lecture-2 electromagnetic wave propagation: wave...

Electromagnetic wave propagation: Wave propagation in lossy dielectrics, plane waves in lossless dielectrics, plane wave in free space,plain waves in good conductors, power and the pointing vector, reflection of a plain wave in a normal incidence.

Lecture-2

Pradeep Singla

Electromagnetic waves

2

2

22

t

EB

t

EEEE

o

02

22

t

EE o

or,

For the electric field E,

i.e. wave equation with v2 = 1/µoPradeep Singla


02

22

t

BB o

Similarly for the magnetic field

i.e. wave equation with v2 = 1/µo

In free space, = o = o ( = 1)

oo

c

1 c = 3.0 X 108 m/s

Pradeep Singla


In a dielectric medium, = n2 and = o = n2 o

n

c

nv

ooo

11

Pradeep Singla

Electromagnetic waves: Phase relations

02

22

t

EE o

02

22

t

BB o

The solutions to the wave equations,

can be plane waves,

ti

o

ti

o

eBB

eEE

rk

rk

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• Using plane wave solutions one finds that, • Consequently

Bit

B

and

EkiE

,BEk

or

t

BE

,

i.e. B is perpendicular to the

Plane formed by k and E !!

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• Since also

• k E and k B ( transverse wave)

• Thus, k, E and B are mutually perpendicular vectors

• Moreover,

00 BkiBEkiE

BBBEkEk ˆˆ

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Thus E and B are in phase since,

requires that

trki

o

trki

o ecBeE

,...2,01 ie

E

B

k

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Electromagnetic waves are transverse waves, but are not mechanical waves (they need no medium to vibrate in).

direction ofpropagation

Therefore, electromagnetic waves can propagate in free space.

At any point, the magnitudes of E and B (of the wave shown) depend only upon x and t, and not on y or z. A collection of such waves is called a plane wave.

y

z

x

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Manipulation of Maxwell’s equations leads to the following plane wave equations for E and B:

These equations have solutions:

You can verify this by direct substitution.

2 2

y y

0 02 2

E E (x,t)=

x t

2 2

z z0 02 2

B B (x,t)=

x t

y maxE =E sin kx - t

z maxB =B sin kx - t

where , ,

2k = =2 f and f = =c.

k

Emax and Bmax in these notes are sometimes written by others as E0 and B0.

Emax and Bmax are the electric and magnetic field

amplitudes

Pradeep Singla

You can also show that

At every instant, the ratio of the magnitude of the electric field to the magnitude of the magnetic field in an electromagnetic wave equals the speed of light.

y zE B

=-x t

max maxE k cos kx - t =B cos kx - t

.

max

max 0 0

E E 1= = =c =

B B k

Pradeep Singla

Summary of Important Properties of Electromagnetic WavesThe solutions of Maxwell’s equations are wave-like with both E and B satisfying a wave equation.



Electromagnetic waves travel through empty space with the speed of light c = 1/(00)½.

Emax and Bmax are the electric and magnetic field amplitudes. Pradeep Singla

Summary of Important Properties of Electromagnetic Waves

The components of the electric and magnetic fields of plane EM waves are perpendicular to each other and perpendicular to the direction of wave propagation. The latter property says that EM waves are transverse waves.

The magnitudes of E and B in empty space are related by

E/B = c.max

max

E E= = =c

B B k

direction ofpropagation

y

z

x

Pradeep Singla

The magnitude S represents the rate at which energy flows through a unit surface area perpendicular to the direction of wave propagation.

Energy Carried by Electromagnetic Waves

Electromagnetic waves carry energy, and as they propagate through space they can transfer energy to objects in their path. The rate of flow of energy in an electromagnetic wave is described by a vector S, called the Poynting vector.*

0

1S= E B

Thus, S represents power per unit area. The direction of S is along the direction of wave propagation. The units of S are J/(s·m2) =W/m2.

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z

x

y

c

SE

B Because B = E/c we can write

These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area.

0

1S= E B

For an EM wave

so

E B =EB

.0

EBS =

.

2 2

0 0

E cBS= =

c

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The time average of sin2(kx - t) is ½, so

2 2

0 0 0

EB E cBS= = =

c

EM waves are sinusoidal.

The average of S over one or more cycles is called the wave intensity I.

2 2

max max max maxaverage

0 0 0

E B E cBI=S = S = = =

2 2 c 2



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Thus,

The magnitude of S is the rate at which energy is transported by a wave across a unit area at any instant:

instantaneous

instantaneous

energypowertimeS= =

area area

average

average

energypowertimeI= S = =

area area

Note: Saverage and <S> mean the same thing!Pradeep Singla

The energy densities (energy per unit volume) associated with electric field and magnetic fields are:

Using B = E/c and c = 1/(00)½ we can write

Energy Density

2

E 0

1u = E

2

2

B

0

1 Bu =

2

2

2220 0

B 0

0 0 0

EE1 B 1 1 1cu = = = = E

2 2 2 2

22

B E 0

0

1 1 Bu =u = E =

2 2Pradeep Singla

For an electromagnetic wave, the instantaneous energy density associated with the magnetic field equals the instantaneous energy density associated with the electric field.

22

B E 0

0

1 1 Bu =u = E =

2 2

22

B E 0

0

Bu=u +u = E =

Hence, in a given volume the energy is equally shared by the two fields. The total energy density is equal to the sum of the energy densities associated with the electric and magnetic fields:

Pradeep Singla

When we average this instantaneous energy density over one or more cycles of an electromagnetic wave, we again get a factor of ½ from the time average of sin2(kx - t).

so we see thatRecall

The intensity of an electromagnetic wave equals the average energy density multiplied by the speed of light.

22

B E 0

0

Bu=u +u = E =

22 max

0 max

0

B1 1u = E =

2 2

2 2

max maxaverage

0 0

E cB1 1S = S = =

2 c 2.S =c u

, 2

E 0 max

1u = E

4,

2

maxB

0

B1u =

4and

Pradeep Singla

Example: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 50 kW. Assuming the wave is radiated equally in all directions above the ground, find the amplitude of the electric and magnetic fields detected by a satellite 100 km from the antenna.

R

Station

SatelliteAll the radiated power passes through the hemispherical surface* so the average power per unit area (the intensity) is

4

-7 2

22 5average

5.00 10 Wpower PI= = = =7.96 10 W m

area 2 R 2 1.00 10 m

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R

Station

Satellite

2

max

0

E1I= S =

2 c

max 0E = 2 cI

-7 8 -7= 2 4 10 3 10 7.96 10

-2 V=2.45 10m

-2

-11maxmax 8

V2.45 10E mB = = =8.17 10 Tc 3 10 m s

Pradeep Singla

Example: for the radio station in the example on the previous two slides, calculate the average energy densities associated with the electric and magnetic field.

2

E 0 max

1u = E

4

2

-12 -2

E

1u = 8.85 10 2.45 10

4

-15

E 3

Ju =1.33 10

m

2

maxB

0

B1u =

4

2

-11

B -7

8.17 101u =

4 4 10

-15

B 3

Ju =1.33 10

m

Pradeep Singla

Momentum and Radiation Pressure

EM waves carry linear momentum as well as energy. When this momentum is absorbed at a surface pressure is exerted on that surface.

If we assume that EM radiation is incident on an object for a time t and that the radiation is entirely absorbed by the object, then the object gains energy U in time t.Maxwell showed that the momentum change of the object is then:

The direction of the momentum change of the object is in the direction of the incident radiation.

incident

Up = (total absorption)

c

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If instead of being totally absorbed the radiation is totally reflected by the object, and the reflection is along the incident path, then the magnitude of the momentum change of the object is twice that for total absorption.

The direction of the momentum change of the object is again in the direction of the incident radiation.

2 Up = (total reflection along incident path)

c

incident

reflected

Pradeep Singla

The radiation pressure on the object is defined as the force per unit area:

From Newton’s 2nd Law (F = dp/dt) we have:

For total absorption,

So

Radiation Pressure

FP =

A

F 1 dpP= =

A A dt

Up=

c

dU1 dp 1 d U 1 SdtP = = = =A dt A dt c c A c

incident

(Equations on this slide involve magnitudes of vector quantities.)Pradeep Singla

This is the instantaneous radiation pressure in the case of total absorption:

SP =

c

For the average radiation pressure, replace S by <S>=Savg=I:

average

rad

S IP = =

c c

Electromagnetic waves also carry momentum through space with a momentum density of Saverage/c2=I/c2. This is not on your equation sheet but you have special permission to use it in tomorrow’s homework, if necessary. Pradeep Singla

Following similar arguments it can be shown that:

rad

IP = (total absorption)

c

rad

2IP = (total reflection)

c

incident

incident

reflected

Pradeep Singla

Example: a satellite orbiting the earth has solar energy collection panels with a total area of 4.0 m2. If the sun’s radiation is incident perpendicular to the panels and is completely absorbed find the average solar power absorbed and the average force associated with the radiation pressure. The intensity (I or Saverage) of sunlight prior to passing through the earth’s atmosphere is 1.4 kW/m2.

3 2 32

WPower =IA= 1.4 10 4.0 m =5.6 10 W=5.6 kWm

Assuming total absorption of the radiation:

32

average -6

rad8

W1.4 10S I mP = = = = 4.7 10 Pamc c 3 10

s

-6 2 -52rad

NF=P A= 4.7 10 4.0 m =1.9 10 Nm Pradeep Singla

New starting equations from this lecture:

0

1S= E B

2 2

max maxaverage

0 0

E cB1 1S = =

2 c 2

max

max 0 0

E E 1= =c =

B B

22

B E 0

0

1 1 Bu =u = E =

2 2

22 max

0 max

0

B1 1u = E =

2 2

U 2 Up = or

c crad

I 2IP = or

c c

, ,

2k = =2 f f = =c

k

There are even more on your starting equation sheet; they are derived from the above! Pradeep Singla

Electromagnetic waves:

Interact with matter in four ways:

Reflection:

Refraction:

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http://apod.gsfc.nasa.gov/apod/image/0103/sunpillar_richard_big.jpg

Scattering:

Diffraction:

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lecture-2 electromagnetic wave propagation: wave...

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