# Chapter 19: Superconductivity¶

## Example 19.1, Page 959¶

In [1]:
#Variable declaration
T_c = 6.2;    # Critical temperature of lead in superconducting state, K
T = 4;        # Temperature at which critical field of lead is to be found out, K
H_c0 = 0.064;    # Critical field for lead at 0 K, MA/m

#Calculation
H_cT = H_c0*(1-(T/T_c)**2);    # Critical field for lead at 4 K, MA/m

#Result
print "The critical field for lead at 4 K = %5.3f MA/m"%H_cT

The critical field for lead at 4 K = 0.037 MA/m


## Example 19.2, Page 959¶

In [2]:
from math import *

#Variable declaration
T_c1 = 4.153;    # Critical temperature of mercury for its one isotope, K
M1 = 200.59;    # Mass of first isotope of mercury, amu
M2 = 204;       # Mass of second isotope of mercury, amu

#Calculation
# From isotopic effect of superconductivity,
# T_c2/T_c1 = sqrt(M1/M2), solving for T_c2
T_c2 = T_c1*sqrt(M1/M2);    # Critical temperature of mercury for second isotope, K

#Result
print "The critical temperature of mercury for its isotope of mass 204 amu = %5.3f K"%T_c2

The critical temperature of mercury for its isotope of mass 204 amu = 4.118 K


## Example 19.3, Page 960¶

In [3]:
#Variable declaration
d = 1e-003;    # Diameter of aluminium wire, m
r = d/2;       # Radius of aluminium wire, m
H_c = 7.9e+003;    # Critical magnetic field for Al, A/m

#Calculation
I_c = 2*3.14*r*H_c;    # Critical current through superconducting aluminium wire, A

#Result
print "The critical current through superconducting aluminium wire = %6.3f A"%I_c

The critical current through superconducting aluminium wire = 24.806 A


## Example 19.4, Page 960¶

In [4]:
#Variable declaration
T_c = 7.18;    # Critical temperature of lead in superconducting state, K
H_c0 = 6.5e+004;    # Critical field for lead at 0 K, A/m
# At T = 4.2 K
T = 4.2;       # Temperature at which critical field of lead is to be found out, K
H_cT = H_c0*(1-(T/T_c)**2);    # Critical field for lead at 4 K, A/m
d = 1e-003;    # Diameter of lead wire, m
I_c = 2*3.14*r*H_cT;    # Critical current through superconducting lead wire, A
J_c = I_c/(3.14*r**2);    # Critical current density for superconducting lead wire, A/Sq. meter
print "The critical current density at %3.1f K = %5.3e A/Sq.m"%(T, J_c)
# At T = 7 K
T = 7;       # Temperature at which critical field of lead is to be found out, K
H_cT = H_c0*(1-(T/T_c)**2);    # Critical field for lead at 4 K, A/m
d = 1e-003;    # Diameter of lead wire, m
I_c = 2*3.14*r*H_cT;    # Critical current through superconducting lead wire, A
J_c = I_c/(3.14*r**2);    # Critical current density for superconducting lead wire, A/Sq. meter
print "The critical current density at %3.1f K = %4.2e A/Sq.m"%(T, J_c)

The critical current density at 4.2 K = 1.710e+08 A/Sq.m
The critical current density at 7.0 K = 1.29e+07 A/Sq.m


## Example 19.5, Page 961¶

In [5]:
#Variable declaration
T1 = 3;    # Initial temperature of lead wire, K
T2 = 7.1;    # Final temperature of lead wire, K
lambda1 = 39.6;    # Initial London penetration depth for lead, mm
lambda2 = 173;    # Final London penetration depth for lead, mm

#Calculations
# As lambda_T = lambda_0*(1-(T/T_c)^4)^(-1/2) so
# (lambda1/lambda2)^2 = (T_c^4 - T2^4)/(T_c^4 - T1^4)
# Solving for T_c
T_c = ((T2**4-T1**4*(lambda1/lambda2)**2)/(1-(lambda1/lambda2)**2))**(1./4);

#Result
print "The critical temperature of lead = %5.3f K"%T_c

The critical temperature of lead = 7.193 K


## Example 19.6, Page 962¶

In [6]:
#Variable declaration
T_c = 7.2;    # Critical temperature of lead in superconducting state, K
T = 5;        # Temperature at which lead loses its superconducting state, K
H_cT = 3.3e+004;    # Critical magnetic field for superconducting lead at 5 K, A/m

#Calculation
# As H_cT = H_c0*(1-(T/T_c)^2), solving for H_c0
H_c0 = H_cT/(1-(T/T_c)**2);    # Critical field for lead at 0 K, A/m

#Result
print "The critical magnetic field for lead at 0 K = %4.2e A/m"%H_c0

The critical magnetic field for lead at 0 K = 6.37e+04 A/m


## Example 19.7, Page 962¶

In [7]:
#Variable declaration
H_c0 = 2e+005;    # Critical field for niobium at 0 K, A/m
H_cT = 1e+005;    # Critical magnetic field for superconducting niobium at 5 K, A/m
T = 8;        # Temperature at which lead loses its superconducting state, K

#Calculation
# As H_cT = H_c0*(1-(T/T_c)^2), solving for T_c
T_c = T/(1-H_cT/H_c0)**(1./2);

#Result
print "The critical temperature for niobium = %6.3f K"%T_c

The critical temperature for niobium = 11.314 K