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| author | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
|---|---|---|
| committer | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
| commit | fffcc90da91b66ee607066d410b57f34024bd1de (patch) | |
| tree | 7b8011d61013305e0bf7794a275706abd1fdb0d3 /Basic_Electrical_Engineering/Chapter5.ipynb | |
| parent | 299711403e92ffa94a643fbd960c6f879639302c (diff) | |
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diff --git a/Basic_Electrical_Engineering/Chapter5.ipynb b/Basic_Electrical_Engineering/Chapter5.ipynb new file mode 100755 index 00000000..3a8210b4 --- /dev/null +++ b/Basic_Electrical_Engineering/Chapter5.ipynb @@ -0,0 +1,841 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5: ELECTROMAGNETISM"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1,Page number: 143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the current necessary to produce a magnetic field inside the solenoid.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "B=20e-03 #Magnitude of magnetic field inside the solenoid(in Tesla)\n",
+ "n=20e02 #Number of turns per cm in the long solenoid\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "abs_per=(4*pi)*(1e-07)\n",
+ "I=B/(abs_per*n)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The required current is %.2f A.\" %(I) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required current is 7.96 A.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2,Page number: 143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the magnetic field near the centre and the ends of a solenoid.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "abs_per=4*pi*(1e-07) #Absolute permeability of free space(in Henry per metre)\n",
+ "I=6.0 #Current carried by the solenoid(in Amperes)\n",
+ "l=50e-02 #Length of solenoid(in metres)\n",
+ "r=1.4e-02 #Radius of the lowest layer(in metres)\n",
+ "turns_per_layer=350 #Number of turns per layer\n",
+ "number_of_layers=4 #Number of layers\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "n=(turns_per_layer*number_of_layers)/l\n",
+ "B_centre=abs_per*n*I\n",
+ "B_end=(abs_per*n*I)/2\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The magnetic field inside the solenois is quite uniform,its strength near the centre on its axis and\" \n",
+ "print \" off its axis is the same and its magnitude is %e T.\" %(B_centre)\n",
+ "print \"(b)The magnetic field strength at the end of the solenoid is %e T.\" %(B_end)\n",
+ "print \"(c)The magnetic field far outside the solenoid near its axis is negligible,compared to the internal field.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The magnetic field inside the solenois is quite uniform,its strength near the centre on its axis and\n",
+ " off its axis is the same and its magnitude is 2.111150e-02 T.\n",
+ "(b)The magnetic field strength at the end of the solenoid is 1.055575e-02 T.\n",
+ "(c)The magnetic field far outside the solenoid near its axis is negligible,compared to the internal field.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3,Page number: 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the force between two long straight parallel wires.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "r=2 #Distance between the parallel current carrying wires(in metres) \n",
+ "abs_per=(4*pi)*(1e-07) #Absolute permeability of free space(in Henry per metre)\n",
+ "I1=80 #Current in the first wire(in Amperes)\n",
+ "I2=30 #Current in the second wire(in Amperes)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "F=(abs_per*I1*I2)/(2*pi*r)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The magnitude of the force between the two wires is %e N/m.\" %(F)\n",
+ "print \"Since the two currents are in the same direction,the force will be attractive.\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of the force between the two wires is 2.400000e-04 N/m.\n",
+ "Since the two currents are in the same direction,the force will be attractive.\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4,Page number: 148 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the force due to current flowing in two straight,parallel wires.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I1=4.0 #Current in the first wire(in Amperes)\n",
+ "I2=6.0 #Current in the second wire(in Amperes)\n",
+ "r=3e-02 #Distance between the parallel current carrying wires(in metres) \n",
+ "abs_per=(4*pi)*(1e-07) #Absolute permeability of free space(in Henry per metre)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "F=(abs_per*I1*I2)/(2*pi*r)\n",
+ "l=15e-02\n",
+ "F_net=F*l\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The net force on 15cm(=0.15 m) section of wire B near its centre is %e N.\" %(F_net)\n",
+ "print \"Since the currents are in opposite directions,the force F_net is repulsive.\"\n",
+ "print \"It means that its direction is normal to wire A away from it.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The net force on 15cm(=0.15 m) section of wire B near its centre is 2.400000e-05 N.\n",
+ "Since the currents are in opposite directions,the force F_net is repulsive.\n",
+ "It means that its direction is normal to wire A away from it.\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5,Page number: 154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the emf produced in a conductor placed in an uniform magnetic field.\"\"\"\n",
+ "\n",
+ "from math import sin,radians\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "B=0.5 #Magnitude of magnetic field(in Tesla)\n",
+ "l=20e-02 #Active length of the conductor(in metres) \n",
+ "v=5 #Velocity of the conductor(in metres per second)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "angle_a=0\n",
+ "e_a=B*l*v*sin(radians(angle_a))\n",
+ "angle_b=90\n",
+ "e_b=B*l*v*sin(radians(angle_b))\n",
+ "angle_c=30\n",
+ "e_c=B*l*v*sin(radians(angle_c))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The emf induced in the straight conductor when its motion is parallel to the magnetic field is %.2f V.\" %(e_a)\n",
+ "print \"(b)The emf induced in the straight conductor when its motion is perpendicular to the magnetic field is %.2f V.\" %(e_b)\n",
+ "print \"(c)The emf induced in the straight conductor when its motion is at an angle 30 degrees to the magnetic field is %.2f V.\" %(e_c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The emf induced in the straight conductor when its motion is parallel to the magnetic field is 0.00 V.\n",
+ "(b)The emf induced in the straight conductor when its motion is perpendicular to the magnetic field is 0.50 V.\n",
+ "(c)The emf induced in the straight conductor when its motion is at an angle 30 degrees to the magnetic field is 0.25 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6,Page number: 154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the emf generated between the wing-tips of an aeroplane.\"\"\"\n",
+ "\n",
+ "from math import sin,radians\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "l=52 #Wing span of the aeroplane(in metres)\n",
+ "B=38e-06 #Magnitude of magnetic field(in Tesla)\n",
+ "v=1100.0*(1000.0/3600.0) #Velocity of the aeroplane(in metres per second)\n",
+ "angle=90 #Angle between the magnetic field vector and the velocity vector(in degrees) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "e=B*l*v*sin(radians(angle))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The emf generated between the wing tips is %.5f V.\" %(e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The emf generated between the wing tips is 0.60378 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7,Page number: 154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the emf developed between the centre and the metallic ring.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "\"\"\" A=(1/2)*R*R*theta; where theta is the angle between the rod and the rod OP(in figure) at time t \"\"\"\n",
+ "B=1.0 #Magnitude of magnetic field(in Tesla)\n",
+ "R=1 #Radius of circular ring(in metres)\n",
+ "f=50 #Frequency of rotation(in revolutions per second) \n",
+ "\n",
+ "\"\"\" B=d(flux)/dt= d(B*A)/dt= (1/2)*B*R*R*d(theta)/dt= (1/2)*B*R*R*ang_freq \"\"\"\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "ang_freq=2*pi*f\n",
+ "e=(1.0/2)*B*R*R*ang_freq\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The emf developed between the centre and the metallic ring is %.2f V.\" %(e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The emf developed between the centre and the metallic ring is 157.08 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8,Page number: 155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the force needed to pull a rectangular loop placed in an uniform magnetic field.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "B=0.5 #Magnitude of magnetic field(in Tesla)\n",
+ "w=3e-02 #Width of the rectangular loop(in metres)\n",
+ "l=10e-02 #Length of the rectangular loop(in metres)\n",
+ "v=1e-02 #Velocity of the rectangular loop(in metre per second)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "e1=B*w*v\n",
+ "t1=l/v\n",
+ "e2=B*l*v\n",
+ "t2=w/v\n",
+ "R=1e-03\n",
+ "F_a=(pow((B*w),2)*v)/R\n",
+ "F_b=(pow((B*l),2)*v)/R\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The induced emf is %e V and the time for which the induced voltage lasts is %.2f secs.\" %(e1,t1)\n",
+ "print \"(b)The induced emf is %e V and the time for which the induced voltage lasts is %.2f secs.\" %(e2,t2)\n",
+ "print \"(c)Because of the gap,no current can flow.Hence there is no heat produced(or no I*I*R losses).\"\n",
+ "print \" If we neglect friction,no force is required to pull the coil.\"\n",
+ "print \"(d)The force required to pull the loop if it has no cut and has a resistance of 1 mill ohm is :\"\n",
+ "print \" For fig (a): F=%e N.\" %(F_a)\n",
+ "print \" For fig (b): F=%e N.\" %(F_b)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The induced emf is 1.500000e-04 V and the time for which the induced voltage lasts is 10.00 secs.\n",
+ "(b)The induced emf is 5.000000e-04 V and the time for which the induced voltage lasts is 3.00 secs.\n",
+ "(c)Because of the gap,no current can flow.Hence there is no heat produced(or no I*I*R losses).\n",
+ " If we neglect friction,no force is required to pull the coil.\n",
+ "(d)The force required to pull the loop if it has no cut and has a resistance of 1 mill ohm is :\n",
+ " For fig (a): F=2.250000e-03 N.\n",
+ " For fig (b): F=2.500000e-02 N.\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9,Page number: 156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the magnetic field due to a line caryying current.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I=90.0 #Current flowing through the line(in Amperes)\n",
+ "x=1.5 #Distance of point of observation from the line(in metres)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "abs_per=(4*pi)/(1e07)\n",
+ "B=(abs_per*I)/(2*pi*x)\n",
+ " \n",
+ "\n",
+ "#Result:\n",
+ "print \"The magnetic field due to the current at a point 1.5 m below the line is %e T.\" %(B)\n",
+ "print \"By applying the right-hand thumb rule,we find that the direction of the magnetic field is from north to south.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnetic field due to the current at a point 1.5 m below the line is 1.200000e-05 T.\n",
+ "By applying the right-hand thumb rule,we find that the direction of the magnetic field is from north to south.\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10,Page number: 156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the force per unit length on a current carrying wire.\"\"\"\n",
+ "\n",
+ "from math import sin,radians\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I=8.0 #Current flowing through the wire(in Amperes)\n",
+ "B=0.15 #Magnitude of uniform magnetic field(in Tesla)\n",
+ "angle=30 #Angle between the direction of current and the uniform magnetic field(in degrees)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Fu=I*B*sin(radians(angle))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The force per unit length of the wire is %.2f N/m.\" %(Fu)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The force per unit length of the wire is 0.60 N/m.\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.11,Page number: 156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the torque experienced by a square coil.\"\"\"\n",
+ "\n",
+ "from math import sin,radians\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I=12.0 #Current flowing through the square coil(in Amperes)\n",
+ "N=20.0 #Number of turns in the square coil\n",
+ "l=10e-02 #Length of each side of the square coil(in metres)\n",
+ "B=0.8 #Magnitude of uniform horizontal magnetic field(in Tesla)\n",
+ "angle=30.0 #Angle made by normal to the plane of the coil with the direction of horizontal magnetic field(in degrees)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "F=I*B*l*N\n",
+ "x=l*sin(radians(angle))\n",
+ "T=F*x\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The torque experienced by the coil is %.3f Nm.\" %(T)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The torque experienced by the coil is 0.960 Nm.\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12,Page number: 156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the net force on a rectangular loop.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I1=15.0 #Current carried by the rectangular loop(in Amperes)\n",
+ "I2=25.0 #Current carried by the straight conductor(in Amperes)\n",
+ "l=25e-02 #Length of the rectangular loop(in metres)\n",
+ "w=10e-02 #Width of the rectangular loop(in metres)\n",
+ "r=2e-02 #Separation between the nearer side of the loop and the conductor(in metres)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "abs_per=(4*pi)/(1e07)\n",
+ "\"\"\"Parts AD and BC do not experience any force,since these conductors are at right angles to the long straight conductor.\n",
+ " \n",
+ " The current I1 in AB and the current I2 in the straight conductor are in the same direction.\n",
+ " Hence,the force F_AB is attractive.\"\"\"\n",
+ "\n",
+ "F_AB=((abs_per*I1*I2)/(2*pi*r))*l\n",
+ "\n",
+ "\"\"\" The current I1 in CD and the current I2 in the straight conductor are in opposite directions.\n",
+ " Hence,the force F_CD is repulsive.\"\"\"\n",
+ "\n",
+ "F_CD=((abs_per*I1*I2)/(2*pi*(r+w)))*l\n",
+ "F_net=F_AB-F_CD\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The net force on the rectangular loop is %.5f mN.\" %(F_net*1000.0) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The net force on the rectangular loop is 0.78125 mN.\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13,Page number: 157"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the location of a point where the resultant magnetic field is zero.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I=10.0 #Current flowing through the wire(in Amperes)\n",
+ "B=2.0e-03 #Horizontal component of Earth's magnetic field(in Tesla)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "\"\"\" To make the resultant magnetic field zero,the magnetic filed produced by the long vertical wire must be equal and opposite \n",
+ " to the horizontal component of Earth's magnetic field.This is possible at a point P,west of the wire. \"\"\"\n",
+ "\n",
+ "abs_per=(4*pi)/(1e07)\n",
+ "x=(abs_per*I)/(2*pi*B)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The distance between the required point from the wire is %e m,to the west of the wire.\" %(x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance between the required point from the wire is 1.000000e-03 m,to the west of the wire.\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.14,Page number: 157"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the force per unit length between two current carrying wires.\"\"\"\n",
+ "\n",
+ "from math import pi \n",
+ "\n",
+ "#Variable Declaration:\n",
+ "I=300.0 #Current flowing through each wire(in Amperes)\n",
+ "r=1.5e-02 #Separation between the wires(in metres)\n",
+ "l=70e-02 #Length of each wire(in metres)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "abs_per=(4*pi)/(1e07)\n",
+ "\"\"\"Since the length of the wires is 70 cm and their separation is only 1.5 cm(i.e, l>>r),\n",
+ " we can conclude that for the given separation the two wires are infinitely long. \"\"\"\n",
+ "F=(abs_per*I*I)/(2*pi*r)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The force per unit length between the two wires is %.2f N/m.\" %(F)\n",
+ "print \"Since the currents in the two wires are in the opposite directions,the force between them will be repulsive.\"\"\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The force per unit length between the two wires is 1.20 N/m.\n",
+ "Since the currents in the two wires are in the opposite directions,the force between them will be repulsive.\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.15,Page number: 158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the counter torque to prevent a coil from turning.\"\"\"\n",
+ "\n",
+ "from math import pi,sin,radians\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N=30.0 #Number of turns in the circular coil\n",
+ "r=8e-02 #Radius of circular coil(in metres)\n",
+ "I=6.0 #Current flowing through the circular coil(in Amperes)\n",
+ "B=1.0 #Magnitude of uniform magnetic field(in Tesla)\n",
+ "angle=60 #Angle made by the field lines with the normal to the coil(in degrees)\n",
+ "\n",
+ " \n",
+ "#Calculations:\n",
+ "\"\"\"The counter torque required to prevent the coil from moving must be equal(and opposite) to the torque developed.\"\"\"\n",
+ "A=pi*r*r\n",
+ "tor=B*I*N*A*sin(radians(angle))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The counter torque that must be applied to prevent the coil from turning is %.2f Nm.\" %(tor) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The counter torque that must be applied to prevent the coil from turning is 3.13 Nm.\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.16,Page number: 158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the maximum voltage induced in a coil.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "A=300e-04 #Area of circular coil(in square-m)\n",
+ "N=25.0 #Number of turns in the circular coil\n",
+ "w=40.0 #Angluar frequency(in radians per second)\n",
+ "B=0.05 #Magnitude of uniform magnetic field(in Tesla)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Em=N*B*A*w\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The maximum value of emf induced in the coil is %.2f V.\" %(Em)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum value of emf induced in the coil is 1.50 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.17,Page number: 158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the induces emf in a circular conducting loop.\"\"\"\n",
+ "\n",
+ "from math import pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "B=0.02 #Magnitude of uniform magnetic field(in Tesla)\n",
+ "r=2e-02 #Radius of the circular loop(in metres)\n",
+ "rate=1.0e-03 #Rate of shrinking of the radius(in metre per second)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "\"\"\" flux=B*A; A=(pi*r*r);\n",
+ " e=d(flux)/dt; \n",
+ " \n",
+ " e=(B*pi*2*r)*(dr/dt); \"\"\"\n",
+ "\n",
+ "e=B*pi*2*r*rate\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The emf induced in the loop at an instant when the radius is 2 cm is %e V.\" %(e) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The emf induced in the loop at an instant when the radius is 2 cm is 2.513274e-06 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |