Page 206 - Engineering Electromagnetics, 8th Edition

188 ENGINEERING ELECTROMAGNETICS

Adding these results, we have
20
H 2 = H 2(x) + H 2(y) =− a z =−6.37a z A/m
π

D7.1. Given the following values for P 1 , P 2 , and I 1 L 1 , calculate H 2 :
(a) P 1 (0, 0, 2), P 2 (4, 2, 0), 2πa z µA·m; (b) P 1 (0, 2, 0), P 2 (4, 2, 3), 2πa z µA·m;
(c) P 1 (1, 2, 3), P 2 (−3, −1, 2), 2π(−a x + a y + 2a z )µA·m.
Ans. −8.51a x + 17.01a y nA/m; 16a y nA/m; 18.9a x − 33.9a y + 26.4a z nA/m

D7.2. A current filament carrying 15 A in the a z direction lies along the entire
√
z axis. Find H in rectangular coordinates at: (a) P A ( 20, 0, 4); (b) P B (2, −4, 4).

Ans. 0.534a y A/m; 0.477a x + 0.239a y A/m

`
7.2 AMPERE’S CIRCUITAL LAW
After solving a number of simple electrostatic problems with Coulomb’s law, we
found that the same problems could be solved much more easily by using Gauss’s
law whenever a high degree of symmetry was present. Again, an analogous procedure
exists in magnetic fields. Here, the law that helps us solve problems more easily is
4
known as Amp` ere’s circuital law, sometimes called Amp`ere’s work law. This law
may be derived from the Biot-Savart law (see Section 7.7).
Amp`ere’s circuital law states that the line integral of H about any closed path is
exactly equal to the direct current enclosed by that path,

H · dL = I (10)

We define positive current as flowing in the direction of advance of a right-handed
screw turned in the direction in which the closed path is traversed.
Referring to Figure 7.7, which shows a circular wire carrying a direct current I,
the line integral of H about the closed paths lettered a and b results in an answer of
I; the integral about the closed path c which passes through the conductor gives an
answer less than I and is exactly that portion of the total current that is enclosed by
the path c. Although paths a and b give the same answer, the integrands are, of course,
different. The line integral directs us to multiply the component of H in the direction
of the path by a small increment of path length at one point of the path, move along
the path to the next incremental length, and repeat the process, continuing until the
path is completely traversed. Because H will generally vary from point to point, and
because paths a and b are not alike, the contributions to the integral made by, say,

4 The preferred pronunciation puts the accent on “circ-.”