{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 81 "PHYSICS RESOURCE PACKETS P
ROJECT                               File: HOWCLOSE.MWS" }}{PARA 0 "" 
0 "" {TEXT -1 110 "Rose-Hulman Institute of Technology                \+
                       Authors: Perry Peters & Greg Williby" }}{PARA 
0 "" 0 "" {TEXT -1 97 "http://www.rose-hulman.edu/~moloney            \+
                       Software: Maple V Release 4" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 308 "A rectangular current lo
op measures 1.50 m by 0.90 m, and carries a current of 10.5 A. At the \+
mid-point of the long (1.50 m) leg inside the loop, how far would we h
ave to be from the wire so that the magnetic field there would be with
in 2% of the field we would calculate assuming wire to be infinitely l
ong?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "So
lution" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 51 "Formula for magnetic field of a finite length w
ire." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "B:=mu[0]*i*(cos(theta)+cos(
phi))/(4*Pi*d);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "B field at dis
tance y from the top horizontal 1.5m segment:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 87 "Bth:=subs(\{d=y,cos(theta)=(3/4)/sqrt((3/4)^2+y^2),co
s(phi)=(3/4)/sqrt((3/4)^2+y^2)\},B);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 46 "B field for the lower horizontal 1.5m segment:" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 34 "Blh:=simplify(subs(y=9/10-y,Bth));" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 43 "B field for the left vertical 0.9m segmen
t:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
97 "Blv:=(subs(\{d=3/4,cos(theta)=y/sqrt(y^2+(3/4)^2),cos(phi)=(9/10-y
)/sqrt((9/10-y)^2+(3/4)^2)\},B));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
44 "B field for the right vertical 0.9m segment:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Brv:=Blv;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Total B field is then:" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 23 "Bloop:=Bth+Blh+Blv+Brv;" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 69 "For a long wire, the magnetic field at a distance
 y from the wire is:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Blong:=mu[0
]*i/(2*Pi*y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "Find y where th
e magnetic field from the loop is within 2% of the magnetic field from
 a \"long\" wire at the same distance y." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "ratio:=expand(Bloop/Blong)=102/100;" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 42 "Solve the above equation for the distance." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "fsolve(ratio,y);" }}}}{MARK "0 2 0
" 97 }{VIEWOPTS 1 1 0 1 1 1803 }