deduce H on the infinite sheet of current
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Deduction of Section 7.4 B – Sadiku's Elements of
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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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
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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