{VERSION 3 0 "IBM INTEL NT" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 2 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "2 D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 256 " " 0 1 48 49 32 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 1 16 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 291 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 292 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 293 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 295 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 296 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 297 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 298 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 299 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 300 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 301 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 16 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 2 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plo t" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 259 12 "Ampere's law" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 375 "Gauss' law rel ates the integral over a closed surface of the electric field vector t o the charge contained in the volume bounded by the surface. Ampere's \+ law represents the equivalent for magnetic fields generated by electri c currents. It is based on the Lorentz force, which forms the basis of understanding the force between two parallel wires that carry electri c currents." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "The law states that the integral of the magnetic field " } {TEXT 258 1 "B" }{TEXT -1 128 " over a closed contour that encloses a \+ current is proportional to the current. The proportionality constant i s the permeability " }{XPPEDIT 18 0 "mu[0]" "6#&%#muG6#\"\"!" }{TEXT -1 38 " which is related to the permittivity " }{XPPEDIT 18 0 "epsilon [0]" "6#&%(epsilonG6#\"\"!" }{TEXT -1 15 " (the product " }{XPPEDIT 18 0 "epsilon[0]*mu[0]=1/c^2" "6#/*&&%(epsilonG6#\"\"!\"\"\"&%#muG6#F( F)*&\"\"\"F)*$%\"cG\"\"#!\"\"" }{TEXT -1 18 " ). In MKSA units " } {XPPEDIT 18 0 "mu[0]" "6#&%#muG6#\"\"!" }{TEXT -1 8 " equals " } {XPPEDIT 18 0 "4Pi/10^7 " "6#*(\"\"%\"\"\"%#PiGF%*$\"#5\"\"(!\"\"" } {TEXT -1 7 " Tm/A ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 102 "The right-hand rule that underlies the Lorentz force d etermines also the orientation of the generated " }{TEXT 257 1 "B" } {TEXT -1 35 " field with respect to the current." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 352 "As with Gauss' law, the \+ usefulness of Ampere's law is restricted by the fact that a high degre e of symmetry must be present, so that simple contours can be found on which the magnetic field is constant. If this is the case, the field \+ can be brought outside the line integral and a simple product of its s trength times the length of the contour results." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 133 "For a straight wire it i s apparent that the right contour is given by concentric circles in th e plane perpendicular to the wire. Thus" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Ampere:=B*2*Pi*r=mu[0 ]*I[0];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'AmpereG/,$*(%\"BG\"\"\"% #PiGF)%\"rGF)\"\"#*&&%#muG6#\"\"!F)&%\"IGF0F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "Outside of the conductor we have for the field:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Brout:=solve(Ampere,B);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%&BroutG,$*&*&&%#muG6#\"\"!\"\"\"&%\" IGF*F,\"\"\"*&%#PiG\"\"\"%\"rG\"\"\"!\"\"#F,\"\"#" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 166 "The strength of the magnetic field decreases with the distance from the wire. As before with Gauss' law we can also use Ampere's law inside the current distribution. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "Suppose the cylindrically shaped wire has radius " }{TEXT 260 1 "R" }{TEXT -1 97 " and the curr ent density inside the wire is constant. What is the magnetic field in side the wire?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "We use a circle with radius " }{TEXT 262 1 "r" }{TEXT -1 2 " <" }{TEXT 261 2 " R" }{TEXT -1 141 ", which encloses a fraction of the current. The constant current density is given as total current d ivided by the total cross-sectional area:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 17 "j:=I[0]/(Pi*R^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%\"jG*&&%\"IG6#\"\"!\"\"\"*&%#PiG\"\"\")%\"RG\"\"#F*!\"\"" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "The amount of current enclosed by \+ a cross-section of radius " }{TEXT 263 2 "r " }{TEXT -1 10 "is called \+ " }{TEXT 19 2 "Ip" }{TEXT -1 19 " (partial current):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Ip:=j*Pi*r^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#IpG*&*&&%\"IG6#\"\"!\"\"\")%\"rG\"\"#\"\"\"F/*$)%\"R G\"\"#F/!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 160 "Ampere's law no w provides a statement about the constant magnetic field strength on t he circle surrounding the enclosed area, which is permeated by the cur rent " }{TEXT 19 2 "Ip" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Ampere:=B*2*Pi*r=mu[0]*Ip;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'AmpereG/,$*(%\"BG\"\"\"%#PiGF)%\"rGF)\"\"#*&*(&%#muG 6#\"\"!F)&%\"IGF1F))F+F,\"\"\"F6*$)%\"RG\"\"#F6!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "We observe that the magnetic field strength in side the conductor grows with the radius (of the circle on which it is constant):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Brin:=solve( Ampere,B);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%BrinG,$*&*(&%#muG6#\" \"!\"\"\"&%\"IGF*F,%\"rGF,\"\"\"*&%#PiG\"\"\")%\"RG\"\"#F0!\"\"#F,\"\" #" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "P1:=plot(subs(mu[0]=1, I[0]=1,R=1,Brin),r=0..1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "P2:=plot(subs(mu[0]=1,I[0]=1,R=1,Brout),r=1..5):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 17 "display(\{P1,P2\});" }}{PARA 13 "" 1 "" {GLPLOT2D 646 235 235 {PLOTDATA 2 "6&-%'CURVESG6$7S7$\"\"!F(7$$\"1nmm;arz@!#<$\" 1.6'4$[7pM!#=7$$\"1LL$e9ui2%F,$\"1&oqqz\"f(['F/7$$\"1nmm\"z_\"4iF,$\"1 :f1#ft@))*F/7$$\"1mmmT&phN)F,$\"1(=qyoD*H8F,7$$\"1LLe*=)H\\5!#;$\"1Uc \\O*4+n\"F,7$$\"1nm\"z/3uC\"FB$\"1$\\&yo:J&)>F,7$$\"1++DJ$RDX\"FB$\"1N $Qg9)y6BF,7$$\"1nm\"zR'ok;FB$\"1$zi#*oI%\\EF,7$$\"1++D1J:w=FB$\"19!p&3 /*f)HF,7$$\"1LLL3En$4#FB$\"1=NK[M=KLF,7$$\"1nm;/RE&G#FB$\"1%3whY5rj$F, 7$$\"1+++D.&4]#FB$\"1@.^mgQ!)RF,7$$\"1+++vB_\")HeRy&pF,7$$\"1LL$e9Ege%FB$\"1$[! 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Magnetic field lines are always closed. One can fo rmulate Gauss' law for magnetism, but given that there is no equivalen t of charges (permanent magnets are always dipoles), the magnetic flux through closed surfaces always vanishes." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 264 24 "Gauss' law for magnetism" }}{PARA 0 "" 0 "" {TEXT -1 345 "We can use Gauss' law \+ for magnetic fields to calculate the magnetic flux through a square wi re loop that is placed between the poles of a permanent magnet (or an \+ electromagnet). The magnet produces a constant field, e.g., of B = 0. 01 T (Tesla). The square wire loop with sides l = 0.01 m is tilted wit h an angle of 20 degrees w.r. to the field." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 125 "We need to find the oriented s urface element and calculate its scalar product with the field. The ve ctorial surface element d" }{TEXT 256 1 "A" }{TEXT -1 115 " is perpend icular to the loop, and thus stands at an angle of theta=90-20 degrees w.r. to the field. Thus, we have:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "theta:=(90-20)*Pi/180;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG,$%#PiG#\"\"(\"#=" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "l:=0.01;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"lG$\" \"\"!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Phi:=B*cos(thet a)*l^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$PhiG,$*&%\"BG\"\"\"-%$co sG6#,$%#PiG#\"\"(\"#=F($F(!\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "B:=0.01;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"BG$\"\"\"!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalf(Phi);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#$\"+G9??M!#;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "The magnetic flux is calculated in Tesla*m^2, a unit which carr ies the name Weber (Wb)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 300 28 "Magnetic field of a solenoid " }}{PARA 0 "" 0 "" {TEXT -1 294 "Ampere's law permits to calculate ap proximately the magnetic field strength inside a solenoid. Many turns \+ are contributing their identical magnetic fields to add up (superposit ion principle for magnetic fields). We consider an ideal solenoid, i.e ., a long coil for which we ignore edge effects." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 437 "Consider a cross section through the length of the solenoid and observe how the top and bottom parts of each turn contribute with their circular B fields to the ove rall magnetic field inside and outside of the coil. The net result is \+ a very strong nearly homogeneous field inside the solenoid, with diver ging field lines at the two ends to close in large loops around the le ngth of the solenoid, i.e., a very weak field outside the coil." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 257 "To calcu late the magnetic field inside the coil we imagine a rectangular loop \+ with one long side inside the solenoid, the other outside with the two short legs making the connection. The solenoid consists of a single l ayer and the rectangular loop encloses " }{TEXT 265 1 "N" }{TEXT -1 34 " turns of wire carrying a current " }{TEXT 266 1 "I" }{TEXT -1 306 ". The line integral along the rectangular loop has to be carried \+ out in a specific direction consistent with the right-hand rule. Only \+ the long leg inside the solenoid contributes to the line integral over the magnetic field: outside the solenoid the field is weak. The short legs are perpendicular to the " }{TEXT 267 1 "B" }{TEXT -1 47 " field , and therefore do not contribute at all." }}{PARA 0 "" 0 "" {TEXT -1 28 "The enclosed current equals " }{TEXT 269 1 "N" }{TEXT -1 1 " " } {TEXT 268 1 "I" }{TEXT -1 16 ", and therefore:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "B:='B': l:='l':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Ampere:=B*l=mu[0]*N*I[0];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'AmpereG/*&%\"BG\"\"\"%\"lGF(*(&%#muG6#\"\"!F(%\"NGF( &%\"IGF-F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "B:=solve(Ampe re,B);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"BG*&*(&%#muG6#\"\"!\"\" \"%\"NGF+&%\"IGF)F+\"\"\"%\"lG!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "This is the magnetic field strength inside the coil." }}{PARA 0 "" 0 "" {TEXT -1 57 "One can also introduce a number of turns per un it length " }{TEXT 270 1 "n" }{TEXT -1 16 ", and eliminate " }{TEXT 271 1 "l" }{TEXT -1 19 " upon substitution " }{TEXT 272 2 "N " }{TEXT -1 2 "= " }{TEXT 273 1 "n" }{TEXT -1 1 " " }{TEXT 274 1 "l" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 267 "Note that the field is homogeneo us inside the coil. This result follows from the fact that the short l egs do not contribute at all to the line integral, i.e., the same resu lt is obtained independent of the location of the contributing leg of \+ length l inside the coil." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 184 "We can use the above finding together with some b asics about conductivity to understand the design of electromagnets. W e consider a single-layer coil with wire of thickness (diameter) " } {TEXT 275 1 "d" }{TEXT -1 13 ". Obviously, " }{TEXT 276 4 "N d " } {TEXT -1 2 "= " }{TEXT 278 2 "l." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "l:=N*d;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"lG*&%\" NG\"\"\"%\"dGF'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "B;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*&*&&%#muG6#\"\"!\"\"\"&%\"IGF'F)\"\" \"%\"dG!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Now suppose a wir e with diameter " }{TEXT 279 1 "d" }{TEXT -1 124 "=0.6 mm can carry at most a current of 30 mA. We can calculate the maximum possible field \+ strength inside the electromagnet:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "evalf(subs(I[0]=0.03,d=0.0006,mu[0]=4*Pi*10^(-7),B)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+3`=$G'!#9" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 246 "The field of 0.06 mT is rather weak. How can we i ncrease it? The amount of current that one can pass through the wire d epends on its cross sectional area, i.e., the square of the diameter. \+ On the other hand the number of turns for a given length " }{TEXT 280 1 "l" }{TEXT -1 109 " decreases with the diameter. We find that the ma ximum value of B is proportional to the diameter as follows:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "subs(I[0]=c1*d^2,B);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*(&%#muG6#\"\"!\"\"\"%#c1GF(%\"dGF(" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "Exercise: an interesting variat ion of the above is the toroidal coil: a torus defined by the big radi us " }{TEXT 281 1 "R" }{TEXT -1 29 " (diameter) and small radius " } {TEXT 282 1 "r" }{TEXT -1 115 " (cross section) has no edge effects du e to the circular hsape. However, the field is not constant inside the coil." }}{PARA 0 "" 0 "" {TEXT -1 45 "Calculate the variation in B wi th the radius " }{TEXT 283 2 "R'" }{TEXT -1 6 " for " }{TEXT 284 1 "R " }{TEXT -1 1 "-" }{TEXT 285 1 "r" }{TEXT -1 5 " < " }{TEXT 286 2 "R '" }{TEXT -1 5 " < " }{TEXT 287 1 "R" }{TEXT -1 1 "+" }{TEXT 288 1 " r" }{TEXT -1 63 " using Ampere's law. Again, ignore the field outside the coil." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 289 26 "A semi-realistic solenoid:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 527 "We can use Maple's graphing capab ilities to draw the magnetic field vectors for a solenoid with a finit e number of turns in a cross section through the center of the coil. W e make use of the result of the dependence of the magnetic field stren gth on the radial distance from the wire obtained above. We use the fa ct that the sign of the field lines around the cross sections of the w ires in the top half is opposite to the bottom half, since the current enters the plane in one half and exits from the plane for the other h alf." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "W e use arbitrary units." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "B:='B':" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 28 "Ampere:=B*2*Pi*r=mu[0]*I[0];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "Brout:=solve(subs(I[0]=1,mu[0]=1,Ampere),B);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%'AmpereG/,$*(%\"BG\"\"\"%#PiGF)%\"rG F)\"\"#*&&%#muG6#\"\"!F)&%\"IGF0F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%&BroutG,$*&\"\"\"F'*&%#PiG\"\"\"%\"rG\"\"\"!\"\"#\"\"\"\"\"#" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "The following procedure takes a po int in the plane with coordinates (" }{TEXT 298 1 "x" }{TEXT -1 2 ", \+ " }{TEXT 299 1 "y" }{TEXT -1 220 ") and performs a sum over the contri butions to the magnetic field at that point that originate from 15 tur ns of the coil. Note that the expression derived from Ampere's law pro vides the field strength at radial distance " }{TEXT 290 1 "r" }{TEXT -1 81 " from the wire. The direction of the field is tangential to the circle of radius " }{TEXT 291 1 "r" }{TEXT -1 164 " at that point. Th e direction of the field is taken into account by either adding positi ve or negative 90 degrees (Pi/2) to the angle of the line connecting p oint (" }{TEXT 293 1 "x" }{TEXT -1 2 ", " }{TEXT 292 1 "y" }{TEXT -1 180 ") with the point where the wire intersects the plane. Maple's arc tangent function with two arguments covers a range of 2 Pi! The proced ure returns a 2dimensional 'vector' with the " }{TEXT 294 1 "x" } {TEXT -1 5 " and " }{TEXT 295 1 "y" }{TEXT -1 53 " components of the t otal magnetic field at location (" }{TEXT 297 1 "x" }{TEXT -1 2 ", " } {TEXT 296 1 "y" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 448 "Imagine that the coil is placed horizont ally: in a cross-sectional diagram the top row of current crossing thr ough the plane corresponds to current going 'into', and the bottom row to current coming 'out' of the plane. Note that the sign of the curre nt has been picked together with the orientation of the winding. To sw itch either of the two one would switch the convention of adding/subtr acting for the top and bottom rows of intersection points." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 264 "The resulting fi eld vector consist of a significant number of 'cancellations': imagine a location directly between two intersections of the wire in the top \+ (or bottom) half and what the effect of circular field lines associate d with the two wires is at that point." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Field:=proc(x,y) local \+ i,B,xi,yi,phi,Bbotx,Bboty,Btopx,Btopy; global Brout;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Btopx:=0: Btopy:=0: #top wires first:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "for i from 1 to 15 do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "xi:=i*0.2: yi:=0.5:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "B:=evalf(subs(r=sqrt((x-xi)^2+(y-yi)^2),Brout));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phi:=arctan(yi-y,xi-x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Btopx:=evalf(Btopx+B*cos(phi+Pi/2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Btopy:=evalf(Btopy+B*sin(phi+Pi/2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Bb otx:=0: Bboty:=0: #bottom wires now:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "for i from 1 to 15 do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "xi: =i*0.2: yi:=-0.5:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "B:=evalf(subs( r=sqrt((x-xi)^2+(y-yi)^2),Brout));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phi:=arctan(yi-y,xi-x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Bbot y:=evalf(Bboty+B*sin(phi-Pi/2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Bbotx:=evalf(Bbotx+B*cos(phi-Pi/2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "od: [Btopx+Bbotx,Btopy+Bboty]; end:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 11 "Field(1,0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$! +-6GBQ!\"*\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plo ts):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "fieldplot(Field(x,y 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The fiel dplot shows also very clearly how the field weakens outside the soleno id." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 301 10 "Exercises:" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 83 "1) Vary the dimensions of the coil, \+ i.e., number, diameter and spacing of windings." }}{PARA 0 "" 0 "" {TEXT -1 140 "2) Do the results change if the top and bottom rows of i ntersections are shifted by one half of the spacing (as is the case in a real coil)?" }}{PARA 0 "" 0 "" {TEXT -1 43 "3) Draw a fieldplot for a single-turn coil." }}{PARA 0 "" 0 "" {TEXT -1 123 "4) Write a proce dure that allows to vary these parameters through the argument list or by modifiaction of global variables." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 217 "In another worksheet a more accurate calculation of the magnetic field is performed by summing the exact m agnetic field components for current loops (Currentloop.mws), and cons tructing magnetic field lines from these." }}}}{MARK "47 0 0" 148 } {VIEWOPTS 1 1 0 1 1 1803 }