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-{
- "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": {}
- }
- ]
-} \ No newline at end of file
generated by cgit (git 2.34.1) at 2025-01-08 20:06:42 +0000