Fundamentals of Electromagnetics with Engineering Applications
Fundamentals of Electromagnetics with Engineering Applications
5th Edition
ISBN: 9780471263555
Author: Stuart M. Wentworth
Publisher: John Wiley & Sons
Question
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Chapter 8, Problem 8.1P

(a)

To determine

The time harmonic electric field of an antenna.

(a)

Expert Solution
Check Mark

Answer to Problem 8.1P

The time harmonic electric field of an antenna is ηoIsrcos2θaϕ .

Explanation of Solution

Given:

The time harmonic magnetic field Hs is Isrcos2θ aθ and the phasor driving current Is is Ioejα .

Concept used:

The expression for time harmonic electric field can be written as,

  Es=ηoar×Hs ..... (1)

Calculation:

Substitute Isrcos2θ aθ for Hs in equation (1).

  Es=ηoar×( I s r cos2θ aθ)=ηoIs cos2θr(ar×aθ)=ηoIsrcos2θaϕ

Therefore, the electric field is ηoIsrcos2θaϕ .

Conclusion:

Thus, the time harmonic electric field of an antenna is ηoIsrcos2θaϕ .

(b)

To determine

The time averaged power density of a wave.

(b)

Expert Solution
Check Mark

Answer to Problem 8.1P

The time averaged power density of a wave is 12ηoIo2r2cos4θ ar .

Explanation of Solution

Concept used:

The expression for time averaged power density can be written as,

  P(r,θ,ϕ)=12Re[Es×Hs] ..... (2)

Calculation:

From part (a), the time harmonic electric field is ηoIsrcos2θaϕ .

The conjugate of time harmonic magnetic field is calculated as,

  Hs=[ I s r cos2θ aθ]=[ I o e jα r cos2θ aθ]=Ioe jαrcos2θ aθ

Substitute ηoIsrcos2θaϕ for Es and Ioejαrcos2θ aθ for Hs in equation (2).

  P(r,θ,ϕ)=12Re[ η o I srcos2θaϕ×( I o e jα r cos 2θ  a θ)]=12Re[ η o I o e jαrcos2θaϕ×( I o e jα r cos 2θ  a θ)]=12Re[ η o I o 2 r 2cos4θ( a ϕ× a θ)]=12ηoIo2r2cos4θ ar

Therefore, the vector for time averaged power density is 12ηoIo2r2cos4θ ar .

Conclusion:

Thus, the time averaged power density of a wave is 12ηoIo2r2cos4θ ar .

(c)

To determine

The radiation resistance of the antenna.

(c)

Expert Solution
Check Mark

Answer to Problem 8.1P

The radiation resistance of the antenna is 950 Ω .

Explanation of Solution

Concept used:

The total power radiated by antenna is,

  Prad=P(r,θ,ϕ)r2sinθdθdϕ ..... (3)

Calculation:

From part (a), the time averaged power density is 12ηoIo2r2cos4θ ar .

Substitute 12ηoIo2r2cos4θ for P(r,θ,ϕ) in equation (3).

  Prad= 1 2 η o I o 2 r 2 cos 4θ  r 2sinθdθdϕ=12ηoIo20π cos4θsinθdθ02πdϕ=12ηoIo20πcos4θsinθdθ[2π]=25ηoπIo2

The relation between power radiation and radiation resistance is shown below.

  Prad=12Io2Rrad

Substitute 25ηoπIo2 for Prad in above equation.

  25ηoπIo2=12Io2RradRrad=45πηo

The value of intrinsic impedance is 120π .

So, the radiation resistance can be written as,

  Rrad=45π(120π)=96π2950 Ω

Therefore, the value of Rrad is 950 Ω .

Conclusion:

Thus, the radiation resistance of the antenna is 950 Ω .

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