ELECTROMAGNETICS
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INSTRUCTOR’S SOLUTIONS MANUAL FOR ELEMENTS OF ELECTROMAGNETICS
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CHAPTER 1 P. E. 1.1 = A(a+) B + 1(,0, 3−) (=5,2, 6−) 6(,2, 3 ) A= +B +3 46 + 9= 7 (bA) B5= − −(5,0, 15 ) (5,2, −6 ) =0(, 2−, 21 ) (c) The component of A along ay is Ay = 0 (dA) B 3= + +(3,0, 9 ) (5,2, −6 ) = (8,2, 3 ) A unit vector parallel to this vector is ( ) ( ) x y z a a a a 0.9117 0.2279 0.3419 6 44 9 ,82 , 3 11 = ± + + ++ = P. E. 1.2 (a) 3 5 p x y z r = a − a + a 8 R y z r = 3a + a (b) The distance vector is (0,3,8) (2, 4,6) 2 2 QR R Q x y z r = r −...
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