Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9780078028151
Author: Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher: Mcgraw-hill Education,
Question
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Chapter 5, Problem 5.1P
To determine

(a)

The total current crossing the plane y=1 in the ay direction in the region 0<x<1,0<z<2.

Expert Solution
Check Mark

Answer to Problem 5.1P

The total current is, I=1.23 MA.

Explanation of Solution

Given Information:

The current density is,

   J=104[sin(2x)e2yax+cos(2x)e2yay] kA/m2.

Calculation:

The current density is given as,

   J=104[sin(2x)e2yax+cos(2x)e2yay] Jay=104cos(2x)e2y

The total current,

   I=S Jn|Sda=0201 J a y | y=1dxdz =0201 10 4 cos( 2x ) e 2y |y=1dxdz =0201104cos(2x)e2dxdz=02[12104e2sin(2x)]01dz =025000e2sin2dz=5000e2sin2[z]02 =10000e2sin2=1230.6 kA =1.23 MA

Conclusion:

The total current is, I=1.23 MA.

To determine

(b)

The total current leaving the region 0<x,y<1,2<z<3 by integrating JdS over the surface of cube.

Expert Solution
Check Mark

Answer to Problem 5.1P

The total current leaving over the surface is zero.

Explanation of Solution

Given Information:

The current density is,

   J=104[sin(2x)e2yax+cos(2x)e2yay] kA/m2.

Calculation:

The current density is given as,

   J=104[sin(2x)e2yax+cos(2x)e2yay] Jax=104sin(2x)e2y Jay=104cos(2x)e2y Jaz=0

The total current leaving over the surface of cube,

   I= 2 3 0 1 J( a x )| x=0 dy dz + 2 3 0 1 J a x | x=1 dy dz+ 2 3 0 1 J( a y )| y=0 dx dz+ 2 3 0 1 J a y | y=1 dx dz

   + 0 1 0 1 J( a z )| z=2 dx dz + 0 1 0 1 J a z | z=3 dx dz

   =0+ 2 3 0 1 10 4 sin( 2 ) e 2y dy dz + 2 3 0 1 10 4 cos( 2x )dx dz+ 2 3 0 1 10 4 cos( 2x ) e 2 dx dz+0+0

   = 10 4 sin2 2 3 [ e 2y 2 ] 0 1 dz + 10 4 ( 1 e 2 ) 2 3 [ sin( 2x ) 2 ] 0 1 dz

   = 10 4 2 ( e 2 1 )sin2+ 10 4 2 ( 1 e 2 )sin2

   =0

Conclusion:

The total current leaving over the surface is zero.

To determine

(c)

The total current leaving the region 0<x,y<1,2<z<3 over the surface of cube using divergence theorem.

Expert Solution
Check Mark

Answer to Problem 5.1P

The total current leaving over the surface is zero, as expected.

Explanation of Solution

Given Information:

The current density is,

   J=104[sin(2x)e2yax+cos(2x)e2yay] kA/m2.

Calculation:

The current density is given as,

   J=104[sin(2x)e2yax+cos(2x)e2yay] Jx=104sin(2x)e2y Jy=104cos(2x)e2y

By using divergence theorem, the total current,

   I=J=Jxx+Jyy =( 10 4sin( 2x) e 2y)x+( 10 4cos( 2x) e 2y)y =2×104cos(2x)e2y+2×104cos(2x)e2y =0

Conclusion:

The total current leaving over the surface is zero, as expected.

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Engineering Electromagnetics

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