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 76P

(a)

To determine

Find reflected and transmitted angles.

(a)

Expert Solution
Check Mark

Answer to Problem 76P

The reflected and transmitted angles are 19.47°_ and 90°_, respectively.

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 magnetic field, the propagation vector is kax+k8az.

Substitute kax+k8az for k in Equation (2).

ak=kax+k8azk2+(k8)2=(13)(ax+8az)

From the given data, an=az.

Substitute (13)(ax+8az) for ak and az for an in Equation (1).

θi=cos1[(13)(ax+8az)az]=cos1(83)19.47°

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

θr=19.47°

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)=μo(9εo)μoεo {μ1=μoε1=9εoμ2=μo for airε2=εo for air}=3

Rearrange the expression to find the angle of transmission θt.

θt=sin1[3sin(θi)]

Substitute 19.47° for θi.

θt=sin1[3sin(19.47°)]90°

Conclusion:

Thus, the reflected and transmitted angles are 19.47°_ and 90°_, respectively.

(b)

To determine

Find the value of k.

(b)

Expert Solution
Check Mark

Answer to Problem 76P

The value of k is 3.333_.

Explanation of Solution

Calculation:

The magnitude of propagation vector is equal to the value of phase constant. Write the expression for phase constant of the medium 1 (dielectric medium).

β1=ωμ1ε1

Rewrite the expression for given data.

β1=ωμo(9εo)

Substitute 109rad/s for ω, 4π×107H/m for μo, and 10936πF/m εo.

β1=(109rad/s)(4π×107H/m)(9)(10936πF/m)=10rad/m

Find the magnitude of the propagation vector.

|k|=k2+(k8)2=3k

As β1=|k|, write the expression.

β1=|k|10rad/m=3k

Simplify the expression for k.

k=103=3.333

Conclusion:

Thus, the value of k is 3.333_.

(c)

To determine

Find the wavelength in the dielectric and in air.

(c)

Expert Solution
Check Mark

Answer to Problem 76P

The wavelength in the dielectric and in air are 0.6283m_ and 1.885m_, respectively.

Explanation of Solution

Calculation:

Consider the expression to find the wavelength.

λ=2πβ        (1)

Modify the expression for wavelength in the dielectric medium.

λ1=2πβ1

Substitute 10rad/m for β1.

λ1=2π10rad/m0.6283m

Write the expression for phase constant of the medium 2 (air).

β2=ωμ2ε2

Rewrite the expression for given data.

β2=ωμoεo

Substitute 109rad/s for ω, 4π×107H/m for μo, and 10936πF/m εo.

β2=(109rad/s)(4π×107H/m)(10936πF/m)=103rad/m

Modify Equation (1) for wavelength in the air (medium 2).

λ2=2πβ2

Substitute 103rad/m for β2.

λ2=2π103rad/m1.885m

Conclusion:

Thus, the wavelength in the dielectric and in air are 0.6283m_ and 1.885m_, respectively.

(d)

To determine

Find the incident electric field.

(d)

Expert Solution
Check Mark

Answer to Problem 76P

The incident electric field is (23.6954ax8.3776az)cos(109tkxk8z)V/m_.

Explanation of Solution

Calculation:

Consider the expression to find the incident electric field for the given data.

Ei=η1H×ak        (2)

Find the intrinsic impedance of medium 1.

η1=μ1ε1

Rewrite the expression for given data.

η1=μo9εo

Substitute  4π×107H/m for μo and 10936πF/m εo.

η1=4π×107H/m9(10936πF/m)=40πΩ

Substitute 40πΩ for η1, 0.2cos(109tkxk8z)ayA/m for H, and (13)(ax+8az) for ak.

Ei=(40πΩ)[0.2cos(109tkxk8z)ayA/m]×[(13)(ax+8az)]=[(40πΩ)(0.2)cos(109tkxk8z)(13)(ay×ax)+(40πΩ)(0.2)cos(109tkxk8z)(83)(ay×az)]V/m[8.3776cos(109tkxk8z)(az)+23.6954cos(109tkxk8z)ax]V/m {ay×ax=azay×az=ax}(23.6954ax8.3776az)cos(109tkxk8z)V/m

Conclusion:

Thus, the incident electric field is (23.6954ax8.3776az)cos(109tkxk8z)V/m_.

(e)

To determine

Find the transmitted and reflected electric fields.

(e)

Expert Solution
Check Mark

Answer to Problem 76P

The transmitted and reflected electric fields are 1357cos(109t3.333x)azV/m_ and (213.3ax+75.4az)cos(109tkx+k8z)V/m_, respectively.

Explanation of Solution

Calculation:

Consider the expression for transmitted electric field.

Et=Eto[cos(θt)axsin(θt)az]cos[109tβ2xsin(θt)+β2zcos(θt)]        (3)

Consider the polarization as parallel polarization.

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=2ηocos(θi)ηocos(θt)+η1cos(θi) {η2=ηo}

As τ=EtoEio, rewrite the expression.

EtoEio=2ηocos(θi)ηocos(θt)+η1cos(θi)

Rearrange the expression for Eto.

Eto=[2ηocos(θi)ηocos(θt)+η1cos(θi)]Eio

Substitute the corresponding parameters and find the value of Eto.

Eto=[2(120πΩ)cos(19.47°)(120πΩ)cos(90°)+(40πΩ)cos(19.47°)][(24π)(3)V/m]1357V/m

Substitute 1357V/m for Eto, 90° for θt, 3.333rad/m for β2 in Equation (3).

Et=1357[cos(90°)axsin(90°)az]cos[109t3.333xsin(90°)+3.333zcos(90°)]V/m=1357cos(109t3.333x)azV/m

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

Er=Ero[cos(θr)ax+sin(θr)az]cos(109tkx+k8z)        (4)

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

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

Rewrite the expression for the given data.

Γ=ηocos(θt)η1cos(θi)ηocos(θt)+η1cos(θi)

As Γ=EroEio, rewrite the expression.

EroEio=ηocos(θt)η1cos(θi)ηocos(θt)+η1cos(θi)

Rearrange the expression for Ero.

Ero=[ηocos(θt)η1cos(θi)ηocos(θt)+η1cos(θi)]Eio

Substitute the corresponding parameters and find the value of Ero.

Ero=[(120πΩ)cos(90°)(40πΩ)cos(19.47°)(120πΩ)cos(90°)+(40πΩ)cos(19.47°)][(24π)(3)V/m]=226.1946V/m

Substitute 226.1946V/m for Ero and 19.47° for θr in Equation (4).

Er=226.1946[cos(19.47°)ax+sin(19.47°)az]cos(109tkx+k8z)V/m(213.3ax+75.4az)cos(109tkx+k8z)V/m

Conclusion:

Thus, the transmitted and reflected electric fields are 1357cos(109t3.333x)azV/m_ and (213.3ax+75.4az)cos(109tkx+k8z)V/m_, respectively.

(f)

To determine

Find the Brewster angle.

(f)

Expert Solution
Check Mark

Answer to Problem 76P

The Brewster angle is 18.43°_.

Explanation of Solution

Calculation:

Consider the expression to find the Brewster angle for the parallel-polarized wave.

tanθB=ε2ε1

Rewrite the expression for the given data.

tanθB=εo9εo=13

Rearrange the expression for Brewster angle.

θB=tan1(13)18.43°

Conclusion:

Thus, the Brewster angle is 18.43°_.

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Chapter 10 Solutions

Elements Of Electromagnetics

Ch. 10.10 - Prob. 11PECh. 10.10 - Prob. 12PECh. 10.11 - Prob. 13PECh. 10 - Prob. 1RQCh. 10 - Prob. 2RQCh. 10 - Prob. 3RQCh. 10 - Prob. 4RQCh. 10 - Prob. 5RQCh. 10 - Prob. 6RQCh. 10 - Prob. 7RQCh. 10 - Prob. 8RQCh. 10 - Prob. 9RQCh. 10 - Prob. 10RQCh. 10 - Prob. 1PCh. 10 - Prob. 2PCh. 10 - Prob. 3PCh. 10 - Prob. 4PCh. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - Prob. 8PCh. 10 - Prob. 9PCh. 10 - Prob. 10PCh. 10 - Prob. 11PCh. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - Prob. 15PCh. 10 - Prob. 16PCh. 10 - Prob. 17PCh. 10 - Prob. 18PCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 25PCh. 10 - Prob. 26PCh. 10 - Prob. 27PCh. 10 - Prob. 28PCh. 10 - Prob. 29PCh. 10 - Prob. 30PCh. 10 - Prob. 31PCh. 10 - Prob. 32PCh. 10 - Prob. 33PCh. 10 - Prob. 34PCh. 10 - Prob. 35PCh. 10 - Prob. 36PCh. 10 - Prob. 37PCh. 10 - Prob. 38PCh. 10 - Prob. 39PCh. 10 - Prob. 40PCh. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Prob. 43PCh. 10 - Prob. 44PCh. 10 - Prob. 45PCh. 10 - Prob. 46PCh. 10 - Prob. 47PCh. 10 - Prob. 48PCh. 10 - Prob. 49PCh. 10 - Prob. 50PCh. 10 - Prob. 51PCh. 10 - Prob. 52PCh. 10 - Prob. 53PCh. 10 - Prob. 54PCh. 10 - Prob. 55PCh. 10 - Prob. 56PCh. 10 - Prob. 57PCh. 10 - Prob. 58PCh. 10 - Prob. 59PCh. 10 - Prob. 60PCh. 10 - Prob. 61PCh. 10 - Prob. 62PCh. 10 - Prob. 63PCh. 10 - Prob. 64PCh. 10 - Prob. 65PCh. 10 - Prob. 66PCh. 10 - Prob. 67PCh. 10 - Prob. 68PCh. 10 - Prob. 69PCh. 10 - Prob. 70PCh. 10 - Prob. 71PCh. 10 - Prob. 72PCh. 10 - Prob. 73PCh. 10 - Prob. 74PCh. 10 - Prob. 75PCh. 10 - Prob. 76PCh. 10 - Prob. 78PCh. 10 - Prob. 79PCh. 10 - Prob. 80PCh. 10 - Prob. 81PCh. 10 - Prob. 82PCh. 10 - Prob. 83PCh. 10 - Prob. 84P
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