Chapter 20 : Magnetic Properties

Example 20.1 Page No 84

In [2]:
b_m=9.27*10**-24   #ampere*m**2  (Bohr Magneton)
Na=6.023*10**23    #atoms/mol  (Avogadro's No.)
d=8.9*10**6        #g/m**3   (density)
uo=4*math.pi*10**-7  #Permitivity of free space
A=58.71             #g/mol  (Atomic weigth of Nickel)
N=d*Na/A            #No. of atoms per cubic meter

M=0.6*b_m*N  #0.6= Bohr Magneton/atom
B=uo*M

print"Saturation Magnetisation is ",M,"A/m"
print"Saturation Flux Density is ",round(B,2),"Tesla"

Saturation Magnetisation is  507834.0 A/m
Saturation Flux Density is  0.64 Tesla

Example 20.2 Page No 88

In [4]:
a=0.839*10**-9   #a is edge length in m
b_m=9.27*10**-24 #ampere*m**2  (Bohr Magneton)
nb=8*4           #8 is no. of Fe++ ions per unit cell
                 #4 is Bohr magnetons per Fe++ ion
M=nb*b_m/a**3  #M is Saturation magnetisation

print"Saturation Magnetisation is ",round(M,0),"A/m"

Saturation Magnetisation is  502278.0 A/m

Design Example 20.1 Page No 88

In [8]:
Ms_Fe=5.25*10**5   #Required saturation Magnetisation
b_m=9.27*10**-24   #ampere*m**2  (Bohr Magneton)
a=0.839*10**-9     #a is edge length in m
M=5*10**5          #From previous question result

nb=Ms_Fe*a**3/b_m
i=8             # No of Divalent ions per unit cell
j=4             #4 is Bohr magnetons per Mn++ ion
n=nb/(i)-j 
         
print"Replacing percent of Fe++ with Mn++ would produce the required saturation magnetisation",round(n*100,2)

Replacing percent of Fe++ with Mn++ would produce the required saturation magnetisation 18.1
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