{VERSION 2 3 "IBM INTEL NT" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Fo nt 0" -1 256 1 {CSTYLE "" -1 -1 "Helvetica" 1 12 255 0 0 1 2 1 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "PHYSICS RESOURCE PACKETS P ROJECT File: loops.MWS" }}{PARA 0 "" 0 "" {TEXT -1 86 "Rose-Hulman Institute of Technology \+ Authors: mjm" }}{PARA 0 "" 0 "" {TEXT -1 97 "http://w ww.rose-hulman.edu/~moloney Software : Maple V Release 4" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "Axial magnetic fields f rom 1, 2, and several loops of current-carrying wire." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "a) Use the starter mat erial and evaluate the formula at the center of the loop. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 159 "b) Make a functi on f1, where you substitute I=1 amp and mu = 4 pi x 10^(-7). Put the c oil center at z=0 give it a radius of 1 m , and plot f1 from z=0 to z= 4. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 212 "c ) At the coil center the slope of Bz is zero and again at z= infinity. Locate the value of z where the slope magnitude is a maximum. [Hint: \+ what does this have to do with the second derivative of the function?] " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "d) P ut a second coil at z=1.2 m and plot Bz for the pair of coils. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "e) Evalua te the 1st and 3rd derivatives of the B field halfway between the two \+ coils" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 "f) Use a loop to create a 'loose' solenoid of 5 coils. Plot axial mag netic field (should have some wiggles in it.)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "g) Estimate the area unde r the graph of B along the axis by using a rectangle. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "restart; with(plots):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "f = Bz @(0,0,z) from lo op centered @ z=c, radius R. This is our basic building block." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "f:= mu/(4*Pi)*I*2*Pi*R^2/(R^2 + (z-c)^2)^(3/2); " }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 49 "check it at center of the loop; want mu I / 2R" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "subs(z=0,c=0,f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "f1:=subs(mu=4*Pi*1e-7,I=1,f);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "Plot this critter to see what it looks li ke (put at z=2 and plot 0 to 4)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "plot(subs(R=1,c=2,f1),z=0..4);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "twoloops:=subs(R=1, c=0,f1) + subs(R=1,c=12/10,f1); # two loops \+ separated by 1.2 units" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "2nd der ivative of f: " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "d2:=diff(f,z,z); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "solution gives two answers f or where Bz is zero for a single loop, half a radius away from loop ce nter to either side" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "sol:=solve(d 2=0,z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "first derivative: " }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "d1:=diff(twoloops,z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "check it at z = 0.6" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "evalf(subs(z=6/10,d1));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "3rd derivative:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "d3: =diff(twoloops,z,z,z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "check i t at z = 0.6" }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 " evalf(subs(z=6/10,d3));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "plot t he axial field of 2 loops, one at zero, one at 1.2." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(twoloops,z=0..1);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 47 "twoa:=subs(R=1, c=1/4,f1) + subs(R=1,c=3/4,f1);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Example of plotting two 'function s' together" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "plot(\{1e7*twoloops, 1e7*twoa\},z=0..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "Example of how to use a software loop to make a group of coils" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "g:=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "fo r k from 1 to 4 do " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "g:=g+subs(R= 1/8,c=k/6,f1); od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "plot (1e6*g,z=-1..2,title=` B(axis) of 'loose' solenoid`);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "This plot is intentionally wiggly; not a \+ very good solenoid." }}{PARA 0 "" 0 "" {TEXT -1 97 "An estimation exer cise would be to guesstimate the area under the curve by means of a a \+ rectangle" }}{PARA 0 "" 0 "" {TEXT -1 48 "By eyeball, try 6.1 height f rom 0.2 to 0.8 = 3.7" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 105 "here's a rectangle with a height of 6.2 and going from 0 to 0.85 . Then we'll plot it with the g function" }}{PARA 0 "" 0 " " {TEXT -1 59 "it needs capitals on Heaviside!! (I found out the hard \+ way)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "rect:= 62/10*(Heaviside(z-1 /100)-Heaviside(z-85/100));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "plot(\{rect,1e6*g\},z=-1..2,title=` Area estimation via rectangle` );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "6.2*0.85;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Lets integrate and see how good this is" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "evalf(int(1e6*g,z=-5..5));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "In the neighborhood" }}}}{MARK "22 0 0" 4 }{VIEWOPTS 1 1 0 1 1 1803 }