Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9780078028151
Author: Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher: Mcgraw-hill Education,
Question
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Chapter 14, Problem 14.13P
To determine

(a)

The Eθs at P(r=100,θ=900,ϕ=300).

Expert Solution
Check Mark

Answer to Problem 14.13P

The value of electric field Eθs=0.2 ej1000πV/m.

Explanation of Solution

Given:

The following information is given:

   r=100,θ=900,ϕ=300.

The radiation field Eθs is given by,

   Eθs= (20r) sinθ ej10πr V/m

Calculation:

Using the condition r=100,θ=90°,ϕ=30° in the formula of Eθs ,

   Eθs=(20100) sin(900) ej1000π V/mEθs=0.2 ej1000π V/m

Conclusion:

The value of electric field Eθs=0.2 ej1000πV/m.

To determine

(b)

The Eθs at P (100 ,90°,30°).

Expert Solution
Check Mark

Answer to Problem 14.13P

The value of electric field Eθs= 0.2 ej1000π ej0.5π V/m.

Explanation of Solution

Given:

The vertical current element location is A(0.1,90°,90°) . The radiation field Eθs is given by,

   Eθs= (20r) sinθ ej10πr V/m

Concept Used:

Due to the change in the position of the vertical current element, there will be practically no change in the distance of P. However, this will enforce a change in the phase term which needs to be included in the overall electric field expression.

Calculation:

Consider two lines are drawn. One line is drawn from the origin to the point P and the other line is being drawn from y = 0.1 to P . Therefore, the difference in their length will be g=0.1sin(30°)=0.05

The overall electric field will be obtained by using the modified result obtained in part (a) with the phase factor due to the path difference. The electric field will be,

   Eθs=0.2 ej1000π ej10π(g) V/mEθs=0.2 ej1000π ej0.5π V/m

Conclusion:

The value of electric field Eθs= 0.2 e-j1000π ej0.5π V/m.

To determine

(c)

The Eθs at P (100 ,900,300).

Expert Solution
Check Mark

Answer to Problem 14.13P

The value of electric field Eθs= 0 V/m.

Explanation of Solution

Given:

The radiation field Eθs is given by,

   Eθs= (20r) sinθ ej10πr V/m

Identical vertical elements are located at A(0.1 ,90°,90°) and B(0.1,90°,270°).

Concept Used:

Due to the presence of the element at A(0.1 ,900,900) , the electric field obtained in part (b) will be present. In addition to this, due to the presence of the element at B(0.1 ,900,2700) , there will be additional path difference which will give rise to some additional phase difference.

Overall electric field can be found by adding the contribution of A(0.1 ,900,900) and B(0.1 ,900,2700).

Calculation:

Since the phase angle of 90° and 270° are 180° apart, the element at B(0.1 ,90°,270°) is essentially located at y=0.1 . This will give rise to a path difference of g'=0.1sin(300)=0.05 . Since this path is longer, the electric field due to the element at B(0.1,90°,270°) will be,

   Eθs=0.2ej1000πej0.5π V/m

Therefore, overall electric field will be,

   Eθs=0.2ej1000π (ej0.5π+e+j0.5π) V/mEθs=0.2ej1000π2cos(0.5π) V/mEθs=0 V/m.

Conclusion:

The value of electric field Eθs= 0 V/m.

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