Exam5_solution - EE321 Exam 5 Solution Outline Problem 1 cm:= 0.01 mm:= 0.001 w:= 1 cm d:= 5 cm ws:= 5 cm d s:= 2 cm g:= 1 mm N:= 100 r:= 1000 7

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

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

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

View Full DocumentRight Arrow Icon
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
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.

This note was uploaded on 02/19/2012 for the course ECE 321 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

Page1 / 5

Exam5_solution - EE321 Exam 5 Solution Outline Problem 1 cm:= 0.01 mm:= 0.001 w:= 1 cm d:= 5 cm ws:= 5 cm d s:= 2 cm g:= 1 mm N:= 100 r:= 1000 7

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