# Ch-22, Superconduting material properties, behaviour and application¶

## example-22.2 page no-660¶

In [8]:
from __future__ import division
#given
#critical  temp of Pb
T0=7.17  #K
#critical field
H0=0.0803  #A/m
#to find the critical field at
T1=3 #K
T2=10  #K
#critical field at T1
Hc1=H0*(1-T1**2/T0**2)  #A/m
#critical field at T2
Hc2=H0*(1-T2**2/T0**2)  #A/m
print " the critical field at 3K temp is %0.3f A/m and at 10K ia %0.3f A/m"%(Hc1,Hc2)

 the critical field at 3K temp is 0.066 A/m and at 10K ia -0.076 A/m


## example-22.3 page no-660¶

In [9]:
from math import pi
#given
#critical magnetic field
Hc=7.9*10**3  #A/m
#diameter of aluminium wire
d=1*10**-3   #m
#critical current is give by
Ic=2*(pi)*d*Hc  #A
print " the critical current which can pass through a long thin superconducting wire of aluminium is %0.3f A"%(Ic)

 the critical current which can pass through a long thin superconducting wire of aluminium is 49.637 A


## example-22.4 page no-667¶

In [10]:
from math import sqrt
#given
sd=11.4
rho=sd*10**3  #kg/m**3
Aw=207.2  #kg/kg-mol
v=1200  #m/s
e=1.602*10**-19  #C/electrons  (charge on an electron)
m=9.1*10**-31  #kg  (mass of electron)
mu0=4*(pi)*10**-7  #H/m  (permeability)
#since lead is type I superconductor, so  London's theory of superconductivity is applicable
#so
ne=2*rho*NA/Aw  #electrons/m**3
#critical current density
Ied=ne*e*v  #A/m**2
#depth of penetration at the surface of lead
dp=sqrt(m/(mu0*ne*e**2))*10**10  #A
print """the electron density is %0.3e electrons/m**3
the critical current density is %0.3e A/m**2
the depth of penetration is %0.3e A"""%(ne,Ied,dp)

the electron density is 6.628e+28 electrons/m**3
the critical current density is 1.274e+13 A/m**2
the depth of penetration is 2.063e+02 A


## example-22.9 page no-673¶

In [11]:
#given
#magnetic field at 0K temp
H0=65*10**3  #A/m
#critical temp
Tc=7.18  #K
#diameter of wire
d=1*10**-3  #m
r=d/2  #m
#area of cross section
A=(pi)*r**2  #m**2
#to find the current density at 4.2 K
#since it is given that Hc is parabolicaly dependent on T, so
T=4.2  #K
Hc=H0*(1-T**2/Tc**2)  #A/m
#critical current
Ic=2*(pi)*r*Hc  #A
#critical current density Jc
Jc=Ic/A  #A/m**2
print " the critical current density of lead is %0.3e A/m**2"%(Jc)

 the critical current density of lead is 1.710e+08 A/m**2


## example-22.10 page no-673¶

In [12]:
from math import sqrt
#given
#critical field at 3K and 14 K are 21 A/m and 10 A/m
T1=7  #K
T2=14  #K
Hc1=21  #A/m
Hc2=10  #A/m
#DETERMINING CRITICAL TEMP
#as we know that H=H0*(1-T**2/Tc**2)
#so we get
#71=H0*(1-7**2/Tc**2)  ----(1)
#10=H0*(1-14**2/Tc**2) --(2)
#dividing 1 and 2 we get
#71/10=(Tc**2-7**2)/(Tc**2-14**2)
#on solving we get
Tc=sqrt(3626/11)  #K
#DETERMINING CRITICAL FIELD AT 0K
H0=Hc1/(1-T1**2/Tc**2)  #A/m
#DETERMINING CRITICAL FIELD AT
T=4.2   #K
Hc=H0*(1-T**2/Tc**2)  #A/m
print " the critical temp is %0.3f K\n, the critical field at 0K is %0.3f A/m and critical field at 4.2 K is %0.3f A/m"%(Tc,H0,Hc)

 the critical temp is 18.156 K
, the critical field at 0K is 24.667 A/m and critical field at 4.2 K is 23.347 A/m


## example-22.11 page no-674¶

In [14]:
from __future__ import division
#given
#depth of penetration at 3K is 39.8 nm and at 7.1 K is 1730 A
T1=3  #K
T2=7.1  #K
dp1=39.6*10**-9  #m
dp2=1730*10**-10  #m
#as we know that depth of penetration and temp are related as
#(dp(T)/dp(T0))=1/(1-t**4/Tc**4)
#so we get
#at 3K
#let dp(T0)=dp0
#dp0=sqrt(dp1**2*(1-T1**4/Tc**4))  -(1)
#also
#dp0=sqrt(dp2**2*(1-T2**4/Tc**4))  .----(2)
#solving 1 and 2 we get
#((Tc**4-81)/(Tc**4-(7.1)**4))=(173)**2/(39.6)**2
#so we get
Tc=(48417.9/18.085)**(1/4)  #K\
#depth of penetration at absolute zero willbe
dp0=sqrt(dp1**2*(1-T1**4/Tc**4))*10**9  #nm
print " critica temp is %0.3f K\n and depth of penetration at critica zero is %0.3f nm"%(Tc,dp0)

 critica temp is 7.193 K
and depth of penetration at critica zero is 38.996 nm