Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9780078028151
Author: Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher: Mcgraw-hill Education,
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Textbook Question
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Chapter 7, Problem 7.1P

  1. Find H in rectangular components at P(2,3,4) if there is a current filament on the z axis carring 8 mA in the az direction. (b) Repeat if the filament is located at x=-1. Y=2. (c) Find H if both filaments are present.

Expert Solution
Check Mark
To determine

(a)

The value of H for given rectangular component and current filament in the az direction.

Answer to Problem 7.1P

The value of H for given rectangular component and current filament in the az direction is Ha=294ax+196ayμA/m

Explanation of Solution

Given:

Rectangular component at P(2,3,4).

Current filament on z-axis carrying 8mA.

Calculation:

Now we can apply Biot-Savart Law, we get;

   Ha= IdL× a R 4π R 2 Ha=Idz a z×[ 2 a x +3 a y +( 4z ) a z ]4π ( z 2 8z+29 ) 3 2 Ha=Idzaz×[2 a y3 a x]4π ( z 2 8z+29 ) 3 2

Now using the integral table, this equation evaluates as

   Ha=I4π[ 2( 2z8 )( 2 a y 3 a x ) 52 ( z 2 8z+29 ) 1/2 ]Ha=I26π(2ay3ax)I=8mAHa=294ax+196ayμA/m

Expert Solution
Check Mark
To determine

(b)

The value of H for given rectangular component and current filament, if it is located at x=1,y=2.

Answer to Problem 7.1P

The value of H for given rectangular component and current filament is Hb=127ax+382ayμA/m

Explanation of Solution

Given:

Current filament carrying 8mA.

Current filament located at x=1,y=2

Calculation:

Now we can apply Biot-Savart Law, we get;

   Hb= IdL× a R 4π R 2 Hb=Idz a z×[ ( 2+1 ) a x +( 32 ) a y +( 4z ) a z ]4π ( z 2 8z+26 ) 3 2 Hb=Idzaz×[3 a y a x]4π ( z 2 8z+26 ) 3 2

Now using the integral table, this equation evaluates as

   I=8mAHb=I4π[ 2( 2z8 )( 3 a y a x ) 40 ( z 2 8z+26 ) 1/2 ]Hb=I20π(3ayax)Hb=127ax+382ayμA/m

Expert Solution
Check Mark
To determine

(c)

The value of H if both filaments are present.

Answer to Problem 7.1P

The value of H if both filaments are present is HT=421ax+578ayμA/m

Explanation of Solution

Given:

Rectangular component at P(2,3,4)

Current filament located at carrying x=1,y=2

Calculation:

For calculating HT when both filaments are present, so we just add the result of HaandHb.

For first filament:

Now we can apply Biot-Savart Law, we get;

   Ha= IdL× a R 4π R 2 Ha=Idz a z×[ 2 a x +3 a y +( 4z ) a z ]4π ( z 2 8z+29 ) 3 2 Ha=Idzaz×[2 a y3 a x]4π ( z 2 8z+29 ) 3 2

Now using the integral table, this equation evaluates as

   Ha=I4π[ 2( 2z8 )( 2 a y 3 a x ) 52 ( z 2 8z+29 ) 1/2 ]Ha=I26π(2ay3ax)I=8mAHa=294ax+196ayμA/m

For second filament:

Now we can apply Biot-Savart Law, we get,

   Hb= IdL× a R 4π R 2 Hb=Idz a z×[ ( 2+1 ) a x +( 32 ) a y +( 4z ) a z ]4π ( z 2 8z+26 ) 3 2 Hb=Idzaz×[3 a y a x]4π ( z 2 8z+26 ) 3 2

Now using the integral table, this equation evaluates as

   I=8mAHb=I4π[ 2( 2z8 )( 3 a y a x ) 40 ( z 2 8z+26 ) 1/2 ]Hb=I20π(3ayax)Hb=127ax+382ayμA/m

   HT=Ha+HbHT=421ax+578ayμA/m

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