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 (visible 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 onto E&M waves and send them at the speed of light, c.  This is the maximum possible speed according to the "Einstein's Theory of Relativity."  Mounting sound waves on electromagnetic waves is called "modulation."

Wave Speed:

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

v = f λ.

For electromagnetic 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 this 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  λ = c/f = {3.00x108/60.0}m = 5,000,000m = 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 15cm or 6.0 inches.  Calculate the frequency of such waves and express it in MHz.   Note that  MHz stands for Mega Hertz or Million Hertz.

Solution:  c = f λ ;  f = c/λ  = {3.00x108/0.15}s-1 = 2.00x109Hz = 2000 MHz.

Example 3:  White light is a mixture of a large number of different electromagnetic waves with wavelengths ranging 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 

Use the values of  εo = 8.85x10-12F/m and  μo = x10-7 Tm/A to verify that  c = 3.00x108 m/s.

 

Energy carried by Electromagnetic Waves:

The energy carried by a mechanical wave is proportional to the square of its amplitude A2.  For  E&M waves, it will be Eo2, or Bo2, or the product 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 Joule per second is watt, we may say that the energy carried by a wave is expressed in watt/m2.   It can be shown that the formula for maximum energy density carried by an electromagnetic wave is either of the three following forms:

I =  c εoEo2    =   cBo2 /μo    =    EoBo/μo

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

I rms  =   (1/2)cεoEo2   =   cBo2 /o   =   EoBo/o

 

Example 4:  A radio set is tuned on 10-kilowatt AM radio station that is 5.0 miles (8.0 km) away.  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 Irms 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 (8000m in 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?  The simple division below will give us the value of  Irms. 

Irms= 10,000watts / [4(3.14 x 80002)m2] = 1.2 x10-5 w/m2  =  12μw/m2.

Irms = (1/2)cεoEo2 , solving for Eo Eo = [2Irms/cεo]1/2  =  9.4 x10-3 V/m.

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

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

 

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) with 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) 3.00x1010cm/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 f is defined as (a) the number of meters per second  (b) the number of ω's 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 (c = 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 = ct   (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 minutes 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, as a result a charge that is on Star, Alpha Centauri, 4 light-years away, will start oscillating (a) 3x108s later   (b) 2yrs 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 for cellular phones is about 2000MHz.  This frequency is (a) 2.0x109Hz  (b) 2.0x106Hz  (c) 2.0x1012Hz.   click here.

23) The λ of E&M waves used for cellular phones (Question 22) 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 is 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 3.00x108m/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 3.00x108m/s  (b) has a speed of 3.00x105km/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 femto-meter 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 the magnet.  Investigate this by drawing a bar-magnet with its field-lines and then once put a closed surface (loop) somewhere in its field-lines and once put the entire magnet inside a closed surface (loop) 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 electromagnetic 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.   When a magnet is moved toward or away from a closed loop of wire such that φm through the loop changes, an induced voltage develops in the loop.  Fig. 2 supports this statement  In Fig. 2, as the bar magnet moves to the right, the changing B induces an electric field E normal to B that results in current I in the wire as sensed by the galvanometer.


In Fig. 3, a varying current is forced into a solenoid by a varying voltage source that causes a varying magnetic field BThe 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 Fig. 3 are circular and the line integral of any closed loop around B results in -dφB/dt.   Faraday's formula can be shown in two forms as shown under Figures 2 & 3.  The equation shown below is the 3rd Maxwell's equation.

 

Ampere-Maxwell Law (4):

The Displacement Current:

 When a capacitor is being charged, we know that there is a varying current in the wires that connects 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 Figure4 and 5 below.  

In Figure 4, Applying Ampere's law to the wire portion can calculate current I in the wire portion for us with no problem.  This can be easily done by finding the line integral of the magnetic field that the current produces around the wire portion.  According to Ampere's law the line integral of B∙dℓ must equal μoI.

     Now, if we decide to apply Ampere's law to the dielectric portion, 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 clarified this shortcoming by using a semispherical surface as shown in Figure 5 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 Figure 4 does cross the flat circular surface, but in Figure 5, it does not cross the semispherical surface.  Does this mean I = 0 in between the plates?  Should we think that the line integral of B∙dℓ equals 0?   Logically this is not true! 

We know that current exists in between the plates but in the form of electric flux.  For clarification, 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.   As electric charge q changes on the plates, φE changes accordingly, and dq/dt becomes:

dq/dt = εodφE /dt   and therefore,   ID = εodφE /dt.

q = εoφE.  (In between the plates)

 

Mathematical Form of Electromagnetic Waves:

By applying Faraday's and Ampere-Maxwell's equations, it is not that difficult to arrive at the following second order differential equation.   The solution to this equation is called the "one-dimensional transverse wave equation" for the propagation of E&M waves in the x-direction at speed c in vacuum:

The solution to these equations are: E = Eosin(kx - ωt)  &  B = Bosin(kx - ωt) where E and B are related by E = cB.  These equations determine the strength of the E and B fields at distance x from the E&M source and at time t.

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 capacitor voltage.

Note that in V = Ed, the quantity E is the uniform electric field in between the plates and d is the the gap for a parallel-Plates capacitor.  For vacuum as the insulator 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 plates Dividing 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 the 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  = B2/μ =o/μo)1/2 EB.    (10)

Now, within every volume Adx in space normal to the direction of wave propagation, the contained energy 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.

Figure 9

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

 the product is:        EB = EoBosin2( kx - ωt). 

The average of this product over one period is (1/2)EoBo. (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 (below) 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 vanes Also, 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.

 

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:

Footnote:

1) Visualize a very big round table that can accommodate at the most exactly 360 chairs around it with exactly 360 people (observers) sitting on them facing the center of the table.  That way every person is using 1o of angular space.  Also, visualize a vertical linear spring hung from the ceiling exactly above the center of this table that has a red glowing marble hanging from it.   Suppose the red marble is pulled down and released to oscillate.  Also, suppose it oscillates at a frequency of say f = 4Hz.   The 360 people can definitely see this oscillating marble.  This means that at least 360 waveforms as shown in Figure 1 keep being originated from the glowing marble and travel horizontally in 360 horizontal directions (each 1o apart) and reach those 360 observers.  If you keep making the table bigger and bigger to accommodate 3600, 36000, 360,000, and so on ....observers, the horizontally propagated waves will have countless horizontal directions because all countless observers will be able to see the oscillating marble.  The countless people around this extremely large table will eventually receive these countless transverse waves going out from the same oscillating charge in countless directions.  The purpose of this footnote is to emphasize the fact that Figure 1 shows only one of such countless directions!

The electron in a hydrogen atom, for example, is spinning within a tiny sphere.  It is not spinning along a single circle (or just on a single plane).  Its plane of rotation keeps changing a large number of times every second (may be of the order of 1014 times per second).  That way, it is like there are 1014 different round tables (each table at a different angle or orientation in space) with a countless number of people around each table watching this oscillating electron!  One may conclude that the effect of the oscillations of the single electron in every hydrogen atom propagates in space in all directions.  The electron in a hydrogen atom does not oscillate just up and down.  The up and down motion shown in Figure 1 for charge q is one possible direction out of almost countless directions.