Question

A cast steel magnetic structure made for a bar of section 8 cm × 2 cm is shown in Fig. . Determine the current that the 500 turn-magnetising coil on the left limb should carry so that a flux of 2 mWb is produced in the right limb. Take μr = 600 and neglect leakage

Answer 1

Since path C and D are in parallel with each other w.r.t. path E (Fig), the m.m.f. across the two is the same

Φ1S1 = Φ2S2

∴ Φ1 x 15/μA = 2 x 25/μA

∴ Φ1 = 10/3 mWb

∴ Φ = Φ1 + Φ2 = 16/3 mWb

Total AT required for the whole circuit is equal to the sum of

(i) that required for path E and (ii) that required for either of the two paths C or D.

Flux density in path E = (16 x 10-3)/(3 x 4 x 10-4) = 40/3 Wb/m2

AT reqd. = (40 x 0.25)/(3 x 4π x 10-7 x 600) = 4, 420

Flux density in path D = (2 x 10-3)/(4 x 10-4) = 5 Wb/m2

AT reqd. = 5/(4π x 10-7 x 600) x 0.25 = 1658

Total AT = 4,420 + 1,658 = 6,078 ;

Current needed = 6078/500 = 12.16 A