{ "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" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.11" } }, "nbformat": 4, "nbformat_minor": 0 }