Bài tập trường điện từ có lời giải chi tiết

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Bài giảng Kỹ thuật điện: Chương 10 - ĐH Bách Khoa TP. HCM

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Đề thi cuối học kỳ I năm học 2018-2019 môn Năng lượng tái tạo - ĐH Bách khoa TP.HCM

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Bài giảng Trường điện từ: Lecture 7 - Trần Quang Việt

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CHAPTER 11.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find:a) a unit vector in the direction of −M + 2N.−M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4)Thusa=(26, 10, 4)= (0.92, 0.36, 0.14)|(26, 10, 4)|b) the magnitude of 5ax + N − 3M:(5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6.c) |M||2N|(M + N):|(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10)= (−580.5, 3193, −2902)1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7, −2, 1):a) Specify the vector A extending from the origin to the point A.A = (4, 3, 2) = 4ax + 3ay + 2azb) Give a unit vector extending from the origin to the midpoint of line AB.The vector from the origin to the midpoint is given byM = (1/2)(A + B) = (1/2)(4 − 2, 3 + 0, 2 + 5) = (1, 1.5, 3.5)The unit vector will bem=(1, 1.5, 3.5)= (0.25, 0.38, 0.89)|(1, 1.5, 3.5)|c) Calculate the length of the perimeter of triangle ABC:Begin with AB = (−6, −3, 3), BC = (9, −2, −4), CA = (3, −5, −1).Then|AB| + |BC| + |CA| = 7.35 + 10.05 + 5.91 = 23.321.3. The vector from the origin to the point A is given as (6, −2, −4), and the unit vector directed from theorigin toward point B is (2, −2, 1)/3. If points A and B are ten units apart, find the coordinates of pointB.With A = (6, −2, −4) and B = 13 B(2, −2, 1), we use the fact that |B − A| = 10, or|(6 − 23 B)ax − (2 − 23 B)ay − (4 + 13 B)az | = 10Expanding, obtain36 − 8B + 49 B 2 + 4 − 83 B + 49 B 2 + 16 + 83 B + 19 B 2 = 100or B 2 − 8B − 44 = 0. Thus B =B=√8± 64−1762= 11.75 (taking positive option) and so212(11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az33311.4. given points A(8, −5, 4) and B(−2, 3, 2), find:a) the distance from A to B.|B − A| = |(−10, 8, −2)| = 12.96b) a unit vector directed from A towards B. This is found throughaAB =B−A= (−0.77, 0.62, −0.15)|B − A|c) a unit vector directed from the origin to the midpoint of the line AB.a0M =(3, −1, 3)(A + B)/2= √= (0.69, −0.23, 0.69)|(A + B)/2|19d) the coordinates of the point on the line connecting A to B at which the line intersects the plane z = 3.Note that the midpoint, (3, −1, 3), as determined from part c happens to have z coordinate of 3. Thisis the point we are looking for.1.5. A vector field is specified as G = 24xyax + 12(x 2 + 2)ay + 18z2 az . Given two points, P (1, 2, −1) andQ(−2, 1, 3), find:a) G at P : G(1, 2, −1) = (48, 36, 18)b) a unit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), soaG =(−48, 72, 162)= (−0.26, 0.39, 0.88)|(−48, 72, 162)|c) a unit vector directed from Q toward P :aQP =(3, −1, 4)P−Q= √= (0.59, 0.20, −0.78)|P − Q|26d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x 2 + 2), 18z2 )|, or10 = |(4xy, 2x 2 + 4, 3z2 )|, so the equation is100 = 16x 2 y 2 + 4x 4 + 16x 2 + 16 + 9z421.6. For the G field in Problem 1.5, make sketches of Gx , Gy , Gz and |G| along the line y = 1, z = 1, for0 ≤ x ≤ 2. We find√G(x, 1, 1) = (24x, 12x 2 + 24, 18), from which Gx = 24x, Gy = 12x 2 + 24,Gz = 18, and |G| = 6 4x 4 + 32x 2 + 25. Plots are shown below.1.7. Given the vector field E = 4zy 2 cos 2xax + 2zy sin 2xay + y 2 sin 2xaz for the region |x|, |y|, and |z| lessthan 2, find:a) the surfaces on which Ey = 0. With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0, with|x|
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