Page 26 - Engineering Electromagnetics, 8th Edition

8 ENGINEERING ELECTROMAGNETICS

[NOTE: Throughout the text, drill problems appear following sections in which
anew principle is introduced in order to allow students to test their understanding of
the basic fact itself. The problems are useful in gaining familiarity with new terms
and ideas and should all be worked. More general problems appear at the ends of the
chapters. The answers to the drill problems are given in the same order as the parts
of the problem.]

D1.1. Given points M(−1, 2, 1), N(3, −3, 0), and P(−2, −3, −4), find:
(a) R MN ;(b) R MN + R MP ;(c) |r M |;(d) a MP ;(e) |2r P − 3r N |.
Ans. 4a x − 5a y − a z ;3a x − 10a y − 6a z ; 2.45; −0.14a x − 0.7a y − 0.7a z ; 15.56

1.5 THE VECTOR FIELD

We have defined a vector field as a vector function of a position vector. In general,
the magnitude and direction of the function will change as we move throughout the
region, and the value of the vector function must be determined using the coordinate
values of the point in question. Because we have considered only the rectangular
coordinate system, we expect the vector to be a function of the variables x, y, and z.
If we again represent the position vector as r, then a vector field G can be
expressed in functional notation as G(r); a scalar field T is written as T (r).
If we inspect the velocity of the water in the ocean in some region near the
surface where tides and currents are important, we might decide to represent it by
avelocity vector that is in any direction, even up or down. If the z axis is taken as
upward, the x axis in a northerly direction, the y axis to the west, and the origin at
the surface, we have a right-handed coordinate system and may write the velocity
vector as v = v x a x + v y a y + v z a z ,or v(r) = v x (r)a x + v y (r)a y + v z (r)a z ; each of
the components v x , v y , and v z may be a function of the three variables x, y, and z.
If we are in some portion of the Gulf Stream where the water is moving only to the
north, then v y and v z are zero. Further simplifying assumptions might be made if
the velocity falls off with depth and changes very slowly as we move north, south,
east, or west. A suitable expression could be v = 2e z/100 a x .Wehaveavelocity of
2 m/s (meters per second) at the surface and a velocity of 0.368 × 2, or 0.736 m/s, at
a depth of 100 m (z =−100). The velocity continues to decrease with depth, while
maintaining a constant direction.

D1.2. Avector field S is expressed in rectangular coordinates as S ={125/
2
2
2
[(x −1) +(y −2) +(z +1) ]}{(x −1)a x +(y −2)a y +(z +1)a z }.(a)Evaluate
S at P(2, 4, 3). (b) Determine a unit vector that gives the direction of S at P.
(c) Specify the surface f (x, y, z)on which |S|= 1.
Ans. 5.95a x + 11.90a y + 23.8a z ;0.218a x + 0.436a y + 0.873a z ;
2
2

(x − 1) + (y − 2) + (z + 1) = 125
2