20075ee1_1_HW5

20075ee1_1_HW5 - CHAPTER 5 5.1. Given the current density J...

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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 . 23MA 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 =0 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 £ cos(2 x ) e − − cos(2 x ) e − 2 ¤ dx dz − 10 4 Z 3 2 Z 1 sin(2) e − 2 y dy dz = 10 4 µ 1 2 ¶ sin(2 x ) ¯ ¯ ¯ 1 (3 − 2) £ 1 − e − 2 ¤ + 10 4 µ 1 2 ¶ sin(2) e − 2 y ¯ ¯ ¯ 1 (3 − 2) = 0 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 £ 2cos(2 x ) e − 2 y − 2cos(2 x ) e − 2 y ¤ = 0 as expected 5.2. A certain current density is given in cylindrical coordinates as J = 100 e − 2 z ( ρ a ρ + a z ) A / m 2 . Find the total current passing through each of these surfaces: a) z = 0, 0 ≤ ρ ≤ 1, in the a z direction: I a = Z S J · d S = Z 2 π Z 1 100 e − 2(0) ( ρ a ρ + a z ) · a z ρ dρ dφ = 100 π where a ρ · a z = 0. b) z = 1, 0 ≤ ρ ≤ 1, in the a z direction: I b = Z S J · d S = Z 2 π Z 1 100 e − 2(1) ( ρ a ρ + a z ) · a z ρ dρ dφ = 100 πe − 2 c) closed cylinder defined by 0 ≤ z ≤ 1, 0 ≤ ρ ≤ 1, in an outward direction: I T = I b − I a + Z 1 Z 2 π 100 e − 2 z ((1) a ρ + a z ) ·...
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This homework help was uploaded on 04/01/2008 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.

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20075ee1_1_HW5 - CHAPTER 5 5.1. Given the current density J...

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