Page 211 - Engineering Electromagnetics, 8th Edition

CHAPTER 7 The Steady Magnetic Field 193

perpendicular to H x . Amp`ere’s circuital law gives
H x1 L + H x2 (−L) = K y L

or
H x1 − H x2 = K y
If the path 3-3 -2 -2-3 is now chosen, the same current is enclosed, and

H x3 − H x2 = K y
and therefore

H x3 = H x1
It follows that H x is the same for all positive z. Similarly, H x is the same for all
negative z. Because of the symmetry, then, the magnetic field intensity on one side
of the current sheet is the negative of that on the other. Above the sheet,
1 (z > 0)
H x = K y
2
while below it
1
H x =− K y (z < 0)
2
Letting a N be a unit vector normal (outward) to the current sheet, the result may be
written in a form correct for all z as
1
H = K × a N (11)
2
If a second sheet of current flowing in the opposite direction, K =−K y a y ,is
placed at z = h, (11) shows that the field in the region between the current sheets is

H = K × a N (0 < z < h) (12)
and is zero elsewhere,

H = 0(z < 0, z > h) (13)

The most difficult part of the application of Amp`ere’s circuital law is the deter-
mination of the components of the field that are present. The surest method is the
logical application of the Biot-Savart law and a knowledge of the magnetic fields of
simple form.
Problem 7.13 at the end of this chapter outlines the steps involved in applying
Amp`ere’s circuital law to an infinitely long solenoid of radius a and uniform current
density K a a φ , as shown in Figure 7.11a.For reference, the result is

H = K a a z (ρ< a) (14a)
H = 0 (ρ> a) (14b)