Calculating electromotive force - practical situation

Discussion in 'Physics' started by elcraft, Sep 24, 2012.

  1. elcraft

    Thread Starter New Member

    Sep 24, 2012

    We know that the generated electromotive force is EMF = dΦ/dt.

    Is it correct for practical calculations if I use EMF = B*S*cosθ/Δt, where Δt is half the period of the 50/60Hz AC frequency current through the machine that produces the field (S the surface and B the magnetic field)?

    Thanks in advance. :)
  2. russ_hensel

    Well-Known Member

    Jan 11, 2009
    That would be an approximation assuming that the surface area is S and at an angle of theta with respect to the surface ( the wire (( or path for the emf ))should form the boundary of the surface). For a solenoid generating the field and a wire enclosing it that may be pretty good. The worst part of the equation may be the attempt to find the derivative or the B field. The output will be a sine wave as I assume the current driving the B field is assumed. The derivative of sin and cosine are easy so why not do the correct derivative, even better is using complex notation where the driver of current and field is exp( i * omega *t ) remember at the end to take the real part. The derivative of e to the x is e to the x so derivatives and integrals are easy.