Exercise 4.3.3: Two materials in series, from the curves

About the further exercises of Section 4.3

Ten further exercises in the spirit of Problems 4.3.1–4.3.3, the applications and the home exercises. They use only

\oint_{C} H \cdot d l = N I \Rightarrow loop \sum Z_{i} Φ_{i} = \sum N I, \oint_{S} B \cdot d S = 0 \Rightarrow node \sum Φ = 0, Z = \frac{l}{μ _{0} μ _{r} S}, L = \frac{N ^{2}}{Z _{eq}}, W = \frac{1}{2} L I^{2} = \int \frac{B ^{2}}{2 μ} d v, F = \frac{B ^{2} S}{2 μ _{0}} per gap, T = \frac{1}{2} i^{2} \frac{d L}{d ϑ}, V_{rms} = 4.44 f N B_{m a x} S .

Where a material is nonlinear the $B$ – $H$ curves of Figure 4.3.19 are used, with the readings given in the statement. Exercises 4.3.1–4.3.4 are about reluctance, flux and inductance, Exercises 4.3.5–4.3.6 about transformers and motors, and Exercises 4.3.7–4.3.10 about recording heads, electromagnets, fringing and leakage. Constant: $μ_{0} = 4 π \times 1 0^{- 7} H/m = 1.2566 \times 1 0^{- 6} H/m$ .

Exercise 4.3.3

Two materials in series, from the curves

cf. Problem 4.3.1

A closed core of uniform cross-section is made of $40 cm$ of sheet steel and $20 cm$ of cast steel joined end to end (no gaps), with a coil of $N = 300$ turns. From the lecture's curves, at $B = 1.2 T$ the sheet steel needs $H = 383 A/m$ and the cast steel $H = 1224 A/m$ . (a) Find the current that gives $B = 1.2 T$ in the core. (b) If a $0.5 mm$ gap is cut in the core, what current is needed for the same $B$ (neglect fringing)? (c) What fraction of the mmf is taken by the cast steel in (a), and by the gap in (b)?

Try it yourself first.