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

Problem 4.3.2

Interactive: Magnetic Circuit Solver

Problem 4.3.2

Three-limb core with two coils

In the magnetic circuit shown in the figure, the magnetic induction field has an intensity of $1 T$ in the air gap and $I_{1} = 10 A$ . Find $I_{2}$ ( $d = 1 mm$ , $l_{1} = 3 m$ , $l_{2} = 1 m$ , $l_{3} = 1.5 m$ , $l_{4} = 1.5 m$ , $N_{1} = 386$ , $N_{2} = 500$ , material is cast steel, cross-sectional areas are assumed to be equal on each branch). (Permeability of air may be taken as $μ_{0} = 1/800000$ .)

Problem 4.3.2. The three readings on the cast-steel curve: H=770\,\mathrm{A/m} at B=1\,\mathrm{T}, B_1=0.43\,\mathrm{T} at H_1=250\,\mathrm{A/m}, and H_2=300\,\mathrm{A/m} at B_2=0.57\,\mathrm{T}. — **Figure 4.3.21.** Problem 4.3.2. The three readings on the cast-steel curve: $H = 770 A/m$ at $B = 1 T$ , $B_{1} = 0.43 T$ at $H_{1} = 250 A/m$ , and $H_{2} = 300 A/m$ at $B_{2} = 0.57 T$ .

Try it yourself first.

Strategy

Three branches meet at the two nodes A (top of the centre limb) and B (bottom): the left branch ( $l_{1}$ , source $N_{1} I_{1}$ ), the centre limb ( $l_{2}$ , source $N_{2} I_{2}$ ) and the right branch ( $l_{3} + l_{4}$ with the gap). All three see the same magnetic potential difference $U_{A B}$ , and $Φ_{1} + Φ_{2} = Φ_{3}$ . The data are given in the right branch, so start there, work out $U_{A B}$ , then the left branch, then the node, then the centre branch.

(b) B_{\mathrm{gap}} against I_2. Problem 4.3.2. (a) The equivalent circuit: the left branch with N_1I_1, the centre limb with N_2I_2 and the gapped right branch all span the nodes A and B. (b) The gap induction as a function of I_2 at I_1=10\,\mathrm{A}, computed from the cast-steel curve: 1\,\mathrm{T} needs 6.8\,\mathrm{A}, and past 1\,\mathrm{T} the curve of the steel is flat, so I_2 rises steeply. — **Figure 4.3.22.** (b) $B_{gap}$ against $I_{2}$ . Problem 4.3.2. (a) The equivalent circuit: the left branch with $N_{1} I_{1}$ , the centre limb with $N_{2} I_{2}$ and the gapped right branch all span the nodes A and B. (b) The gap induction as a function of $I_{2}$ at $I_{1} = 10 A$ , computed from the cast-steel curve: $1 T$ needs $6.8 A$ , and past $1 T$ the curve of the steel is flat, so $I_{2}$ rises steeply.

1. Right branch

$B_{g} = 1 T$ , and with equal areas $B_{3} = B_{4} = 1 T$ in the cast steel too. From the curve, $H = 770 A/m$ at $1 T$ , and in the gap $H_{g} = B_{g} / μ_{0} = 800000 A/m$ . The magnetic potential drop from A to B along this branch is

U_{A B} = H (l_{3} + l_{4}) + H_{g} d = 770 \times 3 + 800000 \times 1 0^{- 3} = 2310 + 800 = 3110 At .

2. Left branch

The same $U_{A B}$ appears across the left branch, which contains the source $N_{1} I_{1} = 386 \times 10 = 3860 At$ and the drop $H_{1} l_{1}$ :

⟹ \Rightarrow N_{1} I_{1} - H_{1} l_{1} = U_{A B} H_{1} = \frac{3860 - 3110}{3} = 250 A/m B_{1} = 0.43 T (curve) .

3. Node

$Φ_{1} + Φ_{2} = Φ_{3}$ and all areas are equal, so $B_{2} = B_{3} - B_{1} = 1.00 - 0.43 = 0.57 T$ , and the curve gives $H_{2} = 300 A/m$ .

4. Centre branch

It also spans $U_{A B}$ :

⟹ N_{2} I_{2} - H_{2} l_{2} = U_{A B} N_{2} I_{2} = 3110 + 300 \times 1 = 3410 At, I_{2} = \frac{3410}{500} .

Answer

I_{2} \approx 6.8 A

Follow the four readings on the cast-steel curve, then raise $B_{gap}$ to 1.1 T and watch $I_{2}$ jump to 9.4 A.

InteractiveLoading simulation…

Check

Loop through the left and right branches: $N_{1} I_{1} = H_{1} l_{1} + H (l_{3} + l_{4}) + H_{g} d = 750 + 2310 + 800 = 3860 At$ . Loop through the centre and right branches: $N_{2} I_{2} = H_{2} l_{2} + 3110 = 3410 At$ . Of the $3410 At$ supplied by coil 2 only $300 At$ are spent in its own limb, the rest is what the gapped right branch demands. The gap, $1 mm$ long, costs $800 At$ , more than a quarter of the whole budget and as much as $1 m$ of cast steel at $1 T$ . Because the cast-steel curve is already bending at $1 T$ , the required $I_{2}$ rises steeply with $B_{g}$ (Figure 4.3.22b): $1.1 T$ would need $9.4 A$ .

Problem 4.3.2: Three-limb core with two coils