# Chapter 14 : Electrical and Magnetic Properties of Materials¶

## Example 14.1 Page No : 443¶

In :

import math

# Variables
l = 100;			#length of wire
p = 2.66*10**(-8);			#resistivity

# Calculation
A = 3*10**(-6);			#cross sectional area
R = p*l/A;			#resismath.tance of an aluminium wire

# Results
print 'resistance of an aluminium wire = %.3e Ohm'%R

resistance of an aluminium wire = 8.867e-01 Ohm


## Example 14.2 Page No : 443¶

In :

import math

# Variables
R_Cu = 1.56;			#Resistivity of pure copper(in micro-ohm-cm)
R_CuNi  =  4.06;			#Resistivity of Cu containing two atomic percent (in micro-ohm-cm)
R_Ni = (R_CuNi-R_Cu)/2;			#Increase in resistivity due to one atomic % Ni

# Calculation
R_CuAg =  1.7;			#resistivity of copper, containing one atomic percent silver (in micro-ohm-cm)
R_Ag = R_CuAg-R_Cu;			#Increase in resistivity due to one atomic % Ag
R_CuNiAg = R_Cu+R_Ni+3*R_Ag;			#Resistivity of copper alloy containing one atomic percent Ni and 3 atomic percent Ag

# Results
print 'Resistivity of copper alloy containing one atomic percent Ni and 3 atomic percent Ag = %.2f micro-ohm-cm'%R_CuNiAg

Resistivity of copper alloy containing one atomic percent Ni and 3 atomic percent Ag = 3.23 micro-ohm-cm


## Example 14.3 Page No : 443¶

In :

import math

# Variables
R_Cu = 1.8*10**(-8);			#resistivity of pure copper at room temperature
R_CuNi = 7*10**(-8);			#resistivity of Cu 4% Ni alloy at room temperature

# Calculation
R_Ni = (R_CuNi-R_Cu)/4;			#resistivity due to Impurity scattering per % of Ni

# Results
print 'resistivity due to impurity scattering per percent of Ni in the Cu lattice = %.1e ohm-meter'%R_Ni

resistivity due to impurity scattering per percent of Ni in the Cu lattice = 1.3e-08 ohm-meter


## Example 14.4 Page No : 455¶

In :

import math

# Variables
C = 10**(-9);			#capacitance(in F)
d = 2*10**(-3);			#distance of separation in a parallel plate condenser
E_o = 8.854*10**(-12);			#dielectric consmath.tant

# Calculation
A = (10*10**(-3))*(10*10**(-3));			#area of parallel plate condenser
#C = E_o*E_r*A/d
E_r = C*d/(E_o*A);			#Relative dielectric constant

# Results
print 'Relative dielectric constant of a barium titanate crystal %.0f'%(E_r)

Relative dielectric constant of a barium titanate crystal 2259


## Example 14.5 Page No : 456¶

In :

import math

# Variables
q = 1.6*10**(-19);			#charge (in C)
d_1 = 0.06			#shift of the titanium ion from the body centre (in Å)
d_2 = 0.08			#shift of the oxygen anions of the side faces (in Å)
d_3 = 0.06			#shift of the oxygen anions of the top and bottom face (in Å)

# Calculation
D_1 = d_1*10**(-10);			#shift of the titanium ion from the body centre (in m)
D_2 = d_2*10**(-10);			#shift of the oxygen anions of the side faces (in m)
D_3 = d_3*10**(-10);			#shift of the oxygen anions of the top and bottom face (in m)
U_1 = 4*q*D_1;			#dipole moment due to two O2– ions on the four side faces(in C-m)
U_2 = 2*q*D_2;			#dipole moment due to one O2– on top and bottom(in C-m)
U_3 = 4*q*D_3;			#dipole moment due to one Ti4+ ion at body centre(in C-m)
U = U_1+U_2+U_3;			#Total dipole moment(in C-m)
V = 4.03*((3.98)**2)*10**(-30);			#volume(in m3)
P = U/V;			#polarization the total dipole moments per unit volume

# Results
print 'polarization = %.2f C/m**2'%P

polarization = 0.16 C/m**2


## Example 14.6 Page No : 478¶

In :

import math

# Variables
V = ((2.87)**3)*10**(-30)			#Volume of unit cell of BCC iron (in m**3)
N = 2.			#Number of atoms in the unit cell

# Calculation
M = 1750.*10**3;			#saturation magnetization of BCC Iron A/m
M_Net = V*M*(1./N)			#net magnetic moment per atom
Bohr_magneton = 9.273*10**(-24);			#Bohr_magneton (magnetic moment) in A/m2
M_moment = M_Net/Bohr_magneton;			#The magnetic moment (in units of U_B)

# Results
print 'The magnetic moment (in units of U_B) = %.1f'%M_moment

The magnetic moment (in units of U_B) = 2.2


## Example 14.7 Page No : 479¶

In :

import math

# Variables
p = 8.90*10**6;			#density of nickel in gm/m3.
N_A = 6.023*10**23;			#Avogadro’s number atoms/mol
At_w = 58.71;			#Atomic weight of Ni in gm/mol

# Calculation
N = p*N_A/At_w;			#number of atoms/m3
U_B = 9.273*10**(-24);			#Bohr_magneton
M_s = 0.60*U_B*N;			#saturation magnetization
pi = 22./7;
U_o = 4*pi*10**(-7);			#magnetic consmath.tant
B_s = U_o*M_s;			#Saturation flux density

# Results
print 'the saturation magnetization = %.1e'%M_s
print 'Saturation flux density = %.2f'%B_s

the saturation magnetization = 5.1e+05
Saturation flux density = 0.64


## Example 14.8 Page No : 479¶

In :

import math

# Variables
#Each cubic unit cell of ferrous ferric oxide contains 8 Fe2+ and 16 Fe3+ ions and
n_b = 32;			#
U_B = 9.273*10**(-24);			#Bohr_magneton

# Calculation
a = 0.839*10**(-9);			#the unit cell edge length in m
V = a**3;			#volume(in m3)
M_s = n_b*U_B/V;			#the saturation magnetization

# Results
print 'the saturation magnetization = %.0e A/m'%M_s

the saturation magnetization = 5e+05 A/m


## Example 14.9 Page No : 479¶

In :
import math

# Variables
#hysteresis loss (Ph) and the induced emf loss (Pe) are proportional to the frequency
#Pe is proportional to the square of the induced emf (Pe)
#Pe + Ph  =  750 W (at 25 Hz)
#4Pe + 2Ph  =  2300 W(at 50Hz)
#solving equation
P_e = 800./2;			#induced emf loss

# Calculation
I_d = 4*P_e;			#The eddy current loss at the normal voltage and frequency

# Results
print 'The eddy current loss at the normal voltage and frequency = %.0f W'%I_d

The eddy current loss at the normal voltage and frequency = 1600 W