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
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
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
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)
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
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
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
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
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