1 Solution to the Drill problems of chapter 05
(Engineering Electromagnetics,Hayt,A.Buck 7th ed)
BEE 4A,4B & 4C
D5.1 (a). J =10ρ2zˆaρ − 4ρ cos2 φˆaφ mA/m2 , P(ρ = 3, φ = 300, z = 2)
⇒ (J)(ρ=3,φ=300,z=2) = 10 × 32 × 2ˆaρ − 4 × 3 × (cos 300)2ˆaφ = (180ˆaρ − 9ˆaφ) mA/ m2
(b). we have I = J · dS, dS = ρdφdzˆaρ ⇒ I = (10ρ2zˆaρ − 4ρ cos2 φˆaφ) · (ρdφdzˆaρ)
⇒ I= (10ρ2z)ρdφdz = 10ρ3 zdzdφ = (10ρ3) 2.8
2 zdz 2π
0 dφ = 3257.20mA = 3.25A
D5.2 (a). we have I = J · dS, dS = ρdφdρˆaz ⇒ I = (−106z1.5ˆaz) · (ρdφdρˆaz) = −106z1.5 20×10−6
0 ρdρ 2π
0 dφ
⇒ I = −106z1.5× | ρ2/2 |20×10−6
0 ×2π = −39.7µA
(b). We have J = ρvv and v = 2 × 106m/s⇒ρv = J/v = −31.622 × 103/2 × 106 = −15.81mC/m3
(c). ρv = −2000C/m3 (at z=0.15m) we have J = ρvv ⇒ v = Jz=0.15/ρv = −58.094 × 103/ − 2000 = 29.04m/s
D5.3 (a). We have vd = −µE ⇒ E = −vd/µ = 2.6 × 10−4V/m also we have J = σE ⇒ J = (6.17 × 107)(2.6 × 10−4)
= 16.5KA/m2
(b). we have J = σE ⇒ J = (6.17 × 107)(1 × 10−3) = 61.7KA/m2
(c). we have J = σV/L = (6.17 × 107)(0.4 × 10−3)/(2.5 × 10−3) = 9.9MA/m2
(d). we have J = I/S = 0.5/(2.5 × 10−3)2 = 80kA/m2
D5.4 (a). we have R = L/σS, L = 1200ft = 1200 × (0.3048) = 365.76m, Diameter = 0.6in
= 0.6 × (0.0254) = 0.01524m ⇒ radius(r) = 7.62 × 10−3m, σ = 5.8 × 107S/m
R = 365.76/(5.8 × 107)(π(7.62 × 10−3)2) = 0.035Ω
(b). we have J = I/S = 50/(π(7.62 × 10−3)2) = 2.74 × 105A/m2
(c). We have V = IR ⇒ V = 50 × (0.035) = 1.75V
(d). We have P = V I ⇒ P = 1.75 × 50 = 87.5W
D5.5 This drill problem is exactly similar to the example 5.2
NOTE: Rest of the drill problems and the topics related to them, of chapter 05 are out of course.
THE END
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This document is prepared in LATEX. (Email: ahmadsajjad01@ciit.net.pk)
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Chapter 05 drill_solution

  • 1.
    1 Solution tothe Drill problems of chapter 05 (Engineering Electromagnetics,Hayt,A.Buck 7th ed) BEE 4A,4B & 4C D5.1 (a). J =10ρ2zˆaρ − 4ρ cos2 φˆaφ mA/m2 , P(ρ = 3, φ = 300, z = 2) ⇒ (J)(ρ=3,φ=300,z=2) = 10 × 32 × 2ˆaρ − 4 × 3 × (cos 300)2ˆaφ = (180ˆaρ − 9ˆaφ) mA/ m2 (b). we have I = J · dS, dS = ρdφdzˆaρ ⇒ I = (10ρ2zˆaρ − 4ρ cos2 φˆaφ) · (ρdφdzˆaρ) ⇒ I= (10ρ2z)ρdφdz = 10ρ3 zdzdφ = (10ρ3) 2.8 2 zdz 2π 0 dφ = 3257.20mA = 3.25A D5.2 (a). we have I = J · dS, dS = ρdφdρˆaz ⇒ I = (−106z1.5ˆaz) · (ρdφdρˆaz) = −106z1.5 20×10−6 0 ρdρ 2π 0 dφ ⇒ I = −106z1.5× | ρ2/2 |20×10−6 0 ×2π = −39.7µA (b). We have J = ρvv and v = 2 × 106m/s⇒ρv = J/v = −31.622 × 103/2 × 106 = −15.81mC/m3 (c). ρv = −2000C/m3 (at z=0.15m) we have J = ρvv ⇒ v = Jz=0.15/ρv = −58.094 × 103/ − 2000 = 29.04m/s D5.3 (a). We have vd = −µE ⇒ E = −vd/µ = 2.6 × 10−4V/m also we have J = σE ⇒ J = (6.17 × 107)(2.6 × 10−4) = 16.5KA/m2 (b). we have J = σE ⇒ J = (6.17 × 107)(1 × 10−3) = 61.7KA/m2 (c). we have J = σV/L = (6.17 × 107)(0.4 × 10−3)/(2.5 × 10−3) = 9.9MA/m2 (d). we have J = I/S = 0.5/(2.5 × 10−3)2 = 80kA/m2 D5.4 (a). we have R = L/σS, L = 1200ft = 1200 × (0.3048) = 365.76m, Diameter = 0.6in = 0.6 × (0.0254) = 0.01524m ⇒ radius(r) = 7.62 × 10−3m, σ = 5.8 × 107S/m R = 365.76/(5.8 × 107)(π(7.62 × 10−3)2) = 0.035Ω (b). we have J = I/S = 50/(π(7.62 × 10−3)2) = 2.74 × 105A/m2 (c). We have V = IR ⇒ V = 50 × (0.035) = 1.75V (d). We have P = V I ⇒ P = 1.75 × 50 = 87.5W D5.5 This drill problem is exactly similar to the example 5.2 NOTE: Rest of the drill problems and the topics related to them, of chapter 05 are out of course. THE END 1 This document is prepared in LATEX. (Email: ahmadsajjad01@ciit.net.pk) 1