Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9781260029963
Author: Hayt
Publisher: MCG
Question
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Chapter 9, Problem 9.12P
To determine

(a)

The displacement current density for given magnetic flux density.

Expert Solution
Check Mark

Answer to Problem 9.12P

   JD=B0k0μ0cos(ωt)sin(k0z)ax

Explanation of Solution

Given:

   B=B0cos(ωt)cos(k0z)ayWb/m2

Medium for magnetic flux density is free space.

Concept used:

Formula for displacement current density is

   ×H=J+JD(JD=displacementcurrentdensity)B=μ0H

Calculation:

Plugging the value B in the formula shown below for magnetic field intensity

   B=μ0H

So,

   H=B0μ0cos(ωt)cos(k0z)ay

Hence,

   ×H=J+JD ( J is zero for a non-conducting medium)

Therefore,

   ×H=JD

On calculating curl of H for displace current density

   JD=( i j k x y z a x a y a z )JD=( i j k x y z 0 B 0 μ 0 cos( ωt )cos( k 0 z ) 0 )JD=B0k0μ0cos(ωt)sin(k0z)ax

To determine

(b)

The electric field intensity.

Expert Solution
Check Mark

Answer to Problem 9.12P

   E=B0k0εμ0ωsin(k0z)sin(ωt)ax volt/m

Explanation of Solution

Given:

   B=B0cos(ωt)cos(k0z)ayWb/m2

Medium for magnetic flux density is free space.

Concept used:

   JD=Dt

Calculation:

Plugging the value JD in the formula shown below for magnetic field intensity and integrating with respect to t

So,

   J Ddt= B 0 k 0 μ 0 cos( ωt)sin( k 0 z)dtD=B0k0μ0sin(k0z)cos( ωt)dtD=B0k0μ0ωsin(k0z)sin(ωt)Volt/m

   D=εE

Hence,

   E=B0k0εμ0ωsin(k0z)sin(ωt)ax volt/m

To determine

(c)

The value of k0.

Expert Solution
Check Mark

Answer to Problem 9.12P

   k0=εμ0ω

Explanation of Solution

Given:

   B=B0cos(ωt)cos(k0z)ayWb/m2

Medium for magnetic flux density is free space.

Concept used:

   ×E=Bt

Calculation:

Plugging the value E in the formula shown above for magnetic flux density

So,

   Bt=( i j k x y z a x a y a z )Bt=( i j k x y z B 0 k 0 ε μ 0 ω sin( k 0 z )sin( ωt ) 0 0 )Bt=B0k02μ0cos(ωt)sin(k0z)ay

On integrating both sides with respect to t

   B t= B 0 k 0 2 μ 0 cos( ωt)sin( k 0 z)dtB=B0k02εμ0ωsin(k0z)sin(ωt)+cc=0(sincenofieldarepreasumedtoexist)B=B0k02εμ0ωsin(k0z)sin(ωt)..................(1)

Comparing magnitudes of (1) to the magnitudes of given B

   B0 = B0k02εμ0ωk0=εμ0ω

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