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authorkinitrupti2017-05-12 18:53:46 +0530
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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": {}
+ }
+ ]
+} \ No newline at end of file