# Chapter 18: Magnetic Properties of Solids¶

## Example 18.1, Page 18.21¶

In [1]:
from math import *

# Given
r = 0.53e-10 # radius of orbit in m
f = 6.6e15 # frequency of revolution in Hz
h = 6.6e-34 # Planck constant in J sec
e = 1.6e-19 # charge on an electron in C
m = 9.1e-31 # mass of electron in kg

#Calculations
M = e * f * pi * r**2
mu = (e * h) / (4 * pi * m)

#Result
print "Magnetic moment is %.3e Am^2\nBohr magneton is %.2e J/T"%(M,mu)

Magnetic moment is 9.319e-24 Am^2
Bohr magneton is 9.23e-24 J/T


## Example 18.2, Page 18.21¶

In [9]:
from math import *

# Given
X = -4.2e-6 # magnetic susceptibility
H = 1.19e5 # magnetic field in A/m
mu_ = 4 * pi * 1e-7 # magnetic permeability of space

#Calculations
I = X * H
B = mu_ * (H + I)
mur = (1 + I/H)

#Results
print "Magnetisation is %.2f A/m\nFlux density is %.3f T\nRelative permeability is %.2f"%(I,B,mur)

Magnetisation is -0.50 A/m
Flux density is 0.150 T
Relative permeability is 1.00


## Example 18.3, Page 18.22¶

In [10]:

# Given
X = 1.2e-5 # magnetic susceptibility of magnesium

#Calculations
p = 100 * X

#Result
print "Percentage increase in magnetic induction is %.4f percent"%p

Percentage increase in magnetic induction is 0.0012 percent


## Example 18.4, Page 18.22¶

In [16]:
from math import *

# Given
X = -0.4e-5 # magnetic susceptibility of material
H = 1e4 # magnetic field in A/m
mu_ = 4 * pi * 1e-7 # magnetic permittivity of space

#Calculations
I = X * H
B = mu_ * (H + I)

#Result
print "Magnetisation is %.2f A/m\nMagnetic flux density is %f T"%(I,B)

Magnetisation is -0.04 A/m
Magnetic flux density is 0.012566 T


## Example 18.5, Page 18.23¶

In [17]:
from math import *

# Given
X = 2.3e-5 # magnetic susceptibility of aluminium
mu_ = 4 * pi * 1e-7 # magnetic permeability of space

#Calculations
mur = 1 + X
mu = mu_ * mur

#Result
print "Permeability of aluminium is %.2e N/A^2"%mu

Permeability of aluminium is 1.26e-06 N/A^2


## Example 18.6, Page 18.23¶

In [18]:
from math import pi

# Given
X = 9.4e-2 # magnetic susceptibility
mu_ = 4 * pi * 1e-7 # magnetic permeability of space

#Calculations
mu_r = 1 + X
mu = mu_ * mu_r

#Results
print "Absolute permeability is %.2e N/A^2\nRelative permeability is %.3f"%(mu,mu_r)

Absolute permeability is 1.37e-06 N/A^2
Relative permeability is 1.094


## Example 18.7, Page 18.23¶

In [19]:
from math import *

# Given
mu = 0.126 # maximum value of the permeability in N/A^2
mu_ = 4 * pi * 1e-7 # magnetic permeability of space

#Calculations
mu_r = mu / mu_
X = mu_r - 1

#Results
print "Magnetic susceptibility is %.f\nRelative permeability is %e"%(X,mu_r)
#Answers differ due to rounding off values

Magnetic susceptibility is 100267
Relative permeability is 1.002676e+05


## Example 18.8, Page 18.24¶

In [20]:
from math import pi

# Given
r = 0.6e-10 # radius of the atom
N = 28e26 # no. of electron in per m^3
mu_ = 4 * pi * 1e-7 # magnetic permeability of space
Z = 2 # atomic no. of helium
m = 9.1e-31 # mass of an electron in kg
e = 1.6e-19 # charge on an electron in C

#Calculations
Chi = -(mu_ * Z * N * r**2 * e**2) / (6 * m)

#Result
print "Diamagnetic susceptibility is %.3e"%Chi

Diamagnetic susceptibility is -1.188e-07


## Example 18.9, Page 18.24¶

In [21]:
from math import pi

# Given
H = 1e3 # magnetisation field in A/m
phi = 2e-5 # magnetic flux in Weber
a = 0.2e-4 # area of cross section in m^2
mu_ = 4 * pi * 1e-7 # magnetic permeability of space

#Calculations
B = phi / a
mu = B / H
X = mu / mu_ - 1

#Results
print "Permeability is %.e N/A^2\nSusceptibility is %.3f"%(mu,X)

Permeability is 1e-03 N/A^2
Susceptibility is 794.775


## Example 18.10, Page 18.24¶

In [10]:

# Given
l = 1 # length of iron rod in m
a = 4e-4 # area in m^2
mu = 50e-4 # permeability of iron in H/m
Phi = 4e-4 # magnetic flux in Weber

#Calculations
B = Phi / a
NI = B / mu

#Result
print "Number of ampere turns is %d A/m"%NI

Number of ampere turns is 200 A/m


## Example 18.11, Page 18.25¶

In [22]:
from math import *

# Given
n = 200 # no. of turns
l = 0.5 # the mean length of iron wire in m
phi = 4e-4 # magnetic flux in Weber
a = 4e-4 # area of cross section in m^2
mu = 6.5e-4 # permeability of iron in wb/Am
mu_ = 4 * pi * 1e-7 # magnetic permeability of space

#Calculations
B = phi / a
N = n / l
I = B / (mu * N)

#Result
print "Current through the winding is %.2f A"%I

Current through the winding is 3.85 A


## Example 18.12, Page 18.25¶

In [23]:
from math import pi

# Given
X = -5.6e-6 # magnetic susceptibility of material
a = 2.55e-10 # lattice constant in m
H = 1e4 # magnetic field in A/m
mu_ = 4 * pi * 1e-7 # magnetic permittivity of space
m = 9.1e-31 # mass of electron in kg
e = 1.6e-19 # charge in an electron in C

#Calculations
N = 2 / a**3
z = 1
R = ((-X * 6 * m) / (mu_ * z * e**2 * N))**(1./2)

#Result
print "Radius of  atom is %.2f A"%(R * 1e10)

Radius of  atom is 0.89 A


## Example 18.13, Page 18.25¶

In [24]:
from math import pi

# Given
N = 6.5e25 # no. of atom per m^3
T = 300 # room temperature in K
mu_ = 4 * pi * 1e-7 # magnetic permittivity of space
k = 1.38e-23 # Boltzmann's constant in J/K
m = 9.1e-31 # mass of electron in kg
e = 1.6e-19 # charge in an electron in C
h = 6.62e-34 # Planck constant in J sec

#Calculations
M = (e * h) / (4 * pi * m)
X = (mu_ * N * M**2) / (3 * k * T)

#Result
print "Susceptibility is %.3e"%X

Susceptibility is 5.642e-07


## Example 18.14, Page 18.26¶

In [25]:
from math import *

# Given
w = 168.5 # molecular weight
d = 4370 # density of material in kg/m^3
H = 2e5 # magnetic field in A/m
T = 300 # room temperature in K
mu_ = 4 * pi * 1e-7 # magnetic permittivity of space]
NA = 6.02e26 # Avogadro no. in per kg
mu_b = 9.24e-24 # Bohr magnetons in Am^2
k = 1.38e-23 # Boltzmann's constant in J/K

#Calculations
N = d * NA / w
X = (mu_ * N * (2 * mu_b)**2) / (3 * k * T)
I = X * H

#Result
print "Magnetisation is %.2f A/m"%I

Magnetisation is 107.89 A/m


## Example 18.15, Page 18.26¶

In [26]:

# Given that
A = 2500 # area of hysteresis loop
m = 10000 # weight in kg
d = 7.5 # density of material in g/cm^3
f = 50 # frequency in Hz

#Calculations
E = f * A * 3600
V = m / d
L = E * V

#Result
print "Total loss of energy per hour is %.e ergs"%L

Total loss of energy per hour is 6e+11 ergs


## Example 18.16, Page 18.27¶

In [16]:

# Given
H = 5e3 # coercivity in A/m
l = 0.10 # length of solenoid in m
n = 50 # no. of turns

#Calculations
N = n / l
i = H / N

#Result
print "Current in solenoid should be %d A"%i

Current in solenoid should be 10 A


## Example 18.17, Page 18.27¶

In [27]:

# Given
l = 0.50 # length of iron rod in m
a = 4e-4 # area of cross section of rod in m^2
mu = 65e-4 # permeability of iron in H/m
fi = 4e-5 # flux in weber

#Calculations
B = fi / a
H = B / mu
N = H * l

#Result
print "Number of turns are %.2f"%N

Number of turns are 7.69


## Example 18.18, Page 18.27¶

In [28]:
from math import *

# Given
H = 600 # magnetic flux in A/m
a = 0.2e-4 # area of cross section of rod in m^2
phi = 2.4e-5 # flux in weber
mu_ = 4*pi * 1e-7 # permeability of space in N/A^2

#Calculations
B = phi / a
mu = B / H
X = mu / mu_ - 1

#Result
print "Permeability is %.3f N/A^2\nSusceptibility is %.f"%(mu,X)

Permeability is 0.002 N/A^2
Susceptibility is 1591


## Example 18.19, Page 18.28¶

In [32]:
from math import *

# Given
X = 9.5e-9 # susceptibility of medium
mu_ = 4*pi * 1e-7 # permeability of space in N/A^2

#Calculations
mu = mu_ * (1 + X)
mu_r = mu / mu_

#Result
print "Relative permeability is 1 + %.3e"%(mu_r -1)

Relative permeability is 1 + 9.500e-09


## Example 18.20, Page 18.28¶

In [37]:

# Given
a = 250. # area of the B-H loop in J/m^3
f = 50. # frequency in Hz
d = 7.5e3 # density of iron in kg/m^3
m = 100. # mass of core in kg

#Calculations
V = m / d
n = 3600 * f
A = a * V * n

#Result
print "Energy loss per hour is %.3e J"%A
#Answer varies due to rounding-off values

Energy loss per hour is 6.000e+05 J


## Example 18.21, Page 18.28¶

In [38]:

# Given
B_max = 1.375 # maximum value of B in Wb/m^2
a = 0.513 # area of the loop in cm^2
k = 1000 # value of 1 cm on x axis in A/m
k_ = 1 # value of 1 cm on y axis in Wb/m^2
B = 1.375 # alternating magnetic flux density in Wb/m^2
v = 1e-3 # volume of specimen in m^3
f = 50 # frequency in Hz

#Calculations
K = a * k * k_
L = K * v * f

#Result
print "Hysteresis loss per sec is %.2f W"%L

Hysteresis loss per sec is 25.65 W