Chapter 6: Magnetic Materials and Circuits

Example 6.1,Page number 2-26

In [1]:
import math

#Given Data:
H=198                  #Magnetizing Force in Ampere per meter
M=2300                 #Magnetization in Ampere per meter
u0=4*math.pi*10**-7    #Permeability in vacuum

#Calculations:
#H=(B/u0)-M
B=u0*(H+M)             #Flux Density
ur=B/(u0*H)            #Relative Permeability

print"Corresponding Flux Density is =",B,"Wb/m^2"
print"Relative Permeability is =",ur

Corresponding Flux Density is = 0.00313907937947 Wb/m^2
Relative Permeability is = 12.6161616162

Example 6.2,Page number 2-26

In [2]:
import math

#Given Data:
x=3.7*10**-3           #Susceptibility at T=300 K
T=300                  #Temperature in kelvin
T1=250                 #Temperature in kelvin
T2=600                 #Temperature in kelvin

#Calculations:
C=x*T                  #Curie's law
ur1=C/T1               #Relative permeability at 250 K
ur2=C/T2               #Relative permeability at 600 K

print"Relative Permeability at 250 K is =",ur1
print"Relative Permeability at 600 K is =",ur2

Relative Permeability at 250 K is = 0.00444
Relative Permeability at 600 K is = 0.00185

Example 6.3,Page number 2-27

In [3]:
import math

#Given Data:
u=0.8*10**-23          #Magnetic dipole moment of an atom in paramagnetic gas in J/T
B=0.8                  #Magnetic field in tesla
K=1.38*10**-23         #Boltzmann constant

#To find Temperature at which Average thermal energy is equal to Magnetic energy 
#i.e. uB=3KT/2
T=2*u*B/(3*K)          #Required temperature

print"Required temperature is =",T,"Kelvin"

Required temperature is = 0.309178743961 Kelvin

Example 6.4,Page number 2-27

In [5]:
import math

#Given Data:
T=27+273           #Temperature in kelvin
B=0.5              #Magnetic field in tesla
C=2*10**-3         #Curie's Constant
u0=4*math.pi*10**-7    #Permeability in vacuum

#  C=u0*M*T/B (Curie's law)
M=C*B/(u0*T)      #Magnetization of material at 300 K

print"Magnetization of material at 300 K is =",M,"A/m"

Magnetization of material at 300 K is = 2.65258238486 A/m

Example 6.5,Page number 2-27

In [7]:
import math

#Given Data:
B=10.9*10**-5            #Horizontal component of B in wb/m^2
u0=4*math.pi*10**-7     #Permeability in free space

H=B/u0                  #Horizontal component of magnetic field
print"Horizontal component of magnetic field is =",H,"Ampere/meter"
print"(Print mistake in unit in book)"

Horizontal component of magnetic field is = 86.7394439851 Ampere/meter
(Print mistake in unit in book)

Example 6.6,Page number 2-28

In [8]:
import math

#Given Data:
u0=4*math.pi*10**-7     #Permeability in vacuum
ur=900                  #Relative permeability of medium
l=2                      #length in meter
A=60*10**-4             #Crosss sectional area of ring in m^2
phi=5.9*10**-3          #flux in weber
n=700                   #Number of turns

#Calculations:
#We know, phi=B*A
B=phi/A            #Flux density
#But, B=u*H
H=B/(u0*ur)        #Magnetic field strength

I=H*l/n            #Required current
print"Current required to produce given flux is =",I,"Ampere"

Current required to produce given flux is = 2.48416445567 Ampere

Example 6.7,Page number 2-28

In [10]:
import math

#Given Data:

u0=4*math.pi*10**-7    #Permeability in vacuum
ur=900                 #Relative permeability of medium
r=25*10**-2            #radius of ring
A=25*10**-4            #Crosss sectional area of ring in m^2
Ag=1*10**-3            #Air gap
phi=2.7*10**-3         #flux in weber
N=400                  #Number of turns

#Calculations:
#We know, phi=B*A
B=phi/A            #Flux density
#But, B=u*H
H=B/(u0*ur)        #Magnetic field strength
L=H*2*math.pi*r+(B*Ag/u0)   #Total amp turns required (iron+air)
I=L/N           #Required current

print"Current required to produce given flux is =",I,"Ampere"

Current required to produce given flux is = 5.89859173174 Ampere

Example 6.8,Page number 2-29

In [11]:
import math

#Given Data:

u0=4*math.pi*10**-7     #Permeability in vacuum
A=0.2*10**-4            #Crosss sectional area of iron bar in m^2
H=1600                  #magnetising field in A/m
phi=2.4*10**-5          #Magnetic flux in weber


#Calculations:
#We know, phi=B*A
B=phi/A            #Flux density
u=B/H              #magnetic permeability
ur=u/u0            #relative permeability
xm=ur-1            #susceptibility of the iron bar

print"magnetic permeability of iron bar is =",u,"N/(A^2)"
print"susceptibility of the iron bar is =",xm

magnetic permeability of iron bar is = 0.00075 N/(A^2)
susceptibility of the iron bar is = 595.831036595

Example 6.9,Page number 2-29

In [13]:
import math

#Given Data:
u0=4*math.pi*10**-7       #Permeability in vacuum
xm=948*10**-11            #susceptibility of the iron bar

#Calculations:
ur=1+xm            #relative permeability
u=u0*ur            #permeability of medium

print"Relative Permeability of medium is =",ur
print"Permeability of medium is =",u,"H/m"

Relative Permeability of medium is = 1.00000000948
Permeability of medium is = 1.25663707335e-06 H/m

Example 6.10,Page number 2-30

In [14]:
import math

#Given Data:
B=2.5                #Magnetic field in tesla
u0=4*math.pi*10**-7    #Permeability in free space
i0=0.7               #current in the core
ri=11*10**-2         #inner radii of core
ro=12*10**-2         #outer radii of core

#Calculations:
r=(ri+ro)/2          #Average radii of core
n=3000/(2*math.pi*r)    #Number of turns

#We know, B=u0*ur*n*i0 .Thus,
ur=B/(u0*n*i0)

print"Relative Permeability of medium is =",ur

Relative Permeability of medium is = 684.523809524

Example 6.11,Page number 2-31

In [15]:
import math

#Given Data:
B=1.0                #Flux density in tesla
u0=4*math.pi*10**-7     #Permeability in free space
i=2.0                #current in the core
n=10*100             #n=N/l i.e. turns per meter

#Calculations:
H=n*i               #Magnetising force produced in wire
print"Magnetising force produced in wire is =",H,"Amp-turn/meter"

#We know that, B=u0(H+I).Thus,
I=B/u0-H            #Magnetisation of material
print"Magnetisation of material is =",I,"Amp-turn/meter"

#u=B/H, i.e. ur*u0=B/H.
ur=B/(u0*H)        #Relative permeability of core
print" Relative Permeability of core is =",ur

Magnetising force produced in wire is = 2000.0 Amp-turn/meter
Magnetisation of material is = 793774.715459 Amp-turn/meter
 Relative Permeability of core is = 397.88735773

Example 6.12,Page number 2-31

In [16]:
import math

#Given Data:
M=40             #Mass of an iron core
D=7.5*10**3      #Density of iron
f=100            #Frequency
A=3800*10**-1    #Loss due to Area of hysterisis loop in J/m^3

#Calculations:
V=M/D            #Volume of iron core
L1=A*V           #Loss of energy in core per cycle
print"Loss of energy in core per cycles is =",L1,"joules"

N=f*60           #Number of cycles per minute
L=L1*N           #Loss of energy per minute

print"Loss of energy per minute is =",L,"joules"

Loss of energy in core per cycles is = 2.02666666667 joules
Loss of energy per minute is = 12160.0 joules

Example 6.13,Page number 2-32

In [17]:
import math

#Given Data:
u0=4*math.pi*10**-7     #Permeability in vacuum
l=30*10**-2             #length in meter
A=1*10**-4          #Crosss sectional area of ring in m^2
phi=2*10**-6        #flux in weber
N=300               #Number of turns
I=0.032             #Current in winding

#Calculations:
#(i):
B=phi/A             #Flux density
print"(i)Flux Density in the ring is =",B,"Wb/m^2"

#(ii):
H=N*I/l            #Magnetic intensity
print"(ii)Magnetic intensity is =",H,"Amp-turn/meter"

#(iii):
u=B/H              #Permeability of ring
print"(iii)Permeability of ring is =",u," Wb/A-m"
ur=u/u0            #Relative permeability of ring
print"Relative Permeability of ring is =",ur

#(iv):
xm=ur-1            #susceptibility of the ring
print"(iv)Magnetic susceptibility of the ring is =",xm

(i)Flux Density in the ring is = 0.02 Wb/m^2
(ii)Magnetic intensity is = 32.0 Amp-turn/meter
(iii)Permeability of ring is = 0.000625  Wb/A-m
Relative Permeability of ring is = 497.359197162
(iv)Magnetic susceptibility of the ring is = 496.359197162

Example 6.14,Page number 2-32

In [18]:
import math

#Given Dta:
M=12*10**3       #Mass of an iron core in grams
D=7.5            #Density of iron in gm/cc
f=50             #Frequency
A=3000           #loss due to Area of hysterisis loop in ergs/cm^3

#Calculations:
V=M/D            #Volume of iron core
L1=A*V           #Loss of energy in core per cycle

L=L1*f*3600      #Loss of energy per hour

print"Loss of energy per hour is =",L,"Erg"

Loss of energy per hour is = 8.64e+11 Erg

Example 6.15,Page number 2-33

In [19]:
import math

#Given Data:
A=0.5*10**3            #Area of B-H loop in Joules per m^3
V=10**-3               #Volume of specimen in m^3
n=50                   #Frequency of a.c.

#Calculations:
H=n*V*A               #Hysteresis power loss

print"Hysteresis power loss is =",H,"Watt"

Hysteresis power loss is = 25.0 Watt

Example 6.16,Page number 2-33

In [22]:
import math

#Given Data:
u0=4*math.pi*10**-7     #Permeability in vacuum
ur=1000             #Relative permeability of medium
V=10**-4            #Volume of iron rod in m^3
n=500               #Number of turns per meter
i=0.5               #Current in windings of solenoid in Amperes

#Calculations:
#We know I=(ur-1)H
#and H=ni , hence
I=(ur-1)*n*i       #Intensity of magnetisation
M=I*V              #Magnetic moment

print"Magnetic moment of the rod is =",M,"A-m^2"

Magnetic moment of the rod is = 24.975 A-m^2

Example 6.17,Page number 2-34

In [23]:
import math

#Given Data:
u0=4*math.pi*10**-7     #Permeability in vacuum
ur=600             #Relative permeability of iron
d=12*10**-2        #mean diameter of ring in m
N=500              #Number of turns
i=0.3              #Current in windings of solenoid in Amperes

#Calculations:
r=d/2              #Radius of ring

B=u0*ur*N*i/(2*math.pi*r)     #Flux densityin the core
print"Flux densityin the core is =",B,"Wb/m^2"

H=B/(u0*ur)         #Magnetic intensity
print"Magnetic intensity is =",H,"Amp-turns/m"

#We know that, B=u0(H+I)
I1=(B-u0*H)/u0      #magnetisation
I2=u0*I1            #Electronic current loop

I=I2/B*100          #Percentage flux density due to electroniuc loop currents
print"Percentage flux density due to electroniuc loop currents is =",I,"percent"

Flux densityin the core is = 0.3 Wb/m^2
Magnetic intensity is = 397.88735773 Amp-turns/m
Percentage flux density due to electroniuc loop currents is = 99.8333333333 percent

Example 6.18,Page number 2-35

In [24]:
import math

#Given Data:

u0=4*math.pi*10**-7   #Permeability in vacuum
ur=900               #Relative permeability of iron ring
d=40*10**-2          #diameter of ring
l=5*10**-3           #air gap in the ring
A=5.8*10**-4         #Crosss sectional area of ring in m^2
phi=1.5*10**-4       #flux in weber
N=600                #Number of turns

#Calculations:
r=d/2                #Radius of ring

#We know, phi=B*A
B=phi/A              #Flux density

#But, B=u*H
H=B/(u0*ur)          #Magnetic field strength

m1=H*ur*l            #amp-turns in air gap
m2=H*2*math.pi*r     #amp-turns by ring
m=m1+m2              #total mmf(amp-turns) required

I=m/N                #Required current
print"Current required to produce given flux is =",I,"Amperes"

Current required to produce given flux is = 2.19395891742 Amperes

Example 6.18.1,Page number 2-38

In [25]:
import math

#Given Data:
u0=4*math.pi*10**-7    #Permeability in vacuum
X=-0.5*10**-5          #Magnetic susceptibility of silicon
H=9.9*10**4            #Magnetic field intensity

#Calculations:

#As, X=I/H. thus,
I=X*H                  #intensity of magnetisation
print"Intensity of magnetisation is =",I

B=u0*(H+I)             #Magnetic flux density
print"Magnetic flux density is =",B,"Wb/ m^2"

Intensity of magnetisation is = -0.495
Magnetic flux density is = 0.124406447047 Wb/ m^2

Example 6.18.2,Page number 2-38

In [27]:
import math

#Given Data:
u0=4*math.pi*10**-7     #Permeability in vacuum
ur=380              #Relative permeability
d=20*10**-2         #diameter of solenoid in m
r=d/2               #radius of ring in m
A=5*10**-4           #Crosss sectional area of ring in m^2
phi=2*10**-3         #flux in weber
N=200                #Number of turns

#Calculations:
l=math.pi*d          #air gap in the ring
S=(l/(u0*ur*A))      #Reluctance of iron ring
print"Reluctance of iron ring is =",S,"Amp-turn/ Wb "

#ohm's law for magnetic circuit is  phi=N*I/S. thus,
I=S*phi/N            #required current
print"Current required to obtain given magnetic flux is =",I,"Amperes"

Reluctance of iron ring is = 2631578.94737 Amp-turn/ Wb 
Current required to obtain given magnetic flux is = 26.3157894737 Amperes

Example 6.18.3,Page number 2-39

In [28]:
import math

#Given Values:
u0=4*math.pi*10**-7     #Permeability in vacuum
ur=1               #Relative permeability of air
r=15*10**-2        #radius of ring in m
A=6*10**-4         #Crosss sectional area of ring in m^2
I=4                #Coil current in amp
N=500              #Number of turns

#Calculations:
m=N*I              #MMF of coil
print"MMF of coil is =",m,"Ampere-turn"

l=2*math.pi*r      #air gap
R=(l/(u0*ur*A))    #Reluctance of iron ring
print"Reluctance of iron ring is =",R,"Ampere-turn/Wb"

phi=m/R            #Magnetic flux
print"Magnetic flux is =",phi,"Weber"

B=phi/A            #Magnetic Flux density
print"Magnetic flux density is =",B,"Weber/m^2"

H=B/(u0*ur)        #Magnetic field intensity
print"Magnetic field intensity is =",H,"Amperes/m"

MMF of coil is = 2000 Ampere-turn
Reluctance of iron ring is = 1250000000.0 Ampere-turn/Wb
Magnetic flux is = 1.6e-06 Weber
Magnetic flux density is = 0.00266666666667 Weber/m^2
Magnetic field intensity is = 2122.06590789 Amperes/m

Example 6.19,Page number 2-36

In [29]:
import math

#Given Data:
u0=4*math.pi*10**-7     #Permeability in vacuum
ur=6*10**-3         #Relative permeability of iron
r=0.5               #radius of ring in m
l=1*10**-2          #air gap in the ring
A=5*10**-4          #Crosss sectional area of ring in m^2
i=5                 #current in ampere
N=900               #Number of turns

#Calculations:
S=(l/(u0*A))+((2*math.pi*r-l)/ur*A)     #Reluctance of iron
print"Reluctance of iron is =",S,"Ampere-turn/Wb"

m=N*i              #mmf produced
print"mmf produced is =",m,"Ampere-turn"

Reluctance of iron is = 15915494.5702 Ampere-turn/Wb
mmf produced is = 4500 Ampere-turn

Example 6.20,Page number 2-36

In [30]:
import math

#Given Data:
H=5*10**3          #coercivity of bar magnet in amp/m
l=10*10**-2        #length of solenoid in m
N=50               #No of turns

#Calculations:

#We know that, H=NI/l ,hence
I=l*H/N            #current through solenoid

print"Current through solenoid is =",I,"Amperes"

Current through solenoid is = 10.0 Amperes

Example 6.21,Page number 2-36

In [31]:
import math

#Given Data:
ur=1200            #Relative permeability of medium
V=10**-3           #volume of iron rod
N=5*10**2          #no of turns per m
i=0.5              #current through solenoid in amp

#Calculations:
x=ur-1             #susceptibility of the ring
H=N*i             #Magnetisisng field

#We know, x=I/H
I=x*H              #magnetisation

#Also, I=M/V , thus
M=I*V              #magnetic moment
print"Magnetic moment is =",M,"Ampere-turn-m^2"

Magnetic moment is = 299.75 Ampere-turn-m^2

Example 6.22,Page number 2-37

In [33]:
import math

#Given Dta:
ur=100            #Relative permeability of medium
l=0.2             #length of iron rod
d=10*10**-3       #diameter of solenoid in m
N=300             #no of turns per m
i=0.5             #current through solenoid in amp
r=d/2             #radius of solenoid

#Calculations:
x=ur-1            #susceptibility of the ring
H=N*i             #Magnetisisng field

#We know, x=I/H
I=x*H             #magnetisation

V=math.pi*(r**2)*l     #volume of iron rod

#Also, I=M/V , thus
M=I*V             #magnetic moment
print"Magnetic moment is =",M,"Ampere-turn-m^2"

Magnetic moment is = 0.233263254529 Ampere-turn-m^2

Example 6.23,Page number 2-38

In [34]:
import math

#Given Data:
l=1.2              #length of circuit in meter
u=7.3*10**-3       #permeability of silicon sheet
A=100              #cross sectional area in cm^2
N=150              #No of turns
B=0.3              #magmetic field in Wb/m^2

#Calculations:

#We know, B=u*H
H=B/u              #Magnetic field strength

m=H*l              #amp-turns in air gap

I1=m/N             #Required current
print"Current required to obtain given magnetic field is =",I1,"Amperes"

I=I1/A             #Required current per unit area
print"Current required per unit area to obtain given magnetic field is =",I,"Amperes"

Current required to obtain given magnetic field is = 0.328767123288 Amperes
Current required per unit area to obtain given magnetic field is = 0.00328767123288 Amperes