Loose Leaf For Engineering Electromagnetics
Loose Leaf For Engineering Electromagnetics
9th Edition
ISBN: 9781260472370
Author: John A. Buck, William H. Hayt
Publisher: McGraw-Hill Education
bartleby

Concept explainers

Question
Book Icon
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 Loose Leaf For 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 Loose Leaf For Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  2along Loose Leaf For Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  3m:math display='block'>Aϕ

is the component of vector Loose Leaf For Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  4along Loose Leaf For 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 Loose Leaf For Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  6m:math display='block'>Bϕ is the component of vector B

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

Calculation:

Substitute 0

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

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

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

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

forLoose Leaf For Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  14Loose Leaf For Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  15for Loose Leaf For Engineering Electromagnetics, Chapter 1, Problem 1.18P , additional homework tip  16and Loose Leaf For 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 Loose Leaf For 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 Loose Leaf 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 Loose Leaf 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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A closed surface is defined in cylindrical coordinates by: 2 < ρ < 5, 0.07 π≤ϕ≤π7 and 1.93≤ z ≤0. Find the the surface area of the sphere
25 Using the coordinate system named, give the vector al point A(2,-1,-3) that extends to B(1,3,4): (a) cartesian; (b) cylindrical; (e) spherical.
Find the surface area of a cone. (Hint: Assume the cone in spherical coordinate system, keepingthe angle θ constant, integrate the differential surface area)
Knowledge Booster
Background pattern image
Electrical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:PEARSON
Text book image
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Text book image
Programmable Logic Controllers
Electrical Engineering
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education
Text book image
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:9780078028229
Author:Charles K Alexander, Matthew Sadiku
Publisher:McGraw-Hill Education
Text book image
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:9780134746968
Author:James W. Nilsson, Susan Riedel
Publisher:PEARSON
Text book image
Engineering Electromagnetics
Electrical Engineering
ISBN:9780078028151
Author:Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:Mcgraw-hill Education,