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diff --git a/Engineering_Physics_Vijaya/chapter5_2.ipynb b/Engineering_Physics_Vijaya/chapter5_2.ipynb new file mode 100755 index 00000000..14018aea --- /dev/null +++ b/Engineering_Physics_Vijaya/chapter5_2.ipynb @@ -0,0 +1,639 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:19dabe1afe46093105a84b4746899bd5b483ca26e3b557510765740ff72179af" + }, + "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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Tc=3.7; #in kelvin\n", + "Hc_0=0.0306; \n", + "T=2\n", + "\n", + "#Calculation\n", + "Hc_2k=Hc_0*(1-((T/Tc)**2));\n", + "Hc_2k=math.ceil(Hc_2k*10**5)/10**5; #rounding off to 5 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "T=4.2; #in kelvin\n", + "Tc=7.18; #in kelvin\n", + "Hc_0=6.5*10**4; #in amp per meter\n", + "D=10**-3\n", + "\n", + "#Calculation\n", + "R=D/2; #radius is equal to half of diameter\n", + "Hc_T=Hc_0*(1-((T/Tc)**2));\n", + "Hc_T=math.ceil(Hc_T*10)/10; #rounding off to 1 decimals\n", + "Ic=2*math.pi*R*Hc_T #critical current is calculated by 2*pi*r*Hc(T)\n", + "Ic=math.ceil(Ic*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"the critical feild in Tesla is\",round(Hc_T));\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "lamda_T=75 #in nm\n", + "T=3.5 \n", + "HgTc=4.12 #in K\n", + "\n", + "#Calculation\n", + "lamda_o=lamda_T*math.sqrt(1-((T/HgTc)**4));\n", + "lamda_o=math.ceil(lamda_o*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "lamda_T1=396 #pentration depth in armstrong\n", + "lamda_T2=1730 #pentration depth in armstrong\n", + "T1=3 #temperature in K\n", + "T2=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\n", + "A=lamda_T2**2\n", + "B=lamda_T1**2\n", + "C=A/B\n", + "C=math.ceil(C*10**4)/10**4; #rounding off to 4 decimals\n", + "X=T1**4\n", + "Y=T2**4\n", + "Y=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\n", + "D=((C*Y)-X)/(C-1)\n", + "D=math.ceil(D*10)/10; #rounding off to 1 decimals\n", + "Tc=D**(1/4)\n", + "Tc=math.ceil(Tc*10**4)/10**4; #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Tc=7.2 #in K\n", + "Ho=6.5*10**3 #in amp per m\n", + "T=5 #in K\n", + "\n", + "#Calculation\n", + "Hc=Ho*(1-((T/Tc)**2))\n", + "Hc=math.ceil(Hc*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Tc=3.5 #in K\n", + "Ho=3.2*10**3 #in amp per m\n", + "T=2.5 #in K\n", + "\n", + "#Calculation\n", + "Hc=Ho*(1-((T/Tc)**2))\n", + "Hc=math.ceil(Hc*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Hc=5*10**3 #in amp per m\n", + "Ho=2*10**4 #in amp per m\n", + "T=6 #in K\n", + "\n", + "#Calculation\n", + "Tc=T/math.sqrt(1-(Hc/Ho))\n", + "Tc=math.ceil(Tc*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Hc=2*10**3 #in amp per m\n", + "R=0.02 #in m\n", + "\n", + "#Calculation\n", + "Ic=2*math.pi*R*Hc\n", + "Ic=math.ceil(Ic*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "M1=199.5 #in a.m.u\n", + "T1=5 #in K\n", + "T2=5.1 #in K\n", + "\n", + "#Calculation\n", + "M2=((T1/T2)**2)*M1\n", + "M2=math.ceil(M2*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "D=3*10**-3 #in meters\n", + "Tc=8 #in K \n", + "T=5 #in K \n", + "Ho=5*10**4\n", + "\n", + "#Calculation\n", + "R=D/2\n", + "Hc=Ho*(1-((T/Tc)**2))\n", + "Ic=2*math.pi*R*Hc\n", + "Ic=math.ceil(Ic*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"critical magnetic feild in amp per m is\",round(Hc));\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "M1=199.5 \n", + "M2=203.4 \n", + "Tc1=4.185 #in K\n", + "\n", + "#Calculation\n", + "Tc2=Tc1*math.sqrt(M1/M2)\n", + "Tc2=math.ceil(Tc2*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "V=8.5*10**-6 #in volts\n", + "e=1.6*10**-19 #in C\n", + "h=6.626*10**-24\n", + "\n", + "#Calculation\n", + "new=2*e*V/h\n", + "new=math.ceil(new*10**5)/10**5; #rounding off to 5 decimals\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#Variable declaration\n", + "p1=1 #in mm\n", + "p2=6 #in mm\n", + "Tc1=5 #in K\n", + "\n", + "#Calculation\n", + "Tc2=Tc1*(p2/p1);\n", + "\n", + "#Result\n", + "print(\"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": [ + " \n", + "#Variable declaration\n", + "Tc=8.7 #in K\n", + "Hc=6*10**5 #in A per m\n", + "Ho=3*10**6 #in A per m\n", + "\n", + "#Calculation\n", + "T=Tc*(math.sqrt(1-(Hc/Ho)))\n", + "\n", + "#Result\n", + "print(\" 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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