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{
"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": [
"\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",
"M=0.6*b_m*N #0.6= Bohr Magneton/atom\n",
"B=uo*M\n",
"\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": [
"\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",
"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": [
"\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",
"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",
"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": {}
}
]
}
|
