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Electromechanical Dynamics (Part 1).0044

Electromechanical Dynamics (Part 1).0044 - A-PDF Split DEMO...

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Circuit Theory material. Thus the only nonzero H occurs in the air gaps g and x, where M = 0, and (1.1.4) becomes B = pH. The use of (1.1.20) with contour (2) in Fig. 2.1.2 shows that the field intensities in the two gaps g are equal in magnitude and opposite in direction. This is expected from the symmetry of the system. Denoting the magnitude of the field intensity in the gaps g as H 1 and the field intensity in gap x by H2, we can integrate (1.1.20) around contour (1) in Fig. 2.1.2 to obtain Hig + Hx = Ni, (a) where H 2 is taken positive upward and H 1 is taken positive to the right. We now use (1.1.21) with a surface that encloses the plunger and passes through the gaps to obtain poH,(2wd) -- p 0 H(2wd) = 0. (b) We combine (a) and (b) to obtain Ni H, = H = - . g+x The flux through the center leg of the core is simply the flux crossing the air gap x and is S= oH 2 (2wd) = 2wdoNi g+z In the absence of leakage flux this same flux links the N-turn winding N times;
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