# Chapter 6 : Magnetic Materials¶

## Example 6.1 Page No : 212¶

In [1]:
import math

# Variables
f = 0.01     #flux in Wb
l = 1.     #mean circumference in m
N = 1000.     #tunrs
Ur = 1000.     #relative permeability

# Calculations
Uo = 4*math.pi*(10**-7)     #permeability of free space in H/m
a = 0.001     # cross section area in m**2
I = (f*l)/(N*Uo*Ur*a)     # current in Amp. math.since f = A*T/(l/Uo*Ur*a)

# Results
print 'Thus Current required is %3.3f Amp'%(I)

Thus Current required is 7.958 Amp


## Example 6.2 Page No : 212¶

In [2]:
import math

# Variables
f = 1.2*(10**-3)     #flux in Wb
l = 1.4     #mean circumference in m
N = 500.     #tunrs

# Calculations
Uo = 4*math.pi*(10**-7)     #permeability of free space in H/m
a = 0.0012     # cross section area in m**2
I = 2     #current in Amp
Ur = (f*l)/(N*I*Uo*a)     #relative permeability

# Results
print 'Thus the relative permeability of iron is %3.2f '%(Ur)

Thus the relative permeability of iron is 1114.08


## Example 6.3 Page No : 213¶

In [3]:
import math

# Variables
l = 0.4     #mean circumference in m
N = 200.     #tunrs
Uo = 4*math.pi*(10**-7)     #permeability of free space in H/m
a = 5*(10**-4)     # cross section area in m**2
I = 6.4     #current in Amp
f = 0.8*(10**-3)     #flux in Wb

# Calculations
fd = f/a     #flux density in Wb/m**2
fi = I*N/l     #Field intensity in AT/m
Ur = (f*l)/(N*I*Uo*a)     #relative permeability

# Results
print 'i) The Flux density is %3.2f Wb/m**2 '%(fd)
print 'ii) The Field intensity is %3.2f AT/m '%(fi)
print 'iii) The Relative permeability of steel is %3.2f '%(Ur)
#The answer to part(iii) has a calculation error in the textbook, hence it doesn't match the answer here.

i) The Flux density is 1.60 Wb/m**2
ii) The Field intensity is 3200.00 AT/m
iii) The Relative permeability of steel is 397.89


## Example 6.4 Page No : 214¶

In [4]:
# Variables
Hl = 250.     #Hysteresis loss per m**3 in J/cycle
V = 1./150     #Volume of specimen in m**3
N = 50.     #No of cycles/sec

# Calculations
E = Hl*V*N     #Energy loss per sec in J
Eh = (E*3600)/1000     #Energy loss per hour in kWh

# Results
print 'Thus Energy loss per hour is %3.2f kWh'%(Eh)

Thus Energy loss per hour is 300.00 kWh


## Example 6.5 Page No : 214¶

In [5]:
# Variables
P = 4.     #no of poles
N = 1600.     # Speed in rpm
f = P*N/120     #Frequency of magnetic reversal
V = 5400.     #volume
d = 7.5     #density

# Calculations
m = (V*d)/1000     #Mass of armature in kg
L = 1.76     #Loss in W/kg
Cl = L*m     #Core loss in Watts

# Results
print 'Thus Frequency of magnetic reversal is %3.2f c/s'%(f)
print ' and Core loss is %3.2f Watts'%(Cl)

Thus Frequency of magnetic reversal is 53.33 c/s
and Core loss is 71.28 Watts


## Example 6.6 Page No : 214¶

In [6]:
# Variables
v = 76300.     #volume in c.c
P = 8.     # no of poles
N = 375.     #rpm
f = P*N/120     #freqency in c/s
Bmax = 12000.     #max. flux density in lines/cm**2
n = 0.002     #(assumed)
d = 7.8     #densityin gm/c.c
l = 1.7     #loss in watts per kg

# Calculations
Hl = n*v*f*(Bmax**1.6)*(10**-7)     #Hysteresis loss in Watts
Al = v*d*l/1000     #Additional loss under particular running conditions
Tl = Hl+Al     #total core loss

# Results
print 'Thus the total core loss is %4.0f Watts'%(Tl)

Thus the total core loss is 2295 Watts


## Example 6.7 Page No : 215¶

In [7]:
# Variables
m = 12000.     #mass in gm
d = 7.5     #density of iron in gm/c.c
Hl = 3000.     #Hysteresis loss per cc in ergs/cycle
N = 50.     #No of cycles per sec

# Calculations
v = m/d     #volume of specimen
E = v*Hl*N     #Energy loss per cc in ergs
Eh = E/(10**10)     #Energy loss per hour in kWh

# Results
print 'Thus the Loss in energy is %3.3f kWh'%(Eh)

Thus the Loss in energy is 0.024 kWh


## Example 6.8 Page No : 215¶

In [8]:
# Variables
m = 10.     #mass in kg
T1 = 20.     #total loss in watts
f1 = 50.     #frequency in c/s
T2 = 35.     #total loss in watts
f2 = 75.     #frequency in c/s

# Calculations
#both have same peak flux density
#total loss = hysteresis loss+ Eddy current loss
#all quantities except frequency are consmath.tant
#so Total loss = Af+Bf**2
#let c1 and c2 be consmath.tants such that total loss = c1*f + c2*f**2
c2 = (T2-(T1*f2/f1))/(f2**2-f1*f2)
c1 = (T1-c2*f1**2)/f1
k = c1/c2     #hysteresis loss/eddy current loss
H50 = T1*k/101     #hysteresis loss at 50 c/s
E50 = T1-H50     #eddy current loss at 50 c/s

# Results
print 'Thus hysteresis loss at 50 c/s is %3.1f Watts '%(H50)
print 'And Eddy current loss at 50c/s is %3.1f Watts'%(E50)

Thus hysteresis loss at 50 c/s is 19.8 Watts
And Eddy current loss at 50c/s is 0.2 Watts