Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 9, Problem 37P
To determine

Find the magnetic field intensity and the value of α.

Expert Solution & Answer
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Answer to Problem 37P

The magnetic field intensity and the value of α are 2.636cos(12πx)sin(1011t331.2y)ax+0.3sin(12πx)cos(1011t331.2y)aymA/m_ and 331.2_, respectively.

Explanation of Solution

Calculation:

Write the generalized form of Maxwell’s third Equation.

×E=Bt=μHt {B=μH}=μHt

Rewrite the expression.

H=1μ(×E)dt=1μo(×E)dt {μ=μo for free space (air)}

Substitute cos(12πx)sin(1011tαy)azV/m for E.

H=1μo|axayazxyz00cos(12πx)sin(1011tαy)|dt=1μo({y[cos(12πx)sin(1011tαy)]}ax{x[cos(12πx)sin(1011tαy)]}ay+0az)dt=1μo{[αcos(12πx)cos(1011tαy)]ax+12πsin(12πx)sin(1011tαy)ay}dt=11011μo{[αcos(12πx)sin(1011tαy)]ax12πsin(12πx)cos(1011tαy)ay}

Simplify the expression.

H=1(1011)(4π×107){[αcos(12πx)sin(1011tαy)]ax+12πsin(12πx)cos(1011tαy)ay}A/m

H=14π×104{[αcos(12πx)sin(1011tαy)]ax+12πsin(12πx)cos(1011tαy)ay}A/m        (1)

Write the generalized form of Maxwell’s fourth Equation.

×H=J+Dt

Rewrite expression for free space.

×H=0+Dt {J=0}=εEt {D=εE}=εoEt {ε=εo}

Rewrite the expression for electric field.

E=1εo(×H)dt        (2)

Find ×H.

×H=14π×104|axayazxyzαcos(12πx)sin(1011tαy)12πsin(12πx)cos(1011tαy)0|=14π×104({z[12πsin(12πx)cos(1011tαy)]}ax{z[αcos(12πx)sin(1011tαy)]}ay+{x[12πsin(12πx)cos(1011tαy)]y[αcos(12πx)sin(1011tαy)]}az)=14π×104(0ax0ay+{(12π)2cos(1011tαy)cos(12πx)+α2cos(12πx)cos(1011tαy)}az)=(12π)2+α24π×104cos(12πx)cos(1011tαy)az

Substitute (12π)2+α24π×104cos(12πx)cos(1011tαy)az for ×H in Equation (2).

E=1εo[(12π)2+α24π×104cos(12πx)cos(1011tαy)az]dt=(12π)2+α24π×104εo(1011)cos(12πx)sin(1011tαy)az=(12π)2+α2(4π×1015)(10936π)cos(12πx)sin(1011tαy)az {εo=10936π}=(9×106)[(12π)2+α2]cos(12πx)sin(1011tαy)azV/m

From the obtained electric field, the magnitude of electric field is (9×106)[(12π)2+α2]V/m.

Equate the obtained magnitude of the electric field to the given magnitude.

(9×106)[(12π)2+α2]V/m=1V/m

Rearrange the expression for α.

α=1(9×106)(12π)2331.2

Substitute 331.2 for α in Equation (1).

H=14π×104{[(331.2)cos(12πx)sin(1011t331.2y)]ax+12πsin(12πx)cos(1011t331.2y)ay}A/m=[2.636cos(12πx)sin(1011t331.2y)ax+0.3sin(12πx)cos(1011t331.2y)ay]mA/m

Conclusion:

Thus, the magnetic field intensity and the value of α are 2.636cos(12πx)sin(1011t331.2y)ax+0.3sin(12πx)cos(1011t331.2y)aymA/m_ and 331.2_, respectively.

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