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If the Coulomb law for the electric field of a point charge represents the underlying ba sis for Gauss' law of electrostatics, the Biot-Savart law plays the sa me role with respect to Ampere's law." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 150 "One can ask the question how the mac roscopic magnetic field generated by a current passing through a wire \+ is composed of infinitesimal contributions d" }{TEXT 257 1 "B" }{TEXT -1 52 " generated by oriented infinitesimal line segments d" }{TEXT 258 1 "l" }{TEXT -1 202 ". One can motivate the Biot-Savart law by per forming the direct calculation for a straight-line wire, and then real ize that the generalization for the magnetic field contribution at dis placement vector " }{TEXT 259 1 "r" }{TEXT -1 24 " from the line segme nt d" }{TEXT 263 1 "l" }{TEXT -1 9 " reads as" }}{PARA 0 "" 0 "" {TEXT -1 1 "d" }{TEXT 262 1 "B" }{TEXT -1 3 " = " }{TEXT 19 27 "mu[0]/ 4 Pi I[0]*crossprod(d" }{TEXT 273 1 "l" }{TEXT 19 1 "," }{TEXT 274 1 " r" }{TEXT 19 5 ")/r^3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 "Note the 1/" }{TEXT 275 1 "r" }{TEXT -1 70 "^2 depende nce of this expression, which upon integration leads to a 1/" }{TEXT 276 1 "r" }{TEXT -1 34 " dependence of the field strength " }{TEXT 277 1 "B" }{TEXT -1 110 ". This is analogous to the electric field gen erated by a charged rod. Note, however that the direction of the " } {TEXT 279 1 "E" }{TEXT -1 28 " field is radial, while the " }{TEXT 278 1 "B" }{TEXT -1 42 " field is orthogonal to the line segment d" } {TEXT 264 1 "l" }{TEXT -1 38 " as well as to the displacement vector" }{TEXT 266 2 " r" }{TEXT -1 37 " (as evident from the cross product). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 160 "Our \+ interest is to make practical use of the Biot-Savart law. We calculate the magnetic field due to a current passing through a finite straight wire of length " }{TEXT 281 1 "L" }{TEXT -1 15 " at a distance " } {TEXT 280 1 "D" }{TEXT -1 22 " away from the middle." }}{PARA 0 "" 0 " " {TEXT -1 29 "We orient the wire along the " }{TEXT 286 1 "x" }{TEXT -1 31 " axis through the origin with -" }{TEXT 287 1 "L" }{TEXT -1 5 " /2 < " }{TEXT 288 1 "x" }{TEXT -1 3 " < " }{TEXT 289 1 "L" }{TEXT -1 55 "/2. Consider the triangle formed from the displacement " }{TEXT 282 1 "x" }{TEXT -1 15 " (location of d" }{TEXT 265 1 "l" }{TEXT -1 65 "=[dx, 0, 0]), the point where the field is to be calculated ([0, \+ " }{TEXT 285 1 "D" }{TEXT -1 50 ", 0]) and the origin. The separation \+ is given as " }{TEXT 19 15 "r=sqrt(x^2+D^2)" }{TEXT -1 32 ", the sine of the angle between " }{TEXT 283 1 "r" }{TEXT -1 6 " and d" }{TEXT 284 1 "l" }{TEXT -1 70 " required for the magnitude of the cross produ ct can be expressed as " }{TEXT 19 14 "sin(theta)=D/r" }{TEXT -1 1 ". " }}{PARA 0 "" 0 "" {TEXT -1 72 "A bit of geometry shows that the foll owing integral is to be calculated:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "r:=sqrt(x^2+D^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%\"rG*$-%%sqrtG6#,&*$)%\"xG\"\"#\"\"\"\"\"\"*$)%\"DGF-F.F/F." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "B:=mu[0]*I[0]/(4*Pi)*Int(D/r *(1/r^2),x=-L/2..L/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"BG,$*&*( &%#muG6#\"\"!\"\"\"&%\"IGF*F,-%$IntG6$*&%\"DG\"\"\"*$),&*$)%\"xG\"\"#F 4F,*$)F3F;F4F,#\"\"$F;F4!\"\"/F:;,$%\"LG#!\"\"F;,$FD#F,F;F,F4%#PiGF@#F ,\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "B0:=value(B);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#B0G,$*&*(&%#muG6#\"\"!\"\"\"&%\"IGF *F,%\"LGF,\"\"\"*(%#PiG\"\"\"-%%sqrtG6#,&*$)F/\"\"#F0F,*$)%\"DGF:F0\" \"%F0F=\"\"\"!\"\"#F,F:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 103 "In th e limit of an infinitely long wire the result previously calculated fr om Ampere's law is obtained:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Blim:=limit(B0,L=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %%BlimG,$*&*&&%#muG6#\"\"!\"\"\"&%\"IGF*F,\"\"\"*&%#PiG\"\"\"%\"DG\"\" \"!\"\"#F,\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 290 10 "Exercise: " } }{PARA 0 "" 0 "" {TEXT -1 233 "Use quantitative numerical examples to \+ investigate the difference beween both expressions: calculate the fiel d strength at distances away from the wire that are fractions (multipl es) of the wire length. How different are the results?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 291 10 "Exercise: " }} {PARA 0 "" 0 "" {TEXT -1 54 "Consider a current-carrying circular loop with radius " }{TEXT 292 1 "R" }{TEXT -1 253 ". Calculate the magneti c field at the center of the loop. Insert numerical values for the rad ius (in the range of 0.1 m) and the current (in the several A range). \+ How does your result compare to the Earth's magnetic field at the surf ace (about 0.04 mT)?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "" 0 "" {TEXT 295 40 "The current loop: calculation of Bz(0,z)" }}{PARA 0 "" 0 "" {TEXT -1 83 "We consider now the following problem: a charge-carrying circular loop with radius " }{TEXT 293 1 " R" }{TEXT -1 126 " produces a magnetic field perpendicular to the plan e of the loop. Calculate the strength of the magnetic field at a dista nce " }{TEXT 294 1 "d" }{TEXT -1 34 " away over the center of the loop ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "We p lace the loop in the " }{TEXT 297 1 "x" }{TEXT -1 1 "-" }{TEXT 296 1 " y" }{TEXT -1 94 " plane with the center at the origin. An analysis of \+ the Biot-Savart law shows that along the " }{TEXT 301 1 "z" }{TEXT -1 84 "-axis the following is true: while there are transverse component \+ contributions in d" }{TEXT 267 1 "B" }{TEXT -1 27 " from individual se gments d" }{TEXT 268 1 "l" }{TEXT -1 18 " (pointing in the " }{TEXT 300 1 "x" }{TEXT -1 1 "-" }{TEXT 299 1 "y" }{TEXT -1 122 " plane), the ir sum vanishes due to symmetry (they all cancel each other). This is \+ apparent from the cross product between " }{TEXT 269 1 "r" }{TEXT -1 6 " and d" }{TEXT 270 1 "l" }{TEXT -1 83 " , and the rotational symmet ry of the problem. Thus, we need to integrate only the " }{TEXT 298 1 "z" }{TEXT -1 84 "-projection of the induced field. The required direc tion cosine can be expressed as " }{TEXT 303 1 "R" }{TEXT -1 1 "/" } {TEXT 302 1 "r" }{TEXT -1 125 ". The integration over the polar angle \+ phi to complete the circular path is trivial due to rotational symmetr y. The vectors " }{TEXT 271 1 "r" }{TEXT -1 6 " and d" }{TEXT 272 1 " l" }{TEXT -1 267 " are orthogonal, i.e., the magnitude of the cross pr oduct involves no angle. These arguments are illustrated with diagrams in first-year physics texts. In the worksheet Currentloop.mws we carr y out the line integral calculation for the case of an arbitrary locat ion (" }{TEXT 304 1 "x" }{TEXT -1 2 ", " }{TEXT 305 1 "y" }{TEXT -1 2 ", " }{TEXT 306 1 "z" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "We have:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "r:=sqrt(R^2+d^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%\"rG*$-%%sqrtG6#,&*$)%\"RG\"\"#\"\"\"\"\"\"*$)%\"dGF-F.F/F." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "B:=mu[0]*I[0]/(4*Pi)*2*Pi*R* (1/r^2)*R/r;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"BG,$*&*(&%#muG6#\" \"!\"\"\"&%\"IGF*F,)%\"RG\"\"#\"\"\"F2*$),&*$F/F2F,*$)%\"dGF1F2F,#\"\" $F1F2!\"\"#F,F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "At distances \+ " }{TEXT 307 1 "d" }{TEXT -1 4 " >> " }{TEXT 308 1 "R" }{TEXT -1 54 " \+ the numerator of this expression can be simplified to" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Blim:=subs(R^2+d^2=d^2,B);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%%BlimG,$*&*(&%#muG6#\"\"!\"\"\"&%\"I GF*F,)%\"RG\"\"#\"\"\"F2*$)*$)%\"dGF1F2#\"\"$F1F2!\"\"#F,F1" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "interface(showassumed=0): as sume(d>0):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Blim:=simplif y(Blim);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%BlimG,$*&*(&%#muG6#\"\" !\"\"\"&%\"IGF*F,)%\"RG\"\"#\"\"\"F2*$)%#d|irG\"\"$F2!\"\"#F,F1" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 198 "This result is important insofar \+ as it describes the magnetic dipole field associated with a current lo op. The expression represents a useful approximation to the magnetic f ield for large distances " }{TEXT 309 1 "d" }{TEXT -1 83 " away from t he source (electromagnetic current loop, or a short permanent magnet). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 434 "Per manent magnetism can be understood as the net result of many aligned m icroscopic spins (electron spin is a quantum phenomenon). One calls th e product of current times the area of the loop that appears in the nu merator the magnetic moment associated with the current loop. It is wo rth remembering that the strength of the magnetic field along the axis behaves as the dipole moment divided by the distance cubed (times mu[ 0]/(2*Pi))." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 310 30 "Maxwell's displacement current" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 245 "Maxwell genera lized Ampere's law to include the situation when a current deposits ch arge (Maxwell displacement current). It is the modified Ampere law tha t includes this current which enters the comprehensive treatment of el ectromagnetic fields." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 311 "To understand the problem consider a capacitor in the process of being charged, i.e., a time-dependent current flows (until the plates are fully charged). The current flows through the wire con necting plate 1, no current flows between the plates, and a current fl ows again from plate 2 into its connecting wire." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "It is possible to conside r two surfaces bounded by the same closed contour around the wire:" }} {PARA 0 "" 0 "" {TEXT -1 88 "surface 1 is defined by the circle, with \+ the wire carrying current I[0] passing through." }}{PARA 0 "" 0 "" {TEXT -1 166 "Let surface 2 be the surface surrounding the capacitor p late (imagine an inflated balloon with the capacitor plate inside, the (missing) open bottom being surface 1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 605 "Clearly no current passes through surface 2, yet it is bounded by the same circular path. It appears th at Ampere's law in its original form gives a contradiction. While it i s true that no usual current can be measured between the capacitor pla tes, it is clear that an electric field is being built up as the capac itor is connected to a voltage source. The current associated with the displacement of charge can be calculated from the total charge deposi ted by the current in the wire. The electric field in the capacitor is homogeneous and the electric flux is related to the field strength, a nd the area:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 19 23 "Phi[e]=E*A=q/epsilon[0]" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Phi[e ] = q/epsilon[0];" "6#/&%$PhiG6#%\"eG*&%\"qG\"\"\"&%(epsilonG6#\"\"!! \"\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "T he displacement current is obtained from the time derivative:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 19 31 "I[d]=epsi lon[0]* diff(Phi[e],t)" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "I[d] = epsilo n[0]*diff(Phi[e],t);" "6#/&%\"IG6#%\"dG*&&%(epsilonG6#\"\"!\"\"\"-%%di ffG6$&%$PhiG6#%\"eG%\"tGF-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "The corrected version of Ampere's law contains a \+ sum of the usual current " }{XPPEDIT 18 0 "I[0];" "6#&%\"IG6#\"\"!" } {TEXT -1 30 " and the displacement current " }{XPPEDIT 18 0 "I[d];" "6 #&%\"IG6#%\"dG" }{TEXT -1 24 " on the right-hand side." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "We can examine the m agnetic field between two circular capacitor plates as they are charge d." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "Sup pose the plates have radius " }{TEXT 311 1 "R" }{TEXT -1 70 ", and we \+ are interested in the magnetic field on a circle with radius " }{TEXT 312 1 "R" }{TEXT -1 62 "/2 (the magnetic field vectors are tangential \+ to this circle)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "We have for the magnetic flux" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 7 "B:='B':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Phi[m]:=B*2*Pi*R/2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%$PhiG6#% \"mG*(%\"BG\"\"\"%#PiGF*%\"RGF*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "The electric field is calculated from the time-varying charge " } {TEXT 314 1 "q" }{TEXT -1 1 "(" }{TEXT 313 1 "t" }{TEXT -1 24 ") and t he plate area as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "E:=q(t )/epsilon[0]/(Pi*R^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"EG*&-%\" qG6#%\"tG\"\"\"*(&%(epsilonG6#\"\"!\"\"\"%#PiG\"\"\")%\"RG\"\"#F*!\"\" " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "We need the electric flux thr ough the area bounded by the circle with radius " }{TEXT 315 1 "R" } {TEXT -1 3 "/2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Phi[e]:= E*Pi*(R/2)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%$PhiG6#%\"eG,$*&-% \"qG6#%\"tG\"\"\"&%(epsilonG6#\"\"!!\"\"#\"\"\"\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "Id:=epsilon[0]*diff(Phi[e],t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#IdG,$-%%diffG6$-%\"qG6#%\"tGF,#\"\"\"\"\" %" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Maxwell:=Phi[m]=mu[0]* Id;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(MaxwellG/*(%\"BG\"\"\"%#PiGF (%\"RGF(,$*&&%#muG6#\"\"!F(-%%diffG6$-%\"qG6#%\"tGF7F(#F(\"\"%" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Bd:=solve(Maxwell,B);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#BdG,$*&*&&%#muG6#\"\"!\"\"\"-%%diff G6$-%\"qG6#%\"tGF3F,\"\"\"*&%#PiG\"\"\"%\"RG\"\"\"!\"\"#F,\"\"%" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "We insert some numerical values:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "simplify(subs(diff(q(t),t )=1*A*s/s,mu[0]=4*Pi*10^(-7)*N/A^2,R=0.5*m,Bd));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%\"NG\"\"\"*&%\"AG\"\"\"%\"mG\"\"\"!\"\"$\"+++++?!# ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "This is the magnetic field \+ strength in Tesla; note that the rate of charging was taken as 1 C/s=1 A." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "We have learned in these lines the following:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 154 "1) for DC currents capacitors \+ are considered to be insulators; this is not true during turn-on and t urn-off, when the plates are being charged/discharged." }}{PARA 0 "" 0 "" {TEXT -1 152 "2) during the charge/discharge times of the plates \+ there is current passing through the wires that connect to the plates, but not in-between the plates." }}{PARA 0 "" 0 "" {TEXT -1 140 "3) ne vertheless, there is a magnetic field associated with the displacement of charge during the process of charging/discharging the plates." }} {PARA 0 "" 0 "" {TEXT -1 352 "4) for an AC current the capictor acts a s a frequency-dependent resistor (impedance); still no current passes \+ in-between the plates (we have a vacuum, or a dielectric between the p lates, and we limit the voltages such that no discharges occur), but t he displacement current causes a magnetic field according to Maxwell's generalization of Ampere's law." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 19 }{VIEWOPTS 1 1 0 1 1 1803 }