Answer to Question 5

The magnetic circuit above is made of a magnetic material of relative permeability 175. It has a square cross section 2.5 cm × 2.5 cm, and the average length is 25 cm. There is a air gap cut into it that is 1.0 mm in length. Calculate: (a) the reluctance of the material; (b) the reluctance of the air gap; (c) the total reluctance. 
(a) Formula:
A = (2.5 × 10^{2} m)^{2} = 6.25 × 10^{4} m^{2}
S_{1} = 0.25 m ÷ (4p × 10^{7} H m^{1} × 175 × 6.25 × 10^{4} m^{2}) = 1.82 × 10^{6} H^{1}
(b) Formula:
S_{2} = 1.0 × 10^{3} m ÷ (4p × 10^{7} H m^{1} × 6.25 × 10^{4} m^{2}) = 1.27 × 10^{6} H^{1}
(c) Formula:
S_{tot} = 1.82 × 10^{6} H^{1} + 1.27 × 10^{6} H^{1} = 3.1 × 10^{6} H^{1} (2 s.f.)
