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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 19 "The Biot-Savart law" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 325 "The Biot
-Savart law serves to calculate magnetic field for situations where Am
pere's law is not useful due to a lack of symmetry. 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 261 1 "l" }{TEXT -1 9 " reads as" }}{PARA 0 "" 0 "" 
{TEXT -1 1 "d" }{TEXT 260 1 "B" }{TEXT -1 3 " = " }{TEXT 19 27 "mu[0]/
4 Pi I[0]*crossprod(d" }{TEXT 271 1 "l" }{TEXT 19 1 "," }{TEXT 272 1 "
r" }{TEXT 19 5 ")/r^3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 11 "Note the 1/" }{TEXT 273 1 "r" }{TEXT -1 70 "^2 depende
nce of this expression, which upon integration leads to a 1/" }{TEXT 
274 1 "r" }{TEXT -1 34 " dependence of the field strength " }{TEXT 
275 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 277 1 "E" }{TEXT -1 28 " field is radial, while the " }{TEXT 
276 1 "B" }{TEXT -1 42 " field is orthogonal to the line segment d" }
{TEXT 262 1 "l" }{TEXT -1 38 " as well as to the displacement vector" 
}{TEXT 264 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 279 1 "L" }{TEXT -1 15 " at a distance " }
{TEXT 278 1 "D" }{TEXT -1 22 " away from the middle." }}{PARA 0 "" 0 "
" {TEXT -1 29 "We orient the wire along the " }{TEXT 284 1 "x" }{TEXT 
-1 31 " axis through the origin with -" }{TEXT 285 1 "L" }{TEXT -1 5 "
/2 < " }{TEXT 286 1 "x" }{TEXT -1 3 " < " }{TEXT 287 1 "L" }{TEXT -1 
55 "/2. Consider the triangle formed from the displacement " }{TEXT 
280 1 "x" }{TEXT -1 15 " (location of d" }{TEXT 263 1 "l" }{TEXT -1 
65 "=[dx, 0, 0]), the point where the field is to be calculated ([0, \+
" }{TEXT 283 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 281 1 "r" }{TEXT -1 6 " and d" }{TEXT 
282 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);" }}}{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);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "B0:=value(B);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 103 "In the limit of an infinitely long wire \+
the result previously calculated from Ampere's law is obtained:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Blim:=limit(B0,L=infinity);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 288 10 "Exercise: " }}{PARA 0 "" 0 "
" {TEXT -1 233 "Use quantitative numerical examples to investigate the
 difference beween both expressions: calculate the field strength at d
istances away from the wire that are fractions (multiples) of the wire
 length. How different are the results?" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT 289 10 "Exercise: " }}{PARA 0 "" 0 "" 
{TEXT -1 54 "Consider a current-carrying circular loop with radius " }
{TEXT 290 1 "R" }{TEXT -1 253 ". Calculate the magnetic field at the c
enter of the loop. Insert numerical values for the radius (in the rang
e of 0.1 m) and the current (in the several A range). How does your re
sult compare to the Earth's magnetic field at the surface (about 0.04 \+
mT)?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT 293 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 291 1 "R" }{TEXT 
-1 126 " produces a magnetic field perpendicular to the plane of the l
oop. Calculate the strength of the magnetic field at a distance " }
{TEXT 292 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 place \+
the loop in the " }{TEXT 295 1 "x" }{TEXT -1 1 "-" }{TEXT 294 1 "y" }
{TEXT -1 94 " plane with the center at the origin. An analysis of the \+
Biot-Savart law shows that along the " }{TEXT 299 1 "z" }{TEXT -1 84 "
-axis the following is true: while there are transverse component cont
ributions in d" }{TEXT 265 1 "B" }{TEXT -1 27 " from individual segmen
ts d" }{TEXT 266 1 "l" }{TEXT -1 18 " (pointing in the " }{TEXT 298 1 
"x" }{TEXT -1 1 "-" }{TEXT 297 1 "y" }{TEXT -1 122 " plane), their sum
 vanishes due to symmetry (they all cancel each other). This is appare
nt from the cross product between " }{TEXT 267 1 "r" }{TEXT -1 6 " and
 d" }{TEXT 268 1 "l" }{TEXT -1 83 " , and the rotational symmetry of t
he problem. Thus, we need to integrate only the " }{TEXT 296 1 "z" }
{TEXT -1 84 "-projection of the induced field. The required direction \+
cosine can be expressed as " }{TEXT 301 1 "R" }{TEXT -1 1 "/" }{TEXT 
300 1 "r" }{TEXT -1 125 ". The integration over the polar angle phi to
 complete the circular path is trivial due to rotational symmetry. The
 vectors  " }{TEXT 269 1 "r" }{TEXT -1 6 " and d" }{TEXT 270 1 "l" }
{TEXT -1 267 " are orthogonal, i.e., the magnitude of the cross produc
t involves no angle. These arguments are illustrated with diagrams in \+
first-year physics texts. In the worksheet Currentloop.mws we carry ou
t the line integral calculation for the case of an arbitrary location \+
(" }{TEXT 302 1 "x" }{TEXT -1 2 ", " }{TEXT 303 1 "y" }{TEXT -1 2 ", \+
" }{TEXT 304 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);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "B:=mu[0]*I[0]/(4*Pi)*2*Pi*R*(1/r^2)*R/r;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 13 "At distances " }{TEXT 305 1 "d" }{TEXT 
-1 4 " >> " }{TEXT 306 1 "R" }{TEXT -1 54 " the numerator of this expr
ession can be simplified to" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "Blim:=subs(R^2+d^2=d^2,B);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 38 "interface(showassumed=0): assume(d>0):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 21 "Blim:=simplify(Blim);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 198 "This result is important insofar as it describes the \+
magnetic dipole field associated with a current loop. The expression r
epresents a useful approximation to the magnetic field for large dista
nces " }{TEXT 307 1 "d" }{TEXT -1 83 " away from the source (electroma
gnetic current loop, or a short permanent magnet). " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 434 "Permanent magnetism ca
n be understood as the net result of many aligned microscopic spins (e
lectron spin is a quantum phenomenon). One calls the product of curren
t times the area of the loop that appears in the numerator the magneti
c moment associated with the current loop. It is worth remembering tha
t the strength of the magnetic field along the axis behaves as the dip
ole moment divided by the distance cubed (times mu[0]/(2*Pi))." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 308 30 "Maxwell's displacement current" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 245 "Maxwell generalized Ampere's l
aw to include the situation when a current deposits charge (Maxwell di
splacement current). It is the modified Ampere law that includes this \+
current which enters the comprehensive treatment of electromagnetic fi
elds." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 311 
"To understand the problem consider a capacitor in the process of bein
g charged, i.e., a time-dependent current flows (until the plates are \+
fully charged). The current flows through the wire connecting plate 1,
 no current flows between the plates, and a current flows again from p
late 2 into its connecting wire." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 91 "It is possible to consider two surfaces b
ounded by the same closed contour around the wire:" }}{PARA 0 "" 0 "" 
{TEXT -1 88 "surface 1 is defined by the circle, with the wire carryin
g current I[0] passing through." }}{PARA 0 "" 0 "" {TEXT -1 166 "Let s
urface 2 be the surface surrounding the capacitor plate (imagine an in
flated balloon with the capacitor plate inside, the (missing) open bot
tom being surface 1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 605 "Clearly no current passes through surface 2, yet it i
s bounded by the same circular path. It appears that Ampere's law in i
ts original form gives a contradiction. While it is true that no usual
 current can be measured between the capacitor plates, it is clear tha
t an electric field is being built up as the capacitor is connected to
 a voltage source. The current associated with the displacement of cha
rge can be calculated from the total charge deposited by the current i
n the wire. The electric field in the capacitor is homogeneous and the
 electric flux is related to the field strength, and 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#/&%$PhiG6#%\"eG*&%\"qG\"\"\"&%(epsilonG6#\"\"!!\"\"" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "The displacement c
urrent is obtained from the time derivative:" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 19 31 "I[d]=epsilon[0]* diff(Phi[e],t)
" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "I[d] = epsilon[0]*diff(Phi[e],t);" 
"6#/&%\"IG6#%\"dG*&&%(epsilonG6#\"\"!\"\"\"-%%diffG6$&%$PhiG6#%\"eG%\"
tGF-" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "T
he corrected version of Ampere's law contains a sum of the usual curre
nt " }{XPPEDIT 18 0 "I[0];" "6#&%\"IG6#\"\"!" }{TEXT -1 30 " and the d
isplacement 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 magnetic field between
 two circular capacitor plates as they are charged." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "Suppose the plates have
 radius " }{TEXT 309 1 "R" }{TEXT -1 70 ", and we are interested in th
e magnetic field on a circle with radius " }{TEXT 310 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 f
or 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;
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "The electric field is calcula
ted from the time-varying charge " }{TEXT 312 1 "q" }{TEXT -1 1 "(" }
{TEXT 311 1 "t" }{TEXT -1 24 ") and the plate area as:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "E:=q(t)/epsilon[0]/(Pi*R^2);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "We need the electric flux through \+
the area bounded by the circle with radius " }{TEXT 313 1 "R" }{TEXT 
-1 3 "/2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Phi[e]:=E*Pi*(
R/2)^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "Id:=epsilon[0]*d
iff(Phi[e],t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Maxwell:=
Phi[m]=mu[0]*Id;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Bd:=sol
ve(Maxwell,B);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "We insert some \+
numerical values:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "simpli
fy(subs(diff(q(t),t)=1*A*s/s,mu[0]=4*Pi*10^(-7)*N/A^2,R=0.5*m,Bd));" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "This is the magnetic field stre
ngth 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 ha
ve 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 turn
-off, when the plates are being charged/discharged." }}{PARA 0 "" 0 "
" {TEXT -1 152 "2) during the charge/discharge times of the plates the
re is current passing through the wires that connect to the plates, bu
t not in-between the plates." }}{PARA 0 "" 0 "" {TEXT -1 140 "3) never
theless, 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 1 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 
1 }{PAGENUMBERS 0 1 2 33 1 1 }