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

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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, 0 < z < 2: This is found through I = S J · n S da = 2 0 1 0 J · a y y =1 dx dz = 2 0 1 0 10 4 cos(2 x ) e 2 dx dz = 10 4 (2) 1 2 sin(2 x ) 1 0 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 = 3 2 1 0 J · ( a y ) y =0 dx dz + 3 2 1 0 J · ( a y ) y =1 dx dz + 3 2 1 0 J · ( a x ) x =1 dy dz = 10 4 3 2 1 0 cos(2 x ) e 0 cos(2 x ) e 2 dx dz 10 4 3 2 1 0 sin(2) e 2 y dy dz = 10 4 1 2 sin(2 x ) 1 0 (3 2) 1 e 2 + 10 4 1 2 sin(2) e 2 y 1 0 (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 = S J · d S = 2 π 0 1 0 100 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 = S J · d S = 2 π 0 1 0 100 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 + 1 0 2 π 0 100 e 2 z ((1) a ρ + a z ) · a ρ (1) dφ dz = 100 π ( e 2 1)+100 π (1 e 2 ) = 0 1
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5.3. Let J = 400sin θ r 2 + 4 a r A / m 2 a) Find the total current flowing through that portion of the spherical surface r = 0 . 8, bounded by 0 . 1 π < θ < 0 . 3 π , 0 < φ < 2 π : This will be I = J ·
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