Electromagnetic Waves Part Iii

Electromagnetic Fields First Semester 2008/2009 Electronic and Comm. Eng. Dept.

(2.3) Sinusoidal time varying fields:  Let all the quantities vary with time in sinusoidal manner. Why?  It is easy to obtain it in laboratory.  The differentiation of sin gives cos and vise versa.  We can convert any function to sin & cos.  To transform the function from time form to complex form, it must be sinusoidal. L

Hint: Solving the function in complex form (algebraic equation ) avoid us the solution of integro-differential equation in time form.

V ∼

I

C

R

1   =   V R + j ω L +  I j ω C  

d i( t ) 1 t V( t ) = R i( t ) + L + ∫ i( t )dt dt C0

Complex representation of field quantities: +j

1. Scalars:

ω ρ

ρ( t ) = ρo cos( ωt + φo ) = Re[ ρo e j( ωt +φo ) ]

= Re[ρo e

jφ o

e

jωt

] = Re[ ρ

e

jωt

]

φ

o

o

ρ = ρ o e jφo ........ Comlex phasor (represented by point in complex domain).

+1

2. Vectors:  E(r, t ) = E ox cos(ωt + φ x ) aˆ x + E oy cos(ωt + φ y ) aˆ y + E oz cos(ωt + φ z ) aˆ z

= Re [ ( E e aˆ + E e aˆ + E = R e [ ( E aˆ + E aˆ + E aˆ ) e ] jφ y

jφ x

ox

x

oy

y

) ]

jφ z jω t ˆ e a e oz z

jω t

x

x

y

y

z

z

 E = E x aˆ x + E y aˆ y + E z aˆ z 3. Derivatives:

ρ( t ) = ρo cos(ωt + φo ) = Re[ ρ ∂ρ( t ) = Re[ (jω) ρ e jωt ] ∂t ∂ 2 ρ( t ) 2 jω t  = Re[ (ω ) ρ e ] 2 ∂t

e jωt ] ∂ ⇒ × ( jω) ∂t ∂2 ⇒ × ( −ω2 ) 2 ∂t

4. Complex parameters of matter ( ε and µ

):

  jψ jψ E = Eo e aˆ1 ; D = Do e aˆ1 ;

Let

E

D

 D D o j( ψD −ψE )  = =  | e − jψε ε e =| ε  E Eo

ε = ε′ − jε′′

Similarly,

µ = µ′ − jµ′′

 Which are the complex permittivity and permeability of the matter

ε′, µ′′ ......

are proportional to the losses.

Or ;

Maxwell’s equations in phasor form (complex domain):     H 1) ∇ × E = − jωµ

      2) ∇ × H = J + jωε E

 3) ∇.D = ρ

 4) ∇.B = 0

 & ∇.J = − jωρ

(2.4) Derivation and solution of wave equations in unbounded media: 2.4.1 Types of media ( according to the values of ε , µ andσ ). 2.4.2 Source-free wave equation. 9.4.2.1 Time form. a. In free-apace ( ε = ε

o

, µ = µo and σ = 0 ).

b. Lossless dielectric ( ε = ε 0 ). c. Lossy dielectric ( ε =ε o ε 0 ).

9.4.2.2 Complex form.

2.4.3 Properties of plane wave.

o

ε

r

, µ = µoµr and σ ≠

r

, µ = µoµr and σ =

(2.4.2) Source-free wave equation :

 Source-free: the solution region does not include any sources ( Jimp = 0 & ρ = 0 ).

(2.4.2.1)Time form: a) In free-apace: ( ε = ε o , µ = µo and σ = 0 )    ∂H 2  ∇ × E = −µo ..... (1) ∂ E 2 ∂ ∇ E − µoεo 2 = 0 t

 ∂E ∇ × H = εo ..... (2) ∂t  ∇.D = 0 ..... (3)  ∇.B = 0 ..... (4)  & ∇.J = 0 ..... (5)

Taking the curl of both sides of eqn. (1) :

  ∂ ∇ × ∇ × E = −µo (∇ × H ) ∂t    ∂ ∂E 2 ∇ (∇.E ) − ∇ E = − µo (εo ) ∂t ∂t 0

∂t

Similarly;

 2  ∂ H 2 ∇ H − µoεo 2 = 0 ∂t

Generally:

 2   ∇ − µoεo ∂  2  ∂ t   2

 E    = 0  H  

Source-free wave equation in free-space (2nd order P.D.E.)

Solution of source-free wave equation in free space:  2  ∂ E 2 ∇ E − µoεo 2 = 0 ......... ( I) ; the solution of this equation has the form: ∂t

    E( r , t ) = E x ( r , t )aˆ x + E y ( r , t )aˆ y + E z ( r , t )aˆ z

For simplicity, let us choose the axes such that the electric field coincides with the x-axis and is function of (z,t) only. I.e.,

 E( r , t ) = E x (z, t )aˆ x

; Substituting in eqn.(I) gives,

∂ 2 E x ( z, t ) ∂ 2 E x ( z, t ) − µoεo =0 2 2 ∂z ∂t Any function f ( z  ct ) is the solution of the above equation because : f ′′(z ct ) − µ oεoc 2f ′′(z ct ) = 0 which is satisfied only when:

1 c= = 3 × 108 m / s µ oεo

So, the wave equation is satisfied by f1(z - ct) and also f2 (z + ct) , therefore Ex can be written as:

Ex=A1 f1(z - ct) +A2 f2 (z + ct)

Wave form f (z + ct) at different times Backward wave…….. f ( z + ct )

f ( z - ct ) …….. Forward wave t1 = 0

-z

z t2 > t1

-z

z t3 > t2

-z

z t4 > t3

-z

z t5 > t4

-z

z

b) Lossless dielectric: ( ε = ε   ∂H ∇ × E = −µ ..... (1) ∂t

o

ε

r

, µ = µoµr and σ = 0 )

 ∂E ∇×H = ε ..... (2) ∂t  ∇.D = 0 ..... (3)  ∇.B = 0 ..... (4)  ∇.J = 0 ..... (5)

Taking the curl of both sides of eqn. (1) :

  ∂ ∇ × ∇ × E = −µ (∇ × H) ∂t    ∂ ∂E 2 ∇ (∇.E ) − ∇ E = −µ (ε ) ∂t ∂t 0

  ∂ E 2 ∇ E−µ ε 2 =0 ∂t

Similarly;

  ∂ H 2 ∇ H−µ ε 2 =0 ∂t 2

Generally:

 2  ∂ ∇ − µ ε  2  ∂ t   2

 E    = 0  H  

As the previous case we can prove that,

1 c v= = .........wave velocity µ ε µrεr

 2 1 ∂  ∴  ∇ − 2 2  v ∂t   2

 E   =0 H  

2

Source-free wave equation in lossless dielectric

c) Lossy dielectric: (dielectric with finite conductivity)( ε =ε  ≠0)  ∂H Similarly; ∇ × E = −µ ..... (1) ∂t  

  ∂E ∇ × H = σE + ε ..... (2) ∂t  ∇.D = 0 ..... (3)  ∇.B = 0 ..... (4)  ∇.J = 0 ..... (5)

o

ε

r

, µ = µoµr and σ

  ∂H ∂ H 2 ∇ H−µ σ −µ ε 2 =0 ∂t ∂t 2

Generally:

 2   ∇ − µ σ∂ − 1 ∂  2 2  ∂ t v ∂ t   2

 E    = 0  H  

Taking the curl of both sides of eqn. (1) :

  ∂ ∇ × ∇ × E = −µ (∇ × H) ∂t

   ∂  ∂E 2 ∇ (∇.E ) − ∇ E = −µ (σE + ε ) ∂t ∂t 0

  2  ∂E ∂ E 2 ∇ E−µ σ −µ ε 2 =0 ∂t ∂t

Source-free wave equation where,

in lossy dielectric

1 v= .........wave velocity µ ε  ∂  E  µ σ    ........damping term ∂t  H 

(2.4.2.2) Complex form (general medium): 

The source-free Maxwell’s equations in complex form (Jimp = 0 & ρ = 0 ) is given  by:    

 =0 ∇.E

H ∇ × E = − jωµ

        ∇ × H = σE + jωε E = (σ + jωε ) E

  =0 ∇.H

 = σ + jωε Let us define the following complex quantities: Z = jωµ & Y Thus, Maxwell’s equations will be;     = 0 ..... (3)    H ..... (1) ∇ × E = −Z ∇. E   (Helmholtz’s equations)      ∇ × H = YE ..... (2) ∇. H = 0 ..... (4)   Taking the curl of both sides of eqn. (1) : Generally;

   )  (∇ × H ∇ × ∇ × E = −Z

    ) 2    ∇(∇.E ) − ∇ E = − Z(YE 0

   =0 2   ∇ E − ZY E Similarly;

   =0 2    ∇ H − ZY H

(∇

2

 +K

2

)

E   =0    H  

where,  = − Z Y  ....... Comlex wave number K

 = −Z Y  = − ( jωµ  )( σ + jωε ) K = − jγ = β − jα Phase shift const.

Attenuation const.

γ = α + jβ ...... Propagation Constant.

Solution of source-free wave equation (complex form):   2 2   ∇ E + K E = 0 ..... (1) One of the solutions of this equation has the form:

  - jK . r E = A1 e ........ (2) where,

 K ....... propagation vector.   x aˆ x + K  yaˆ y + K  zaˆ z = K  kˆ ; K  = β - jα .... complex wave number K=K

 r ....... position vector.

 r = x aˆ x + y aˆ y + z aˆ z = r rˆ   - jK . r  - jK (kˆ . r )  - j(β - jα ) (kˆ . r) = A1 e ⇒ From eqn. (2): E = A1 e = A1 e

   e - α (kˆ . r ) e - jβ (kˆ . r ) E = A 1 Attenuation

Phase angle

A wave propagating a distance ξ in an kˆ arbitrary direction has the equation of wave front,ˆ 

z

 r

β (k.r ) = β (r cos ψ) = β ξ

ψ

Then, for the chosen surface all points have the same phase β ξ (which represent a plane). So, this wave is called “ plane

wave ”.

  - α ξ - jβ ξ E = A1 e e

kˆ ξ

y

x

Note: “ for any source-free wave equation, the solution is always a plane wave ”

Plane wave Source

Solution check:   − jK .r  − j( K x x + K y y + K z z ) E = A1 e = A1e   ∂ 2 ∂2 ∂ 2   ∇ E =  2 + 2 + 2  E ∂y ∂z   ∂x   xx+K  yy+ K  z z) − j( K 2 2 2    = A1 (-jK x ) + (-jK y ) + (-jK z ) e 2

[

]

 − j( K x x + K y y + K z z )  2 2 2 2      = − [K x + K y + K z ] [A1e ] = − K E ; which satisfies eqn.(1). Another possible solution is:

  + jK . r E = A2 e ........ (3)

; which will also satisfies eqn.(1).

Generally, the solution of eqn. (1) can be written as:

  − jK . r  + jK . r E = A1 e + A2 e Forward wave Backward wave

(2.4.3) Properties of plane wave : Definition: Plane wave is a class of waves having both electric and magnetic fields contained in one plane called the wave front which is perpendicular to the direction of propagation. 1. Operator ∇ :    − jK . r

E = E0 e

eˆ

 x+ K  y+ K  z) − j( K   ∂ ∂ ∂    x y z ∇.E =  aˆ x + aˆ y + aˆ z  .  E 0 e eˆ  ∂y ∂z     ∂x   )aˆ + (− jK  )aˆ + (− jK  )aˆ . E = ( − jK

[

x

x

y

y

z

z

]

    x aˆ x + K  y aˆ y + K  z aˆ z ) . E = − jK.E .... (1) = − j( K Similarly,

   ∇ × E = − jK × E ... (2)

From eqns. (1) and (2):

  kˆ ∇ ≡ − jK = − jK

 : 2. Intrinsic impedance η From Maxwell’s first equation ,

From Maxwell’s second equation,

     ∇ × E = −ZH     ˆ   − jK k × E = −ZH

(

     ∇ × H = YE     ˆ   − jK k × H = YE

)

(

   jK   ˆ H= k ×E = Z   Y  ˆ = k ×E  Z

(

Let

η =

Y   Z  ˆk × E  Z

)

)

(

)

(

Y       E = − jK  =− Z ˆk × H ˆk × H   Y Y   Z  × kˆ = H  Y

(

)

)

Z =| η | ∠ϕη ….. is the intrinsic impedance of the medium. Y

(

   = 1 kˆ × E  then, H  η

)

and

(

  E = η H × kˆ

Example: In free space (ε o , µ o and σ = 0)

 = η = Z /Y

)

kˆ …direction of propagation.

 = jω εo then, Z = jω µo , Y

4π ∗ 10 −7 jω µo / jω εo = µ o / ε o = = 120π Ω = ηo −9 10 / 36π

  3. Relations between E, H and kˆ :

 E

From Maxwell’s 3rd. equation ,

 ∇.D = 0  ∇.E = 0

(ρ = 0)

  kˆ . E = 0 − jK

 kˆ . E = 0

(

)

 H kˆ

 E ⊥ kˆ

   1 ˆ    and H = k ×E H ⊥ E ⊥ kˆ  η   e − jK Z aˆ , A  = A e jψo and η = | η | ∠ϕ  . Find: Example: Let E = A x o η    H , E(z, t) and H(z, t).   jωt  E (z, t ) = Re[ E e ] jψ o − α z - j β z E = Ao e e e aˆ x  − αz   E ( z , t ) = A e cos(ωt - β z + ψ o ) aˆ x o  1  H=

( kˆ × E )  η

similarly,

  A o jψ − αz - jβz - jϕη A o − αZ o H ( z , t ) = e cos(ωt - βz + ψ o − ϕη ) aˆ y H= e e e e aˆ y | η | | η |

4. Phase velocity ( Vp ): “The velocity of points in the wave moving at constant phase” or ;

“ it is the velocity of equiphase plane (wave front) ”.

Const. = (ωt - βz + ψ o − ϕη )

f(z,t)

Differentiate w.r.t. time ,

0 = (ω - β

dz ) dt

 vp

dz ω vp = = dt β

z

5. Wavelength ( λ ):

λ

“The distance between two successive points having the same phase”.

ωt - βz + ψ o - ϕη = ωt - β(z + λ) + ψ o - ϕη + 2π 6. Depth of penetration ( δ ) : (skin depth) “The distance that the wave travel through the medium until its amplitude is reduced by (1/e) ”.  | E |= A oe − αz 1

A o e − α δ = A o e −1

δ=

α

β λ= 2π

or,

2π λ= β

 |E| Ao Ao e δ

z

7. Electric and magnetic energy densities (we and wm): a) Time form:

1  1   1 2 w e = εE.E = D.E = ε | E | .........[ J / m3 ] 2 2 2  2 1   1   1 w m = µH.H = B.H = µ | H | .......[ J / m3 ] 2 2 2 The average values (denoted by < >) of the above quantities are given by:

< wm

1T > = ∫ w m (t, z) dt T0

1T < w e > = ∫ w e (t, z) dt T0

b) Complex form:

1   * w e = ε E.E .......[J/m3 ] 2

1   * < w e > = ε E.E 4

 1    *...... [J/m 3 ]  m = µ H.H w 2

1   *  m > = µ H.H <w 4

The ( * ) stands for the complex conjugate.

 8. The Poynting vector ( S ) : a) Time form:

   S = E × H = S kˆ  1 T < S > = ∫ S( t , z) dt T0

where, where,

 S ….. Poynting vector S ….. Power density [W/m2]. kˆ ….. Direction of propagation.  < S > ….. Average power density.

b) Complex form:

 1   * S= E× H 2

  1   * < S > = Re[S] = Re[E × H ] 2

W/m2

 9. Group velocity (Vg ) : It is defined as the ratio between the average power flow density to the total average electromagnetic energy.

  < S> Vg = < we > + < wm >

Example:

  Eo cos(ωt − β z + ψ o ) aˆ y Let E(z, t ) = E o cos(ωt − β z + ψ o ) aˆ x , H(z, t ) = η

  Find, S, < S > , w e , < w e > , w m and < w m > .

   E o2 S = E×H = cos 2 (ωt − βz + ψo ) aˆ z η  1 T E o2 E o2 2 <S >= ∫ cos (ωt - βz + ψo ) dt aˆ z = aˆ z T0 η 2η 1  2 ε 2 w e = ε | E | = E o cos 2 (ωt − βz + ψ o ) 2 2 1 Tε 2 1 < w e > = ∫ E o cos 2 (ωt − βz + ψo ) dt = εE o2 To2 4  2 µ E o2 1 wm = µ| H | = cos 2 (ωt − βz + ψ o ) 2 2 2 η 1 T µ E o2 1 E o2 2 < wm > = ∫ cos (ωt − βz + ψo ) dt = µ 2 2 To2 η 4 η

   T n 2 2 Note: for sinusoidal E and H; S is f (sin or cos ) then, ∫ (sin 2 or cos 2 ) dt = T / 2 0

(2.5) plane wave in different media: Relation between wave parameters in different media (summary):

Z = jωµ

 = σ + jωε Y

 = − Z Y  = − jγ = β - jα K

γ = α + jβ

....... Comlex wave number

................ Propagation constant.

β ....... Phase shift constant [rad/m].

α ....... Attenution constant [Np/m] (Nepers/meter). ω vp = ....... Phase velocity [m/s] . β

λ=

2π ...... Wavelength [m] . β

1 δ = ...... Skin depth (Penetration depth) [m]. α

 =| η | ∠ϕη ........ Interinsic impedance [Ω]. η = Z / Y

(2.5.1) Free space (εo , μo and σ = 0) :

Z = jω µo

 = jω εo Y

;

 = − Z Y  = ω µoεo = ω / c K

vp = ω / β = c ;

λ=

α=0

;

2π 2π 2π c = = = ; β ω / c 2πf / c f

 = µ o / ε o = 120π η = Z / Y

β = ω /c

δ = 1/ α = ∞

| η |= 120π ; ϕη = 0

(2.5.2) Lossless dielectric ( ε , μ and σ = 0) :

Z = jω µ

 = jω ε Y

;

 = − Z Y  = ω µ ε= ω / v K

vp = ω / β = v =

c ; µrεr

 = µ/ε η = Z / Y

α=0 2π 2π v vp λ= = = = β ω/ v f f

; ;

β = ω /v δ = 1/ α = ∞

| η |= µ / ε ; ϕη = 0

(2.5.3) Good dielectric (ε = ε′ - jε′′ , μo and σ = 0) ( ε′′ < < ε′ ) : Usually the magnetic response is very weak compared to the dielectric response in most materials for wave propagation, so, In such materials μ=μo, so, Assume that μ is real value

Z = jω µ

;

 = jωε = jω(ε′ − jε′′) Y

 = − Z Y  = ω µ(ε′ − jε′′) = ω µε′ 1 − jε′′ / ε′ = ω µε′ ( 1 − jε′′ / 2ε′ ) K α and β are found by taking the real and imaginary parts of jk

ε′′  ωε′′ µ  α = ω µε′  = 2 ε′  2ε′ 

;

β = ω µε′

1 vp = ω / β = ; µε′

Z η = = Y

vp 2π 2π λ= = = β ω µε′ f

1 2 ε′ δ= = α ωε′′ µ

jω µ µ 1 µ ε′′  = = 1 + j  ′ ′ ′ ′ ′ jω(ε − jε ) ε 1 − j(ε′′ / ε′) ε 2ε′ 

µ µ  ε′′  | η |= 1+   ≈ ε′ ε′  2ε′  2

 ε′′  ϕη = tan  ≈0 ′  2ε  −1

;

(ε′ > > ε′′)

(ε′ > > ε′′)

(2.5.4) Dielectric with losses (ε = ε′ - jε′′ , μo and σ = 0) ( ε′′ > > ε′ ) :

Z = jω µ

;

 = jωε = jω(ε′ − jε′′) Y

K = − Z Y  = ω µ(ε′ − jε′′) = ω − jµε′′ 1 + j ε′  = ω µε′′  1 + j ε′′  1 − j  2ε′  2   ε′′  

α = ω µε′′ / 2 ( 1 − ε′ / 2ε′′ )

;

β = ω µε′′ / 2 ( 1 + ε′ / 2ε′′ )

ω 2 1 2  ε′  vp = = = 1 −  β µε′′ ( 1 + ε′/2ε′′ ) µε′′  2ε′′  vp 2π 2π λ= = = β ω / vp f

Z η = = Y

1 1 2  ε′  δ= = 1 +  α ω µε′′  2ε′′ 

;

jω µ = jω(ε′ − jε′′)

µ 1 µ ε′   1 + j  = 1 − j   − jε′′ 1 + j(ε′ / ε′′) ε′′  2ε′′  2 

µ µ  ε′  | η |= 1+  ;  ≈ ε′′ ε′′  2ε′′  2

ε′   ϕη = tan  1 −   ε′′  −1

(ε′ > > ε′′ )

( ε , μ and σ ) ( ω ε> σ )

(2.5.5) Dielectric with finite conductivity  = σ + jω ε Z = jω µ ; Y

:

 = − Z Y  = − jω µ(σ + jω ε) = ω2µ ε(1 − jσ / ω ε) = ω µ ε(1 − jσ / 2ω ε) K

 σ α = ω µ ε  2ω vp = ω / β =

 σ µ = ε 2 ε

1 ; µ ε

β=ω µ ε

;

2π v p λ= = β f

1 2 ε δ= = α σ µ

;

Z jω µ µ 1 µ σ  η = = = = 1 + j   Y (σ + jω ε) ε 1 − j(σ / ω ε) ε 2ω ε µ  σ | η |= 1+  ε  2ω

2

µ   ≈ ε ε

<<1

;

 σ  ϕη = tan    2ω ε −1

(2.5.6) Good conductor ( ε , μ and σ ) ( σ > > ω ε) : A good conductor has high conductivity and large conduction current.

Loss tangent = ε” / ε’ >> 1 Ratio of the conduction current density to the displacement current density in a conducting material is given by σ / ωε >> 1

Z = jω µ

;

 = σ + jω ε Y

jω ε ωµ σ   = − Z Y  = − jω µ(σ + jω ε) = − jω µ σ K 1 + ≈ (1 − j)   σ  2  <<1

ωµ σ α= 2

;

ωµ σ β= 2

− j = 1∠ − 90 1 (1 − j ) 1∠ − 90 = 1∠ − 45 = 2

2ω vp = ω / β = ; µ σ η =

Z jω µ = = Y (σ + jω ε)

ωµ | η |= σ

2π v p λ= = β f

1 2 δ= = α ωµ σ

;

jω µ 1 ωµ = (1 + j) σ 1 − ( jω ε/ σ) 2σ << 1

;

ϕη = tan −1 (1) = 45

(2.5.7) Perfect conductor ( ε , μ and σ → ∞ ):

Z = jω µ

;

 =∞ Y

 = − Z Y  =∞ K

vp = ω / β = 0 ;

Z η = =0 Y

α =β= ∞ λ=

2π =0 ; β

| η |= 0

δ=

;

1 =0 α

ϕη = 0

(9.6) Power, energy and Poynting theorems: (9. 6.1) Poynting theorem (time form) : (in 1884 by John H. Poynting) To find the power flow associated with the electromagnetic wave, it is necessary to develop a power theorem for the electromagnetic field, known as the Poynting Theorem

   S = E × H = S kˆ

 S ….. Poynting vector

Take the scalar product of both sides:

       ∇.S = ∇. E × H = H. ∇ × E − E. ∇ × H

(

)

(

)

(

)

   ∂H  ∂E   = −µ H . − ε E. − E.J ∂t ∂t

       ∂H  ∂E   − E .  J + ε  = H .  − µ ∂t  ∂t   

  µ ∂H 2 ε ∂E 2   ∇. E × H = − − −E.J 2 ∂t 2 ∂t     ∂ 1 1 2 2 − ∇. E × H =  µH + ε E  + E . J ∂t  2 2 

(

)

(

)

Note:

( )

dx 1 d 2 x = x dt 2 dt

Integrating both sides over volume V,

(

)

    ∂ 1 1 2 2 − ∫ ∇. E × H dv = ∫  µH + ε E dv + ∫ E . Jdv ∂t v  2 2  v v

  J = σE

Using Gauss’ theorem:

(

)

   ∂ 1 1 2 2 2 − ∫ E × H .dS = µ H + ε E  dv + ∫ σE dv ∫ ∂t v  2 2  s v

Poynting theorem (time form)

(

)

 ∂ 1   1  2 −∫ E ×H . dS = µ H + ε E 2 dv + ∫ σE 2 dv ∫ ∂t v  2 2  s v L.H.S. :

    S = E × H = S kˆ where, S ….. Poynting vector and S ….. Power density [W/m2].      P = ∫ E × H . dA = ∫ S . dA where, P….. Total power crossing certain area [W]. A

(

)

R.H.S. :

A

(-ve ..means power flowing into the volume)

1 1   2 Wm = ∫ µH dv = ∫ µ(H.H)dv = ∫ w mdv v2 v2 v

.. Magnetic energy [J].

1 2 1  We = ∫ εE dv = ∫ ε(E.E)dv = ∫ w edv v2 v2 v

.. Electric energy [J].

W = Wm + We

… Total electromagnetic energy [J].

∂W … Rate of change (increase) of electromagnetic energy with time (Power) ∂t [W]. Ploss = ∫ σE 2 dv … Total power loss in the volume V [W]. v

Classification of media Dielectric media Free space:

ε o , µ o and σ = 0

Dissipative media Dielectric with finite conductivity:

ε , µ and σ ; (ω ε> σ) Lossless dielectric: ε , µ and σ = 0

Good dielectric:

ε = ε′ − jε′′ , µ and σ = 0 ; (ε′′ < < ε′) Dielectric with losses:

ε = ε′ − jε′′ , µ and σ = 0 ; (ε′′ > > ε′)

Good conductor:

ε , µ and σ ; (σ > > ω ε) Perfect conductor:

ε , µ and σ ; (σ → ∞ )

(2.4) Derivation and solution of wave equations in unbounded media: of media ( according to the values of ε , µ andσ ): (2.4.1)Types 1. Free space:

ε =ε

o

, µ = µo and σ = 0 .

2. Homogeneous media: ε , µ and σ are not functions of position. 3. Inhomogeneous media: ε and/or µ and/or σ are functions of position.   ε = ε(r ) and/or µ = µ(r ) and/or

 σ = σ (r )

4. Isotropic media: ε , µ and σ are not functions of direction.

  D = εE 5. Anisotropic media: ε and/or µ and/or σ are functions of direction.

 ε xx ε xy ε xz     ε = ε yx ε yy ε yz  and    ε zx ε zy ε zz   

 Dx   ε xx ε xy ε xz   Ex       D y  =  ε yx ε yy ε yz   E y        D z   ε zx ε zy ε zz   E z 

6. Linear media: ε , µ and σ are constants.

7. Non linear media: ε and/or µ and/or σ are functions of the applied field (source).    ε = ε(E) and/or µ = µ(H ) and/or σ = σ(E )

8. Dispersive media: µ ε =and/or and/or σσ= are ε = ε(f ) and/or µ(f ) µand/or σ(f )functions of frequency. 9. Lossy media: µ =complex µ′ − jµ′′ and/or σ ≠ 0. ε = ε′ − εjε′′and/or µ are 10. Simple media (linear isotropic): ε , µ and σ are constants.