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Series Magnetic CircuitsAuthor: E.E. Kimberly A magnetic circuit may be made up of sections of different reluctances in series, just as an electric circuit may be made up with different resistances in series. Fig. 74 (a) shows a rod of iron of uniform quality, but of various crosssections in series. The analogous electric circuit is shown in (b). The different component lengths of the magnetic circuit may be of different materials, and therefore may have different permeabilities, just as the electric circuit may be made up of sections of iron, copper, aluminum, etc., every one of which has different resistivity. There is, however, an important difference. The permeability of any section (being a complex function of flux density in that section) cannot be known unless the flux density is knoAvn. Since the resistance of a metallic conductor does not change (constant temperature being assumed) when the current density in it changes, Ohm's Law may be applied readily regardless of current magnitude. A comparison of the electric, magnetic, and hydraulic circuits is shown in Fig. 75.
Although the sum of the differences of magnetic potential in a magnetic circuit is equal to the total applied difference in magnetic potential, the actual percentage division of difference of potential among the several sections of Fig. 74 (a) will change as the flux density is changed. Therefore, it is not possible to calculate directly the flux density in any section of Fig. 74 (a) when a knoAvn difference of magnetic potential is applied to the whole. Such a problem can be solved by trial and error only. However, it is possible to calculate accurately the total mmf necessary to produce any specific flux density in any part, as is demonstrated in Example 71, which follows.
Example 71.  Determine the number of ampereturns necessary to produce a magnetic field of 1,000,000 lines of flux in the structure of Fig. 76 (a). '
Solution.  The first step is to find the number of ampereturns for part a of the magnetic circuit and the number for part 6. For part a,
From curve a in Fig. 76 (6) corresponding to the kind of steel in section a in (a), a flux density of 12,730 lines per sq cm requires 11.5 ampereturns per centimeter. NIa = ampturns per centimeter X centimeters in part a NIa = 11.5 X 157 = 1805 ampturns for part a For part b,
From curve b in (b) corresponding to the kind of steel in section b in (a), a flux density of 19,890 lines per sq cm requires 33 ampereturns per centimeter.
For the airgap, The flux density in the airgap (fringing being ignored) is the same as that in part 6, or 19,890 lines per sq cm. Hence, and The total number of ampereturns is
When it is required to find the number of ampereturns necessary to produce a given flux density in air and the dimensions are in inches instead of centimeters, it is convenient to use the converted equation for ampereturns. Thus,
(78) where B = flux density, in lines per square inch; I = length of gap, in inches.


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