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 ◦