hw3_solution

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 × =
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 +
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online