1.8Cast-steelCross-sectionalpap1 = 0.5 mm.area -3 cm?1.6l4 = 10 cmSilicon ironCast iron1.4Mild steel1.2Cross-sectionalarea 32 ст?1.00.8Cast ironCross-sectionalarea-1.4 cm?= 0.1 mm0.6(a)0.40.2A10002000 3000 4000 500060007000Magnetic field strength, H(A/m)(b)Figure (1)Silicon Ironuɔ S7 = "1Flux density, B(T) Q1//Consider the magnetic circuit shown in Figure (1-a). The circuit consists of thefollowing parts:Section A: made of silicon iron with a cross-sectional area of 1.4 cm? and a lengthof 10 cm.Section B: made of cast iron with a cross-sectional area of 3 cm? and a length of 25cm.Two air gaps, both of a cross-sectional area of 2 cm. The length of the first gap is0.5 mm and the second gap is 0.1 mm. If a coil was uniformly wound on section B,find the following:(a) The required m.m.f. to produce a flux of 210 uWb in the air gaps.(b) The required current to produce this m.m.f., if the number of turns for the coilwas 3000.(c) The relative permeabilities of silicon iron and cast iron under the aboveconditions..Make use of the B-H graph in Figure (1-b) in your solution.

Section A : silicon iron cross-sectional area = 1.4 cm2 = 1.4×10-4 m2 length = 10 cm = 0.1 m…

Answered: Q1//Consider the magnetic circuit shown…

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