Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 4.3.1

Interactive: Magnetic Circuit Solver

Problem 4.3.1

Two C-cores with two air gaps

What should be the width of the air gap, namely $d$ , such that the magnetic induction field in this gap will have an intensity of $0.5 T$ ( $l_{1} = 0.5 m$ , $l_{2} = 1 m$ , $N_{1} = 500$ , $N_{2} = 200$ , $I_{1} = 10 A$ , $I_{2} = 7 A$ , $μ_{r 1}$ sheet steel, $μ_{r 2}$ cast iron, cross-sectional areas are assumed to be equal on each branch)? (Permeability of air may be taken as $μ_{0} = 1/800000$ .)

Problem 4.3.1. The B–H curves of sheet steel, cast steel and cast iron, with the readings at B=0.5\,\mathrm{T}: H_1=60\,\mathrm{A/m} for the sheet steel and H_2=1500\,\mathrm{A/m} for the cast iron. — **Figure 4.3.19.** Problem 4.3.1. The $B$ – $H$ curves of sheet steel, cast steel and cast iron, with the readings at $B = 0.5 T$ : $H_{1} = 60 A/m$ for the sheet steel and $H_{2} = 1500 A/m$ for the cast iron.

Try it yourself first.

Strategy

The two C-cores and the two gaps form one closed flux tube, so the same flux $Φ$ , and since all cross-sections are equal the same $B = 0.5 T$ , is carried by the sheet steel, the cast iron and both gaps. Ampère's law round the mean path then has four terms on the left and the two magnetomotive forces on the right. The iron terms are found by reading $H$ at $0.5 T$ from the two curves, and the gap term is $H_{g} = B / μ_{0}$ .

1. Equivalent circuit

In series: $Z_{1}$ (sheet steel, $l_{1}$ ), $Z_{g}$ , $Z_{2}$ (cast iron, $l_{2}$ ) and $Z_{g}$ again, with the two sources $N_{1} I_{1}$ and $N_{2} I_{2}$ driving the flux the same way round (Figure 4.3.20a). The flux crosses the gap twice.

2. Loop rule

H_{1} l_{1} + H_{2} l_{2} + 2 \frac{B}{μ _{0}} d = N_{1} I_{1} + N_{2} I_{2} = 500 \times 10 + 200 \times 7 = 6400 At .

3. Read the curves at B = 0.5 T

Sheet steel: $H_{1} = 60 A/m$ , so $H_{1} l_{1} = 60 \times 0.5 = 30 At$ . Cast iron: $H_{2} = 1500 A/m$ , so $H_{2} l_{2} = 1500 \times 1 = 1500 At$ . In the gaps $H_{g} = B / μ_{0} = 0.5 \times 800000 = 4 \times 1 0^{5} A/m$ , and the flux crosses a gap twice: $2 H_{g} d = 8 \times 1 0^{5} d$ .

4. Solve for d

⟹ 8 \times 1 0^{5} d = 6400 - 30 - 1500 = 4870 d = 6.09 \times 1 0^{- 3} m .

Answer

d \approx 6.1 mm

(b) B_{\mathrm{gap}} against d at the fixed mmf. Problem 4.3.1. (a) The equivalent circuit: the sheet steel, the cast iron and the two gaps in series with the two mmf sources. (b) The gap induction obtained for every gap width d with F_m=6400\,\mathrm{At} (computed from the curves of Figure 4.3.19): the required 0.5\,\mathrm{T} is reached at d=6.1\,\mathrm{mm}, and however small the gap is made, the cast iron saturates and B cannot pass about 0.65\,\mathrm{T}. — **Figure 4.3.20.** (b) $B_{gap}$ against $d$ at the fixed mmf. Problem 4.3.1. (a) The equivalent circuit: the sheet steel, the cast iron and the two gaps in series with the two mmf sources. (b) The gap induction obtained for every gap width $d$ with $F_{m} = 6400 At$ (computed from the curves of Figure 4.3.19): the required $0.5 T$ is reached at $d = 6.1 mm$ , and however small the gap is made, the cast iron saturates and $B$ cannot pass about $0.65 T$ .

The tool reads both curves at 0.5 T and returns $d = 6.09$ mm. Switch to 'Given d → find B' and shrink $d$ to see the cast-iron ceiling near 0.65 T.

InteractiveLoading simulation…

Check

Back into the loop: $30 + 1500 + 8 \times 1 0^{5} \times 6.09 \times 1 0^{- 3} = 30 + 1500 + 4870 = 6400 At$ . Units: $H$ in A/m times $l$ in m gives A, and $(B / μ_{0}) d$ is $T m/ (H/m) = Wb/H = A$ . Of the $6400 At$ , $4870 At$ ( $76 %$ ) are spent on $12 mm$ of air and only $1530 At$ on $1.5 m$ of iron, and $1500 At$ of those on the cast iron, whose permeability at $0.5 T$ is $μ = B / H = 0.5/1500 = 3.3 \times 1 0^{- 4} H/m$ , i.e. $μ_{r} \approx 270$ , against $μ_{r} \approx 6700$ for the sheet steel. The cast iron also sets a ceiling (Figure 4.3.20b): its curve flattens near $0.65 T$ , so however small $d$ is made, the gap induction cannot exceed about $0.65 T$ with this mmf.

Problem 4.3.1: Two C-cores with two air gaps