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diff --git a/Materials_science_and_engineering_an_introduction/CH20.ipynb b/Materials_science_and_engineering_an_introduction/CH20.ipynb new file mode 100644 index 00000000..c410a6ac --- /dev/null +++ b/Materials_science_and_engineering_an_introduction/CH20.ipynb @@ -0,0 +1,157 @@ +{
+ "metadata": {
+ "name": "CH20"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 20 : Magnetic Properties"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 20.1 Page No 84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Calculation of saturation magnetisation and flux density for Nickel\n",
+ "\n",
+ "#Given\n",
+ "b_m=9.27*10**-24 #ampere*m**2 (Bohr Magneton)\n",
+ "Na=6.023*10**23 #atoms/mol (Avogadro's No.)\n",
+ "d=8.9*10**6 #g/m**3 (density)\n",
+ "uo=4*math.pi*10**-7 #Permitivity of free space\n",
+ "A=58.71 #g/mol (Atomic weigth of Nickel)\n",
+ "N=d*Na/A #No. of atoms per cubic meter\n",
+ "\n",
+ "#Calculation\n",
+ "# M is saturation magnetisation\n",
+ "M=0.6*b_m*N #0.6= Bohr Magneton/atom\n",
+ "#B = Saturation Flux Density\n",
+ "B=uo*M\n",
+ "\n",
+ "#Result\n",
+ "print\"Saturation Magnetisation is \",M,\"A/m\"\n",
+ "print\"Saturation Flux Density is \",round(B,2),\"Tesla\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Saturation Magnetisation is 507834.0 A/m\n",
+ "Saturation Flux Density is 0.64 Tesla\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 20.2 Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Calculation of saturation magnetisation of Fe3O4\n",
+ "\n",
+ "#Given\n",
+ "a=0.839*10**-9 #a is edge length in m\n",
+ "b_m=9.27*10**-24 #ampere*m**2 (Bohr Magneton)\n",
+ "nb=8*4 #8 is no. of Fe++ ions per unit cell\n",
+ " #4 is Bohr magnetons per Fe++ ion\n",
+ "#Calculation\n",
+ "M=nb*b_m/a**3 #M is Saturation magnetisation\n",
+ "\n",
+ "print\"Saturation Magnetisation is \",round(M,0),\"A/m\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Saturation Magnetisation is 502278.0 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Design Example 20.1 Page No 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Design Example 20.1: Designing a cubic mixed-ferrite magnetic material\n",
+ "\n",
+ "#Given\n",
+ "Ms_Fe=5.25*10**5 #Required saturation Magnetisation\n",
+ "b_m=9.27*10**-24 #ampere*m**2 (Bohr Magneton)\n",
+ "a=0.839*10**-9 #a is edge length in m\n",
+ "M=5*10**5 #From previous question result\n",
+ "\n",
+ "#Calculation\n",
+ "nb=Ms_Fe*a**3/b_m\n",
+ "i=8 # No of Divalent ions per unit cell\n",
+ "j=4 #4 is Bohr magnetons per Mn++ ion\n",
+ "n=nb/(i)-j \n",
+ " \n",
+ "#Result\n",
+ "print\"Replacing percent of Fe++ with Mn++ would produce the required saturation magnetisation\",round(n*100,2)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Replacing percent of Fe++ with Mn++ would produce the required saturation magnetisation 18.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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