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Unformatted text preview: CHAPTER 5 5.1. Given the current density J = 10 4 [sin ( 2 x)e 2 y a x + cos ( 2 x)e 2 y a y ] kA / m 2 : a) Find the total current crossing the plane y = 1 in the a y direction in the region 0 < x < 1, < z < 2: This is found through I = Z Z S J n S da = Z 2 Z 1 J a y y = 1 dx dz = Z 2 Z 1 10 4 cos ( 2 x)e 2 dx dz = 10 4 ( 2 ) 1 2 sin ( 2 x) 1 e 2 = 1 . 23 MA b) Find the total current leaving the region 0 < x,x < 1, 2 < z < 3 by integrating J dS over the surface of the cube: Note first that current through the top and bottom surfaces will not exist, since J has no z component. Also note that there will be no current through the x = 0 plane, since J x = 0 there. Current will pass through the three remaining surfaces, and will be found through I = Z 3 2 Z 1 J ( a y ) y = dx dz + Z 3 2 Z 1 J ( a y ) y = 1 dx dz + Z 3 2 Z 1 J ( a x ) x = 1 dy dz = 10 4 Z 3 2 Z 1 h cos ( 2 x)e cos ( 2 x)e 2 i dx dz 10 4 Z 3 2 Z 1 sin ( 2 )e 2 y dy dz = 10 4 1 2 sin ( 2 x) 1 ( 3 2 ) h 1 e 2 i + 10 4 1 2 sin ( 2 )e 2 y 1 ( 3 2 ) = c) Repeat part b , but use the divergence theorem: We find the net outward current through the surface of the cube by integrating the divergence of J over the cube volume. We have J = J x x + J y y = 10 4 h 2 cos ( 2 x)e 2 y 2 cos ( 2 x)e 2 y i = as expected 5.2. Let the current density be J = 2 cos 2 a sin 2 a A / m 2 within the region 2 . 1 < < 2 . 5, < < . 1 rad, 6 < z < 6 . 1. Find the total current I crossing the surface: a) = 2 . 2, 0 < < . 1, 6 < z < 6 . 1 in the a direction: This is a surface of constant , so only the radial component of J will contribute: At = 2 . 2 we write: I = Z J d S = Z 6 . 1 6 Z . 1 2 ( 2 ) cos 2 a a 2 ddz = 2 ( 2 . 2 ) 2 ( . 1 ) Z . 1 1 2 ( 1 + cos 2 )d = . 2 ( 2 . 2 ) 2 1 2 ( . 1 ) + 1 4 sin 2 . 1 = 97 mA b) = . 05, 2 . 2 < < 2 . 5, 6 < z < 6 . 1 in the a direction: In this case only the component of J will contribute: I = Z J d S = Z 6 . 1 6 Z 2 . 5 2 . 2 sin 2 = . 05 a a d dz = ( . 1 ) 2 2 2 2 . 5 2 . 2 = 7 mA 60 5.2c. Evaluate J at P( = 2 . 4 , = . 08 ,z = 6 . 05 ) : J = 1 (J ) + 1 J = 1 ( 2 2 cos 2 ) 1 ( sin 2 ) = 4 cos 2 2 cos 2 . 08 = 2 . 0 A / m 3 5.3. Let J = 400 sin r 2 + 4 a r A / m 2 a) Find the total current flowing through that portion of the spherical surface r = . 8, bounded by . 1 < < . 3 , 0 < < 2 : This will be I = Z Z J n S da = Z 2 Z . 3 . 1 400 sin (. 8 ) 2 + 4 (. 8 ) 2 sin d d = 400 (. 8 ) 2 2 4 . 64 Z . 3 . 1 sin 2 d = 346 . 5 Z . 3 . 1 1 2 [1 cos ( 2 ) ] d = 77 . 4 A b) Find the average value of J over the defined area. The area is Area = Z 2 Z . 3 . 1 (. 8 ) 2 sin d d = 1 . 46 m 2 The average current density is thus J avg = ( 77 . 4 / 1 . 46 ) a r = 53 . a r A / m 2 ....
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This note was uploaded on 06/10/2008 for the course EE 325 taught by Professor Brown during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Brown
 Electromagnet

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