Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9781260029963
Author: Hayt
Publisher: MCG
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Question
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Chapter 1, Problem 1.18P
To determine

(a)

The value of S=E×H and convert the value of S in rectangular coordinates.

Expert Solution
Check Mark

Answer to Problem 1.18P

The value of S=E×H is [(AB r 2)sin2θ]r^ and the rectangular expression for S is

   (ABr2)sin2θ(sinθcosϕax+sinθsinϕay+cosθaz).

Explanation of Solution

Given:

The value of E is E=(Ar)sinθaθ.

The value of H is H=(Br)sinθaϕ

The expression for S

is Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  1

Concept used:

Write the expression for cross product of two vector.

   A×B=|r^θ^ϕ^ A r A θ A ϕ B r B θ B ϕ| ...... (1)

Here, r is the radius, θ is angle between z axis and line drawn from origin, ϕ is the angle between x axis and line drawn from origin, Ar is the component of vector A along r^ , Aθ

is the component of vector Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  2along Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  3m:math display='block'>Aϕ

is the component of vector Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  4along Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  5m:math display='block'>Br is the component of vector B along r^ , Bθ is the component of vector B

along Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  6m:math display='block'>Bϕ is the component of vector B

along Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  7

Calculation:

Substitute 0

for Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  8m:math display='block'>(Ar)sinθ

for Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  9m:math display='block'>0

forEngineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  10m:math display='block'>0

for Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  11Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  12for Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  13m:math display='block'>(Br)sinθ

forEngineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  14Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  15for Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  16and Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  17for B in equation (1).

   E×H=| r ^ θ ^ ϕ ^ 0 ( A r )sinθ 0 0 0 ( B r )sinθ|=r^[(( A r )sinθ)(( B r )sinθ)0]θ^[(0)(( B r )sinθ)0]+ϕ^[0(0)(( A r )sinθ)]=( AB r 2 )sin2θr^θ^(0)+ϕ^(0)=[( AB r 2 )sin2θ]r^ Substitute [(AB r 2)sin2θ]r^ for E×H in given equation.

   S=[(AB r 2)sin2θ]r^

Write the expression of x component for S.

Sx=S.ax ...... (2)

Here, Sx

is Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  18component for S.

Substitute [(AB r 2)sin2θ]r^ for S in equation (2).

   Sx=[( AB r 2 )sin2θ]r^.ax=[( AB r 2 )sin2θ]ar.ax=[( AB r 2 )sin2θ](sinθcosϕ)            {ar.ax=sinθcosϕ}

Simplify further.

Sx=(ABr2)sin3θcosϕ ...... (3)

Write the expression of y component for S.

Sy=S.ay ...... (4)

Here, Sy is y

component for Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  19

Substitute [(AB r 2)sin2θ]r^ for S in equation (4).

   Sy=[( AB r 2 )sin2θ]r^.ay=[( AB r 2 )sin2θ]ar.ay=[( AB r 2 )sin2θ](sinθsinϕ)            {ar.ay=sinθsinϕ}

Simplify further.

Sy=(ABr2)sin3θsinϕ ...... (5)

Write the expression of z component for S.

Sz=S.az ...... (6)

Here, Sz is z

component for Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  20

Substitute [(AB r 2)sin2θ]r^ for S in equation (6).

   Sz=[( AB r 2 )sin2θ]r^.az=[( AB r 2 )sin2θ]ar.az=[( AB r 2 )sin2θ](cosθ)            {ar.az=cosθ}

Simplify further.

Sz=(ABr2)sin2θcosθ ...... (7)

So the rectangular expression for S.

   S=(( AB r 2 ) sin3θcosϕ)ax+(( AB r 2 ) sin3θsinϕ)ay+(( AB r 2 ) sin2θcosθ)az=( AB r 2 )sin2θ(sinθcosϕax+sinθsinϕay+cosθaz)

Conclusion:

Thus,the value of S=E×H is [(AB r 2)sin2θ]r^ and the rectangular expression for S is

   (ABr2)sin2θ(sinθcosϕax+sinθsinϕay+cosθaz).

To determine

(b)

The value of S along x , y and z axes.

Expert Solution
Check Mark

Answer to Problem 1.18P

Thevalue of S along x axis is (ABr2)sin3θcosϕ , the value of S along y axis is (ABr2)sin3θsinϕ and the value of S along z axis is (ABr2)sin2θcosθ.

Explanation of Solution

Calculation:

From equation (3) the value of Sx is

   Sx=(ABr2)sin3θcosϕ

So the value of S along x axis is (ABr2)sin3θcosϕ.

From equation (5) the value of Sy is

   Sy=(ABr2)sin3θsinϕ

So the value of S along y axis is (ABr2)sin3θsinϕ.

From equation (7) the value of Sz is

   Sz=(ABr2)sin2θcosθ

So the value of S along z axis is (ABr2)sin2θcosθ.

Conclusion:

Thus, the value of S along x axis is (ABr2)sin3θcosϕ , the value of S along y axis is (ABr2)sin3θsinϕ and the value of S along z axis is (ABr2)sin2θcosθ.

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