{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 9: Electromagnetic induction
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 102
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the current flowing in the conductor when \n",
"#(a) its ends are open-circuited, \n",
"#(b) its ends are connected to a load of 20 ohm resistance.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"l = 0.3;# in m\n",
"v = 4;# in m/s\n",
"B = 1.25;# in Tesla\n",
"R = 20;# in ohms\n",
"u0 = 4*math.pi*1E-7;\n",
"\n",
"#calculation:\n",
"E = B*l*v\n",
"I2 = E/R\n",
"\n",
"#Results\n",
"print \"\\n\\nResult\\n\\n\"\n",
"print \"\\n (a)If the ends of the conductor are open circuited \"\n",
"print \"no current will flow even though \",E,\" V has been induced.\\n\"\n",
"print \"\\n (b)From Ohms law, I = \",I2,\" Ampere\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
"Result\n",
"\n",
"\n",
"\n",
" (a)If the ends of the conductor are open circuited no current will flow even though 1.5 V has been induced.\n",
"\n",
"\n",
" (b)From Ohms law, I = 0.075 Ampere"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2, page no. 103
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"l = 0.075;# in m\n",
"E = 9;# in Volts\n",
"B = 0.6;# in Tesla\n",
"R = 20;# in ohms\n",
"u0 = 4*math.pi*1E-7;\n",
"\n",
"#calculation:\n",
"v = E/(B*l)\n",
"\n",
"#Results\n",
"print \"\\n\\nResult\\n\\n\"\n",
"print \"\\n velocity v = \",v,\" m/s\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
"Result\n",
"\n",
"\n",
"\n",
" velocity v = 200.0 m/s"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3, page no. 103
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#find the magnitude of the induced e.m.f. in each case.\n",
"from __future__ import division\n",
"import math\n",
"\n",
"#initializing the variables:\n",
"l = 0.02;# in m\n",
"b = 0.02;# in m\n",
"v = 15;# in m/s\n",
"R = 20;# in ohms\n",
"Phi = 5E-6;# in Wb\n",
"u0 = 4*math.pi*1E-7;\n",
"a1 = 90;# in degrees\n",
"a2 = 60;# in degrees\n",
"a3 = 30;# in degrees\n",
"\n",
"#calculation:\n",
"A = l*b\n",
"B = Phi/A\n",
"E90 = B*l*v*math.sin(a1*math.pi/180)\n",
"E60 = B*l*v*math.sin(a2*math.pi/180)\n",
"E30 = B*l*v*math.sin(a3*math.pi/180)\n",
"\n",
"#Results\n",
"print \"\\n\\nResult\\n\\n\"\n",
"print \"\\n Induced e.m.f. at angles 90\u00b0, 60\u00b0, 30\u00b0 are \",(E90/1E-3),\" V, \",round((E60/1E-3),2),\" V, \"\n",
"print \"(E30/1E-3),\" V respectively\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
"Result\n",
"\n",
"\n",
"\n",
" Induced e.m.f. at angles 90\u00c2\u00b0, 60\u00c2\u00b0, 30\u00c2\u00b0 are 3.75 V, 3.25 V, 1.875 V respectively"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 103
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#determine the e.m.f. induced between its wing tips\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"s = 36;# in m\n",
"v = 400;# in km/h\n",
"u0 = 4*math.pi*1E-7;\n",
"B = 40E-6;# in Tesla\n",
"\n",
"#calculation:\n",
"v0 = v*5/18\n",
"E = B*s*v0\n",
"\n",
"#Results\n",
"print \"\\n\\nResult\\n\\n\"\n",
"print \"\\n Induced e.m.f. = \",E,\" V\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
"Result\n",
"\n",
"\n",
"\n",
" Induced e.m.f. = 0.16 V"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 105
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine the e.m.f. induced in a coil\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"N = 200;# no. of turns\n",
"dt = 0.050;# change of time in sec\n",
"u0 = 4*math.pi*1E-7;\n",
"dPhi = 0.025;# change of flux in Wb\n",
"\n",
"#calculation:\n",
"E = -1*N*dPhi/dt\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\nResult\\n\\n\"\n",
"print \"\\n Induced e.m.f. = \",E,\" V\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
"Result\n",
"\n",
"\n",
"\n",
" Induced e.m.f. = -100.0 V"
]
}
],
"prompt_number": 5
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 105
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Find the average e.m.f. induced.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"N = 150;# no. of turns\n",
"dt = 0.040;# change of time in sec\n",
"u0 = 4*math.pi*1E-7;\n",
"dPhi = 800E-6;# change of flux in Wb\n",
"\n",
"#calculation:\n",
"#Since the flux reverses, the flux changes from C400 \u03bcWb to \u0003400 \u03bcWb, a total change of flux of 800 \u03bcWb\n",
"E = -1*N*dPhi/dt\n",
"\n",
"#Results\n",
"print \"\\n\\nResult\\n\\n\"\n",
"print \"\\n Induced e.m.f. = \",E,\" V\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
"Result\n",
"\n",
"\n",
"\n",
" Induced e.m.f. = -3.0 V"
]
}
],
"prompt_number": 6
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8, page no. 105
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the e.m.f. induced in a coil\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"L = 12;# in Henry\n",
"u0 = 4*math.pi*1E-7;\n",
"dIdt = 4;# change of current with change in time in A/s\n",
"\n",
"#calculation:\n",
"E = -1*L*dIdt\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\nResult\\n\\n\"\n",
"print \"\\n Induced e.m.f. = \",E,\" V\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
"Result\n",
"\n",
"\n",
"\n",
" Induced e.m.f. = -48 V"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 9, page no. 106
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#find inductance of the coil.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"E = 1500;# in Volts\n",
"dt = 0.008;# Change of time in sec\n",
"dI = 4;# change of current in A/s\n",
"\n",
"#calculation:\n",
"L = abs(E)*dt/dI\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n Inductance L= \",L,\" H\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" Inductance L= 3.0 H"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 10, page no. 107
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#How much energy is stored in the magnetic field of the inductor?\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"L = 8;# in Henry\n",
"I = 3;# in Amperes\n",
"\n",
"#calculation:\n",
"W = L*I*I/2\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n Energy stored, W = \",W,\" J\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" Energy stored, W = 36.0 J"
]
}
],
"prompt_number": 9
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 11, page no. 107
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the coil inductance\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"I = 4;# in Amperes\n",
"N = 800;#turns\n",
"Phi = 0.005;# in Wb\n",
"\n",
"#calculation:\n",
"L = N*Phi/I\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n Inductance L = \",L,\" H\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" Inductance L = 1.0 H"
]
}
],
"prompt_number": 10
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 12, page no. 107
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate (a) the inductance of the coil, \n",
"#(b) the energy stored in the magnetic field, and \n",
"#(c) the average e.m.f. induced if the current falls to zero in 150 ms.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"I1 = 3;# in Amperes\n",
"I2 = 0;# in Amperes\n",
"dt = 0.150;# in secs\n",
"N = 1500;#turns\n",
"Phi = 0.025;# in Wb\n",
"\n",
"#calculation:\n",
"L = N*Phi/I1\n",
"W = L*I1*I1/2\n",
"dI = I1 - I2\n",
"E = -1*L*dI/dt\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)Inductance L = \",L,\" H\\n\"\n",
"print \"\\n (b)energy stored W = \",W,\" J\\n\"\n",
"print \"\\n (c)e.m.f. induced = \",E,\" V\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)Inductance L = 12.5 H\n",
"\n",
"\n",
" (b)energy stored W = 56.25 J\n",
"\n",
"\n",
" (c)e.m.f. induced = -250.0 V"
]
}
],
"prompt_number": 11
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 13, page no. 108
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the flux linking the coil and the e.m.f. induced in the coil when the current collapses to zero in 20 ms\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"I1 = 2;# in Amperes\n",
"I2 = 0;# in Amperes\n",
"dt = 0.020;# in secs\n",
"N = 750;#turns\n",
"L = 3;# in Henry\n",
"\n",
"#calculation:\n",
"Phi = L*I1/N\n",
"dI = I1 - I2\n",
"E = -1*L*dI/dt\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)Flux = \",Phi,\" Wb\\n\"\n",
"print \"\\n (b)e.m.f. induced = \",E,\" V\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)Flux = 0.008 Wb\n",
"\n",
"\n",
" (b)e.m.f. induced = -300.0 V"
]
}
],
"prompt_number": 12
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14, page no. 108
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the mutual inductance between two coils\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"dI1dt = 200;# change of current with change in time in A/s\n",
"N = 2;# no. of coils\n",
"E2 = 1.5;# in Volts\n",
"\n",
"#calculation:\n",
"M = abs(E2)/dI1dt\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n mutual inductance, M = \", M,\" H\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" mutual inductance, M = 0.0075 H"
]
}
],
"prompt_number": 13
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 15, page no. 109
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the steady rate of change of current in one coil to induce an e.m.f. of 0.72 V in the other.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"M = 0.018;# in Henry\n",
"N = 2;# no. of coils\n",
"E2 = 0.72;# in Volts\n",
"\n",
"#calculation:\n",
"dI1dt = abs(E2)/M\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n rate of change of current dI1/dt = \", dI1dt,\" A/s\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" rate of change of current dI1/dt = 40.0 A/s"
]
}
],
"prompt_number": 14
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16, page no. 109
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate (a) the average induced e.m.f. in the second coil, \n",
"#(b) the change of flux linked with the second coil if it is wound with 500 turns.\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"M = 0.2;# in Henry\n",
"I1 = 10;# in Amperes\n",
"I2 = 4;# in Amperes\n",
"dt = 0.010;# in secs\n",
"N = 500;# turns\n",
"\n",
"#calculation:\n",
"dI1dt = (I1 -I2)/dt \n",
"E2 = -1*dI1dt*M\n",
"dPhi = abs(E2)*dt/N\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)Induced e.m.f. E2 = \", E2,\" V\\n\"\n",
"print \"\\n (b)change of flux = \", dPhi,\" Wb\\n\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)Induced e.m.f. E2 = -120.0 V\n",
"\n",
"\n",
" (b)change of flux = 0.0024 Wb"
]
}
],
"prompt_number": 15
}
],
"metadata": {}
}
]
}