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exam5_solution

exam5_solution - EE321 Exam 5 Solution Outline Problem 1...

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EE321 Exam 5 Solution Outline Problem 1 cm 0.01 = mm 0.001 = w 1 cm = d 5 cm = w s 5 cm = d s 2 cm = g 1 mm = N 100 = μ r 1000 = μ 0 4 π 10 7 - = F e 25 = A w d = R 45 d s w 2 + μ 0 μ r A = R 45 39.789 10 3 × = R 56 w s w + μ 0 μ r A = R 56 95.493 10 3 × = R 81 w 2 μ 0 μ r A = R 81 7.958 10 3 × = R 12 w s w + μ 0 μ r A = R 12 95.493 10 3 × = R iron 2 R 81 R 12 + 2 R 45 + R 56 + = R iron 286.479 10 3 × =
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L N 2 R iron 2 g w d μ 0 + = F e 1 2 i 2 N 2 R iron 2 g w d μ 0 + 2 2 w d μ 0 = attractive ( ) i 1 w d μ 0 w d μ 0 F e ( 29 1 2 R iron w d μ 0 2 g + N = i 4.348 10 0 × = Problem 2 i as 5 = T e 50 = T e 4 sin 2 θ rm ( 29 i as 2 = θ rm asin T e 4 i as 2 1 2 = Looking at the graph - this is the unstable answer. The stable solution is θ rm 261.799 10 3 - × = θ rm π 2 θ rm - = θ rm 1.309 10 0 × =
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Problem 3 v dc 100 = v fsw 2 = v fd 1 = i mn 9.95 = i mx 10.05 = d 0.9 = ω rm 10000 2 π 60 = T sw 1 10000 =
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