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//CHAPTER 6- MAGNETIC CIRCUITS
//Example 2
disp("CHAPTER 6");
disp("EXAMPLE 2");
//VARIABLE INITIALIZATION
dr=25/100; //diameter of steel ring in m
ds=3/100; //diameter of circular section in m
lg=1.5/1000; //length of air-gap in m
N=700; //number of turns
mu0=4*%pi*10^(-7); //absolute permeability in Henry/m
I=2; //in Amperes
//SOLUTION
//solution (i)
mmf=N*I;
disp(sprintf("(i) MMF is %d AT", mmf));
//solution (ii)
netMMF=(mmf-(0.35*mmf)); //mmf taken by iron path is 35% of total mmf
b=(mu0*netMMF)/lg; //phi=b*area, r=lg/(mu0*area) & mmf=phi*r => mmf=(b*lg)/mu0 => b=(mmf*mu0)/lg
disp(sprintf("(ii) The flux density of the air gap is %E Wb/m^2", b));
//solution (iii)
ar=%pi*((ds/2)^2); //area of cross-section of circular section
phi=ar*b;
disp(sprintf("(iii) The magnetic flux is %E Wb",phi));
//solution (iv)
rt=mmf/phi;
disp(sprintf("(iv) The total reluctance is %E AT/wb",rt));
//solution (v)
rg=lg/(mu0*ar); //reluctance of air gap
rs=rt-rg; //reluctance of steel
lr=%pi*dr; //circumference of ring
mur=lr/(mu0*rs*ar);
disp(sprintf("(v) The relative permeability of the steel ring is %E",mur));
//solution (vi)
disp(sprintf("(vi) Reluctance of steel is %E AT/Wb",rs));
//END
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