329sum11hw1sol - ECE 329 Homework 1 Solution Due 5PM 1 The...

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ECE 329 Homework 1 — Solution Due: June 17, 2011, 5PM 1. The 3D vectors are ! A =3ˆ x y - z, ! B x y - ˆ ! C x - y +3 , where x , y and z constitute an orthogonal set of unit vectors, we can calculate the following as shown below:- a) The vector ! D = ! A + ! B x y - z x y - ˆ z =4ˆ x +2ˆ y - z. b) The vector ! A + ! B - 4 ! C x y - z x y - ˆ z - x +8ˆ y - 12ˆ z =10ˆ y - 15ˆ c) The vector magnitude ± ± ± ! A + ! B - 4 ! C ± ± ± = | 10ˆ y - 15ˆ z | = 10 2 +15 2 = 325 . d) The unit vector ˆ u = ± A +2 ± B - ± C | ± A +2 ± B - ± C | = x +5ˆ y - z 4 2 +5 2 +7 2 = x +5ˆ y - z 90 (Here, ˆ u is along vector ! A +2 ! B - ! C ). e) The dot product ! A · ! B =(3ˆ x y - z ) · x y - ˆ z )=3ˆ x · ˆ x + ˆ y · ˆ y z · ˆ z = 3+1+2=6 . f) The cross product ! B × ! C = ± ± ± ± ± ± ˆ x ˆ y ˆ z 11 - 1 1 - 23 ± ± ± ± ± ± =(3 - 2)ˆ x - (3+1)ˆ y +( - 2 - 1)ˆ z = ˆ x - y - 2. Let ! J = z 2 x y z ) A / m 2 denote the electrical current density Feld (current ±ux per unit area) in a region of space represented in Cartesian coordinates. The magnitude of ! J is z 2 3 amperes (A) per unit area and its direction is x y z ) 3 . The total current ±ux ¸ S ! J · d ! S out of a closed surface S enclosing a cubic volume V =1 m 3 (Fgure below) can be calculated by adding the surfaces integrals I i = ´ S i ! J · d ! S i taken over each of the six faces of the cube. (1 , 1 , 1) (0 , 0 , 0) x y z d ± S 2 d ± S 5 d ± S 1 d ± S 3 d ± S 6 d ± S 4 1
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Face S 1 , the integral is I 1 = ˆ S 1 ! J (1 ,y,z ) · d ! S 1 = 1 ˆ 0 1 ˆ 0 z 2 x y z ) · ˆ xdydz = 1 ˆ 0 1 ˆ 0 z 2 dydz = ± z 3 3 ² 1 0 = 1 3 A .
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329sum11hw1sol - ECE 329 Homework 1 Solution Due 5PM 1 The...

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