Chapter 11 Classification of magnetic materials

Example 11.1 Page no 614

In [7]:
#Given
e=1.6*10**-19
f=6.8*10**15
n=1
r=0.53*10**-10

#Calculation
import math
I=e*f
M=n*I*math.pi*r**2

#Result
print"Equivalent magnetic moment is", round(M*10**24,1)*10**-24,"Am**2"

Equivalent magnetic moment is 9.6e-24 Am**2

Example 11.2 Page no 615

In [18]:
#Given
E=240
R=474.0
r=12.5*10**-2
N=500
ur=5000

#Calculation
import math
I=E/R
I1=2*math.pi*r
H=(N*I)/I1
u=4*math.pi*10**-7
B=u*ur*H

#Result
print"(i) The magnetising force is", round(H,0),"AT/m"
print"(ii) The magnetic flux density is",round(B,2),"Wb/m**2"

(i) The magnetising force is 322.0 AT/m
(ii) The magnetic flux density is 2.03 Wb/m**2

Example 11.3 Page no 615

In [24]:
#Given
r1=11
r2=12
B=2.5                     #T
a=3000
I=0.70                     #A

#Calculation
import math
r=((r1+r2)/2.0)*10**-2
n=a/(2*math.pi*r)
ur=B*2*math.pi*r/(4*math.pi*10**-7*a*I)

#Result
print"Relative permeability is", round(ur,1)

Relative permeability is 684.5

Example 11.4 Page no 616

In [35]:
#Given
I=0.5                              #m
N=500
I1=0.15                              #A
a=5000

#Calculation
import math
H=(N*I1)/I
B=4*math.pi*10**-7*H
B1=B*a
I3=(B1-(H*4*math.pi*10**-7))/(4.0*math.pi*10**-7)

#Result
print round(I3*10**-5,1),"*10**5 A/m"

7.5 *10**5 A/m

Example 11.5 Page no 616

In [46]:
#Given
B=0.6
H=360.0

#Calculation
u=B/H
x=(u-1*4*math.pi*10**-7)/(4.0*math.pi*10**-7)

#Result
print"(i) Permeability is",round(u*10**3,2),"*10**-3 T/A m"
print"(ii) Susceptibility of the material is",round(x,0)

(i) Permeability is 1.67 *10**-3 T/A m
(ii) Susceptibility of the material is 1325.0

Example 11.6 Page no 616

In [52]:
#Given
M=8.0*10**22                  #Am**2
R=64*10**5                    #m

#Calculation
import math
I=(3*M)/(4.0*math.pi*R**3)

#Result
print"Earth's magnetisation is", round(I,1),"A/m"

Earth's magnetisation is 72.9 A/m

Example 11.7 Page no 616

In [63]:
#given
N=1800
l=0.6
I=0.9                           #A
ur=500
n1=6.02*10**26
a=55.85
y=7850

#Calculation
n=N/l
H=n*I
I1=(ur-1)*H
B=4*math.pi*10**-7*ur*H
x=(y*n1)/a
X=I1/x

#Result
print"Average magnetic moment per iron atom is", round(X*10**23,2)*10**-23," A m**2"

Average magnetic moment per iron atom is 1.59e-23  A m**2

Example 11.8 Page no 617

In [70]:
#Given
M=8.4                         #g
d=7200.0
f=50                              #Hz
E=3.2*10**4
t=30*60.0

#Calculation
V=M/d
P=E/t
E1=P/(V*f)

#Result
print"Energy dissipated per unit volume is", round(E1,0),"J/m**3/cycle"

Energy dissipated per unit volume is 305.0 J/m**3/cycle

Example 11.9 Page no 617

In [73]:
#Given
H=4*10**3                         #A/m
a=60
b=0.12

#Calculation
n=a/b
I=H/n

#Result
print"Current should be sent through the solenoid is", I,"A"

Current should be sent through the solenoid is 8.0 A

Example 11.10 Page no 617

In [79]:
#Given
x=1.68*10**-4
T1=293
T2=77.4

#Calculation
x1=(x*T1)/T2

#Result
print"Susceptibility is", round(x1*10**4,2),"*10**-4"

Susceptibility is 6.36 *10**-4

Example 11.11 Page no 617

In [113]:
#Given
l=10**-6                     #m
d=7.9                       #g
a=6.023*10**23
n=55.0
M1=9.27*10**-24

#Calculation
V=l**2
M=V*d
N=(a*M)/n
Mmax=N*M1
Imax=Mmax/V*10**-4

#Result
print"Number of iron atom is",round(N*10**-10,2),"*10**10 atoms"
print"Magnetisation of the dipole is",round(Imax*10**5,0),"*10**5 A/m"
print"Maximum possible dipole moment is",round(Mmax*10**13,0)*10**-13,"A m**2"

Number of iron atom is 8.65 *10**10 atoms
Magnetisation of the dipole is 8.0 *10**5 A/m
Maximum possible dipole moment is 8e-13 A m**2