electromagnetic relay question

Discussion in 'Homework Help' started by exidez, Nov 13, 2008.

  1. exidez

    Thread Starter Member

    Aug 22, 2008
    26
    0
    this question has been bothering me as i do not have any solutions or answers! See attached jpg for the question.

    It states the air gap has a flux intensity of 0.8 Telsa
    so from the graph i can get H to be around 400 At/m for the cast steel arm

    the length of the air gap is 1.5mm which is 1.5x10^-3 meters
    H* l = N I
    (H*L)/N = I
    (400*1.5x10^-3)/1000 = 0.6 mA

    is this correct? is all the other information useless??

    i can easily work out part b. thats not problem
     
  2. exidez

    Thread Starter Member

    Aug 22, 2008
    26
    0
    i found the solution!

    you have to add all the components
    NI = ƩΦR = ƩHL = ƩH(B/μ)

    so the answer is
    NI = HL (cast steel) + HL (sheet steel) + H(B/μ) (air gap)
    NI = (0.12*500) + (0.06*250) + (1.5x10^-3 * [ 0.8 / 4∏*10^-7 ] )
    NI = 1030
    1030 / 1000 = I = 1.03 A

    it always seems so simple AFTER you know the answer!
     
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