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# chap5 - CHAPTER 5 5.1 Given the current density J = − 10...

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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 ² 1 2 ³ sin ( 2 x) ± ± ± 1 ( 3 − 2 ) h 1 − e − 2 i + 10 4 ² 1 2 ³ 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 φ cos 2 φ a ρ − ρ sin 2 φ a φ A / m 2 within the region 2 . 1 < ρ < 2 . 5, < φ < . 1 rad, 6 < z < 6 . 1. Find the total current I crossing the surface: a) ρ = 2 . 2, 0 < φ < . 1, 6 < z < 6 . 1 in the a ρ direction: This is a surface of constant ρ , so only the radial component of J will contribute: At ρ = 2 . 2 we write: I = Z J · d S = Z 6 . 1 6 Z . 1 2 ( 2 ) cos 2 φ a ρ · a ρ 2 dφdz = 2 ( 2 . 2 ) 2 ( . 1 ) Z . 1 1 2 ( 1 + cos 2 φ)dφ = . 2 ( 2 . 2 ) 2 ´ 1 2 ( . 1 ) + 1 4 sin 2 φ ± ± ± . 1 µ = 97 mA b) φ = . 05, 2 . 2 < ρ < 2 . 5, 6 < z < 6 . 1 in the a φ direction: In this case only the φ component of J will contribute: I = Z J · d S = Z 6 . 1 6 Z 2 . 5 2 . 2 − ρ sin 2 φ ± ± φ = . 05 a φ · a φ dρ dz = − ( . 1 ) 2 ρ 2 2 ± ± ± 2 . 5 2 . 2 = − 7 mA 60 5.2c. Evaluate ∇ · J at P(ρ = 2 . 4 ,φ = . 08 ,z = 6 . 05 ) : ∇ · J = 1 ρ ∂ ∂ρ (ρJ ρ ) + 1 ρ ∂J φ ∂φ = 1 ρ ∂ ∂ρ ( 2 ρ 2 cos 2 φ) − 1 ρ ∂ ∂φ (ρ sin 2 φ) = 4 cos 2 φ − 2 cos 2 φ ± ± ± . 08 = 2 . 0 A / m 3 5.3. Let J = 400 sin θ 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 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 400 sin 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 . 4 A 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 . 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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chap5 - CHAPTER 5 5.1 Given the current density J = − 10...

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