Answer to Question 8 The diagram shows a piece of magnetic material of relative permeability 130.   The dimensions are shown on the diagram.  Calculate: (a) The Magnetomotive force; (b) The reluctance for Loop 1; (c) The reluctance for Loop 2; (d) The total reluctance; (e) The total flux; (f) The flux in Loops 1 and 2. (g) Are the fluxes consistent?  Explain your answer. (a) Fm = 250 turns × 2.5 A = 625 A turns   (b) Formula: A = 1.5 × 10-2 m × 2.0 × 10-2 m = 3.0 × 10-4 m2 m = 4p × 10-7 H m-1 × 130 = 1.634 × 10-4 H m-1   S1  = 0.35 m ÷ (1.634 × 10-4 H m-1 × 3.0 × 10-4 m2) = 7.140 × 106 H-1 = 7.1 × 106 H-1 (2 s.f.)   (c) Formula: A = 1.0 × 10-2 m × 2.0 × 10-2 m = 2.0 × 10-4 m2   S2  = 0.30 m ÷ (1.634 × 10-4 H m-1 × 2.0 × 10-4 m2) = 9.180 × 106 H-1 = 9.2 × 106 H-1 (2 s.f.)     (d) Formula: (Stot)-1 = (7.140 × 106 H-1 )-1  + (9.180 × 106 H-1 )-1 = 2.490 × 10-7 H   Stot = 4.016 × 106 H-1 = 4.0 × 106 H-1 (2 s.f.)   (e) Formula:    Ftot  = 625 A turns  ÷ 4.016 × 106 H-1  = 1.56 × 10-4 Wb = 1.6 × 10-4 Wb (2 s.f.)   (f) F1  = 625 A turns  ÷ 7.140 × 106 H-1 = 8.75× 10-5 Wb = 8.8× 10-5 Wb (2 s.f.)       F2  = 625 A turns  ÷ 9.180 × 106 H-1 = 6.81× 10-5 Wb = 6.8× 10-5 Wb (2 s.f.)   (g) Ftot  = 8.75× 10-5 Wb + 6.81× 10-5 Wb = 1.556 × 10-4 Wb      The calculation shows that the answers are consistent.