Electrical Machine Question

Discussion in 'Homework Help' started by athrun300, Jan 24, 2011.

  1. athrun300

    Thread Starter New Member

    Jan 24, 2011
    Q2. The magnetic circuit shown in Figure Q2 is made of high permeability steel so that its reluctance can be negligible. The dimensions are as follows: N =1200, g = 2 mm, l = 15 cm, and i = 10 Adc.
    (i) Derive an expression for flux density in the air gap, Bg, as a function of dimensions and magnetic field quantities.
    (ii) Derive an expression for the inductance, L, in the two air gaps.
    (iii) Find the force, Fm, on the moving part using (i) and (ii).
    (iv) If the maximum flux density, Bmax, is to be limited to 1.6 T, determine Fm(max).

    Pls help,,,,,,,
  2. amilton542

    Active Member

    Nov 13, 2010
    for the flux density of an air gap this formula may help you B= 4 x 3.142 to the power of 10 minus 7 x NI then divide by the air gap (g)
    Last edited: Jan 25, 2011