# CHAPTER15 : MAGNETIC CIRCUIT COMPUTATIONS¶

## Example E01 : Pg 634¶

In :
# a
import math
phi = 6.*10.**-4.; # given magnetic flux (in Wb)
A = 0.001; #  cross sectional area (in meter square)
B = phi/A ; #
Ha = 10.; # magnetic field intensity of material a needed to establish   the given magnetic flux
Hb = 77.; #  magnetic field intensity of material b
Hc = 270.; #  magnetic field intensity of material c
La = 0.3; # arc length of material a (in meters)
Lb = 0.2; # arc length of material b (in meters)
Lc = 0.1; # arc length of material c (in meters)

F = Ha*La + Hb*Lb + Hc*Lc; # magnetomotive force
print '%s' %("a")
print '%s %.2f' %("magnetomotive force needed to establish a flux of 6*10**-4(in At) = ",F)

# b
N = 100.; # no. of turns
I = F/N; # current in amps
print '%s' %("b")
print '%s %.2f' %("current that must be made to flow through the coil(in amps) = ",I)

# c
MU0 = 4.*math.pi*10.**-7.;
MUa = B/Ha; # permeability of material a
MUb = B/Hb; # permeability of material b
MUc = B/Hc; # permeability of material c

MUra = MUa/MU0; # relative permeability of material a
MUrb = MUb/MU0; # relative permeability of material b
MUrc = MUc/MU0; # relative permeability of material c

Ra = Ha*La/phi; # reluctance of material a
Rb = Hb*Lb/phi; # reluctance of material b
Rc = Hc*Lc/phi; # reluctance of material c

print '%s' %("c")
print '%s %.2f' %("relative permeability of material a = ",MUra)
print '%s %.2f' %("relative permeability of material b = ",MUrb)
print '%s %.2f' %("relative permeability of material c = ",MUrc)
print '%s %.2f' %("reluctance of material a = ",Ra)
print '%s %.2f' %("reluctance of material b = ",Rb)
print '%s %.2f' %("reluctance of material c = ",Rc)

a
magnetomotive force needed to establish a flux of 6*10**-4(in At) =  45.40
b
current that must be made to flow through the coil(in amps) =  0.45
c
relative permeability of material a =  47746.48
relative permeability of material b =  6200.84
relative permeability of material c =  1768.39
reluctance of material a =  5000.00
reluctance of material b =  25666.67
reluctance of material c =  45000.00


## Example E03 : Pg 637¶

In :
import math
mu0 = 4.*math.pi*10.**-7.;
A = 0.0025; # cross sectional area of the coil
# dimensions of the coil (in meters)
Lg = 0.002; # air gap length (in meters)
Lbd = 0.025;
Lde = 0.1;
Lef = 0.025;
Lfk = 0.2;
Lbc = 0.175;
Lcab = 0.5;

Lbghc = 2.*(Lbd + Lde + Lef + (Lfk/2.)) - Lg;# length of the ferromagnetic material involved here

phig = 4.*10.**-4.; # air gap flux (in Wb)
Bg = phig/A ; # air gap flux density (in tesla)
Hg = Bg/mu0 ; # feild intensity of the air gap
mmfg = Hg*Lg ; # mmf produced in the air gap (in At)

Bbc = 1.38 ; # flux density corresponding to cast steel

Hbghc = 125.; # field intensity corresponding to flux density of 0.16T in the steel
mmfbghc = Hbghc*Lbghc ; #  mmf corresponding to bghc

mmfbc = mmfg + mmfbghc ; # mmf across path bc
Hbc = mmfbc/Lbc;
phibc = Bbc*A ; # flux produced in bc

phicab = phig + phibc; # total fiux existing in leg cab
Bcab = phicab/0.00375; # flux density
Hcab = 690.;
mmfcab = Hcab*Lcab; # mmf in leg cab

mmf = mmfbc + mmfcab ; # mmf produced by the coil

print '%s %.2f' %("mmf produced by the coil(in At) = ",mmf)

mmf produced by the coil(in At) =  661.90


## Example E05 : Pg 646¶

In :
# b
import math
mu0 = 4.*math.pi*10.**-7. ;
# plunger magnet dimensions (in meters)
x = 0.025;
h = 0.05;
a = 0.025;
g = 0.00125;

mmf = 1414.; # (in At)

F = math.pi*a*mu0*(mmf**2.)*(h**2.)*(1./(x + h)**2.)/g; # magnitude of the force
print '%s %.2f' %("magnitude of the force (in Newtons) = ",F)

magnitude of the force (in Newtons) =  70.16