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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 H1 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) -- p0H(2wd) = 0.
We combine (a) and (b) to obtain H, = H Ni
=- (b) g+x . The flux through the center leg of the core is simply the flux crossing the air gap x and is S= oH (2wd) = 2wdoNi 2 g+z In the absence of leakage flux this same flux links the N-turn winding N times; that is, when we evaluate SB-nda over a surface enclosed by the wire of the N-turn winding, we obtain the flux linkage A as aN = ~2w g+x (c) Note that because A is a linear function of i the system is electrically linear and we can write (c) as A = L(x)i, (d) where L(x) 2wdpN 2 g+x (e) When we assume that the current i and displacement x are specified functions of time, we can use (d) with (2.1.13) to evaluate the terminal voltage as 2wdpoN 2 di
g+z dt 2wdPoN 2i dx
(g +x)2 dt The first term is the transformer voltage that will exist if z is fixed and i is varying. The second term is the speed voltage that will exist if i is constant and x is varying. Example 2.1.2. As a second example, consider the system in Fig. 2.1.3 which has two electrical terminal pairs and the mechanical displacement is rotational. This system consists _ ...
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- Fall '11