Question

6) Figure Q4(b) shows a coils of 200 turns. The coil is uniformly wound around a wooden ring with a mean circumference of 600mm and cross-section area of 500mm%. The current flowing into the call is 4A. (Permeability of free space fle = 4ex 10' Hm'. Calculate: L. the magnetic field strength (2 marks) flux density (1 marks) total flux (1 marks) turn I meter Figure Q4 (b) A magnetic circuit made of cast steel is arranged as in Figure Q4(cl). All limbs have a cross section area of 2x102m². A coil of 300 turns is wound on the left limb. The effective length of the section is given as led = lea=0.25 m; lda = 0.2 m; ldea = 0.35 m and l (loc) 0.25x10 m. Neglect any magnetic leakage and fringing effect. Used the BH curve characteristic given in Figure Q4@). i Draw the equivalent Ampere's Circuital Law model. (2 marks) Determine the current required to establish a flux of 6x 10 Wb in the air gap limb.

Answer 1

R o Docs By Ampero's uireital law, $ Ho di = Lenclosed Now, inside the toroid, we have amperianloop A By equation (d) we have budi = NI te Ny no. of twins - Amperian loop A from (2) we have 14 x 2TV = 2008 (4) P1 = 4x103 600x103 (2) 11 Magnetic field cMagnehe flux demaity IŠ 15 Moln 18 (2) I (20= 4TX 10 - 7 x 4 x 103 or al = 800 strength 3 20 = 1B1= 16.75*164 Tesla and Total fluxe sectional area B lacross = 16.75*10-4 x500x109 = 8.375 x 10-10 Wb 25