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

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  • 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