# Ch-21, Magnetic material properties, behaviour and application¶

## example-21.1 page no-621¶

In [14]:
from __future__ import division
from math import pi
#given
#magnetic field
H=2400  #A/m
#susceptibilty
kie=1500
#part(a)
#relative permeability is given by
mur=1+kie
#part(b)
#intensity of magnetization
M=kie*H  #A/m
#part(c)
#permeability
mu0=4*(pi)*10**-7
#remanance
B=mu0*mur*H  #T
print " the realative permeability is %0.f \n, the intensity of magnetisation is %.f A/m \n and the remanance is %0.3f"%(mur,M,B)

 the realative permeability is 1501
, the intensity of magnetisation is 3600000 A/m
and the remanance is 4.527


## example-21.2 page no-621¶

In [15]:
from __future__ import division
#given
#relative permeability of superalloy
mur=200000
mu0=4*(pi)*10**-7  #henry/m
#intensity of magnetisation
M=6000  #A/m
#magnetic field is given by
H=M/(mur-1)  #A/m
#strength of magnet
B=mu0*mur*H  #tesla
print "The strength of magnet is %0.3f T"%(B)

The strength of magnet is 0.008 T


## example-21.3 page no-621¶

In [17]:
#given
#magnetic moment is 0.6 times bohr magneton and we know that beta is 9.27*10**-24 Am**2
beta=9.27*10**-24  #A/m**2
M=0.6*beta  #A/m**2
#attice constant
a=0.35*10**-9  #m
#no of atoms per unit cell is given by
Ne=4
#saturation magnetisation for FCC unit cell is given by
Ms=Ne*M/a**3  #A/m
print "he saturation magnetisation is %0.3f A/m"%(Ms)

he saturation magnetisation is 518903.790 A/m


## example-21.4 page no-632¶

In [5]:
#given
#refer to fig-21.6
#width of loop
#W=(OA+OB)
#so
W=80*2  #A/m
#height of loop
#H=(OC+OD)  #wb/m**2
#so
H=0.15*2  #Wb/m**2
#area of loop
A=W*H  #T A/m or J
print "The energy loss per ubitvolume of the magnetic material during one cycle is %.f J"%(A)

The energy loss per ubitvolume of the magnetic material during one cycle is 48 J


## example-21.5 page no-633¶

In [6]:
#given
#volume
V=0.01  #m**3
#frequency
f=50  #Hz
#area of loop
A=600   #J/m**2
#as we know that
#A=mu*Bmax**16
#also power loss is given by
#P=mu*Bmax**16*f*V
#so
P=A*f*V  #watt
print " the power loss due to hysteresis is %.f W"%(P)

 the power loss due to hysteresis is 300 W


## example-21.6 page no-633¶

In [18]:
#given
#hysteresis loss is
W1=300  #W
#max flux density is
Bmax1=0.9  #Wb/m**2
Bmax2=1.1  #Wb/m**2
#frequency
f1=50  #Hz
f2=40  #Hz
#we know that
#W=nu*(Bmax)**1.7*f*V
#s0
#W1=nu*(Bmax1)**1.7*f1*V   ---------(1)
#W2=nu*(Bmax2)**1.7*f*V   -------(2)
#from one and 2 we get
W2=W1*(Bmax2)**1.7*f2/((Bmax1)**1.7*f1)  #W
print "The hysteresis loss at 40 Hz frequency is %0.3f W"%(W2)

The hysteresis loss at 40 Hz frequency is 337.572 W


## example-21.7 page no 634¶

In [19]:
#given
#flux density
Bm=1.10  #Wb/m**2
#frequency
f=50  #Hz
#thickness of sheet
t=0.5*10**-3  #m
#resistivity
rho=30*10**-8  #per ohm m
#density
rhodash=7800  #kg/m**3
#mass
m=1  #kg
#volume of material
V=m/rhodash  #m**3
#and
k=1.11
#hysteresis loss in each cycle
Wh=380  #W s/m**3
#loss per kg of specimen is given by
We=4/3*(Bm*f*t*k)**2*V/(rho)   #watt/kg
print " the loss is %0.3f watt/kg"%(We)

 the loss is 0.531 watt/kg


## example-21.8 page no-634¶

In [20]:
#given
#frequency
f=50  #Hz
#mass
m=50  #kg
#density
rho=7500  #kg/m**3
#volume of material
V=m/rho  #m**3
#hysteresis loop area
A=150  #m**2
#scale factor
#1 cm=0.008 Wb/m**2 on y-axis and 1cm=20 A/m on x-axis
#energy lost during each cycle
E=A*0.008*20*10**4  #J/m**3
#poer loss due to hysteresis
P=E*f*V  #J/s
#energy lost in one hour
Wh=P*(60*60)  #J
print " the energy lost in one hour is %0.3e J"%(Wh)

 the energy lost in one hour is 2.880e+08 J


## example-21.11 page no-650¶

In [22]:
#given
#frequency
f=50  #Hz
#eddy current loss in transformer
We=100  #W
#to find eddy current loss at frequencies
f1=60  #Hz
f2=100  #Hz
#as we know that
#We is directly proportional to f**2
We1=f1**2*We/f**2  #W
#similarly
We2=f2**2*We/f**2  #W
print " the eddy current loss at 60 Hz is %0.f W \n and at 100 Hz is %.f W"%(We1,We2)

 the eddy current loss at 60 Hz is 144 W
and at 100 Hz is 400 W


## example-21.15 page no-651¶

In [23]:
#given
#length of wire
l=250*10**-3  #m
#no of turns
N=400
#current
I=15  #A
#permeability in vaccum
mu0=1.2457*10**-6  #H/m
#relative permeability
mur=1
#magnetic field strength
H=N*I/l  #AT/m
#flux density is
B=mu0*mur*H  #Wb/m**2
print " the magnetic field strength is %.f AT/m and flux density is %0.3f Wb/m"%(H,B)

 the magnetic field strength is 24000 AT/m and flux density is 0.030 Wb/m