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//CHAPTER 6- MAGNETIC CIRCUITS
//Example 4
disp("CHAPTER 6");
disp("EXAMPLE 4");
//VARIABLE INITIALIZATION
di=10; //diameter of iron ring in cm
dr=1.5; //diameter of iron rod in cm
mui=900; //relative permeability of rod
mu0=4*%pi*10^(-7); //absolute permeability in Henry/m
lg=5/10; //length of air-gap in cm
N=400; //number of turns
I=3.4; //current through the winding in Amperes
//SOLUTION
li=(di*%pi)-lg; //length of iron path
area=((dr^2)*%pi)/4; //area of iron cross-section
//solution (a)
mmf=(4*%pi*N*I)/10; //in gilberts, since 1 AT=(4*pi)/10
mmf=round(mmf); //to round off the value
disp(sprintf("(a) MMF is %d Gilberts",mmf));
//solution (b)
//tot reluctance = iron reluctance + air gap reluctance(mur=1 for air)
totR=(li/(area*mu0*mui))+(lg/(area*mu0*1));
disp(sprintf("(b) The total reluctance is %E Gilberts/Maxwell",totR));
//solution (c)
phi=mmf/totR;
disp(sprintf("(c) The flux in the circuit is %f Maxwell",phi));
//solution (d)
b=phi/area;
disp(sprintf("(d) The flux density in the circuit is %f Gauss",b));
//Answers of (b), (c) & (d) are different because absolute permeability is not included in (b)
//END
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