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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 18:Magnetism"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex18.1:pg-901"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The force on the wire is F= 0.000958363 N\n"
]
}
],
"source": [
" #Example 18_1\n",
"import math \n",
" \n",
"#To find the force on the wire\n",
"b=2*10**-4 #Units in T\n",
"i=20 #Units in A\n",
"l=0.3 #Units in meters\n",
"theta=53 #Units in degrees\n",
"thetaa=math.sin(theta*math.pi/180) #Units in Radians \n",
"f=b*i*l*thetaa #Units in N\n",
"print \"The force on the wire is F=\",round(f,9),\" N\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex18.2:pg-902"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The magnitude of magnetic field is B= 0.2609 T\n"
]
}
],
"source": [
" #Example 18_2\n",
" \n",
" \n",
" #To find the magnitude of the magnetic field\n",
"m=1.67*10**-27 #Units in Kg\n",
"v=10**6 #Units in meters/sec\n",
"r=4*10**-2 #Units in Meters\n",
"q=1.6*10**-19 #Units in C or eV\n",
"b=(m*v)/(r*q) #Units in T\n",
"print \"The magnitude of magnetic field is B=\",round(b,4),\" T\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex18.3:pg-903"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The magnetic field exerts a force of q*v*B upwards on the particle.\n",
"The particle doesnot deflect because the two forces are equal\n",
"Hence v=(E/B)\n",
"A particle with this speed will pass through the region of the crossfields and undeflected\n"
]
}
],
"source": [
" #Example 18_3\n",
" \n",
" \n",
"#To show that the particles does not deflect from its straight line path\n",
"print \"The magnetic field exerts a force of q*v*B upwards on the particle.\\nThe particle doesnot deflect because the two forces are equal\\nHence v=(E/B)\\nA particle with this speed will pass through the region of the crossfields and undeflected\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex18.4:pg-903"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of B is= 1.2 G\n"
]
}
],
"source": [
" #Example 18_4\n",
" \n",
" \n",
"#To calculate the value of B at a radial distance of 5 cm\n",
"u=4*math.pi*10**-7 #Units in T m/A\n",
"i=30 #Units in A\n",
"r=0.05 #Units in Meters\n",
"b=(u*i)/(2*math.pi*r) #Units in T\n",
"b=b*10**4 #Units in G\n",
"print \"The value of B is=\",round(b,2),\" G\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex18.5:pg-904"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The magnetic moment of Hydrogen atom is=\n",
"9.319e-24\n",
"A meters**2\n"
]
}
],
"source": [
" #Example 18_5\n",
" \n",
" \n",
"#To find the magnetic moment of hydrogen atom\n",
"r=0.53*10**-10 #Units in meters\n",
"a=math.pi*r**2 #Units in meters**2\n",
"q=1.6*10**-19 #Units in C\n",
"f=6.6*10**15 #Units in sec**-1\n",
"i=q*f #Units in A\n",
"u=i*a #Units in A meter**2\n",
"print \"The magnetic moment of Hydrogen atom is=\"\n",
"print round(u,27)\n",
"print \"A meters**2\"\n"
]
}
],
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"display_name": "Python 2",
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},
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"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
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|
