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deduce H on the infinite sheet of current

deduce H on the infinite sheet of current

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Deduction of Section 7.4 B – Sadiku's Elements of Electromagnetics 7th Edition – please refer to the 2 images for it.

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

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B. Infinite Sheet of Current Consider an infinite current sheet in the z = 0 plane. If the sheet has a uniform current density K = K,a, A/m as shown in Figure 7.11, applying Ampère's law to the rectangular closed path 1-2-3-4-1 (Amperian path) gives dl (7.21a) enc qx = *1 = 1P :He To evaluate the integral, we first need to have an idea of what H is like. To achieve this, we regard the infinite sheet as comprising filaments; dH above or below the sheet due to a pair of filamentary currents can be found by using eqs. (7.14) and (7.15). As evident in Figure 7.11(b), the resultant dH has only an x-component. Also, H on one side of the sheet is the negative of that on the other side. Owing to the infinite extent of the sheet, the sheet can be
Transcribed Image Text:B. Infinite Sheet of Current Consider an infinite current sheet in the z = 0 plane. If the sheet has a uniform current density K = K,a, A/m as shown in Figure 7.11, applying Ampère's law to the rectangular closed path 1-2-3-4-1 (Amperian path) gives dl (7.21a) enc qx = *1 = 1P :He To evaluate the integral, we first need to have an idea of what H is like. To achieve this, we regard the infinite sheet as comprising filaments; dH above or below the sheet due to a pair of filamentary currents can be found by using eqs. (7.14) and (7.15). As evident in Figure 7.11(b), the resultant dH has only an x-component. Also, H on one side of the sheet is the negative of that on the other side. Owing to the infinite extent of the sheet, the sheet can be
regarded as consisting of such filamentary pairs so that the characteristics of H for a pair are the same for the infinite current sheet, that is, (H,a H = z > 0 (7.21b) -H,a, z<0 where H, is yet to be determined. Evaluating the line integral of H in eq. (7.21a) along the closed path in Figure 7.11(a) gives fa a - C, d1 H· dl 2. 3 — 0(—а) + (-н,)(-b) + о(a) + Н.(6) (7.21c) 2H,b 1 From eqs. (7.21a) and (7.21c), we obtain H, = K. Substituting H, in eq. (7.21b) gives 2 Ka. z>0 H = (7.22) K,a z< 0 In general, for an infinite sheet of current density K A/m, 1 кха, (7.23) H = where a, is a unit normal vector directed from the current sheet to the point of interest.
Transcribed Image Text:regarded as consisting of such filamentary pairs so that the characteristics of H for a pair are the same for the infinite current sheet, that is, (H,a H = z > 0 (7.21b) -H,a, z<0 where H, is yet to be determined. Evaluating the line integral of H in eq. (7.21a) along the closed path in Figure 7.11(a) gives fa a - C, d1 H· dl 2. 3 — 0(—а) + (-н,)(-b) + о(a) + Н.(6) (7.21c) 2H,b 1 From eqs. (7.21a) and (7.21c), we obtain H, = K. Substituting H, in eq. (7.21b) gives 2 Ka. z>0 H = (7.22) K,a z< 0 In general, for an infinite sheet of current density K A/m, 1 кха, (7.23) H = where a, is a unit normal vector directed from the current sheet to the point of interest.
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