Prof. Rakhesh Singh Kshetrimayum

Solutions to selected problems from

Exercise 5

Prof. Rakhesh Singh Kshetrimayum

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

Exercise 5.1

� Since

� the direction of wave propagation

ˆ ˆx yk β

+ =

r

( ) 2

0ˆ ˆ

x yj

E E x y eβ

+ −

= − +r

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

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

ˆ ˆ

2

x yk β

+ =

Exercise 5.1

0

0

0 0

2 2

ˆ ˆ ˆ

0

x y x yj j

x y z

EEE j H H

j j x y z

e eβ β

ωµωµ ωµ

+ + − −

∇× ∂ ∂ ∂∇× = − ⇒ = =

− − ∂ ∂ ∂

−

rr r r

( ) 2

0ˆ ˆ

x yj

E E x y eβ

+ −

= − +r

3/23/2018Electromagnetic 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 yj j j j

j

e e

E e e e ex y z

j z z x y

E je

j

β β β β

β

ωµ

β

ωµ

+ + + + − − − −

−

−

∂ ∂ ∂ ∂

= − − + + − ∂ ∂ ∂ ∂

−=

−2 2 2 2

0 0

0 0

2 2ˆ ˆ ˆ

2

x y x y x y x yj j jj j

e z E e z E e zj

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 2ˆ ˆ ˆRe Re

2 2 2

x y x yj j

avg

y xES E H E x y e E e z

β β

η η

+ + − +

+

= × = − + × =

r r r

3/23/2018Electromagnetic 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

( ) ( )22tzftzf ±∂

=±∂ ωβ

εµωβ

( )tzf ωβ ±

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

( ) ( )2002

t

tzf

z

tzf

∂

±∂=

∂

±∂ ωβεµ

ωβ

( ) ( )

∂

±∂

∂

∂=

∂

±∂

∂

∂⇒

t

tzf

tz

tzf

z

ωβεµ

ωβ00

Exercise 5.2

( ) ( )( )

( ) ( )( )

( )( )

( ) ( )( )

±∂

±∂±

∂

∂=

±∂

±∂

∂

∂⇒

±∂

±∂

∂

±∂

∂

∂=

±∂

±∂

∂

±∂

∂

∂

00

00

tz

tzf

ttz

tzf

z

tz

tzf

t

tz

ttz

tzf

z

tz

z

ωβ

ωβωεµ

ωβ

ωββ

ωβ

ωβωβεµ

ωβ

ωβωβ

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

( )( )

( )

( )( )

( ) ( )( )

±∂

±∂=

±∂

±∂⇒

±∂±

∂=

±∂∂⇒

2

22

002

22

00

tz

tzf

tz

tzf

tzttzz

ωβ

ωβωεµ

ωβ

ωββ

ωβωεµ

ωββ

Exercise 5.2

( ) ( )cv p =⇒=⇒=⇒

00

2

2

2

00

2 1

εµβ

ωωεµβ

3/23/2018Electromagnetic 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

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

Exercise 5.4

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

( )0ˆ ˆ( )

j ya E E jx z e

β−= +r

( ) tjyjeexjzEE

ωβ−+= ˆˆr

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� Taking the real part of the wave( ) tjyj

eexjzEE += ˆˆ0

( )

+−+−= xytzytEE ˆ

2cosˆcos0

πβωβω

r

Exercise 5.4

� Observing at y=0

( ) ( )( )xtztEE ˆsinˆcos0 ωω −=r

x

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

z

x

Time t=∆t

Time t=0

LHCP

Exercise 5.4

� Obviously EP

� Putting time dependence

( ) ( ){ }0ˆ ˆ( ) 2 1 3

j yb E E j x z j e

β−= + + +r

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

� Taking real part

( ) tjyjj

jeexezeEE

ωβ−

+=

−−

ˆ5ˆ10 2

1tan

3tan

0

11r

( ) ( )( )xytzytEE ˆ6.26cos5ˆ6.71cos1000

0 +−++−= βωβωr

Exercise 5.4� Observing at y=0

( ) ( )( )xtztEE ˆ6.26cos5ˆ6.71cos1000

0 +++= ωωr

x

3/23/2018Electromagnetic 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

( )0ˆ ˆ( )

j zc H H x jy e

β+= −r

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

� Observing at z=0

( )

++

−+= ztxyztHH βω

πβω cosˆˆ

2cos0

r

( ) ( )( )txytHH ωω cosˆˆsin0 +=r

Exercise 5.4

x Time t=0

( ) ( )( )txytHH ωω cosˆˆsin0 +=r

3/23/2018Electromagnetic 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

sin52

cos5,cos3

2222

=+=+⇒

−=

+==

ttEE

ttEtE

zy

zy

ωω

ωπ

ωω

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

� Maximum field magnitude equals 5 V/m

( ) ( ) 1sincos259

22 =+=+⇒ ttEE

zy ωω

Exercise 5.5

� yzx forms a right handed system

z

( ) ( )ztytE ˆsin5ˆcos3 ωω −=r

3/23/2018Electromagnetic 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

3/23/2018Electromagnetic 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

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

� The average Poynting vector reduces at the rate of

� inside the solid ground

297901922.02

==ε

µσα

ze

α2−

Exercise 5.9� Hence, for depth into the ground

�where the average power of the EM wave has been reduced by 1%

� to that of the surface is given by

( ) ( )α

αα

2

01.0ln01.0ln201.0

2 −=⇒=−⇒=−zze

z

3/23/2018Electromagnetic Field Theory by Rakhesh Singh Kshetrimayum19

� For f=10kHz,

� the required depth is approximately 163.89m

� whereas for f=100MHz,

� it is approximately 7.72934m

( )α

α2

01.0ln201.0 −=⇒=−⇒= zze

Exercise 5.9

� (c) In order to image an object at a depth of 50 m,

� for roundtrip of the radar signals we need a distance of 100m

� From the knowledge of the above part of the section (100 MHz won’t be able to get that far),

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(100 MHz won’t be able to get that far),

�we may assume that the solid ground should

� behave as a good conductor for the problem at hand

Exercise 5.9

� Then for a power reflected back from the object

�which has dropped to just 1% of the original power,

�we have,

( ) ( ){ }22201.0ln401.0ln201.0 =⇒=−⇒=−

zfzfez µσπµσπα

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( ) ( ){ }22201.0ln401.0ln201.0 =⇒=−⇒=−

zfzfez µσπµσπα

( ){ } kHzz

f 86.2601.0ln4

1 2

2≅=⇒

πµσ

Exercise 5.9

� Hence the wavelength at this frequency is approximately 1.117 km,

� therefore, we may not be able to image at any small object at this depth

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