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Unformatted text preview: 172 MAXWELL'S EQUATIONS AND BOUNDARY CONDITIONS (3103). One concludes that a surface charge is ' con
' d d on the interface by the normal components ofE if at least on: reg1;):ng bath
m um h hand no free. surface charge exists at the inter ace 1
on the Qt er a the special proportion 61/61 = 01/0;
event and oflittlc importance.
(3134») yields the special case in which £1 and £2 are given by ductive. I _
regions are nonconductlve (a1 = 02 H 0) or (t?!) are
is true among the region parameters, presuma rt
For both regions noneonducttve, putting p, = m o 61113"1 _ 621;": = 03 orjust Dnl _ DnZ : 0 a) = 02 = 0 a result agreeing with (343) for the nonconductive case. '  r  ' If: h parat g tW 3 I I [)(‘tt'l [Illllt‘ ll (3 lf‘lr'u llvﬁ ldW l0! dwth LLlll L'le tit an In llaf I‘ E l! O 1501.! UP“ (Ullduc [HR 16 10115. i; Cl ldllze the 163111‘ l0] 0116 LolldU(UU1tY IllUCll 1.31 C! that].
l g p g . ' ' _ dar
[he 0tgimme the J vectors tilted by amounts 91 and 61 as Show" m (a)' rhf boun y Condition (3432) for dc becomes “7.1:an (I) urrent flux 239/ (a) Region 12w” Region 1:071) Region 2:“; >> 0'1) Region 2: {02)
(d) Region 21(62 =10 ‘71) m (c) (1 Ref: JCUO 0‘ cut: Iltb b LIE. lt’S Ol urrent flux I' a ‘lltJIl [a = “)0 .
‘ P C CfI' L. I 2 1 l] C . ( ) m
(6) current ﬂux for (113310“. 2 conductive. Constraint {0 tangential ﬂOW at Inlet lace lOI region 1 nonconductive. PROBLEMS 173 while the boundary condition involving tangential components is obtained from (3—79),
with J = 0E a: a: (2) From the geometry, the tilt angles obey _ Jrl tan 3] _ tan 9; = & nl jnl The latter combines with (l) and (2), whereupon inserting the expression for tan 61
obtains the refractive law [an 91 = 2 tan 61 (3138)
0'2 The analogy with the refractive laWS (376) and (380) for B and E might be noted.
For an example in which a; = 1001, the refractive effects of direct current streamlines at an interface are shown typically in (b) of the accompanying ﬁgure. For :12 » or, the
near perpendicularity of the current ﬂux occurs in regions 1, as noted in (r). Ifcri were reduced to zero, then I, = O, constraining the current ﬂow in region 2 to paths tan
gential to the conductor—insulator boundary as in (d), a result evident from the insertion of},l = j” = 0 into the boundary conditions (4133) and {4134}. REFERENCES ELLIOTT, R. S. Electromagnetirr. New York: McGrawiﬂill, 1966.
JAVID, M., and P. M. BROWN. FitM Analysis and Electromagnelicj. New York: McGraw—Hill, 1963. _IORDAN, E. C., and K. G. BALMAIN. Electromagnetic Waves and Radiating Systems, 2nd ed., Engle
wood Cliffs, NJ; Prentice—Hall, [963: LORRAIN, P., and D. R. Conson. Elcrtramagnetic Fields and Warm. San Francisco: Freeman, 1970. Rsirz, R., and F. J. MILFORD. Foundations qf Electromagnets": Themy. Reading, Mass: Addison~
Wesley, New York: 1960. PROBLEMS
SECTION 31 31. Pure copper, with a free (outer orbit) electron concentration of about 10‘29 electrons/m3,
has the conductivity 0 = 5.8 x If.)7 mho/m at room temperature (Table 3—3). (a) Find the
mobility of the free electrons in copper. (b) Express the free electron charge density in coulombs
per cubic millimeter for this material. (c) Find the drift velocity of the electrons for the unit
applied electric ﬁeld E = a, V/m. What is the corresponding volume current density in this
specimen? Sketch the vectors depicting 9‘, J, and E in the sample. (Explain from physical reasoning why u, and E are in opposite directions, although J and E are in the same sense.)
{Answerz (c) —3.ﬁnx min/sec] 32. Find the current density (expressed in A/cmz) in the. following conductors, possessing only
negative electronic charge carriers under the given conditions. (a) The average drift velocity
is —n,4.5 mm/sec and the charge carriers have the density 2 x If)“ electrons/m3. (b) The
volume density of electronic charge carriers is —3.5 x 108 C/m’ and the carrier average drift
velocity is 4.2 mm/see, with E = 101, V/m within the conductor. What is the conductivity of
the region in the latter case? [Answer: (a) 3‘1440 A/cm2 (b) 1,147 A/ctn’. 0.147 MU/m] 174 MAXWELL's EQUATIONS AND BOUNDARY CONDITIONS SECTION 32 33. At some particular temperature, helium gas has 1025 atomsj'm’ and is measured to have
the dielectric susceptibility of 1.5 x 10“. What is its electric polarizatiOn ﬁeld P for the applied
ﬁeld E = 103 V/m? What is the charge density p“ and the average displacement cl of the
nucleus relative to the electron cloud for the given E? What is 6,? [Answer: p, — 3.2 x 10‘ Cfm’] 34. At low frequencies, the measured relative permittivity of water is 81 (Table 33). What
is then its electric susceptibility? What electric ﬁeld E must be applied to produce, at the sinusoidal
frequency a), the polarization ﬁeld P = 1,10 sin to! ,uC/mz in a water sample? (Ex ress E in
kV/m.) Firfd the corresponding electric displacement density D, expressed in [AC/m . Without
using ﬁeld values, form appropriate ratios to determine by what factor the magnitude of D is larger than that of P; similarly, compare P with EOE. 35. (a) To make the electric polarization density P and the applied ﬁeld EOE exactly the
same in a material, what must its relative permittivity be? (b) What is the relative permittivity
of a material ifP has 10 times the value of 50E therein? (c) H!) has 10 times the strength of
£013 in a material, what is its relative permittivity? (d) If D is 10P in some region, what is its
6,? [Answers (b) 11 (d) 1.111] 36. The same electric ﬁeld, E = 103az V/m, is applied to the following regions having the
dielectric susceptibilities: (a) zero (what sort ofregion is this?); (b) 103; (c) l; (d) 103. Deter
mine the relative permittivity, the applied ﬁeld 60E, the electric polarization ﬁeld P, and the
electric displacement ﬁeld D for each region. SECTION 32A. 57. The following E ﬁelds are given to exist in some block of polyethylene. for which E, = 2.26
(from Table 33): (a) 1,103.13 sin to! V/m; (b) 31,1031: sin tot V/m; (c) n,(103/r1) sin cot V/m.
Find the ﬁelds EOE, P, D, the polarization (bound) charge density p,, and the volume
polarization (bound) current density J, for each applied E ﬁeld. [Answerz (c) p,=0,
J, = 1,.(1 1.14/r2)cu cos cu! nAfmz] 38. Corresponding to the electric polarizatitm ﬁeld P = 11,10 sin to! pG/m2 of Problem 34,
ﬁnd the polarization (bound) current density J, at the frequencies: (a) 1 kHz; (b) 1 MHz. SECTION 323 _ 39. Apply the Gauss—Maxwell integral law (336) to a vanishing volume element Ar: in a
dielectric region, to rederive its differential form (324). [Hint Divide (336) by An and consider
the meaning of each ratio as As —a 0.] 310. Making use of the divergence theorem, show how the differential expression (321) can
be manipulated to yield the integral form (338). Explain the physical meaning of this result. SECTION 320 311. The coaxial, circular cylindrical conductor pair (coaxial line) of great length and with
the dimensions shown contains a homogeneous dielectric sleeve with the permittivity £. Assume
the static surface charges totaling 1 Qon every axial length if of the inner and outer conductors
respectively. (a) Making use of the symmetry and Gauss's law (337), determine for each region
between the conductors the D and the E ﬁelds. (b) Determine P in the dielectric region. By use
of the criterion (321), determine whether there is any volume density of excess polarization
(hound) charge, of density pp, within the dielectric. (c) Making use of the appropriate boundary
conditions, ﬁnd the free charge densities on the conductor surfaces at p = a and d, as well as the
surface polarization (bound) densities at p = b and c. (d) Letting a = 2 mm, b = 4 mm, c =
8 mm, d = 1 cm, Q]! = 10'2 ,uC/m, and e, = 2.26 (polyethylene), ﬁnd the values oil. and p,
at the conductor surfaces at p = a and d. Find also D, P, and E at the surface ,0 = bl (just within
the dielectric), comparing their values with those at p = b— (just outside the dielectric).
[Answen (d) 13(6) = 90 kV/ln, p,(a) = 0.796 ,uC/mz, E(b+) = 19.9 kV/in] PROBLEMS 175 sin PROBLEM 31 1 312. IAssume that the region a < p < d between the coaxial conductors of Problem 311 is
ﬁlled With a Single, inhomogeneous dielectric material for which the permittivity is €(p) a func
tion ofonly p. (a) Make use of the symmetry and Gauss's law (3—37) to establish the fuiictional
dependence of e on p required to make E between the conductors independent of ,0. Express
the answer such that 6(p) has the value c, at the outer radius ,0 = d. What is then E? (b) Find
both the polarization density ﬁeld P and the volume density p of polarization (bound) charge
for this choice ofe(p). (c) Note that this nonuniform design of the dielectric region provides a
way to avmd high electric ﬁelds in a coaxial conﬁguration, thus reducing the possibility of di
electric breakdown. Suggest how the nonuniform permittivity conditions of this problem might
be met approxtmately, using, say, three or four diﬂerent but homogeneous dielectric materials. [Answer: (a) €(p) = Elsi/p (b) p, = QED/21:6,dtp] The concentric, spherical conductor pair is separated by two dielectric shells of permit I
tivmes 61 and 62 as shown, the interface between them appearing at r = 5. Assuming the static
surface charges totaling i Qon the inner (r = a) and outer (r = r) conductor surfaces respec
tively, answer the following. (a) Use Gauss’s law (337) and the symmetry to deduce D and E
Withinthe two regions. Both these ﬁelds are normal to the interface at r = [1. Which boundary
condition in Table 32 is applicable at this interface? (b) Find the expression for P in each region
From (321),_deduce whether there is a polarization (bound) charge density ,0 within either
dielectric region. (c) Employ the proper boundary conditions to ﬁnd the free liiurface charge
density p, on the conductor surfaces, as well as the surface polarization charge density p at r = b.
(d) With a = 1m, 11: 1.02 m, r I: 1.05 m, 6,, = 2.26, 6,: =1 (air), and Q= 0.1 ,uép ﬁnd the
values of E at the radii b' and 15*. as well as on thehconductor surfaces 1* = a and c. .‘iketch E
versus r from a to c. [Answer (a) E” = Q/47telr’ (d) iE',,(a) = 398 kV/m] r PROBLEM 313 176 MAXWELL‘S EQUATIONS AND BOUNDARY CONDITIONS C" N 33 _ v v
:54 gased on a pillbox construction suggested by Figure 34, prove the boundary condition ' ' f an interface.
(3 50} concerning the continuity of the normal components of B to either side O 34 ‘ ‘ 
:Elfnglzinning with the force (351) acting on each edge of the current loop of Figure .5 7(a), Fill in the remaining details to prove (354). . . . tar
316 Prove that the net magnetic force 1",, acting on anarbitrary, closed, thiiyﬁfaﬂii‘igmy‘):
circtiit carrying the uniform current I and immersed in a uruform B ﬁeld, is zero. [ iii . ' ' l
(352) about the circuit path {, noting that f B x «if can be written 3 x See also Examp e
16.] ‘ n
' ‘  located in the z = 0 plane as in
 h d mensions of the square current loop, ‘
3717. 3E;(l:‘infotatiarlge scale by assuming each side to be 2 meters‘long (side (1 lgcated :I.
Figure. ihe 1*! = 0 plane etc}. The current ﬂows clockwise when looking in the +1: intent;g .
j _ gsiriime the current lbop to be immersed in a magnetic ﬁeld having only the CDmpOﬂTEJ‘cgé
£fitJetch the system. Determine the force JF, due to the magnetic ﬁeld actizfgﬂonBansy :1 men—y
. 1 of the side !.. as well as the total force on (1. showing that F, a: 3.! y y “tin ,
; a"italhat the net magnetic force on the loop is zero. Show also that the differenittia torqhue ri lug
0: current element of t,. relative to the moment arm R = a; + o,o measure roir: t 1;: tgrim;
iindT = 3,133 23:11:: and that the total torque on (I is zero. (b)dR{epeat (a). afucfll‘lrhgl the (ma:
I' . Sh wh the forces on the sides {I an ‘ are zero, n
that only lihiislopﬁpscdiie tool; , isylyiRZIBI. Find the total torque, due to B, pnly,Il)3yﬁatialoglyé
:l‘luihree components of]! present, what is the total torquez on the loop. (all e 111:: u
mzignetic moment ofthis ﬁnite sized current loop as In = In: :31) tzﬁashgvyhtiiziagﬁgg‘tzamomtlm
' ‘ ed 'n {b is e uivalem to (354), = m x . I _ I
:finrthiﬁn clirreni 100‘; of sides 2n = 10 cm and carrying 10 A, immersed 1n the ﬁeld
B = 0.3.; + 0.4., + 0.5., bem2? ' ' ’ ‘ “I 3  d with the steady
" tizable material has 5.3 x 10 atomsj'ni , an ‘
318. ‘Aﬁgfduliluraifyi‘n'i applied to its interior, there results the averagedipole mgjiiient
magnltifa: Iii—2“ Ami. (a) Find the density of magnetiZation M. the magnetic Siliscepti 1121',
title—relative permeability, and the permeability of this material. (b) Find 3 in t is mater: .
[Answerz M = 39,600 Aim, pi = 12.4 nilHm] . . . . d
' ' ‘ ' " ' ‘ lar s ecimen of magnetic material is measure
. Th ma netic SUSCCpUblllty of a particu p _ A I ‘ B
tsdlge 59. Viliatgis the magnetic polarization M and the magnetic intensity H, if the ﬁeld in the material is 0.0lax bemz? 320. Given the following volume magnetization ﬁelds M within certair;I regions of magnetic materials, ﬁnd the volume densities J. produced by the bound Currents t eretn. (a) 150”, (b) 1,200,351I (c) n.520 (cylindrical) (d) aglotlr cos 9 (e) 1,160,“: [Answen (b) 0 (d) “‘200 cm 6] h l l t' hip (3 56) relating the
‘ ’  l to show how t e cur re 3 ions  , :3;etifrtiiiioriziiiiik:ii:iiiiliﬁraltbztlifinagnetization current density J“, is transformed to the integral relation (36? )1. SECTION 341‘ _ z A I
322 Show that the magnetization current denstty j”, = —a,l[} Afrn as:fociat‘edt writ; $1:
bourid currents in the sample of Example 33 yields, from an appropriatie :ﬁiatastfmtglgrogmn
d rrent ﬂow of [06’ A through any fixed it cross section 0 I .
ihiaiaifiglhnnsiv: by use of the line integral of (367}. Sketch the system, appropriately labeled. PROBLEMS 177
SECTION 343 323. Employ a suitable sketch, showing how the quantity n I: I'll,
boundary condition (372}, specifies the tangmtia! component of the
in both magnitude and direction. used in the magneticﬁeld
surface current density L 324.. Apply the appropriate boundary condition in answering the following. (a) An airto
perfectconductor interface is at z = 0. the region z > El being air. With H = 150:, Afm in the
air region, what is the surface current density on the perfect conductor? How much total current
I ﬂows in a 20cmwide xdirected strip of this conductor surface? Sketch this system showing
H, J,, and a few current ﬂux lines. (b) Find the current density on the conductor surface of
(a), this time assuming H = 301,, + 40.3, Afrn. Sketch this system. (c) Suppose in the geometry
of Figure l19(a} that the long, straight wire shown is a perfect conductor, and that surface
currents totaling 1 ﬂow on the conductor surface ,0 = a. The B ﬁeld for p > a is still given
correctly by (164}. Use this ﬁeld to deduce the surface current density J, on the wire. Formulate
a vector integral relationship between I and L, showing a related sketch. 325. What two simultaneous boundary conditions are being satisﬁed by the magnetic ﬁeld
refraction expression (376}? Establish that, if region 1 is air and region 2 is iron with pr, = if)4
(a case of high contrast in permeabilities), the tilt angle 3, Of Bl from the normal in region is very small for most values of 92. For example, ﬁnd 91 if 91 = 0, 45°, 89°, and then 39.9”.
How far from the normal must 6'; be if 91 is to become as large as 10“? Sketch this example. 326. The toroidal iron core of rectangular cross section partly ﬁlls the closely Wound toroidal
coil of it turns and carrying the direct current I as shown. (it) Use the righthand rule (thumb
in the sense of I) to establish the direction of H inside the winding. (b) Use the static form
of Ampere's law (366) to deduce H at any radius p within the winding, and determine B for
the two regions. Which boundary condition for magnetic ﬁelds (Table 32) is being satisﬁed
at the airiron interface? (c) From H deduce expressions for the magnetization density ﬁeld M
in the [W0 regions. Sketch ﬂux plots showing (in side views) the relative densities of H, Ema,
and M in the two regions. assuming it, >> 1 for the iron. (d) Find J," within the iron as well
as j”, on the four sides of the iron core. Sketch representatEVe vectors or ﬂuxes depicting these
quantities. {e} If a = 1 cm, 6 = 1.5 cm, (‘2 2 cm, :1: 1 cm, it, =1000, it = 100 turns, .f= 100 mA, find the values ofH and B at p = a+ and {9— (just within the iron], at p = bI» and
,0: c. 327. As a simple exercise in applying boundary conditions, an air space (region 1) deﬁned
for all e > U and a magnetic substrate with p, = 4 (region 2) occurring for all e < 0 are separated
by the inﬁnite plane interface at e = 0. The constant, static magnetic ﬁeld in region 1 is given
to be .3, = 0.3a, + 0.4, + 0.50Jr Wit/nil. Sketch 31 (shown for convenience at the origin) and
the normal unit vector In at the interface (its direction taken as going from region 2 to region 1).
(a) Make use of the boundary conditions (Table 32), concerning the continuity of appropriate
tangential or normal ﬁeld components at the interface, to deduce the vector ﬁelds H1, 3,,
and H, in the two regions, as well as the ﬁeld magnitudes. {Leave H expressions in terms of PROBLEM 326 5 ye. mansram o ownsiuno mu munmur UUNDI'I'IUNS 2: M
3: W
PROBLEM 328 the symbolic 11..) (b) By use of the deﬁnition of n' 3. ﬁnd the angles 9. and 9, between ss
andI(orﬂ)inthetworegiom.(LabelO,onthesketch.)Oieckyouramesslryuseof
(376). [Amm: (a) a,  1.2., + 1.6:, + 0.5, bem' (b) a,  76°] 3”. Averylong, nonmagneticconductorm l) afradiuscearriesthestadccurtentlas
shown. Thecontluctorissunounded byaeylindrical sleeveoi'nomconducting magnetic material
withathicknasumdingﬁnmp—ampbanddtepmnuhiﬁtyn.1hemundingregim
isair. (a) Makeuseof and Ampére‘a law (366) toﬁndllandlinthe three regions.
(Iabelthecloaedlincscmployedin thepmoﬁdepictingﬂinthepmperﬁnsecneachline.)
(b) Find thehlﬁeldinthemagneticregion. [fl628A,¢ l cth l.5ust,p,6forthe
magnetic sleeve, sketch h", B‘. and M‘ versus ,9 for this system. Comment on the continuity
(or otherwise) of these tangential ﬁelds at the interﬁces. (c) By use of (356) and (373b), ﬁnd
the volume magnetization current densityl“I and the bound sufﬁce current densities J_ within
andcn themagneticaleeve. SECTION 35 . 329. Twominﬁninresimair (region l)ﬁirz>ﬂ,andadideenic (regions,th
e  tilen) ibrz < 0, arcseparated by the interﬁsce ate  0. In the airregicn, the constant electric
ﬁeldEl  —l5n,+2m,+30a,Wmisgiven.Sketchl,foreonvenienoeattheorigin.(a)F'md
D and E for both regions, making use of boundary conditions (Table 32). (Leave to explicitly
intheDettpmcslions.) (b) Fuidthereﬁacdonangluﬂ, and9,&omthemrmalinbolhregions,
makingusecfthedeﬁnitionofn°lifnisdirectcdﬁomngion2toregion LUsethereﬁ'action
law (380) at a check. [Answen E: It —l§n,+ 20;,4 7.5.. Villa, 92 I 73.30"] CTION 37
3!”. Provedqu (390a) and (390b) brtheattcnuation oonstantaand thephase
constant 3 associated with uniform plane waves in an unbounded, [any region. 381. Asurneuniformplanewavestobetravelingattheﬁequencyf lWMHzinaloy
region having the constitutive parameters is u 110, e  6:0, 0  10'3 mbofm. (a) By direct sub
stitution into (338), determine the value of the complex pmpagation onnstant associated with
the waves, expressing 7 in its complex rectangular form denoted by (389). From this result
infer the values of the wave attenuation constant and phase constant. (b) Find the attenuation
constant and the constant by use of (3903) and (390b). [Answen a 0.76l prm,
)3 I 5.187 radfm] 342. ReputhblemSSl,thistisneasunﬁngthepammetersofthelmsyrcgiontobenp0,
e: l.8£o. er lOmhofm, and in which uniformplane
f 10 6111. [A'nswen y: 597.? +ﬁ60.5 m"] 383. Making use ofthe freespace parametersppo, (6., and 60. show that the
«missions (SNa}. (390b) and (3998) reducetotheﬁ'eehspaoetuultstl  OJ I 500112118),
and q = no of(2130b). 34¢. Provedlatthcpenmﬁonoflhreeskhtdepthsbyapbnewaveinmamndncﬁw
region produces an amplitude reduction to 5% of the reference value. Show that six skin depths yields 0.25%. mounts 179 I345. Giventheelecuicﬁcldplanewaveaolution 391!) ' ' prepay.“ comm
1s deﬁned by (Retail:wa substitution into the Mixwulcrnwelﬁut (333) 3.1.9 cone: magnetic utionbecomes{3Nc,iftho’mnmc' ‘ 'peﬂance 'tleﬁned
by (399a). [Hulk Show that the coefﬁcient mini}: reduce:II to 8' 1:" W m q u as; Showthatthcuprusinnforinuinsicwaveimpcdanceﬁ,dcﬁnedby(397)as wave penetrate before teaching 5% of its surface value’ Perﬁarm this cal ula ' ‘
lotvradioﬁ'equendesHOItIla intheVLFrange l. mummgc ‘umnmm
tltecosutantse,8landa(40lm atmmtl‘ezmueﬁﬂzmul’ mumIns
undersea radio communication, basedouyour results. SECTION 38 349. Uaethemlm' (390 ‘ .
conduct(forum 0/095) 1), ' 90h 99" m we'll”, a[alilllttlliletoa good mgionattlusﬁequencyisaboummm.Useasketchcl‘therealtinseelecuiclield ' .mns' ' ' I
tithe zongIthocapiainthenseaningofHepth ofpeneostion.” (1:) ﬁnd thetaaunt: this cavealabelsngtton the sketchofpart (b). Comparethiswavelength ‘
Intimregionassunsingnowthatitismmpictelyloale ...
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This note was uploaded on 11/22/2010 for the course ECSE ecse 351 taught by Professor Dennis during the Winter '07 term at McGill.
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