The magnetic circuit shown in figure below is built up of solid iron cross section, 1x1cm2. Each air gap is 1 mm wide. For this type of air gap, the effective cross-section of the air may be taken as 1.2 x that of iron path to allow for fringing. Two different grades of iron are used. When a coil having 2000 turns is wound on part B, and a current of 0.5A is applied to it, the relative permeability of part A is measured as 500, and of part B as 1000.
a) Calculate the Reluctance of part A
b) Calculate the Reluctance of part B
Ra = 0.2 / [1 x 10^-4 x 4pi x 10^-7 x 500] = 3.18 x 10^6 At/W
Rb = 0.38 / [1 x 10^-4 x 4pi x 10^-7 x 1000] = 3.02 x 10^6 At/W
The above is a provided solution for parts (a) and (b). Please explain (comprehensively) how these equations were formed and particularly, where did the 0.38 come from when calculating the reluctance of part B?
The magnetic circuit with two air gaps is given and the reluctance or the circuit is calculated by calculating the series and parallel combinations of the reluctances.