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Please assist with non graded problem. Deduction of Section 7.4 A – Sadiku's Elements of Electromagnetics 7th Edition – please refer to the first image for it. Question: I need help to deduce H on the infinite line current (section 7.4 A). Please comment each passage and expand every possible equation, as I am having deep trouble understading where does it all come from. Thank you!

Please assist with non graded problem. Deduction of Section 7.4 A – Sadiku's Elements of Electromagnetics 7th Edition – please refer to the first image for it. Question: I need help to deduce H on the infinite line current (section 7.4 A). Please comment each passage and expand every possible equation, as I am having deep trouble understading where does it all come from. Thank you!

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Please assist with non graded problem.

Deduction of Section 7.4 A – Sadiku's Elements of Electromagnetics 7th Edition – please refer to the first image for it.

Question: I need help to deduce H on the infinite line current (section 7.4 A). Please comment each passage and expand every possible equation, as I am having deep trouble understading where does it all come from.

Thank you!

A. Infinite Line Current Consider an infinitely long filamentary current I along the z-axis as in Figure 7.10. To determine H at an observation point P, we allow a closed path to pass through P. This path, on which Ampère's law is to be applied, is known as an Amperian path (analogous to the term "Gaussian surface"). We choose a concentric circle as the Amperian path in view of eq. (7.14), which shows that H is constant provided p is constant. Since this path encloses the whole current I, according to Ampère's law, I = | Hgas p dp as = H4 p do = H4 · 2mp or I H = 2πρ (7.20) as expected from eq. (7.14).
Transcribed Image Text:A. Infinite Line Current Consider an infinitely long filamentary current I along the z-axis as in Figure 7.10. To determine H at an observation point P, we allow a closed path to pass through P. This path, on which Ampère's law is to be applied, is known as an Amperian path (analogous to the term "Gaussian surface"). We choose a concentric circle as the Amperian path in view of eq. (7.14), which shows that H is constant provided p is constant. Since this path encloses the whole current I, according to Ampère's law, I = | Hgas p dp as = H4 p do = H4 · 2mp or I H = 2πρ (7.20) as expected from eq. (7.14).
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