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{
"metadata": {
"name": "chapter5"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": "Superconductivity"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.1, Page number 148"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the critical feild at 2K\n\n#importing modules\nimport math\n\n#Variable declaration\nTc=3.7; #in kelvin\nHc_0=0.0306; \nT=2\n\n#Calculation\nHc_2k=Hc_0*(1-((T/Tc)**2));\nHc_2k=math.ceil(Hc_2k*10**5)/10**5; #rounding off to 5 decimals\n\n#Result\nprint(\"the critical feild at 2K in tesla is\",Hc_2k);\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the critical feild at 2K in tesla is', 0.02166)\n"
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.2, Page number 149\n"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the critical current\n\n#importing modules\nimport math\n\n#Variable declaration\nT=4.2; #in kelvin\nTc=7.18; #in kelvin\nHc_0=6.5*10**4; #in amp per meter\nD=10**-3\n\n#Calculation\nR=D/2; #radius is equal to half of diameter\nHc_T=Hc_0*(1-((T/Tc)**2));\nHc_T=math.ceil(Hc_T*10)/10; #rounding off to 1 decimals\nIc=2*math.pi*R*Hc_T #critical current is calculated by 2*pi*r*Hc(T)\nIc=math.ceil(Ic*10**2)/10**2; #rounding off to 2 decimals\n\n#Result\nprint(\"the critical feild in Tesla is\",round(Hc_T));\nprint(\"the critical current in Amp is\",Ic);\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the critical feild in Tesla is', 42759.0)\n('the critical current in Amp is', 134.34)\n"
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.3, Page number 149\n"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the pentration depth at 0k\n\n#importing modules\nimport math\n\n#Variable declaration\nlamda_T=75 #in nm\nT=3.5 \nHgTc=4.12 #in K\n\n#Calculation\nlamda_o=lamda_T*math.sqrt(1-((T/HgTc)**4));\nlamda_o=math.ceil(lamda_o*10**2)/10**2; #rounding off to 2 decimals\n\n#Result\nprint(\"the pentration depth at 0k is\",lamda_o);",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the pentration depth at 0k is', 51.92)\n"
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.4, Page number 150"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the critical magnitude\n\n#importing modules\nimport math\n\n#Variable declaration\nlamda_T1=396 #pentration depth in armstrong\nlamda_T2=1730 #pentration depth in armstrong\nT1=3 #temperature in K\nT2=7.1 #temperature in K\n\n#Calculation\n#lamda_T2**2=lamda_0**2*(((Tc**4-T2**4)/Tc**4)**-1)\n#lamda_T1**2=lamda_0**2*(((Tc**4-T1**4)/Tc**4)**-1)\n#dividing lamda_T2**2 by lamda_T1**2 = (Tc**4-T1**4)/(Tc**4-T2**4)\n#let A=lamda_T2**2 and B=lamda_T1**2\nA=lamda_T2**2\nB=lamda_T1**2\nC=A/B\nC=math.ceil(C*10**4)/10**4; #rounding off to 4 decimals\nX=T1**4\nY=T2**4\nY=math.ceil(Y*10**2)/10**2; #rounding off to 2 decimals\n#C*((TC**4)-Y)=(Tc**4)-X\n#C*(Tc**4)-(Tc**4)=C*Y-X\n#(Tc**4)*(C-1)=(C*Y)-X\n#let Tc**4 be D\n#D*(C-1)=(C*Y)-X\nD=((C*Y)-X)/(C-1)\nD=math.ceil(D*10)/10; #rounding off to 1 decimals\nTc=D**(1/4)\nTc=math.ceil(Tc*10**4)/10**4; #rounding off to 4 decimals\n\n#Result\nprint(\"the pentration depth at 0k is\",Tc);",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the pentration depth at 0k is', 7.1932)\n"
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.5, Page number 150"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the critical feild at 5K\n\n#importing modules\nimport math\n\n#Variable declaration\nTc=7.2 #in K\nHo=6.5*10**3 #in amp per m\nT=5 #in K\n\n#Calculation\nHc=Ho*(1-((T/Tc)**2))\nHc=math.ceil(Hc*10**2)/10**2; #rounding off to 2 decimals\n\n#Result\nprint(\"the critical magnetic feild at 5K in amp per m is\",Hc)\n\n# answer given in the book is wrong",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the critical magnetic feild at 5K in amp per m is', 3365.36)\n"
}
],
"prompt_number": 33
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.6, Page number 151"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the critical feild at 2.5K\n\n#importing modules\nimport math\n\n#Variable declaration\nTc=3.5 #in K\nHo=3.2*10**3 #in amp per m\nT=2.5 #in K\n\n#Calculation\nHc=Ho*(1-((T/Tc)**2))\nHc=math.ceil(Hc*10**2)/10**2; #rounding off to 2 decimals\n\n#Result\nprint(\"the critical magnetic feild at 5K in amp per m is\",Hc)\n\n#answer in the book is wrong",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the critical magnetic feild at 5K in amp per m is', 1567.35)\n"
}
],
"prompt_number": 45
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.7, Page number 151"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the transition temperature\n\n#importing modules\nimport math\n\n#Variable declaration\nHc=5*10**3 #in amp per m\nHo=2*10**4 #in amp per m\nT=6 #in K\n\n#Calculation\nTc=T/math.sqrt(1-(Hc/Ho))\nTc=math.ceil(Tc*10**2)/10**2; #rounding off to 2 decimals\n\n#Result\nprint(\"the critical magnetic feild at 5K in amp per m is\",Tc)\n\n#answer in the book is wrong",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the critical magnetic feild at 5K in amp per m is', 6.93)\n"
}
],
"prompt_number": 66
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.8, Page number 152"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the critical current\n\n#importing modules\nimport math\n\n#Variable declaration\nHc=2*10**3 #in amp per m\nR=0.02 #in m\n\n#Calculation\nIc=2*math.pi*R*Hc\nIc=math.ceil(Ic*10**2)/10**2; #rounding off to 2 decimals\n\n#Result\nprint(\"the critical current is\",Ic)\n\n#answer in the book is wrong",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the critical magnetic feild at 5K in amp per m is', 251.33)\n"
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.9, Page number 152"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the isotopic mass of M2\n\n#importing modules\nimport math\n\n#Variable declaration\nM1=199.5 #in a.m.u\nT1=5 #in K\nT2=5.1 #in K\n\n#Calculation\nM2=((T1/T2)**2)*M1\nM2=math.ceil(M2*10**3)/10**3; #rounding off to 3 decimals\n\n#Result\nprint(\"the isotopic mass of M2 is\",M2)",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the isotopic mass of M2 is', 191.754)\n"
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.10, Page number 152"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the critical magnetic feild and critical current\n\n#importing modules\nimport math\nfrom __future__ import division\n\n#Variable declaration\nD=3*10**-3 #in meters\nTc=8 #in K \nT=5 #in K \nHo=5*10**4\n\n#Calculation\nR=D/2\nHc=Ho*(1-((T/Tc)**2))\nIc=2*math.pi*R*Hc\nIc=math.ceil(Ic*10**3)/10**3; #rounding off to 3 decimals\n\n#Result\nprint(\"critical magnetic feild in amp per m is\",round(Hc));\nprint(\"critical current in amp is\",Ic);\n\n#answer in the book is wrong",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('critical magnetic feild in amp per m is', 30469.0)\n('critical current in amp is', 287.162)\n"
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.11, Page number 153"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the critical temperature\n\n#importing modules\nimport math\n\n#Variable declaration\nM1=199.5 \nM2=203.4 \nTc1=4.185 #in K\n\n#Calculation\nTc2=Tc1*math.sqrt(M1/M2)\nTc2=math.ceil(Tc2*10**3)/10**3; #rounding off to 3 decimals\n\n#Result\nprint(\"the critical temperature is\",Tc2)",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the critical temperature is', 4.145)\n"
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.12, Page number 154"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#To calculate the EM wave frequency\n\n#importing modules\nimport math\nfrom __future__ import division\n\n#Variable declaration\nV=8.5*10**-6 #in volts\ne=1.6*10**-19 #in C\nh=6.626*10**-24\n\n#Calculation\nnew=2*e*V/h\nnew=math.ceil(new*10**5)/10**5; #rounding off to 5 decimals\n\n#Result\nprint(\"EM wave generated frequency in Hz is\",new)\n\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('EM wave generated frequency in Hz is', 0.41051)\n"
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.13, Page number 154"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#to calculate the critical temperature at 6mm presure of Hg\n\n#Variable declaration\np1=1 #in mm\np2=6 #in mm\nTc1=5 #in K\n\n#Calculation\nTc2=Tc1*(p2/p1);\n\n#Result\nprint(\"the critical temperature in K is\",round(Tc2))",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "('the critical temperature in K is', 30.0)\n"
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example number 5.14, Page number 154\n"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#to calculate maximum critical temperature\n\n#Variable declaration\nTc=8.7 #in K\nHc=6*10**5 #in A per m\nHo=3*10**6 #in A per m\n\n#Calculation\nT=Tc*(math.sqrt(1-(Hc/Ho)))\n\n#Result\nprint(\" maximum critical temperature in K is\",T)\n\n#answer given in the book is wrong",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "(' maximum critical temperature in K is', 7.781516561699267)\n"
}
],
"prompt_number": 15
},
{
"cell_type": "code",
"collapsed": false,
"input": "",
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}
|
