# Chapter 6: MAGNETIC CIRCUITS¶

## Example 6.1,Page number: 167¶

In [1]:
#Question:
"""Finding the magnetic field strength and total flux passing through a coil."""

from math import pi

#Variable Declaration:
N=200                 #Number of turns in the coil
l=0.60                #Length(Circumference) of the wooden ring(in metres)
A=500e-06             #Cross-sectional area of the ring(in square-metres)
I=4                   #Current through the coil(in Amperes)

#Calculations:
H=(N*I)/l
rel_per=1
per=(4*pi)*(1e-07)
B=per*rel_per*H
flux=B*A

#Result:
print "(a)The magnetic field strength is %.2f A/m." %(H)
print "(b)The flux density is %.2f micro T." %(B*1000000)
print "(c)The total flux is %.5f micro Wb." %(flux*1000000)

(a)The magnetic field strength is 1333.33 A/m.
(b)The flux density is 1675.52 micro T.
(c)The total flux is 0.83776 micro Wb.


## Example 6.2,Page number: 168¶

In [2]:
#Question:
"""Finding the magnetomotive force(mmf) required to produce flux."""

from math import pi

#Variable Declaration:
flux=0.015            #Flux across the air-gap(in Webers)
l=2.5e-03             #Length of the air-fap(in metres)
A=200e-04             #Effective area of the air-gap(in square-metres)

#Calculations:
B=flux/A
abs_per=(4*pi)*(1e-07)
H=B/abs_per
mmf=H*l

#Result:
print "The magneto motive force(mmf) required is %.2f At." %(mmf)

The magneto motive force(mmf) required is 1492.08 At.


## Example 6.3,Page number: 168¶

In [3]:
#Question:
"""Finding the reluctance of a mild-steel ring."""

from math import pi

#Variable Declaration:
N=200                       #Number of turns in the coil
abs_per=(4*pi)*(1e-07)      #Absolute permeability of free space
rel_per=380                 #Relative permeability of mild-steel
l=400e-03                   #Length(Circumference) of the mild-steel ring(in metres)
A=500e-06                   #Cross-sectional area of the mild-steel ring(in square-metres)
flux=800e-06                #Flux in the ring(in Webers)

#Calculations:
Rel=l/(abs_per*rel_per*A)
mmf=flux*Rel
I=mmf/N

#Result:
print "(a)The reluctance of the ring is %6e A/Wb." %(Rel)
print "(b)The magnetising current is %.2f A." %(I)

(a)The reluctance of the ring is 1.675315e+06 A/Wb.
(b)The magnetising current is 6.70 A.


## Example 6.4,Page number: 171¶

In [4]:
#Question:
"""Finding the current in the coil to produce a flux density of 0.9 T in the air gap."""

from math import pi

#Variable Declaration:
B=0.90                #Flux Density in the air gap(in Tesla)
N=4000                #Number of turns in the core
l_core=220e-03        #Length of the core(in metres)
A_core=50e-06         #Cross-sectional area of the core(in square-metres)
H_core=820            #Magnetic field intensity of the core(in Ampere per metre)
l_gap=1e-03           #Length of the air-gap(in metres)
A_gap=50e-06          #Cross-sectional area of the air gap(in square-metres)

#Calculations:
mmf_core=H_core*l_core
abs_per=(4*pi*(1e-07))
H_gap=B/abs_per
mmf_air_gap=H_gap*l_gap
Total_mmf=mmf_core+mmf_air_gap
I=Total_mmf/N

#Result:
print "The magnetisation current is %.5f A." %(I)

The magnetisation current is 0.22415 A.


## Example 6.5,Page number: 173¶

In [9]:
#Question:
"""Finding the magnetising current required to produce a flux across the air gap."""

from math import pi

#Variable Declaration:
w=40e-03               #Width of the core(in metres)
d=50e-03               #Depth of the core(in metres)
lg=2.0e-03             #Length of the air gap(in metres)
Ag=2500e-06            #Area of the air gap(in square metres)
N=800                  #Number of turns in the coil
flux=2.50e-03          #Flux across the air gap(in Webers)
lf=1.2                 #Leakage Factor
abs_per=(4*pi)*(1e-07) #Absolute permeability of free space(Henry per metre)
lc=600e-03             #Length of the core(in metres)

#Calculations:
Ac=w*d
Bg=flux/Ag
Hg=Bg/abs_per
mmf_g=Hg*lg
eff_Ac=0.92*Ac
flux_c=flux*lf
Bc=flux_c/eff_Ac
Hc=4000.0
mmf_c=Hc*lc
mmf=mmf_c+mmf_g
I=mmf/N

#Result:
print "The magnetising current required to produce a flux of 0.0025 Wb across the air gap is %.2f A." %(I)

The magnetising current required to produce a flux of 0.0025 Wb across the air gap is 4.99 A.


## Example 6.6,Page number: 174¶

In [4]:
#Question:
"""Finding the current in the coil."""

#Variable Declaration:
lA=0.3                #Length of silicon steel material(in metres)
lB=0.2                #Length of low-carbon mild steel material(in metres)
lC=0.1                #Length of cast iron material(in metres)
N=100                 #Number of turns in the exciting coil
A=0.001               #Cross-sectional area(in square-metres)
flux=600e-06          #Flux in the coil(in Webers)
abs_per=(4*pi)*(1e-07)#Absolute permeability of free space(in Henry per metre)

#Calculations:
B=flux/A
B_A=B
B_C=B
B_C=B
"""From magnetisation characteristics of ferromagnetic materials,"""
H_A=20.0
H_B=700.0
H_C=2500.0
"""According to Kirchoff's magnetomotive force law(KML),the total mmf required, tot_mmf= mmf_A+mmf_B+mmf_C."""
tot_mmf=(H_A*lA)+(H_B*lB)+(H_C*lC)
I=tot_mmf/N
rel_A=B/(abs_per*H_A)
rel_B=B/(abs_per*H_B)
rel_C=B/(abs_per*H_C)
R_A=(H_A*lA)/flux
R_B=(H_B*lB)/flux
R_C=(H_C*lC)/flux

#Result:
print "(a)The mmf for setting up a flux of 600 micro Weber is %.2f At." %(tot_mmf)
print "(b)The current in the coil is %.2f A." %(I)
print "(c)The relative permeability of:\n   Material A: %.2f\n   Material B: %.2f\n   Material C: %.2f.\n" %(rel_A,rel_B,rel_C)
print "   The reluctances are :\n   R_A=%.2f At/Wb \n   R_B=%.2f At/Wb \n   R_C=%.2f At/Wb." %(R_A,R_B,R_C)

(a)The mmf for setting up a flux of 600 micro Weber is 396.00 At.
(b)The current in the coil is 3.96 A.
(c)The relative permeability of:
Material A: 23873.24
Material B: 682.09
Material C: 190.99.

The reluctances are :
R_A=10000.00 At/Wb
R_B=233333.33 At/Wb
R_C=416666.67 At/Wb.


## Example 6.7,Page number: 175¶

In [15]:
#Question:
"""Finding the current in the exciting coil."""

from math import pi

#Variable Declaration:
N=300                 #Number of the turns in the exciting coil
flux=600e-06          #Flux in the air gap(in Webers)
lg=1e-03              #Length of the air gap(in metres)
lc=40e-02             #Mean length of the core(in metres)
A=4e-04               #Cross sectional area of the core(in square metres)
abs_per=(4*pi)*1e-07  #Absolute permeability of free space(in Henry per metre)

#Calculations:
B=flux/A
Hg=B/abs_per
mmf_g=Hg*lg
mmf_c=3000*lc
mmf=mmf_g+mmf_c
I=mmf/N

#Result:
print "The current in the exciting coil to set up a flux of 600 micro Weber in the air gap is %.2f A." %(I)

The current in the exciting coil to set up a flux of 600 micro Weber in the air gap is 7.98 A.


## Example 6.8,Page number: 175¶

In [10]:
#Question:
"""Finding the current in the exciting coil."""

#Variable Declaration:
l1=10e-02             #Length of first side of the magnetic circuit(in metres)
l2=18e-02             #Length of second side of the magnetic circuit(in metres)
l3=18e-02             #Length of third side of the magnetic circuit(in metres)
A1=6.25e-04           #Cross sectional area of l1 path(in square-metre)
A2=3.00e-04           #Cross sectional area of l2 path(in square-metre)
A3=3.00e-04           #Cross sectional area of l3 path(in square-metre)
lg=2e-03              #Length of air gap(in metres)
rel_per=800.0         #Relative permeability of core material
N=600                 #Number of turns
flux=100e-06          #Flux in the air gap(in Webers)
abs_per=(4*pi)*1e-07  #Absolute permeability of free space(in Farad per metre)

#Calculations:
Bg=flux/A1
Hg=Bg/abs_per
mmf_g=Hg*lg
B1=Bg
H1=B1/(rel_per*abs_per)
mmf_1=H1*l1
flux2=flux/2.0
B2=flux2/A2
H2=B2/(rel_per*abs_per)
mmf_2=H2*l2
tot_mmf=mmf_g+mmf_1+mmf_2
I=tot_mmf/N

#Result:
print "The current in the 600 turn exciting coil is %.3f A." %(I)

The current in the 600 turn exciting coil is 0.501 A.


## Example 6.9,Page number: 176¶

In [16]:
#Question:
"""Finding the value of exciting current."""

#Variable Declaration:
A_B=0.01              #Area of cross section of limb B(in square metre)
A_C=0.02              #Area of cross section of limb C(in square metre)
l_B=1e-03             #Length of air gap in limb B(in metres)
l_C=2e-03             #Length of air gap in limb C(in metres)
flux_B=1.5e-03        #Flux in limb B(in Webers)
N=500                 #Number of turns in the coil
abs_per=(4*pi)*(1e-07)#Absolute permeability of free space(in Henry per metre)

#Calculations:
"""The mmf across parallel paths must be same.
(R_B*flux_B)=(R_C*flux_C);"""
flux_C=flux_B*(l_B/l_C)*(A_C/A_B)
flux=flux_B+flux_C
R_B=l_B/(abs_per*A_B)
R_C=l_C/(abs_per*A_C)
R_net=1.0/((1.0/R_B)+(1.0/R_C))
I=(flux*R_net)/N

#Result:
print "The flux in limb A is %e Wb." %(flux)
print "The current in the exciting coil is %e A." %(I)

The flux in limb A is 3.000000e-03 Wb.
The current in the exciting coil is 2.387324e-01 A.


## Example 6.10,Page number: 177¶

In [21]:
#Question:
"""Finding the ampere turns required to produce a flux."""

from math import pi

#Variable Declaration:
abs_per=(4*pi)*(1e-07) #Absolute permeability of free space(in Henry per metre)
D=21e-02               #Mean diameter of the ring(in metres)
A=10e-04               #Cross sectional area(in square metre)
flux=0.8e-03           #Flux to be produced(in Webers)
lg=0.4e-03             #Length of the air gap(in metres)
rel_i=166.0            #Relative permeability of iron
rel_s=800.0            #Relative permeability of steel

#Calculations:
B=flux/A
li=(pi*D)/2.0
ls=(pi*D)/2.0
mmf_g=(B/abs_per)*lg
mmf_i=(B/(abs_per*rel_i))*li
mmf_s=(B/(abs_per*rel_s))*ls
mmf=mmf_g+mmf_i+mmf_s

#Result:
print "The total ampere turns required is %.2f At." %(mmf)

The total ampere turns required is 1782.21 At.


## Example 6.11,Page number: 177¶

In [3]:
#Question:
"""Finding the ampere turns of the coil wound on the central limb."""

#Variable Declaration:
flux=1e-03            #Flux in the central limb(in Webers)
Ac=8e-04              #Area of the central limb(in square metres)
As=5e-04              #Area of each side limb(in square metres)
l=0.15                #Length of the central limb(in metres)
lg=0.001              #Length of the air gap(in metres)
abs_per=(4*pi)*1e-07  #Absolute permeability of free space(in Henry per metre)
l1=0.34               #Length of the part ABCD(in metres)

#Calculations:
B=flux/Ac
"""Corresponding to this value of B, H is 500 At/m from the table."""
mmf_DG=500*l
Hg=1.25/abs_per
mmf_g=Hg*lg
B1=(flux/2.0)/As
"""Corresponding to this value of B, H is 200 At/m from the table."""
mmf_1=200*l1
mmf_tot=mmf_DG+mmf_g+mmf_1

#Result:
print "The required ampere turns of the coil wound on the central limb is %.2f At." %(mmf_tot)

The required ampere turns of the coil wound on the central limb is 1137.72 At.