Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 10, Problem 75P

(a)

To determine

Find incidence angle.

(a)

Expert Solution
Check Mark

Answer to Problem 75P

The incidence angle is 36.87°_.

Explanation of Solution

Calculation:

Consider the expression to find the incidence angle.

cos(θi)=akan

Rearrange the expression for incidence angle.

θi=cos1(akan)        (1)

Consider the expression to find ak.

ak=k|k|        (2)

Here,

k is the propagation vector.

From the given electric field, the propagation vector is 4ay+3az.

Substitute 4ay+3az for k in Equation (2).

ak=4ay+3az42+32=4ay+3az5

From the given data, an=ay.

Substitute (4ay+3az5) for ak and ay for an in Equation (1).

θi=cos1[(4ay+3az5)ay]=cos1(45)36.87°

Conclusion:

Thus, the incidence angle is 36.87°_.

(b)

To determine

Find the time-average power in air.

(b)

Expert Solution
Check Mark

Answer to Problem 75P

The time-average power in air is 106.1ay+79.58azmW/m2_.

Explanation of Solution

Calculation:

Consider the expression to find the time-average power in air.

Pave=Eo22ηoak        (3)

Here,

Eo is the magnitude of electric field.

ηo is the intrinsic impedance in the air, which is 120πΩ.

Find Eo from the given data.

Eo=82+(6)2V/m=10V/m

Substitute 10V/m for Eo, 120πΩ for ηo, and (4ay+3az5) for ak in Equation (3).

Pave=(10V/m)22(120πΩ)(4ay+3az5)106.1ay+79.58azmW/m2

Conclusion:

Thus, the time-average power in air is 106.1ay+79.58azmW/m2_.

(c)

To determine

Find the reflected and transmitted electric fields.

(c)

Expert Solution
Check Mark

Answer to Problem 75P

The reflected and transmitted electric fields are (1.518ay+2.02az)sin(ωt+4y3z)V/m_ and (1.879ay5.968az)sin(ωt9.539y3z)V/m_, respectively.

Explanation of Solution

Calculation:

Consider the expression for reflected electric field for the given data.

Er=Ero[sin(θr)ay+cos(θr)az]sin(ωtkrr)        (4)

The reflection angle is equal to incidence angle. Therefore, θr=θi.

θr=36.87°

Consider the expression of reflection coefficient in the parallel polarization of oblique incidence.

Γ=η2cos(θt)η1cos(θi)η2cos(θt)+η1cos(θi)        (5)

Here,

η1 is the intrinsic impedance of the medium 1 (air),

η2 is the intrinsic impedance for medium 2 (dielectric half-space),

θt is the angle of transmission, and

θi is the angle of incidence.

Rewrite Equation (5) for the given data.

Γ=ηo2cos(θt)ηocos(θi)ηo2cos(θt)+ηocos(θi)

As Γ=EroEio, rewrite the expression.

EroEio=ηo2cos(θt)ηocos(θi)ηo2cos(θt)+ηocos(θi)

Rearrange the expression for Ero.

Ero=[ηo2cos(θt)ηocos(θi)ηo2cos(θt)+ηocos(θi)]Eio        (6)

Consider the expression to find the angle of transmission θt.

sin(θt)sin(θi)=μ1ε1μ2ε2

Rewrite the expression for the given data.

sin(θt)sin(θi)=μoεo(2μo)(2εo) {μ1=μoε1=εoμ2=2μo for half-space dielectricε2=2εo for half-space dielectric}=0.5

Rearrange the expression to find the angle of transmission θt.

θt=sin1[0.5sin(θi)]

Substitute 36.87° for θi.

θt=sin1[0.5sin(36.87°)]=17.46°

Substitute 10V/m for Eio, 120πΩ for ηo, 36.87° for θi, and 17.46° for θt in Equation (6).

Ero={(120πΩ2)cos(17.46°)(120πΩ)cos(36.87°)(120πΩ2)cos(17.46°)+(120πΩ)cos(36.87°)}(10V/m)2.53V/m

From the given Figure, find kr.

kr=kryay+krzaz=krcos(θr)ay+krsin(θr)az=(5)cos(36.87°)ay+(5)sin(36.87°)az=4ay+3az

From the given Figure, find krr.

krr=(0ax4ay+3az)(xax+yay+zaz)=4y+3z

Substitute 2.53V/m for Ero, 36.87° for θr, and 4y+3z for krr in Equation (4).

Er=(2.53V/m)[sin(36.87°)ay+cos(36.87°)az]sin[ωt(4y+3z)](1.518ay+2.02az)sin(ωt+4y3z)V/m

Consider the expression for transmitted electric field.

Et=Eto[sin(θt)aycos(θt)az]sin(ωtktr)        (7)

Consider the expression of transmission coefficient in the parallel polarization of oblique incidence.

τ=2η2cos(θi)η2cos(θt)+η1cos(θi)

Rewrite the expression for the given data.

τ=2(ηo2)cos(θi)ηo2cos(θt)+ηocos(θi)=2ηocos(θi)ηocos(θt)+2ηocos(θi)

As τ=EtoEio, rewrite the expression.

EtoEio=2ηocos(θi)ηocos(θt)+2ηocos(θi)

Rearrange the expression for Eto.

Eto=[2ηocos(θi)ηocos(θt)+2ηocos(θi)]Eio

Substitute 10V/m for Eio, 120πΩ for ηo, 36.87° for θi, and 17.46° for θt.

Eto=[2(120πΩ)cos(36.87°)(120πΩ)cos(17.46°)+2(120πΩ)cos(36.87°)](10V/m)6.265V/m

From the given Figure, find kt.

kt=ktyay+ktzaz

kt=ktcos(θt)ay+ktsin(θt)az        (8)

Find kt.

kt=β2=ωμ2ε2=ω(2μo)(2εo)=2ωμoεo

Find ki.

ki=β1=ωμ1ε1=ω(μo)(εo)=ωμoεo

Find the value of kt.

ktki=2ωμoεoωμoεo=2

Simplify the expression.

kt=2ki=2β1=2(5) {β1=5rad/m}=10

Substitute 10 for kt and 17.46° for θt in Equation (8).

kt=10cos(17.46°)ay+(10)sin(17.46°)az9.539ay+3az

From the given Figure, find ktr.

ktr=(0ax+9.539ay+3az)(xax+yay+zaz)=9.539y+3z

Substitute 6.265V/m for Eto, 17.46° for θt, and 9.539y+3z for ktr in Equation (7).

Et=(6.265V/m)[sin(17.46°)aycos(17.46°)az]sin[ωt(9.539y+3z)](1.879ay5.968az)sin(ωt9.539y3z)V/m

Conclusion:

Thus, the reflected and transmitted electric fields are (1.518ay+2.02az)sin(ωt+4y3z)V/m_ and (1.879ay5.968az)sin(ωt9.539y3z)V/m_, respectively.

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Chapter 10 Solutions

Elements Of Electromagnetics

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