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Diffstat (limited to 'Engineering_Physics/chapter5_2.ipynb')
| -rw-r--r-- | Engineering_Physics/chapter5_2.ipynb | 407 |
1 files changed, 362 insertions, 45 deletions
diff --git a/Engineering_Physics/chapter5_2.ipynb b/Engineering_Physics/chapter5_2.ipynb index 66e0fe37..14018aea 100644 --- a/Engineering_Physics/chapter5_2.ipynb +++ b/Engineering_Physics/chapter5_2.ipynb @@ -1,6 +1,7 @@ { "metadata": { - "name": "chapter5" + "name": "", + "signature": "sha256:19dabe1afe46093105a84b4746899bd5b483ca26e3b557510765740ff72179af" }, "nbformat": 3, "nbformat_minor": 0, @@ -11,25 +12,47 @@ "cell_type": "heading", "level": 1, "metadata": {}, - "source": "Superconductivity" + "source": [ + "Superconductivity" + ] }, { "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.1, Page number 148" + "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", + "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" + "text": [ + "('the critical feild at 2K in tesla is', 0.02166)\n" + ] } ], "prompt_number": 4 @@ -38,19 +61,45 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.2, Page number 149\n" + "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", + "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" + "text": [ + "('the critical feild in Tesla is', 42759.0)\n", + "('the critical current in Amp is', 134.34)\n" + ] } ], "prompt_number": 19 @@ -59,19 +108,39 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.3, Page number 149\n" + "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);", + "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" + "text": [ + "('the pentration depth at 0k is', 51.92)\n" + ] } ], "prompt_number": 29 @@ -80,19 +149,58 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.4, Page number 150" + "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);", + "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" + "text": [ + "('the pentration depth at 0k is', 7.1932)\n" + ] } ], "prompt_number": 44 @@ -101,19 +209,41 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.5, Page number 150" + "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", + "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" + "text": [ + "('the critical magnetic feild at 5K in amp per m is', 3365.36)\n" + ] } ], "prompt_number": 33 @@ -122,19 +252,41 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.6, Page number 151" + "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", + "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" + "text": [ + "('the critical magnetic feild at 5K in amp per m is', 1567.35)\n" + ] } ], "prompt_number": 45 @@ -143,19 +295,41 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.7, Page number 151" + "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", + "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" + "text": [ + "('the critical magnetic feild at 5K in amp per m is', 6.93)\n" + ] } ], "prompt_number": 66 @@ -164,19 +338,40 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.8, Page number 152" + "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", + "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" + "text": [ + "('the critical magnetic feild at 5K in amp per m is', 251.33)\n" + ] } ], "prompt_number": 2 @@ -185,19 +380,39 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.9, Page number 152" + "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)", + "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" + "text": [ + "('the isotopic mass of M2 is', 191.754)\n" + ] } ], "prompt_number": 4 @@ -206,19 +421,45 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.10, Page number 152" + "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", + "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" + "text": [ + "('critical magnetic feild in amp per m is', 30469.0)\n", + "('critical current in amp is', 287.162)\n" + ] } ], "prompt_number": 22 @@ -227,19 +468,39 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.11, Page number 153" + "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)", + "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" + "text": [ + "('the critical temperature is', 4.145)\n" + ] } ], "prompt_number": 3 @@ -248,19 +509,41 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.12, Page number 154" + "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", + "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" + "text": [ + "('EM wave generated frequency in Hz is', 0.41051)\n" + ] } ], "prompt_number": 24 @@ -269,19 +552,35 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.13, Page number 154" + "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))", + "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" + "text": [ + "('the critical temperature in K is', 30.0)\n" + ] } ], "prompt_number": 14 @@ -290,19 +589,37 @@ "cell_type": "heading", "level": 2, "metadata": {}, - "source": "Example number 5.14, Page number 154\n" + "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", + "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" + "text": [ + "(' maximum critical temperature in K is', 7.781516561699267)\n" + ] } ], "prompt_number": 15 @@ -310,7 +627,7 @@ { "cell_type": "code", "collapsed": false, - "input": "", + "input": [], "language": "python", "metadata": {}, "outputs": [] |