Electromagnetic Waves Part Ii

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

References •

William H. Hayt, “Engineering Electromagnetic,” McGraw-Hill, 1989.

•

Plonsey & Collin, “ Principles and Applications of Electromagnetic Fields,” McGraw-Hill, 1962.

•

F.T. Ulaby, “Fundamentals of Applied Electro-magnetic,” Prentice-Hall, 1997.

Derivation and solution of wave equation

How Maxwell’s equations used to show wave equation? Consider the electric and magnetic fields in a region does not include any sources, which called:

Source-free

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

(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. 2.4.2.1 Time form. a. In free-apace ( ε = ε o , µ = µo and σ = 0 ). b. Lossless dielectric ( ε = ε o ε r , µ = µoµr and σ = 0 ). c. Lossy dielectric ( ε =ε o ε r , µ = µoµr and σ ≠ 0 ).

2.4.2.2 Complex form.

2.4.3 Properties of plane wave.

a) In free-apace: ( ε = ε o , µ = µo and σ = 0 ) (2.4.2.1)Time form:

  ∂H ∇ × E = −µo ..... (1) ∂t

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

To derive the wave equation for the electric field: Take the curl of both sides of eqn. (1) :   ∂ ∇ × ∇ × E = −µo (∇ × H ) ∂t

Using vector relationship:    ∂ ∂E 2 ∇ (∇.E ) − ∇ E = − µo (εo ) ∂t ∂t 0

Similarly;

  ∂ E 2 ∇ E − µoεo 2 = 0 ∂t 2

Generally:

 2  ∂ ∇ − µo εo  2   ∂ t   2

 2  ∂ H 2 ∇ H − µoεo 2 = 0 ∂t

 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 fact that the experimentally determined speed of light is also 3x108 m / s. So the light is just a form of electromagnetic wave propagation

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)

b) Lossless dielectric: ( ε = ε o ε r , µ = µoµr and σ = 0 )   ∂H Similarly; ∇ × E = −µ ..... (1)  ∂t 2

 ∂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 − µε 2 = 0 ∂t

 ∂ H ∇ H − µε 2 = 0 ∂t 2

Generally:

 2  ∂  ∇ − µε  2  ∂t   2

 E    = 0  H  

As the previous case we can prove that,

1 c v= = .........wave velocity µε 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 o ε r , µ = µoµr and σ ≠ 0 )   ∂H Similarly; ∇ × E = −µ ..... (1) ∂t   

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

2  ∂ H ∂ H ∇ 2H − µσ − µε 2 = 0 ∂t ∂t

Generally:

 2   ∇ − µσ ∂ − 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 − µσ − µε 2 = 0 ∂t ∂t

Source-free wave equation where,

in lossy dielectric

1 v= .........wave velocity µε  ∂  E  µσ    ........damping term ∂t  H 