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{
"metadata": {
"name": "",
"signature": "sha256:9b85f177bb11398b4f5d182de9fe8e9e406f1c070a747a1b0c1a3320e4f20360"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Ch-22, Superconduting material properties, behaviour and application"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-22.2 page no-660"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"#given\n",
"#critical temp of Pb\n",
"T0=7.17 #K\n",
"#critical field\n",
"H0=0.0803 #A/m\n",
"#to find the critical field at \n",
"T1=3 #K\n",
"T2=10 #K\n",
"#critical field at T1\n",
"Hc1=H0*(1-T1**2/T0**2) #A/m\n",
"#critical field at T2\n",
"Hc2=H0*(1-T2**2/T0**2) #A/m\n",
"print \" the critical field at 3K temp is %0.3f A/m and at 10K ia %0.3f A/m\"%(Hc1,Hc2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" the critical field at 3K temp is 0.066 A/m and at 10K ia -0.076 A/m\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-22.3 page no-660"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi\n",
"#given\n",
"#critical magnetic field\n",
"Hc=7.9*10**3 #A/m\n",
"#diameter of aluminium wire\n",
"d=1*10**-3 #m\n",
"#critical current is give by\n",
"Ic=2*(pi)*d*Hc #A\n",
"print \" the critical current which can pass through a long thin superconducting wire of aluminium is %0.3f A\"%(Ic)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" the critical current which can pass through a long thin superconducting wire of aluminium is 49.637 A\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-22.4 page no-667"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import sqrt\n",
"#given\n",
"#specific density of lead\n",
"sd=11.4\n",
"#density of lead\n",
"rho=sd*10**3 #kg/m**3\n",
"#atomic weight of lead\n",
"Aw=207.2 #kg/kg-mol\n",
"#velocity of sound in lead\n",
"v=1200 #m/s\n",
"NA=6.023*10**26 #particles/kg-mol (avogadro's number)\n",
"e=1.602*10**-19 #C/electrons (charge on an electron)\n",
"m=9.1*10**-31 #kg (mass of electron)\n",
"mu0=4*(pi)*10**-7 #H/m (permeability)\n",
"#since lead is type I superconductor, so London's theory of superconductivity is applicable\n",
"#so\n",
"ne=2*rho*NA/Aw #electrons/m**3\n",
"#critical current density\n",
"Ied=ne*e*v #A/m**2\n",
"#depth of penetration at the surface of lead\n",
"dp=sqrt(m/(mu0*ne*e**2))*10**10 #A\n",
"print \"\"\"the electron density is %0.3e electrons/m**3\n",
"the critical current density is %0.3e A/m**2\n",
"the depth of penetration is %0.3e A\"\"\"%(ne,Ied,dp)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the electron density is 6.628e+28 electrons/m**3\n",
"the critical current density is 1.274e+13 A/m**2\n",
"the depth of penetration is 2.063e+02 A\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-22.9 page no-673"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#given\n",
"#magnetic field at 0K temp \n",
"H0=65*10**3 #A/m\n",
"#critical temp \n",
"Tc=7.18 #K\n",
"#diameter of wire\n",
"d=1*10**-3 #m\n",
"#radius of wire\n",
"r=d/2 #m\n",
"#area of cross section\n",
"A=(pi)*r**2 #m**2\n",
"#to find the current density at 4.2 K\n",
"#since it is given that Hc is parabolicaly dependent on T, so\n",
"T=4.2 #K\n",
"Hc=H0*(1-T**2/Tc**2) #A/m\n",
"#critical current\n",
"Ic=2*(pi)*r*Hc #A\n",
"#critical current density Jc\n",
"Jc=Ic/A #A/m**2\n",
"print \" the critical current density of lead is %0.3e A/m**2\"%(Jc)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" the critical current density of lead is 1.710e+08 A/m**2\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-22.10 page no-673"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import sqrt\n",
"#given\n",
"#critical field at 3K and 14 K are 21 A/m and 10 A/m\n",
"T1=7 #K\n",
"T2=14 #K\n",
"Hc1=21 #A/m\n",
"Hc2=10 #A/m\n",
"#DETERMINING CRITICAL TEMP\n",
"#as we know that H=H0*(1-T**2/Tc**2)\n",
"#so we get\n",
"#71=H0*(1-7**2/Tc**2) ----(1)\n",
"#10=H0*(1-14**2/Tc**2) --(2)\n",
"#dividing 1 and 2 we get\n",
"#71/10=(Tc**2-7**2)/(Tc**2-14**2)\n",
"#on solving we get\n",
"Tc=sqrt(3626/11) #K\n",
"#DETERMINING CRITICAL FIELD AT 0K\n",
"H0=Hc1/(1-T1**2/Tc**2) #A/m\n",
"#DETERMINING CRITICAL FIELD AT \n",
"T=4.2 #K\n",
"Hc=H0*(1-T**2/Tc**2) #A/m\n",
"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)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" the critical temp is 18.156 K\n",
", the critical field at 0K is 24.667 A/m and critical field at 4.2 K is 23.347 A/m\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"example-22.11 page no-674"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"#given\n",
"#depth of penetration at 3K is 39.8 nm and at 7.1 K is 1730 A\n",
"T1=3 #K\n",
"T2=7.1 #K\n",
"dp1=39.6*10**-9 #m\n",
"dp2=1730*10**-10 #m\n",
"#as we know that depth of penetration and temp are related as\n",
"#(dp(T)/dp(T0))=1/(1-t**4/Tc**4)\n",
"#so we get\n",
"#at 3K\n",
"#let dp(T0)=dp0\n",
"#dp0=sqrt(dp1**2*(1-T1**4/Tc**4)) -(1)\n",
"#also\n",
"#dp0=sqrt(dp2**2*(1-T2**4/Tc**4)) .----(2)\n",
"#solving 1 and 2 we get\n",
"#((Tc**4-81)/(Tc**4-(7.1)**4))=(173)**2/(39.6)**2\n",
"#so we get\n",
"Tc=(48417.9/18.085)**(1/4) #K\\\n",
"#depth of penetration at absolute zero willbe\n",
"dp0=sqrt(dp1**2*(1-T1**4/Tc**4))*10**9 #nm\n",
"print \" critica temp is %0.3f K\\n and depth of penetration at critica zero is %0.3f nm\"%(Tc,dp0)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" critica temp is 7.193 K\n",
" and depth of penetration at critica zero is 38.996 nm\n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}
|
