# Mathcad - 04-14 - turns 1 Problem 4-14: Solution lcs := 0.7...

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turns 1 Problem 4-14: Solution l cs 0.7 m := l ci 0.2 m := N 300 turns := A cs 0.0025 m 2 := A ci 0.00625 m 2 := This problem must be solved iteratively or graphically. We will do it graphically. Since we have to read values from a graph, it is easier to use vectors to solve this problem. Start by defining a vector for the flux density in the iron. Due to the differences in area, the steel flux density will be higher by the ratio of the iron area to the steel area (flux is the same in both materials) B ci 0 .03 .06 .09 .12 .15 .18 .21 .24 .27 .30 T := B cs A ci A cs B ci := B cs 0 0 1 2 3 4 5 6 7 8 9 10 0 0.075 0.15 0.225 0.3 0.375 0.45 0.525 0.6 0.675 0.75 T = For each value of flux density in the materials, we can read a value for the field intensity H ci 0 75 150 225 300 375 450 525 700 850 1000 amp

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## This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University.

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Mathcad - 04-14 - turns 1 Problem 4-14: Solution lcs := 0.7...

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