Engineering Electromagnetics 9th Edition.

CHAPTER 1 – 9th Edition 1.1. If A represents a vector two units in length directed due west, B represents a vector three units in length directed due north, and A + B = C − D and 2B − A = C + D, find the magnitudes and directions of C and D. Take north as the positive y direction: With north as positive y, west will be -x. We may therefore set up: C + D = 2B − A = 6ay + 2ax and C − D = A + B = −2ax + 3ay Add the equations to find C = 4.5ay (north), and then D = 2ax + 1.5ay (east of northeast). 1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2). a) Find the unit vector in the direction of (A − B): First A − B = (ax + 2ay + 3as) − (2ax + 3ay − 2as) = (−ax − ay + 5as) whose magnitude is |A − B| = (−a − a + 5a ) ∙ (−a – a + 5a ) 1/2 = , 1 + 1 + 25 = , x y s 3 3 = 5.20. The unit vector is therefore x y s aAB = (−ax − ay + 5as)/5.20 b) find the unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of A and B: The midpoint is located at Pmp = [1 + (2 − 1)/2, 2 + (3 − 2)/2, 3 + (−2 − 3)/2)] = (1.5, 2.5, 0.5) The unit vector is then a = (1.5ax + 2.5ay + 0.5as) = (1.5a + 2.5a + 0.5a )/2.96 mp , (1.5)2 + (2.5)2 + (0.5)2 x y s 1.3. The vector from the origin to the point A is given as (6, −2, −4), and the unit vector directed from the origin toward point B is (2, −2, 1)/3. If points A and B are ten units apart, find the coordinates of point B. With A = (6, −2, −4) and B = 1B(2, −2, 1), we use the fact that |B − A| = 10, or |(6 − 2B)ax − (2 − 2B)ay − (4 + 1B)as| = 10 3 3 3 Expanding, obtain 36 − 8B + 4B 2 + 4 − 8B + 4B 2 + 16 + 8B + 1B 2 = 100 9 3 9 3 9 , or B 2 − 8B − 44 = 0. Thus B = 8± 64−176 = 11.75 (taking positive option) and so B = 2 (11.75)a − 2 (11.75)a + 1 (11.75)a = 7.83a – 7.83a + 3.92a 3 x 3 y 3 s x y s 3 a1 a5 a6 30° a2 a3 a4 F × G = = a − a + ( − ) a = ( 2 − 2 ) a 1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. D, etermine the unit vector in rectangular components that lies in the xy plane, is tangent to the circle at ( general direction of increasing values of x: 3, −1, 0), and is in the A unit vector tangent to this circle in the general increasing x direction is t = +aø. y components are tx = aø ∙ ax = − sin ø, and ty = aø ∙ ay = cos ø. At the point ( , Its x and 3, −1), ø = 330◦ , and so t = − sin 330◦ax + cos 330◦ay = 0.5(ax + , 3ay). 1.5. An equilateral triangle lies in the xy plane with its centroid at the origin. One vertex lies on the positive y axis. a) Find unit vectorsthat are directed from the origin to the three vertices: Referring to the figure,the easy one is a1 = ay. Then, a2 will have negative x and y components, and can be constructed as a2 = G(−ax – tan 30◦ ay) where G = (1 + tan 30◦ ) 1/2 = 0.87. So finally a2 = −0.87(ax + 0.58ay). Then, a3 isthe x same as a2, but with the x component reversed: a3 = 0.87(ax − 0.58ay). b) Find unit vectors that are directed from the origin to the three sides, intersecting these at right angles: These will be a4, a5, and a6 in the figure, which are in turn just the part a results, oppositely directed: a4 = −a1 = −ay, a5 = −a3 = −0.87(ax − 0.58ay), and a6 = −a2 = +0.87(ax + 0.58ay). 1.6. Find the acute angle between the two vectors A = 2ax + ay + 3as and B = ax − 3ay + 2as by using the definition of: a) the dot product: First, A ∙ B = 2 − 3 + 6 = 5 = AB cos ϴ, where A , 2 2 2 , , , , = 2 + 1 + 3 = 14 and where B = 1 2 + 3 2 + 2 2 = 14. Therefore cos ϴ = 5/14, so that ϴ = 69.1 ◦ . b) the cross product: Begin with jax ay asj A × B = j 2 1 3 j = 11ax − ay − 7as j 1 −3 2 j , j , j , and then |A × B, | = 112 + 1 2 + 7 2 = 171. So now, with |A × B| = AB sin ϴ = 171, find ϴ = sin−1 171/14 = 69.1 ◦ 1.7. Given the field F = xax + yay. If F ∙ G = 2xy and F × G = (x 2 − y 2 ) as, find G: Let G = g1 ax + g2 ay + g3 as Then F ∙ G = g1x + g2y = 2xy, and jax ay asj j x y 0 j g3 y x g3 x y g2 x g1 y s x y s jg1 g2 g3 j j j From the last equation, it is clear that g3 = 0, and that g1 = y and g2 = x. This is confirmed in the F ∙ G equation. So finally G = yax + xay. 4 J J J J J x y s 3 )( 3 t 1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between two vectors by finding the angle between A = 3ax −2ay +4as and B = 2ax +ay −2as. Does this ambiguity exist when the dot product is used? We use the relation A × B = |A||B|sin ϴn. With the given vectors we find , L 2a + a M , , A × B = 14a + 7a = 7 5 y s = 9 + 4 + 16 4 + 1 + 4 sin ϴ n y s , 5 z ______¬ ±n where n is identified as shown; we see that n can be positive or negative, as sin ϴ can be positive or negative. This apparent sign ambiguity is not the real problem, however, as we really want , t h e m, a g n i , t u d e of the angle anyway. Choosing the positive sign, we are left with sin ϴ = 7 5/( 29 9) = 0.969. Two values of ϴ (75.7 ◦ and 104.3 ◦ ) satisfy this equation, and hence the real ambiguity. In using the do , t product, we find A ∙ B = 6 − 2 − 8 = −4 = |A||B| cos ϴ = 3 , 29 cos ϴ, or cos ϴ = −4/(3 29) = −0.248 Ù ϴ = −75.7 ◦ . Again, the minus sign is not important, as we care only about the angle magnitude. The main point isthat only one ϴ value results when using the dot product, so no ambiguity. 1.9. A field is given as Find: G = 25 (xa + ya ) (x 2 + y 2) x y a) a unit vector in the direction of G at P(3, 4, −2): Have Gp = 25/(9 + 16) × (3, 4, 0) = 3ax + 4ay, and |Gp| = 5. Thus aG = (0.6, 0.8, 0). b) the angle between G and ax at P: The angle is found through aG ∙ ax = cos ϴ. So cos ϴ = (0.6, 0.8, 0) ∙ (1, 0, 0) = 0.6. Thus ϴ = 53◦ . c) the value of the following double integral on the plane y = 7: 4 2 G ∙ aydsdx 0 0 4 2 25 ( 4 2 25 4 350 0 0 x 2 + y 2 xax + yay) ∙ aydsdx = 0 0 x 2 + 49 × 7 dsdx = J0 x 2 + 49 dx = 350 × 1 an 7 −1 4 − 0 7 = 26 1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle between any two diagonals of a cube, where each diagonal connects diametrically opposite corners, and passes through the center of the cube: Assuming a side length, b, two diagonal vectors would be A = b(ax + ay + as) and B = b(a – a + a ). Now use A ∙ B = |A||B| cos ϴ, or b 2 , b , b) cos ϴ Ù cos ϴ = 1/3 Ù ϴ = 70.53◦ . This result (in magnitude) is the same for any two diagonal vectors. (1 − 1 + 1) = ( J 5 A 1.11. Given the points M(0.1, −0.2, −0.1), N(−0.2, 0.1, 0.3), and P (0.4, 0, 0.1), find: a) the vector RMN : RMN = (−0.2, 0.1, 0.3) − (0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4). b) the dot product RMN ∙ RMP : RMP = (0.4, 0, 0.1) − (0.1, −0.2, −0.1) = (0.3, 0.2, 0.2). RMN ∙ RMP = (−0.3, 0.3, 0.4) ∙ (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05. c) the scalar projection of RMN on RMP : RMN ∙ aRMP = (−0.3, 0.3, 0.4) ∙ , (0.3, 0.2, 0.2) 0.09 + 0.04 + 0.04 0.05 , 0.17 = 0.12 d) the angle between RMN and RMP : 0 R ∙ R 1 H I 0.05 ϴM = cos−1 MN MP = cos−1 , , = 78◦ |RMN ||RMP | 0.34 0.17 1.12. Write an expression in rectangular componentsfor the vector that extendsfrom (x1 , y1 ,s1 ) to (x2 , y2 ,s2 ) and determine the magnitude of this vector. The two points can be written as vectors from the origin: A1 = x1ax + y1ay + s1as and A2 = x2ax + y2ay + s2as The desired vector will now be the difference: A12 = A2 − A1 = (x2 − x1 )ax + (y2 − y1 )ay + (s2 − s1 )as whose magnitude is |A12| = , 12 ∙ A12 = (x2 − x1 ) 2 + (y2 − y1 ) 2 + (s2 − s1 ) 2 1/2 1.13. a) Find the vector component of F = (10, −6, 5) that is parallel to G = (0.1, 0.2, 0.3): F|| = F ∙ G G = (10, −6, 5) ∙ (0.1, 0.2, 0.3)(0.1, 0.2, 0.3) = (0.93, 1.86, 2.79) G |G| 2 0.01 + 0.04 + 0.09 b) Find the vector component of F that is perpendicular to G: FpG = F − F||G = (10, −6, 5) − (0.93, 1.86, 2.79) = (9.07, −7.86, 2.21) c) Find the vector component of G that is perpendicular to F: G = G − G = G − G ∙ F F = (0.1, 0.2, 0.3) − 1.3 (10, −6, 5) = (0.02, 0.25, 0.26) pF ||F |F| 2 100 + 36 + 25