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### Transcript of Solutions to selected problems from Exercise 5 Prof. Rakhesh Singh Kshetrimayum Solutions to...

• Prof. Rakhesh Singh Kshetrimayum

Solutions to selected problems from

Exercise 5

Prof. Rakhesh Singh Kshetrimayum

3/23/20181 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum

• Exercise 5.1

� Since

� the direction of wave propagation

ˆ ˆx y k β

+  =  

r

( ) 20 ˆ ˆ x y

j

E E x y e β

+  −  

 = − + r

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum2

� the corresponding magnetic field can be obtained from Maxwell’s curl equation

ˆ ˆ

2

x y k β

+  =  

 

• Exercise 5.1

0

0

0 0

2 2

ˆ ˆ ˆ

0

x y x y j j

x y z

EE E j H H

j j x y z

e e β β

ωµ ωµ ωµ

+ +    − −   

   

∇× ∂ ∂ ∂ ∇× = − ⇒ = =

− − ∂ ∂ ∂

r r r r

( ) 20 ˆ ˆ x y

j

E E x y e β

+  −  

 = − + r

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum3

2 2

2 2 2 2

0

0

0

0

0

ˆ ˆ ˆ

2

x y x y x y x y j j j j

j

e e

E e e e e x y z

j z z x y

E j e

j

β β β β

β

ωµ

β

ωµ

   

+ + + +        − − − −       

       

    ∂ ∂ ∂ ∂ 

= − − + +   − ∂ ∂ ∂ ∂  

  

− =

− 2 2 2 2

0 0

0 0

2 2 ˆ ˆ ˆ

2

x y x y x y x y j j jj j

e z E e z E e z j

β β ββ β

ωµ η

+ + + +        − − −       

          − −  + = =    −   

• Exercise 5.1

� time-average power flow per unit area

( ) ( ) ( ) 2

* 2 2 0

0 0

0 0

ˆ ˆ1 1 2 ˆ ˆ ˆRe Re

2 2 2

x y x y j j

avg

y xE S E H E x y e E e z

β β

η η

+ +    − +   

      + 

= × = − + × =    

r r r

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum4

• Exercise 5.2

� For any function to be a solution of the wave equation, it must satisfy the following equation

( ) ( )22 tzftzf ±∂ =

±∂ ωβ εµ

ωβ

( )tzf ωβ ±

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum5

( ) ( ) 2002

t

tzf

z

tzf

±∂ =

±∂ ωβ εµ

ωβ

( ) ( )   

  

±∂

∂ =

  

  

±∂

∂ ⇒

t

tzf

tz

tzf

z

ωβ εµ

ωβ 00

• Exercise 5.2

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) 

  

±∂

±∂ ±

∂ =

  

±∂

±∂

∂ ⇒

  

  

±∂

±∂

±∂

∂ =

  

  

±∂

±∂

±∂

00

00

tz

tzf

ttz

tzf

z

tz

tzf

t

tz

ttz

tzf

z

tz

z

ωβ

ωβ ωεµ

ωβ

ωβ β

ωβ

ωβωβ εµ

ωβ

ωβωβ

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum6

( ) ( )

( )

( ) ( )

( ) ( ) ( ) 

 

  

±∂

±∂ =

  

  

±∂

±∂ ⇒

 

 

±∂ ±

∂ =

 

 

±∂∂ ⇒

2

2 2

002

2 2

00

tz

tzf

tz

tzf

tzttzz

ωβ

ωβ ωεµ

ωβ

ωβ β

ωβ ωεµ

ωβ β

• Exercise 5.2

( ) ( ) cv p =⇒=⇒=⇒ 00

2

2

2

00

2 1

εµβ

ω ωεµβ

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum7

� This wave has phase velocity equal to speed of light

� Hence, any function f(βz±ωt) is a solution of the wave equation

• Exercise 5.3

� Already solved in the class

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum8

• Exercise 5.4

� Note that zxy form a right handed system, putting in time dependence

( )0 ˆ ˆ( ) j y

a E E jx z e β−= +

r

( ) tjyj eexjzEE ωβ−+= ˆˆ r

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum9

� Taking the real part of the wave ( ) tjyj eexjzEE += ˆˆ0

( )  

  

  

  

 +−+−= xytzytEE ˆ

2 cosˆcos0

π βωβω

r

• Exercise 5.4

� Observing at y=0

( ) ( )( )xtztEE ˆsinˆcos0 ωω −= r

x

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum10

z

x

Time t=∆t

Time t=0

LHCP

• Exercise 5.4

� Obviously EP

� Putting time dependence

( ) ( ){ }0 ˆ ˆ( ) 2 1 3 j yb E E j x z j e β−= + + + r



3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum11

� Taking real part

( ) tjyj j

j eexezeEE

ωβ−  

  

 

 

 +=

− −

ˆ5ˆ10 2 1

tan 3tan

0

1 1r

( ) ( )( )xytzytEE ˆ6.26cos5ˆ6.71cos10 000 +−++−= βωβω r

• Exercise 5.4 � Observing at y=0

( ) ( )( )xtztEE ˆ6.26cos5ˆ6.71cos10 000 +++= ωω r

x

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum12

z

x

Time t=∆t Time t=0

RHEP

• Exercise 5.4

� In that case, yx (-z) forms a right handed system

� Putting in time dependence and taking real part

( )0 ˆ ˆ( ) j z

c H H x jy e β+= −

r

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum13

� Observing at z=0

( ) 

  

 ++

  

 −+= ztxyztHH βω

π βω cosˆˆ

2 cos0

r

( ) ( )( )txytHH ωω cosˆˆsin0 += r

• Exercise 5.4

x Time t=0

( ) ( )( )txytHH ωω cosˆˆsin0 += r

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum14

y

x

Time t=∆t

Time t=0

LHCP

• Exercise 5.5

� equation of the ellipse of polarization

( ) ( )

( ) ( ) 1sincos

sin5 2

cos5,cos3

22 22

=+=+⇒

−= 

  

 +==

tt EE

ttEtE

zy

zy

ωω

ω π

ωω

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum15

� Maximum field magnitude equals 5 V/m

( ) ( ) 1sincos 259

22 =+=+⇒ tt EE zy ωω

• Exercise 5.5

� yzx forms a right handed system

z

( ) ( )ztytE ˆsin5ˆcos3 ωω −= r

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum16

y

z

Time t=∆t

Time t=0

LHEP

• Exercise 5.9 � (a) Loss tangent of a medium is given by � At frequency f=10 kHz,

� loss tangent is approximately equal to 899 which is greater than 100 and

� hence the solid ground acts as a good conductor

ωε

σ

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum17

� hence the solid ground acts as a good conductor � At frequency f=100 MHz,

� loss tangent is approximately equal to 0.0899 which is closer to 0.01 and

� hence the solid ground may act as a bad conductor or good dielectric

• Exercise 5.9

� For good conductor,

� and good dielectric

014049629.0== µσπα f

3/23/2018 Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum18

� The average Poynting vector reduces at the rate of

� inside the solid ground

297901922.0 2

== ε

µσ α

z e