.MCAD 304020000 1 74 133 0
.CMD PLOTFORMAT
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.CMD FORMAT  rd=d ct=10 im=i et=3 zt=300 pr=3 mass length time charge temperature tr=0 vm=0
.CMD SET ORIGIN 0
.CMD SET TOL 0.001000000000000
.CMD SET PRNCOLWIDTH 8
.CMD SET PRNPRECISION 4
.CMD PRINT_SETUP 1.200000 1.218750 1.200000 1.200000 0
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.CMD UNITS U=1
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.TXT 2 1 43 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}{\f1\fcharset0\fnil Times New Roman;}}
\plain\cf1\fs20 \pard {\cf2\f1\b COPYRIGHT MIRON KAUFMAN 1997}}
.TXT 4 20 42 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}{\f1\fcharset0\fnil Britannic Bold;}}
\plain\cf1\fs20 \pard {\cf2\f1\fs36\b Physics B PHY232/235\par 
Computer Project\par Laboratory #3a}}
.TXT 11 -20 45 0 0
Cg a71.000000,71.000000,491
{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 We 
study the motion of a particle of charge q and mass m in a magnetic 
field }{\cf2\b B}{\cf2  and in an electric field }{\cf2\b E}{\cf2 .  
The force acting on the particle is given by the Lorentz formula: }{
\cf2\b F}{\cf2  = q}{\cf2\b E}{\cf2  + q}{\cf2\b v}{\cf2 x}{\cf2\b B }{
\cf2 where}{\cf2\b  v }{\cf2 is the velocity vector. For small enough 
velocities, v << c, we can use Newtonian mechanics to describe the 
motion of the charged particle: m}{\cf2\b a}{\cf2  = }{\cf2\b F}{\cf2  
.  Lorentz formula and Newton's second law give: d}{\cf2\b v}{\cf2 /dt 
= (q/m)}{\cf2\b E}{\cf2  + (q/m)}{\cf2\b v}{\cf2 x}{\cf2\b B. \par }{
\cf2 First we input the mass and charge of the particle, proton for example.}}
.EQN 13 2 3 0 0
{0:q}NAME:1.6*(10)^(-19)
.EQN 0 18 56 0 0
{0:m}NAME:1.67*(10)^(-27)
.TXT 6 -20 119 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 Next 
we input the magnetic field }{\cf2\b B}{\cf2  and the electric field }{
\cf2\b E.  \par }{\cf2 In this computation we will study the case of 
no electric field}{\cf2\b  E}{\cf2  = 0.}}
.EQN 8 0 51 0 0
{0:B}NAME:({3,1}ö0.0001ö0ö0)
.EQN 0 12 85 0 0
{0:E}NAME:({3,1}ö0ö0ö0)
.TXT 6 -12 57 0 0
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\fonttbl{\f0\fcharset0\fnil Arial;}{\f1\fcharset2\fnil Symbol;}}\plain
\cf1\fs20 \pard {\cf2 The Larmor (cyclotron) angular velocity }{\cf2\f1 w}{
\cf2  and period are T:}}
.EQN 5 1 46 0 0
{0:\w}NAME:({0:q}NAME*|({0:B}NAME))/({0:m}NAME)
.EQN 0 15 59 0 0
{0:T}NAME:(2*{0:\p}NAME)/({0:\w}NAME)
.EQN 5 -1 111 0 0
{0:T}NAME={0}?_n_u_l_l_
.TXT 5 -15 61 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 We 
will compute the position and velocity of the particle 1000 times 
during a period.}}
.EQN 5 2 60 0 0
{0:\Dt}NAME:({0:T}NAME)/(1000)
.EQN 4 0 8 0 0
{0:n}NAME:0;5000
.EQN 3 0 6 0 0
({0:t}NAME)[({0:n}NAME):{0:n}NAME*{0:\Dt}NAME
.TXT 5 0 63 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 The 
initial velocity is }{\cf2\b v}{\cf2\b\dn 0}{\cf2  and the initial 
position is}{\cf2\b  r}{\cf2\b\dn 0}{\cf2 .}}
.EQN 6 0 64 0 0
({0:v}NAME)[(0):({3,1}ö(10)^(5)ö0ö(10)^(6))
.EQN 0 13 67 0 0
({0:r}NAME)[(0):({3,1}ö0ö0ö0)
.TXT 15 -15 110 0 0
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\fonttbl{\f0\fcharset0\fnil Arial;}{\f1\fcharset0\fnil Times New Roman;}}
\plain\cf1\fs20 \pard {\cf2\f1\b COPYRIGHT MIRON KAUFMAN 1997}}
.TXT 5 2 81 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}{\f1\fcharset2\fnil Symbol;}}\plain
\cf1\fs20 \pard {\cf2 We integrate numerically the equation of motion }{
\cf2\b a}{\cf2  = }{\cf2\b F}{\cf2 /m by using }{\cf2\b a}{\cf2  = d}{
\cf2\b v}{\cf2 /dt = (}{\cf2\b v}{\cf2\b\dn n+1}{\cf2\b  }{\cf2 - }{
\cf2\b v}{\cf2\b\dn n}{\cf2 )/}{\cf2\f1 D}{\cf2 t and }{\cf2\b  v}{\cf2  
= (}{\cf2\b r}{\cf2\b\dn n+1}{\cf2  - }{\cf2\b r}{\cf2\b\dn n}{\cf2 )/}{
\cf2\f1 D}{\cf2 t.  This implementation is called the Euler method and 
its accuracy depends on the smallness of the time interval:  }{\cf2\f1 D}{
\cf2 t << T. }}
.EQN 7 1 70 0 0
({0:v}NAME)[({0:n}NAME+1):({0:v}NAME)[({0:n}NAME)+{0:\Dt}NAME*((({0:q}NAME)/({0:m}NAME))*({0:v}NAME)[({0:n}NAME){50}{0:B}NAME+({0:q}NAME)/({0:m}NAME)*{0:E}NAME)
.EQN 4 0 71 0 0
({0:r}NAME)[({0:n}NAME+1):({0:r}NAME)[({0:n}NAME)+{0:\Dt}NAME*({0:v}NAME)[({0:n}NAME)
.TXT 5 -2 82 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 To 
graph the trajectory,  I use the 3d scatter plot available in Mathcad 6.}}
.EQN 10 3 75 0 0
({3,1}ö({0:v.z}NAME)[({0:n}NAME)ö({0:v.y}NAME)[({0:n}NAME)ö({0:v.x}NAME)[({0:n}NAME)):({0:v}NAME)[({0:n}NAME)
.EQN 0 21 77 0 0
({3,1}ö({0:z}NAME)[({0:n}NAME)ö({0:y}NAME)[({0:n}NAME)ö({0:x}NAME)[({0:n}NAME)):({0:r}NAME)[({0:n}NAME)
.EQN 12 -25 120 0 0
{0:x}NAME,{0:y}NAME,{0:z}NAME{3 5 3 45 35 0 69 33 0 1 1 1 4 -1 1 0 1 1 1 4 -1 1 0 1 1 1 4 -1 1 22 4194368 65408 1 100 2 NO-TITLE}{57}
2 5 21 21 0 1 1.5 7
1 0 4 3 1 4 1 4
3 1 2 0.1
.TXT 39 0 83 0 0
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{\rtf\ansi \deff0{\colortbl;\red0\green64\blue128;\red0\green0\blue128;}{
\fonttbl{\f0\fcharset0\fnil Arial;}}\plain\cf1\fs20 \pard {\cf2 So 
when E = 0 the trajectory is a helix.  To see what happens in the 
presence of the electric field click to get the following file: }{\cf2
\b\ul\link1 Phyblab3b.mcd}{\cf2 \par }{\cf2\fs24\b \par }}
.ATT
.ATT_END
.ATT
.LINK file:C:\WINMCAD\mirmcd\Phyblab3b.MCD
.ATT_END