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diff --git a/Engineering_Physics/chapter5_2.ipynb b/Engineering_Physics/chapter5_2.ipynb new file mode 100644 index 00000000..66e0fe37 --- /dev/null +++ b/Engineering_Physics/chapter5_2.ipynb @@ -0,0 +1,322 @@ +{ + "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": {} + } + ] +}
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