Question - A 1000-turn coil is wound on the central limb of the cast steel sheet frame as shown in Fig. bellow where dimensions are in cm. A total flux of 1.2 mWb is required to be in the gap. Find the current required in the magnetizing coil. Assume gap density is uniform and all lines pass straight across the gap. Following data is given : B (T) 0.2 0.45 0.75 11.2 900 1100 0.1 H(AT/m) 300 500 700 80 4x4cm

Step 1 Calculation of flux densities T(no. of turns) = 1000α = 1.2 mWb through air gap and central limb* flux density in either parallel path is half of that in the central path becauseflux divides into 2 equal parts at 'A'→ The two (equal ) parallel paths are Let ABCD and AXD across central path AD∴ y/2 = 1.2 mWb2 = 0.6 mWb through yokes* corresponding flux density are (for air gap & central limb) = y/A (area)                                                = 1.2 x 10-34 x 4x 10+4                                                = 0.75 Wb /m2∴By (air gap) = Bc(Central gap)                            = 0.75 Wb/m2Flux density for yolkes (By) = 0.75/2 = 0.375 Wb/m2     Step 2 mmf calculation for central and air gapsfor MMF calculations(a) for Air gap , for Bg of 0.75 Wb/m2Hg = Bgμ0 = 0.754π x 10-7 = 59.68 x 104 AT/mHg = 59.68 x 104 x 10-3      = 596.8 ampere turns .....(1)since ATg = Hgx Ig                                      Ig = 0.1 x 10-2(given)b) For central limbATc = Hcx Ichere Hc = 700 AT/m   (from given data table)∴ ATc = 7900 x 0.24                                     (Ic = 0.24  from fig.)= 168 ampere turns ......(2) Step 3 Calculation of coil current Calculation of coil current,* The yolkes are working at a flux density of 0.375 corresponding 'H'value= 500from given data table, interception can be done for accuracy∴AT required = 500 x 0.8 = 400 AT ...(3)Thus total AT required = (1) +(2)+(3)                                         = 596.8+168+400                                        = 1164.8Current required = 1164.81000=1.164 Ampere  ...