chap5o hayt

chap5o hayt - Chapter 5 – Odd-Numbered 5.1 Given the...

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Unformatted text preview: Chapter 5 – Odd-Numbered 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 dxdz = Z 2 Z 1 − 10 4 cos(2 x ) e − 2 dxdz = − 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 dxdz + Z 3 2 Z 1 J · ( a y ) ¯ ¯ ¯ y =1 dxdz + 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 ¤ dxdz − 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 5.3. 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 < . 3 π , 0 < φ < 2 π : This will be I = Z Z J · n ¯ ¯ ¯ S da = Z 2 π Z . 3 π . 1 π 400sin θ ( . 8) 2 + 4 ( . 8) 2 sin θ dθ dφ = 400( . 8) 2 2 π 4 . 64 Z . 3 π . 1 π sin 2 dθ = 346 . 5 Z . 3 π . 1 π 1 2 [1 − cos(2 θ )] dθ = 77 . 4A b) Find the average value of J over the defined area. The area is Area = Z 2 π Z . 3 π . 1 π ( . 8) 2 sin θ dθ dφ = 1 . 46m 2 The average current density is thus J avg = (77 . 4 / 1 . 46) a r = 53 . a r A / m 2 . 54 5.5. Let J = 25 ρ a ρ − 20 ρ 2 + 0 . 01 a z A / m 2 a) Find the total current crossing the plane z = 0 . 2 in the a z direction for ρ < . 4: Use I = Z Z S J · n ¯ ¯ ¯ z = . 2 da = Z 2 π Z . 4 − 20 ρ 2 + . 01 ρdρdφ = − µ 1 2 ¶ 20ln( . 01 + ρ 2 ) ¯ ¯ ¯ . 4 (2 π ) = − 20 π ln(17) = − 178 . 0A b) Calculate ∂ρ v /∂t : This is found using the equation of continuity: ∂ρ v ∂t = −∇ · J = 1 ρ ∂ ∂ρ ( ρJ ρ ) + ∂J z ∂z = 1 ρ ∂ ∂ρ (25) + ∂ ∂z µ − 20 ρ 2 + . 01 ¶ = 0 c) Find the outward current crossing the closed surface defined by ρ = 0 . 01, ρ = 0 . 4, z = 0, and z = 0 . 2: This will be I = Z . 2 Z 2 π 25 . 01 a ρ · ( − a ρ )( . 01) dφdz + Z . 2 Z 2 π 25 . 4 a ρ · ( a ρ )( . 4) dφdz + Z 2 π Z . 4 − 20 ρ 2 + . 01 a z · ( − a z ) ρdρdφ + Z 2 π Z . 4 − 20 ρ 2 + .....
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chap5o hayt - Chapter 5 – Odd-Numbered 5.1 Given the...

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