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+//Finding flux
+//Example 15.26(pg. 409)
+clc
+clear
+D=0.3//diameter of anchor ring in m
+l=%pi*D//length of iron ring in m
+N=400//number of turns on the iron ring
+a=0.0012//area of cross section of iron path in m^2
+Ur=1000//relative permeability
+Uo=4*%pi*(10^-7)//permeability
+I=2//current in A
+phi=(N*I)/(l/(Uo*Ur*a))//flux through iron path in WB
+phi1=phi/(10^-3)//flux in mWb
+printf('Thus flux through iron path is %2.2f mWb',phi1)