Page 221 - Engineering Electromagnetics, 8th Edition

CHAPTER 7 The Steady Magnetic Field 203

Figure 7.16 The sum of the closed line integrals
about the perimeter of every S is the same as the
closed line integral about the perimeter of S because
of cancellation on every interior path.

because every interior wall is covered once in each direction. The only boundaries
on which cancellation cannot occur form the outside boundary, the path enclosing S.
Therefore we have

(∇× H) · dS (30)
H · dL ≡
S
where dL is taken only on the perimeter of S.
Equation (30) is an identity, holding for any vector field, and is known as Stokes’
theorem.

EXAMPLE 7.3
A numerical example may help to illustrate the geometry involved in Stokes’ theorem.
Consider the portion of a sphere shown in Figure 7.17. The surface is specified by r =
4, 0 ≤ θ ≤ 0.1π,0 ≤ φ ≤ 0.3π, and the closed path forming its perimeter is com-
posed of three circular arcs. We are given the field H = 6r sin φa r +18r sin θ cos φa φ
and are asked to evaluate each side of Stokes’ theorem.
Solution. The first path segment is described in spherical coordinates by r = 4, 0 ≤
θ ≤ 0.1π, φ = 0; the second one by r = 4,θ = 0.1π, 0 ≤ φ ≤ 0.3π; and the third
by r = 4, 0 ≤ θ ≤ 0.1π, φ = 0.3π. The differential path element dL is the vector
sum of the three differential lengths of the spherical coordinate system first discussed
in Section 1.9,

dL = dr a r + rdθ a θ + r sin θ dφ a φ