{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 12: Magnetic Properties of Materials" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.1, Page 603" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "H = 5e+3; # Coercivity of a bar magnet, A/m\n", "L = 0.1; # Length of the solenoid, m\n", "N = 50; # Turns in solenoid\n", "n = 500; # Turns/m\n", "\n", "#Calculations\n", "# Using the relation\n", "I = H/n; # where I is the current through the solenoid\n", "\n", "#Result\n", "print \"The current through the solenoid is = %2d A\"%I\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The current through the solenoid is = 10 A\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.2, Page 603" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "n = 500; # Number of turns wound per metre on the solenoid\n", "i = 0.5; # Current through the solenoid, A\n", "V = 1e-03; # Volume of iron rod, per metre cube\n", "mu_r = 1200; # Relative permeability of the iron\n", "\n", "#Calculations&Results\n", "H = n*i; # Magnetic intensity inside solenoid, ampere-turn per metre\n", "# As B = mu_o * (H + I) => I = B/mu_o - H \n", "# But B = mu_o * mu_r * H and solving for I \n", "I = (mu_r - 1) * H;\n", "print \"The Intensity of magnetisation inside the solenoid, I = %5.3e A/m\"%I\n", "M = I * V; # Magnetic moment of the rod, ampere metre square \n", "print \"The magnetic moment of the rod, M = %3d ampere metre square\"%M\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Intensity of magnetisation inside the solenoid, I = 2.998e+05 A/m\n", "The magnetic moment of the rod, M = 299 ampere metre square\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.3, Page 604" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "n = 300; # Number of turns wound per metre on the solenoid\n", "i = 0.5; # Current through the solenoid, A\n", "V = 1e-03; # Volume of iron rod, per metre cube\n", "mu_r = 100; # Relative permeability of the iron\n", "\n", "#Calculations&Results\n", "H = n*i; # Magnetic intensity inside solenoid, ampere-turn per metre\n", "# As, I = (B-mu_o* H)/mu_o\n", "#But, B= mu * H = mu_r * mu_o * H and I = (mu_r-1)* H\n", "I = (mu_r-1)*n*i; \n", "print \"The Intensity of magnetisation inside the solenoid, I = %5.3e A/m\"%I\n", "l = 0.2; #length of the rod,m\n", "r = 5e-3; #radius of the rod,m\n", "V = 1.57e-5; #V=%pi*r^2*l where the volume of the rod having radius r and length,m\n", "M = I * V ; # Magnetic moment of the rod, ampere metre square\n", "print \"The magnetic moment of the rod, M = %5.3f ampere metre square\"%M\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Intensity of magnetisation inside the solenoid, I = 1.485e+04 A/m\n", "The magnetic moment of the rod, M = 0.233 ampere metre square\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.4, Page 605" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "B = 0.0044; # Magnetic flux density, weber/meter square\n", "mu_o = 4*pi*1e-07; # Relative permeability of the material, henery/m \n", "I = 3300; # Magnetization of a magnetic material, A/m\n", "\n", "#Calculations&Results\n", "#B = mu_o*(I+H), solving for H \n", "H = (B/mu_o)- I; # Magnetizing force ,A/m\n", "print \"The magnetic intensity,H = %3d A/m\"%H\n", "# Relation between intensity of magnetization and relative permeability \n", "mu_r = (I/H)+1; #substitute the value of I and H \n", "print \"The relative permeability, mu_r = %5.2f\"%mu_r\n", "\n", "#Incorrect answers in the textbook" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The magnetic intensity,H = 201 A/m\n", "The relative permeability, mu_r = 17.38\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.5, Page 605" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "mu_o = 4*pi*1e-07; # Magnetic permeability of the free space, henery/m\n", "mu_r = 600;\n", "mu = mu_o*mu_r; # Magnetic permeability of the medium, henery/m\n", "n = 500; # Turns in a wire\n", "i = 0.3; # Current flows through a ring,amp\n", "r = 12e-02/2; # Mean radius of a ring, m\n", "\n", "#Calculations&Results\n", "B = mu_o*mu_r*n*i/(2*pi*r);\n", "print \"The magnetic flux density = %2.1f weber/meter-square\"%B\n", "H = B/mu; # Magnetic intensity, ampere-turns/m\n", "print \"The magnetic intensity = %5.1f ampere-turns/m\"%H\n", "mui = B -mu_o\n", "per = mui/i*100\n", "print \"The percentage magnetic flux density due to electronic loop currents = \",per,\"%\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The magnetic flux density = 0.3 weber/meter-square\n", "The magnetic intensity = 397.9 ampere-turns/m\n", "The percentage magnetic flux density due to electronic loop currents = 99.999581121 %\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.6, Page 606" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "M_i = 4.5; # Intial value of total dipole moment of the sample\n", "H_i = 0.84; # External magnetic field, tesla\n", "T_i = 4.2; # Cooling temerature of the sample, K\n", "H_f = 0.98; # External magnetic field, tesla\n", "T_f = 2.8; # Cooling temerature of the sample, K\n", "\n", "#Calculations\n", "# According to the curie's law, Mf/Mi = (Hf/Hi)*(Ti/Tf)\n", "M_f = M_i*H_f/H_i*T_i/T_f;\n", "\n", "#Result\n", "print \"The total dipole moment of the sample = %5.3f joule/tesla\"%M_f\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The total dipole moment of the sample = 7.875 joule/tesla\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.7, Page 606" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "mu_o = 4*pi*1e-07; # Magnetic permeability of free space, henry/m\n", "n = 1e+29; # Number density of atoms of iron, per metre cube \n", "p_m = 1.8e-23; # Magnetic moment of each atom, ampere-metre square \n", "k_B = 1.38e-23; # Boltzmann constant, J/K\n", "B = 0.1; # Magnetic flux density, weber per metre square\n", "T = 300; # Absolute room temperature, K\n", "l = 10e-02; # Length of the iron bar, m\n", "a = 1e-04; # Area of cross-section of the iron bar, metre square\n", "\n", "#Calculations&Results\n", "V = l*a; # Voluem of the iron bar, metre cube\n", "chi = n*p_m**2*mu_o/(3*k_B*T);\n", "print \"The paramagnetic susceptibility of a material = %5.3e\"%chi\n", "pm_mean = p_m**2*B/(3*k_B*T); # Mean dipole moment of an iron atom, ampere metre-square\n", "P_m = n*pm_mean; # Dipole moment of the bar, ampere metre-square\n", "I = n*p_m; # Magnetization of the bar in one domain, ampere/metre\n", "M = I*V; # Magnetic moment of the bar, ampere metre-square\n", "print \"The dipole moment of the bar = %5.3e ampere metre-square\"%P_m\n", "print \"The magnetization of the bar in one domain = %3.1e ampere/metre\"%I\n", "print \"The magnetic moment of the bar = %2d ampere metre-square\"%M\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The paramagnetic susceptibility of a material = 3.278e-03\n", "The dipole moment of the bar = 2.609e+02 ampere metre-square\n", "The magnetization of the bar in one domain = 1.8e+06 ampere/metre\n", "The magnetic moment of the bar = 18 ampere metre-square\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.8, Page 607" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "A = 500; # Area of the B-H loop, joule per metre cube\n", "n = 50; # Total number of cycles, Hz\n", "m = 9; # Mass of the core, kg\n", "d = 7.5e+3; # Density of the core, kg/metre cube\n", "t = 3600; # Time during which the energy loss takes place, s\n", "\n", "#Calculations\n", "V = m/d; # Volume of the core, metre cube\n", "E = n*V*A*t; # Hystersis loss of energy per hour, joule\n", "\n", "#Result\n", "print \"The hystersis loss per hour = %5.2eJ\"%E\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The hystersis loss per hour = 1.08e+05J\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.9, Page 607" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "n = 50; # Total number of cycles per sec, Hz\n", "V = 1e-03; # Volume of the specimen, metre cube\n", "t = 1; # Time during which the loss occurs, s\n", "A = 0.25e+03; # Area of B-H loop, joule per metre cube\n", "\n", "#Calculations\n", "E = n*V*A*t; # Energy loss due to hysteresis, J/s\n", "\n", "#Result\n", "print \"The hystersis loss = %4.1f J/s\"%E\n", "#incorrect answer in the textbook" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The hystersis loss = 12.5 J/s\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.10, Page 608" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "e = 1.6e-19; # Charge on anlectron, C\n", "m = 9.1e-31; # Mass of the electron, kg\n", "r = 5.1e-11; # Radius of the electronic orbit, m\n", "B = 2.0; # Applied magnetic field, weber per metre-square\n", "\n", "#Calculations\n", "delta_pm = e**2*r**2*B/(4*m);\n", "\n", "#Result\n", "print \"The change in the magnetic dipole moment of the electron = %3.1e A-metre square\"%delta_pm\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The change in the magnetic dipole moment of the electron = 3.7e-29 A-metre square\n" ] } ], "prompt_number": 21 } ], "metadata": {} } ] }