Magnetic Circuits Problems And Solutions Pdf May 2026

So: [ \mathcalR_eq, branches = \frac(\mathcalR_o + 2\mathcalR_y)2 = \frac530.5 + 132.62 = 331.55 \ \textkA-t/Wb ] Wait – (2\mathcalR_y = 132.6), so (\mathcalR_o + 2\mathcalR_y = 530.5+132.6 = 663.1). Half of that is kA-t/Wb.

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)). magnetic circuits problems and solutions pdf

Flux: [ \Phi = \frac4001.725\times 10^6 \approx 0.232 \ \textmWb ] Yokes have (A=6\ \textcm^2), but they carry (\Phi_c)