# Chapter 12: Magnetic Properties of Materials¶

## Example 12.1, Page 603¶

In [1]:
#Variable declaration
H = 5e+3;   # Coercivity of a bar magnet, A/m
L = 0.1;   # Length of the solenoid, m
N = 50;   # Turns in solenoid
n = 500;   # Turns/m

#Calculations
# Using the relation
I = H/n;   # where I is the current through the solenoid

#Result
print "The current through the solenoid is = %2d A"%I

The current through the solenoid is = 10 A


## Example 10.2, Page 603¶

In [2]:
#Variable declaration
n = 500;   # Number of turns wound per metre on the solenoid
i = 0.5;   # Current through the solenoid, A
V = 1e-03;   # Volume of iron rod, per metre cube
mu_r = 1200;    # Relative permeability of the iron

#Calculations&Results
H = n*i;    # Magnetic intensity inside solenoid, ampere-turn per metre
# As B = mu_o * (H + I) => I = B/mu_o -  H
# But B = mu_o * mu_r * H and solving for I
I = (mu_r - 1) * H;
print "The Intensity of magnetisation inside the solenoid, I = %5.3e A/m"%I
M = I * V;    # Magnetic moment of the rod, ampere metre square
print "The magnetic moment of the rod, M = %3d ampere metre square"%M

The Intensity of magnetisation inside the solenoid, I = 2.998e+05 A/m
The magnetic moment of the rod, M = 299 ampere metre square


## Example 10.3, Page 604¶

In [4]:
#Variable declaration
n = 300;   # Number of turns wound per metre on the solenoid
i = 0.5;   # Current through the solenoid, A
V = 1e-03;   # Volume of iron rod, per metre cube
mu_r = 100;    # Relative permeability of the iron

#Calculations&Results
H = n*i;    # Magnetic intensity inside solenoid, ampere-turn per metre
# As, I = (B-mu_o* H)/mu_o
#But, B= mu * H = mu_r * mu_o * H and I = (mu_r-1)* H
I = (mu_r-1)*n*i;
print "The Intensity of magnetisation inside the solenoid, I = %5.3e A/m"%I
l = 0.2;   #length of the rod,m
r = 5e-3;   #radius of the rod,m
V = 1.57e-5;   #V=%pi*r^2*l where the volume of the rod having radius r and length,m
M = I * V ;    # Magnetic moment of the rod, ampere metre square
print "The magnetic moment of the rod, M = %5.3f ampere metre square"%M

The Intensity of magnetisation inside the solenoid, I = 1.485e+04 A/m
The magnetic moment of the rod, M = 0.233 ampere metre square


## Example 12.4, Page 605¶

In [5]:
from math import *

#Variable declaration
B = 0.0044;   # Magnetic flux density, weber/meter square
mu_o = 4*pi*1e-07;   # Relative permeability of the material, henery/m
I = 3300;   # Magnetization of a magnetic material, A/m

#Calculations&Results
#B = mu_o*(I+H), solving for H
H = (B/mu_o)- I;   # Magnetizing force ,A/m
print "The magnetic intensity,H = %3d A/m"%H
# Relation between intensity of magnetization and relative permeability
mu_r = (I/H)+1;   #substitute the value of I and H
print "The relative permeability, mu_r = %5.2f"%mu_r

#Incorrect answers in the textbook

The magnetic intensity,H = 201 A/m
The relative permeability, mu_r = 17.38


## Example 12.5, Page 605¶

In [16]:
from math import *

#Variable declaration
mu_o = 4*pi*1e-07;   # Magnetic permeability of the free space, henery/m
mu_r = 600;
mu = mu_o*mu_r;     # Magnetic permeability of the medium, henery/m
n = 500;   # Turns in a wire
i = 0.3;   # Current flows through a ring,amp
r = 12e-02/2;   # Mean radius of a ring, m

#Calculations&Results
B = mu_o*mu_r*n*i/(2*pi*r);
print "The magnetic flux density = %2.1f weber/meter-square"%B
H = B/mu;   # Magnetic intensity, ampere-turns/m
print "The magnetic intensity = %5.1f ampere-turns/m"%H
mui = B -mu_o
per = mui/i*100
print "The percentage magnetic flux density due to electronic loop currents = ",per,"%"

The magnetic flux density = 0.3 weber/meter-square
The magnetic intensity = 397.9 ampere-turns/m
The percentage magnetic flux density due to electronic loop currents =  99.999581121 %


## Example 12.6, Page 606¶

In [17]:
#Variable declaration
M_i = 4.5;   # Intial value of total dipole moment of the sample
H_i = 0.84;   # External magnetic field, tesla
T_i = 4.2;   # Cooling temerature of the sample, K
H_f = 0.98;   # External magnetic field, tesla
T_f = 2.8;   # Cooling temerature of the sample, K

#Calculations
# According to the curie's law, Mf/Mi = (Hf/Hi)*(Ti/Tf)
M_f = M_i*H_f/H_i*T_i/T_f;

#Result
print "The total dipole moment of the sample = %5.3f joule/tesla"%M_f

The total dipole moment of the sample = 7.875 joule/tesla


## Example 12.7, Page 606¶

In [18]:
from math import *

#Variable declaration
mu_o = 4*pi*1e-07;   # Magnetic permeability of free space, henry/m
n = 1e+29;   # Number density of atoms of iron, per metre cube
p_m = 1.8e-23;   # Magnetic moment of each atom, ampere-metre square
k_B = 1.38e-23;   # Boltzmann constant, J/K
B = 0.1;    # Magnetic flux density, weber per metre square
T = 300;   # Absolute room temperature, K
l = 10e-02; # Length of the iron bar, m
a = 1e-04;  # Area of cross-section of the iron bar, metre square

#Calculations&Results
V = l*a;    # Voluem of the iron bar, metre cube
chi = n*p_m**2*mu_o/(3*k_B*T);
print "The paramagnetic susceptibility of a material = %5.3e"%chi
pm_mean = p_m**2*B/(3*k_B*T); # Mean dipole moment of an iron atom, ampere metre-square
P_m = n*pm_mean;  # Dipole moment of the bar, ampere metre-square
I = n*p_m;  # Magnetization of the bar in one domain, ampere/metre
M = I*V;    # Magnetic moment of the bar, ampere metre-square
print "The dipole moment of the bar = %5.3e ampere metre-square"%P_m
print "The magnetization of the bar in one domain = %3.1e ampere/metre"%I
print "The magnetic moment of the bar = %2d ampere metre-square"%M

The paramagnetic susceptibility of a material = 3.278e-03
The dipole moment of the bar = 2.609e+02 ampere metre-square
The magnetization of the bar in one domain = 1.8e+06 ampere/metre
The magnetic moment of the bar = 18 ampere metre-square


## Example 12.8, Page 607¶

In [19]:
#Variable declaration
A = 500;   # Area of the B-H loop, joule per metre cube
n = 50;   # Total number of cycles, Hz
m = 9;   # Mass of the core, kg
d = 7.5e+3;   # Density of the core, kg/metre cube
t = 3600;   # Time during which the energy loss takes place, s

#Calculations
V = m/d;   # Volume of the core, metre cube
E = n*V*A*t;   # Hystersis loss of energy per hour, joule

#Result
print "The hystersis loss per hour = %5.2eJ"%E

The hystersis loss per hour = 1.08e+05J


## Example 12.9, Page 607¶

In [20]:
#Variable declaration
n = 50;   # Total number of cycles per sec, Hz
V = 1e-03;   # Volume of the specimen, metre cube
t = 1;   # Time during which the loss occurs, s
A = 0.25e+03;   # Area of B-H loop, joule per metre cube

#Calculations
E = n*V*A*t;    # Energy loss due to hysteresis, J/s

#Result
print "The hystersis loss = %4.1f J/s"%E
#incorrect answer in the textbook

The hystersis loss = 12.5 J/s


## Example 12.10, Page 608¶

In [21]:
#Variable declaration
e = 1.6e-19;   # Charge on anlectron, C
m = 9.1e-31;   # Mass of the electron, kg
r = 5.1e-11;   # Radius of the electronic orbit, m
B = 2.0;   # Applied magnetic field, weber per metre-square

#Calculations
delta_pm = e**2*r**2*B/(4*m);

#Result
print "The change in the magnetic dipole moment of the electron = %3.1e A-metre square"%delta_pm

The change in the magnetic dipole moment of the electron = 3.7e-29 A-metre square