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+{
+"cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 31: Induction and Inductance"
+ ]
+ },
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.1: Sample_Problem_1.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"i = 1.5 //in A\n",
+"D = 3.2*10^-2 //in meter\n",
+"N = 220/10^-2 //in turns/m\n",
+"n = 130\n",
+"d = 2.1*10^-2 //in meter\n",
+"deltaT = 25*10^-3 //in s\n",
+"uo = 4*%pi*10^-7 //in SI unit\n",
+"\n",
+"//Sample Problem 31-1\n",
+"printf('**Sample Problem 31-1**\n')\n",
+"A = %pi*(d/2)^2\n",
+"deltaPhi = uo*N*i*A\n",
+"E = n*deltaPhi/deltaT\n",
+"printf('The emf induced is equal to %eV', E)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.2: Sample_Problem_2.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"r = 0.20 //in meter\n",
+"t = poly(0, 't')\n",
+"B = 4.0*t^2 + 2.0*t + 3.0\n",
+"E = 2.0 //in Volts\n",
+"R = 2 //in Ohm\n",
+"\n",
+"//Sample Problem 31-2a\n",
+"printf('**Sample Problem 31-2a**\n')\n",
+"t = 10 //in sec\n",
+"flux = B*%pi*r^2/2\n",
+"Et = derivat(flux)\n",
+"E1 = horner(Et, t)\n",
+"printf('The Emf induced is equal to %fV\n', E1)\n",
+"\n",
+"//Sample Problem 31-2b\n",
+"printf('\n**Sample Problem 31-2b**\n')\n",
+"I = (E1-E)/R\n",
+"printf('The induced current is equal to %fA', I)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.3: Sample_Problem_3.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"t = poly(0, 't')\n",
+"//B = 4*t^2*x^2\n",
+"W = 3.0 //in meter\n",
+"H = 2.0 //in meter\n",
+"t1 = 0.10 //in sec\n",
+"\n",
+"//Sample Problem 31-3\n",
+"printf('**Sample Problem 31-3**\n')\n",
+"flux = integrate('4*x^2*H', 'x', 0, W)\n",
+"E = derivat(flux*t^2)\n",
+"E1 = horner(E, t1)\n",
+"printf('The induced emf is equal to %fV', E1)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.4: Sample_Problem_4.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"R = 8.5*10^-2 //in meter\n",
+"Rb = 0.13 //in T/s\n",
+"r = 5.2*10^-2 //in meter\n",
+"\n",
+"//Sample Problem 31-4a\n",
+"printf('**Sample Problem 31-4a**\n')\n",
+"//Using Faraday's law\n",
+"Rf = Rb*%pi*r^2\n",
+"E = Rf/(2*%pi*r)\n",
+"printf('The induced electric field is equal to %eV/m\n', E)\n",
+"\n",
+"//Sample Problem 31-4b\n",
+"printf('\n**Sample Problem 31-4b**\n')\n",
+"r = 12.5*10^-2 //in meter\n",
+"Rf = Rb*%pi*R^2\n",
+"E = Rf/(2*%pi*r)\n",
+"printf('The induced electric field is equal to %eV/m', E)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.5: Sample_Problem_5.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"R = 9.0 //in Ohm\n",
+"L = 2*10^-3 //in Henery\n",
+"E = 18 //in Volts\n",
+"\n",
+"//Sample Problem 31-5a\n",
+"printf('**Sample Problem 31-5a**\n')\n",
+"//As soon as switch is closed the inductor will act like current barrier\n",
+"Io = E/R\n",
+"printf('The current as soon as qwitch is closed is equal to %1.2fA\n', Io)\n",
+"\n",
+"//Sample Problem 31-5b\n",
+"printf('\n**Sample Problem 31-5b**\n')\n",
+"//After long time inductor will act like short circuit\n",
+"Req = R/3\n",
+"If = E/(R/3)\n",
+"printf('The current through the battery after long time will be %1.2fA', If)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.6: Sample_Problem_6.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"L = 53*10^-3 //in H\n",
+"R = 0.37 //in Ohm\n",
+"\n",
+"//Sample Problem 31-6\n",
+"printf('**Sample Problem 31-6**\n')\n",
+"//i = io(1-e^(t/T))\n",
+"//ln2 = t/T\n",
+"T = L/R\n",
+"t = T*log(2)\n",
+"printf('The time taken to rach the current to half of its stedy state value is %fs', t)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.7: Sample_Problem_7.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"L = 53*10^-3 //in H\n",
+"R = 0.35 //in Ohm\n",
+"V = 12 //in Volts\n",
+"\n",
+"//Sample Problem 31-7a\n",
+"printf('**Sample Problem 31-7a**\n')\n",
+"i = V/R //in steady state\n",
+"E = 1/2*L*i^2\n",
+"printf('The Energy stored in the inductor in steady state is %fJ\n', E)\n",
+"\n",
+"//Sample Problem 31-7b\n",
+"printf('\n**Sample Problem 31-7b**\n')\n",
+"Et = E/2\n",
+"//hence It = Io/sqrt(2)\n",
+"f = log(1-1/sqrt(2)) //the number of times of time constant\n",
+"printf('After t=%1.1fT, the energy stored in the inductor will be half of tis steady state value', f)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.8: Sample_Problem_8.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"a = 1.2*10^-3 //in meter\n",
+"b = 3.5*10^-3 //in meter\n",
+"i = 2.7 //in Amp\n",
+"l = 1 //in meter(say)\n",
+"uo = 4*%pi*10^-7\n",
+"\n",
+"//Sample Problem 31-8\n",
+"printf('**Sample Problem 31-8**\n')\n",
+"B = uo*i/(2*%pi) //divided by r\n",
+"Ul = B^2/(2*uo) //divided by r^2\n",
+"//Energy as a funtion of r\n",
+"U = Ul*2*%pi*l //divided by r by r\n",
+"Energy = integrate('U/r', 'r', a, b)\n",
+"printf('Energy per unit length is equal to %1.2eJ/m', Energy)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 31.9: Sample_Problem_9.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Given that\n",
+"N1 = 1200 //turns\n",
+"N2 = N1\n",
+"R2 = 1.1*10^-2 //in meter\n",
+"R1 = 15*10^-2 //in meter\n",
+"uo = 4*%pi*10^-7\n",
+"\n",
+"//Sample Problem 31-9\n",
+"printf('**Sample Problem 31-9**\n')\n",
+"//let's assume\n",
+"i = 1 //in amp\n",
+"B1 = uo*N1*i/(2*R1)\n",
+"phi2 = B1*%pi*R2^2*N2\n",
+"M = phi2/i\n",
+"printf('The mutual inductance of the two coil is equal to %1.2eH', M)"
+ ]
+ }
+],
+"metadata": {
+ "kernelspec": {
+ "display_name": "Scilab",
+ "language": "scilab",
+ "name": "scilab"
+ },
+ "language_info": {
+ "file_extension": ".sce",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "scilab",
+ "version": "0.7.1"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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