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2. A unit normal vector from ān21 = (6ā× +2ā, −3ā₂)/7. If region 2, where μ = 2μ。, to Ĥ₁ =10㸠+ā, +12ā, A/m and region 2, where μ = μ₁, is H₂ = H₂xª× −5㸠+4ā₁A/m, determine: a) H2x b) The surface current density K on the interface. c) The angles B₁ and B₂ make with the normal to the interface. Answers: (a) 5.833 (b) 4.86ā, -8.64ā, +3.95ā₂ A/m² (c) 76.27° and 77.62°.

2. A unit normal vector from ān21 = (6ā× +2ā, −3ā₂)/7. If region 2, where μ = 2μ。, to Ĥ₁ =10㸠+ā, +12ā, A/m and region 2, where μ = μ₁, is H₂ = H₂xª× −5㸠+4ā₁A/m, determine: a) H2x b) The surface current density K on the interface. c) The angles B₁ and B₂ make with the normal to the interface. Answers: (a) 5.833 (b) 4.86ā, -8.64ā, +3.95ā₂ A/m² (c) 76.27° and 77.62°.

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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2. A unit normal vector from ān21 = (6ā× +2ā, −3ā₂)/7. If region 2, where μ = 2μ。, to Ĥ₁ =10㸠+ā, +12ā, A/m and region 2, where μ = μ₁, is H₂ = H₂xª× −5㸠+4ā₁A/m, determine: a) H2x b) The surface current density K on the interface. c) The angles B₁ and B₂ make with the normal to the interface. Answers: (a) 5.833 (b) 4.86ā, -8.64ā, +3.95ā₂ A/m² (c) 76.27° and 77.62°.
Transcribed Image Text:2. A unit normal vector from ān21 = (6ā× +2ā, −3ā₂)/7. If region 2, where μ = 2μ。, to Ĥ₁ =10㸠+ā, +12ā, A/m and region 2, where μ = μ₁, is H₂ = H₂xª× −5㸠+4ā₁A/m, determine: a) H2x b) The surface current density K on the interface. c) The angles B₁ and B₂ make with the normal to the interface. Answers: (a) 5.833 (b) 4.86ā, -8.64ā, +3.95ā₂ A/m² (c) 76.27° and 77.62°.
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