hw3_solution

# hw3_solution - EE321 Spring 08 Homework 3 Problem 11 From...

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EE321, Spring 08 Homework 3 Problem 11 From minimum cost solution in class N 260 := d 8.3788 10 2 := g 13.069 10 3 := w 1.8361 10 2 := d s 8.9184 10 2 := w s 8.9184 10 2 := μ 0 4 π 10 7 := i4 0 := L des 510 3 := Recomputing the reluctance R 2g μ 0 wd 2w s + () 2d s + 2500 μ 0 + := R 13.609 10 6 × = a) To find the inductance L N 2 R := L 4.967 10 3 × = b) To find the flux density in the magnetic material Φ Ni R = BA = so B m Ni Rw d := B m 496.731 10 3 × = c) The flux density in the air gap is the same (areas assumed equal) B g B m := B g 496.731 10 3 × = d) The field intensity in the magnetic material is H m B m 2500 μ 0 := H m 158.114 10 0 × = e) The field intensity in the air gap is H g B g μ 0 := H g 395.286 10 3 × =

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f) The percentage difference between this calculated inductance and the design value is LL des L des 100 655.764 10 3 × = percent Problem 12 From given E integral equation E 1 2 μ H H v = Using results from Problem 12 E g 1 2 μ 0 H g 2 2g w d () := E g 3.948 10 0 × = E m 1 2 2500 μ 0 H m 2 2w s +
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hw3_solution - EE321 Spring 08 Homework 3 Problem 11 From...

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