An iron ring of mean length 50 cm and cross-sectional area 10 cm² has a relative permeability of 800. It is wound with 500 turns. Calculate the current required to produce a flux of 0.8 mWb. Neglect leakage and fringing.
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MMF and current. [ F = \Phi R_total = (0.5 \times 10^-3) \times (2.2281 \times 10^6) = 1114.05 \text At ] [ I = \fracFN = \frac1114.051000 = 1.114 \text A ] An iron ring of mean length 50 cm
Desired flux (\Phi_des = 1.2 \ \textmWb) with (NI = 250 \ \textA-turns) (since (0.5 \times 500)). Neglect leakage and fringing
Total Reluctance ($\mathcalR total$): $$\mathcalR total = \mathcalR_i + \mathcalR_g = 1,590,986 + 3,183,098 = 4,774,084 \text AT/Wb$$
Electromagnetics Education Lab Date: April 2026
Mistake: Desired flux is (1.2\ \textmWb) – that’s higher than actual? No, problem says: after fault, measured flux = 0.8 mWb at same current. So with fault: [ \mathcalR total,fault = \frac2500.8\times 10^-3 = 312.5 \ \textkA-t/Wb ] Without fault, if no gap: (\mathcalR iron \approx 497\ \textkA-t/Wb) – but that would give even lower flux? Contradiction.