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{
"metadata": {
"name": "",
"signature": "sha256:c4f4158f2a13ac09e07abe135472c68abbdd09915f4866e5eedc3ddf9b9f42f5"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"9: Superconductivity"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 9.1, Page number 9.3"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"Tc = 3.7 #critical temperature(K)\n",
"Hc_0 = 0.0306 #critical field(T)\n",
"T = 2 #temperature(K)\n",
"\n",
"#Calculation\n",
"Hc_2 = Hc_0*(1-(T/Tc)**2) #critical field(T)\n",
"Hc_2 = math.ceil(Hc_2*10**5)/10**5 #rounding off to 5 decimals\n",
"\n",
"#Result\n",
"print \"critical field at 2K is\",Hc_2,\"T\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"critical field at 2K is 0.02166 T\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 9.2, Page number 9.5"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"T = 4.2 #temperature(K)\n",
"d = 1 #diameter(mm)\n",
"Tc = 7.18 #critical temperature(K)\n",
"H0 = 6.5*10**4 #critical field(A/m)\n",
"\n",
"#Calculation\n",
"d = d*10**-3 #diameter(m)\n",
"Hc = H0*(1-((T/Tc)**2)) #critical field at 2K(A/m)\n",
"ic = math.pi*d*round(Hc); #critical current(A)\n",
"ic = math.ceil(ic*10**2)/10**2; #rounding off to 2 decimals\n",
"\n",
"#Result\n",
"print \"critical current for lead is\",ic,\"A\"\n",
"print \"answer given in the book differs due to rounding off errors\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"critical current for lead is 134.34 A\n",
"answer given in the book is wrong\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 9.3, Page number 9.13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"lamda_T = 750 #penetration depth of mercury(Armstrong)\n",
"T = 3.5 #temperature(K)\n",
"Tc = 4.12 #critical temperarure(K)\n",
"\n",
"#Calculation\n",
"lamda_0 = lamda_T*((1-(T/Tc)**4))**(1/2) #penetration depth(Armstrong)\n",
"\n",
"#Result\n",
"print \"penetration depth at 0K is\",int(lamda_0),\"armstrong\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"penetration depth at 0K is 519 armstrong\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 9.4, Page number 9.13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"T1 = 3 #temperature(K)\n",
"T2 = 7.1 #temperature(K)\n",
"lamda_T1 = 396 #penetration depth(armstrong)\n",
"lamda_T2 = 1730 #penetration depth(armstrong)\n",
"\n",
"#Calculation\n",
"A = (((lamda_T2/lamda_T1)**2)*T2**4) - T1**4\n",
"B = ((lamda_T2/lamda_T1)**2)-1\n",
"Tc = (A/B)**(1/4) #critical temperature(K)\n",
"Tc = math.ceil(Tc*10**4)/10**4; #rounding off to 4 decimals\n",
"\n",
"#Result\n",
"print \"critical temperature for lead is\",Tc,\"K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"critical temperature for lead is 7.1932 K\n"
]
}
],
"prompt_number": 4
}
],
"metadata": {}
}
]
}
|
