{ "metadata": { "name": "", "signature": "sha256:bc4f145530554f550ad02c41bb92fc1d8f9fdac80b4d0ae7b519b9e02a3d9b4f" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter18:Magnetic Materials" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex18.1:pg-346" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Example 18.1 : magnetization and flux density\n", "import math\n", "#given data :\n", "mu0=4*math.pi*10**-7;\n", "H=10**4;# in A/m\n", "Xm=3.7*10**-3;# room temperature\n", "mu_r=1+Xm;\n", "B=mu0*mu_r*H;\n", "M=Xm*H;\n", "print \"{:.2e}\".format(B),\"= the flux density,B(Wb/m**2) \"\n", "print M,\"= magnetization,M(A/m) \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "1.26e-02 = the flux density,B(Wb/m**2) \n", "37.0 = magnetization,M(A/m) \n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex18.2a:pg-350" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Example 18.2.a : saturation magnetization\n", " \n", "#given data :\n", "mu_b=9.27*10**-24;# A.m**2\n", "p=8.9; # in g/cm**3\n", "Na=6.023*10**23;# avogadro's number\n", "A=58.71; # in g/mol\n", "n=((p*Na)/A)*10**6;\n", "Ms=0.60*mu_b*n;\n", "print \"{:.1e}\".format(Ms),\"= saturation magnetization,Ms(A/m) \"\n", "# the answe ris slightly different in textbook due to approximation" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "5.1e+05 = saturation magnetization,Ms(A/m) \n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex18.2b:pg-350" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Example 18.2.b : saturation flux density\n", "import math\n", "#given data :\n", "\n", "mu0=4*math.pi*10**-7;\n", "mu_b=9.27*10**-24;# A.m**2\n", "p=8.9; # in g/cm**3\n", "Na=6.023*10**23;# avogadro's number\n", "A=58.71; # in g/mol\n", "n=((p*Na)/A)*10**6;\n", "Ms=0.60*mu_b*n;\n", "Bs=mu0*Ms;\n", "print round(Bs,2),\"= saturation flux density,Bs(tesla) \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "0.64 = saturation flux density,Bs(tesla) \n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex18.3:pg-351" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Example 18.3 : magnetic moment\n", "import math \n", "#given data :\n", "\n", "mu0=4*math.pi*10**-7;\n", "mu_b=9.27*10**-24;# A.m**2\n", "p=8.9; # in g/cm**3\n", "Na=6.023*10**23;# avogadro's number\n", "A=58.71; # in g/mol\n", "n=((p*Na)/A)*10**6;\n", "Bs=0.65;#in Wb/m**2\n", "Ms=Bs/mu0;\n", "m_mu_b=Ms/n;\n", "print round(m_mu_b,26),\" is saturation magnetisation,m_mu_b(A.m**2) \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "5.67e-24 is saturation magnetisation,m_mu_b(A.m**2) \n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex18.4:pg-355" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Example 18.4 : power loss\n", " \n", "#given data :\n", "V=0.01;#in m**3\n", "f=50;# in Hz\n", "area=600;#in jm**-1\n", "Wh=area*V*f;\n", "print Wh,\"= power loss,Wh(watts) \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "300.0 = power loss,Wh(watts) \n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex18.5:pg-355" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Example 18.4 : los of energy\n", " \n", "#given data :\n", "mass=10.0;# in kg\n", "energy_loss=250.0;# in J/m**2\n", "#energy loss at the rate of 50 cycles/s\n", "E=energy_loss*50.0;# in J/m**3\n", "E_loss=E*3600.0;#in J/m**3\n", "D=7500.0;#density in kg/m**3\n", "Volume=mass/D;\n", "energy_loss_per_hour=E_loss/Volume;\n", "print \"{:.1e}\".format(energy_loss_per_hour),\"= energy_loss_per_hour(J/hour) \"\n", "\n", "# answer is incorrect in textbook" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "3.4e+10 = energy_loss_per_hour(J/hour) \n" ] } ], "prompt_number": 5 } ], "metadata": {} } ] }