Magnetic Circuits Problems And Solutions Pdf Direct

Flux density in yokes = same as center limb area? Yokes have (A=6\ \textcm^2), but they carry (\Phi_c)? No – yokes carry the outer branch flux? Actually each yoke segment carries (\Phi_o) if symmetric. Check: At top yoke, flux from center splits: half to left outer, half to right outer. So yoke carries (\Phi_o). [ B_yoke = \frac0.4845\times 10^-36\times 10^-4 = 0.8075 \ \textT ] Desired flux (\Phi_des = 1.2 \ \textmWb) with (NI = 250 \ \textA-turns) (since (0.5 \times 500)).

Center limb: [ \mathcalR_c = \frac0.1(4\pi\times 10^-7)(1000)(6\times 10^-4) \approx 132.6 \ \textkA-t/Wb ] Each outer limb: [ \mathcalR_o = \frac0.2(4\pi\times 10^-7)(1000)(3\times 10^-4) \approx 530.5 \ \textkA-t/Wb ] Yoke (each, two yokes in series effectively for each flux path): [ \mathcalR y = \frac0.05(4\pi\times 10^-7)(1000)(6\times 10^-4) \approx 66.3 \ \textkA-t/Wb ] Total for one outer path (center → yoke → outer limb → yoke → center): [ \mathcalR outer, total = \mathcalR_c + 2\mathcalR_y + \mathcalR_o ] [ = 132.6 + 2(66.3) + 530.5 = 795.7 \ \textkA-t/Wb ] But careful: The two outer paths are after the center limb. magnetic circuits problems and solutions pdf

Ah – critical insight: If the core originally had , its reluctance is 497 kA-t/Wb. Then flux would be (250/497k \approx 0.503 \ \textmWb), not 1.2 mWb. So the “desired” 1.2 mWb must have come from a different core or higher current. The problem as written is inconsistent – an excellent teaching point: always check if numbers make physical sense . Flux density in yokes = same as center limb area

Flux density: [ B = \frac\PhiA = \frac1.005\times 10^-35\times 10^-4 = 2.01 \ \textT ] Good – below saturation for typical iron. Solution 2 – With Air Gap (a) Core reluctance same as above: (\mathcalR_c \approx 398 \ \textkA-turns/Wb) Gap reluctance: [ \mathcalR g = \fracl_g\mu_0 A = \frac0.001(4\pi\times 10^-7)(5\times 10^-4) \approx 1.592 \times 10^6 \ \textA-turns/Wb ] Total reluctance: [ \mathcalR total = 3.98\times 10^5 + 1.592\times 10^6 = 1.99 \times 10^6 \ \textA-turns/Wb ] Actually each yoke segment carries (\Phi_o) if symmetric

The center limb carries (\Phi_c). That flux splits into two paths, each with total reluctance (\mathcalR_branch = \mathcalR_o + 2\mathcalR_y). The center limb reluctance is in series with the parallel combination of the two branch reluctances.

Hint: By symmetry, the two outer limbs carry equal flux. A DC relay has a magnetic circuit that should produce (\Phi = 1.2 \ \textmWb) at (I = 0.5 \ \textA) with (N = 500). After years of use, the measured flux is only (0.8 \ \textmWb) at the same current. You suspect an unexpected air gap has developed (e.g., due to corrosion or mechanical wear).