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+clc;
+N=1000; // Number of turns
+f=1*10^-3; // flux in central limb
+Ac=8*10^-4; // Area of central limb
+Ao=4*10^-4; // Area of outer limb
+lg=2*10^-3; // length of air gap
+lc=0.15; // length of central limb in metre
+lo=0.25; // length of outer limb in metre
+uo=4*%pi*10^-7; // permeability of free space
+disp('case a');
+// for ur=infinity, reluctance offered by cast steel is zero
+Rl1=lg/(uo*Ao); // reluctance offered by outer limb
+Rl2=lg/(uo*Ac); // reluctance offered by central limb
+// Assuming magnetic circuit as a close circuit, applying KVl in one of loop gives
+I=(f*(Rl2+(Rl1/2)))/N;
+printf('Coil current for first case is %f A\n',I);
+disp('case b');
+ur=6000; // relative permability
+Rlc1=(lc+lo)/(uo*ur*Ao); // reluctance of outer steel core (including the top)
+Rlc2=(lc)/(uo*ur*Ac); // reluctance offered by central steel core
+r=(Rlc1+Rl1)/2; // resultant of outer reluctance
+// By kVL we get
+I=(f*(Rlc2+Rl2+r))/N;
+printf('Coil current for second case is %f A\n',I);