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Transformer, flux linkage and emf short question

  1. Nov 5, 2012 #1
    http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s05_qp_4.pdf1. The problem statement, all variables and given/known data

    6b

    2. Relevant equations

    flux linkage = BA
    B=F/iL

    3. The attempt at a solution
    from my understanding, if B=F/iL , if the current is zero, shouldnt B be infinite,hence phi infinite or something? As seen in the 1st graph. Why does the graph follow a sine graph instead of phi(max)sin(wt)?

    Also, can anyone explain why the phase difference is 90 deg instead of 180? i thought the EMF curve would just follow a cos curve because of the negative sign of faraday's law, hence 180 phase difference.
     
  2. jcsd
  3. Nov 5, 2012 #2
    Apply Ampere's law along a path inside the core where constant H is assumed to follow.

    There is a derivative involved. If the flux is represented as a phasor ψejwt (w = frequency), then E = -dψ/dt = -jwψejwt. The j means there is a 90 degree relationship between the two. You could also see this by taking the derivative of flux assuming it is a sin and you'd end up with a cos, but phasors (and diagram) helps to keep all the relative phases straight.
     
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