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)
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)
#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)
#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)
#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)
#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)
#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)
#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)
#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)
#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)