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1₁ = 2A P₁ = -4āx - 4āz, |p₁|=√32 Ø₁ = ₁ × ₁ = x (x + (az-ax) āØ₁ = A₁ :āx √32 PARTMEN √32 āz جامعاروف كلية الهندسة DNI قسم المنصبية الكهر CAL 2 2√32 (32 x)]——(āz- az - āx)]-(az - āx) 12 H₂ ·āØ₂ 2πP₂ P₂ = -3āy+6āz,|p₂|=√45, ap₂ = √ay+az ². āØ₂ = ax x (āy+z)=(-āz-2ay) H₂ = 2√5 (2ay+az)]=[-2ay - äz] H = H₂ + H₂ -ax 12 π 45m ay +(-az āz 8 45. H = -0.04āx -0.085ãy-0.002az A/m Isl

1₁ = 2A P₁ = -4āx - 4āz, |p₁|=√32 Ø₁ = ₁ × ₁ = x (x + (az-ax) āØ₁ = A₁ :āx √32 PARTMEN √32 āz جامعاروف كلية الهندسة DNI قسم المنصبية الكهر CAL 2 2√32 (32 x)]——(āz- az - āx)]-(az - āx) 12 H₂ ·āØ₂ 2πP₂ P₂ = -3āy+6āz,|p₂|=√45, ap₂ = √ay+az ². āØ₂ = ax x (āy+z)=(-āz-2ay) H₂ = 2√5 (2ay+az)]=[-2ay - äz] H = H₂ + H₂ -ax 12 π 45m ay +(-az āz 8 45. H = -0.04āx -0.085ãy-0.002az A/m Isl

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Ex2:-Two infinite line current ₁ = 2ay A at x =z=4 and I₂ = 4āx at y-3, z =-6. Calculat A² Hat (0,0,0). Chuy Solution - Xx At (0,0,0) then use H=¹₁ãØ 2πρι 4₁ = 2A P₁ = -4āx - 4āz, |p₁|=√32 az :āx √32 √32 :āØ₁ =āl₁ xap₁ = ay x āØ₁ = (az- (az - āx) :: H₂ = 2²³/2₂(az - āx)]—¹—(āz – āx) 2√32 32 12 H₂ -āØ₂ 2πP₂ Ã₂ = −3āy+6āz, | Ã₂|=√45, ap₂ = √y+√āz :āØ₂ = ax x ( ( ngày + vasaz) = vao (-az – 2āy) H₂ = 2√45√³ (2ãy + āz)]—[-2āy — āz] -ax 12 H = H₂ + 12 ay + (-)āz I 45T 8 H = -0.04āx -0.085āy-0.002āz A/m How solve the question? PARTMEN a Pi جامعار- كلية الهندسة AL EN dut Ang NI 5.5
Transcribed Image Text:Ex2:-Two infinite line current ₁ = 2ay A at x =z=4 and I₂ = 4āx at y-3, z =-6. Calculat A² Hat (0,0,0). Chuy Solution - Xx At (0,0,0) then use H=¹₁ãØ 2πρι 4₁ = 2A P₁ = -4āx - 4āz, |p₁|=√32 az :āx √32 √32 :āØ₁ =āl₁ xap₁ = ay x āØ₁ = (az- (az - āx) :: H₂ = 2²³/2₂(az - āx)]—¹—(āz – āx) 2√32 32 12 H₂ -āØ₂ 2πP₂ Ã₂ = −3āy+6āz, | Ã₂|=√45, ap₂ = √y+√āz :āØ₂ = ax x ( ( ngày + vasaz) = vao (-az – 2āy) H₂ = 2√45√³ (2ãy + āz)]—[-2āy — āz] -ax 12 H = H₂ + 12 ay + (-)āz I 45T 8 H = -0.04āx -0.085āy-0.002āz A/m How solve the question? PARTMEN a Pi جامعار- كلية الهندسة AL EN dut Ang NI 5.5
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