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Unformatted text preview: 3:32 we g (a) Find H in rectangular components at P(2, 3, 4) if there is a current
ﬁlament on the z axis carrying 8'rnA in the aZ direction. (b) Repeat if the
ﬁlament is locatedat x = —l, y z: 2. (c) Find E if both ﬁlaments are present. Two Semiinﬁnite ﬁlaments on the z axis lie in the regions 5—00 < z < a
and a < z ,< 00.. Each carries a current I in the aZ direction. (a) Calculate H
as a function of p and d at z = O. (b) What value of a will cause the
magnitude of H at p = 1, z = O, to be onehalf the value obtained for an , maﬂsesnt? AEurrent sheet K = Sax A/m ﬂows in theregion —2 <. y < 2. in the plane ‘_
z=O.CalculateHatP(0,0,3).l' Aw” “Lgg Aim \M_wn_.wnwmﬂw—_n 74  v. H _ V ,_ ’ _ .1 _ _ :
Assume that there is a region with cylindrical symmetry in which the .
conductivity is given by a = 1.56‘150PkS/m. An electric ﬁeld of 303z V/rn ‘
is present. (a) Find J. .(b) Find the total current crossing the surface ,0 < pg,
2 = 0; all 45. (0) Make use of Ampere’s circuita—l law to ﬁnd H. < g: ,2 Given the ﬁeld H = 20/02 21¢ A/rn: (a) Determine the current density J. (b) Integrate J over the circular surface ,0 = l, 0 < <25 < 2n, z :0, to
determine the tqtal eunenthnassing through that iiithe a; direction. "
(c) Find the total current once more; this time by a line integral around the
circularpathp = 1,0 < 1,23 < 27:,z=0. ' i\ V .Whenlx, y, and z are posrnve and less than in certain magnetic ﬁeld ' mtensrty may be expressed as H = [xzyz/(y + lﬂax + 3262222}, —
[xyz2 / (y +V 1)]az. Find the total current in the ax direction that crosses the
stupx = 2,15 31 5 4, 3 5 z 5 4, by'amethod utilizing: (a) a surface (Z?) a closed line integral. ® The magnetic ﬁeld intensity is'giv'e'n'm": certain regibﬁ'oi’épéeé‘aé n [(x '+ 2y)/2:2]ay + (2/z)az A/m. (a) Find V x H. (17) Find J. (c) Use J to ﬁnd
‘ the total current passing through the surface 2 = 4, 1 5 x s 2, 3 5 z 5 5,
i in the a; direction. (d) Show that the same result is obtained using the other '
__side of Stokes’ theorem. ' A long, straight, nonmagnetic conductor of 0.2 mm radius carries a '
' ‘ uniférmly distributed current of 2 A dc. (a) Find J within the conductor.
(17) Use AmPEre’s circuital law to ﬁnd H and B Within the conductor.
(6) Show that V x H = J within the conductor. (0!) Find H and B outside the, ’
conductor. (e) Show that V x H = J outside the conductor. ' §“ The cylindrical shell deﬁned by 1 cm é < l.4 'ccnsists 5f nonmagnetic conducting material and carries a total current of 50 A in the az .
direction. Find the total magnetic ﬂux crossing the plane it; = 0,0 < z < 1:
(a) 0 < p < 1.2 cm; ((7) 1.0 cm <_ p < 1.4 cm;v(c) 1.4 cm < p < 20 cm. @ Assume that A = 5043232 W'b/m in a certain region of free space. (a) Find
and B. (27) Find J. (cl Use J to ﬁnd the total current crossing'the surface O 5 p.51, 0 5 ('15 k 221’, z = 0. (d) Use the value of [1.15 at p =_ 1 to calCulate‘g
fHdeorp=1,z=O.'  . A point charge, Q = ——O.3 [LC’andm = 3 x 10"16 kg, is moving through
"W/ the ﬁeld E = 30212 V/m. Use Eq. (1) and Newton’s laws to develop the
appropriate diﬁerential equations and solve them1 subject to the initial
conditions at t = 0, v = 3 x 105aI m/s at the origin. At t = 3 us, ﬁnd:
(a) the position P(x, 3:, z) of the charge; (b) the velocity v; (c) the kinetic
energy of the charge. _ ~ A point charge for which Q: 2 x 10"16 C and m = 5 x 10‘26 kg is moving
in the combined ﬁelds E = 100:1: — 20an + 30032 V/m and B = —33x +
2ay —— az mT. If the charge velocity at t = O is v(0) = (23K —— 333, — 43;)105 m/s: (a) give the unit vector shoWing the direction in which the ’ charge is accelerating at t = 0; (27) ﬁnd the kinetic energy of the charge at
t = O.  @ éiyen‘a material for which Xm = 3:1 and Withjn WhjCh B = OAyaz T, ﬁnd: @H; (b) u; (0) gr; (d) M; (e) J; (f) 11;; (3) Jr Let [id = 2 region 1, deﬁned by 2:: + 3y —— 4z > 1, while 11,; = 5
in region 2 where 2x + By — 4z_ < 1. Ingregion 1, H1 = 503Jr — 302), +
2021Z A/m. Find: (a) Hm; (19) Ha; (c) Ha; (d) HNZ'; (e) 61, the angle between ELEPWNZI; (f) 92’ the angle heme“ H2 and aNZI' . ‘ g (a) 4.833,. — 7.242.y + 9.66aZ A/m ' ' ‘ ' ‘ “ ‘ " £9m‘5 ‘ (b) 54.83ax 7 22.7633, + 10.3421z A/m / mg (1:; (a) ~29421,r + 196a,, nA/m
(c) —42lax + 578a,. uA/m (g) (a) (b) 1/45 @ (a)45e_150pazkA/.m2 1
' (b) 12.6[1 — (1 + 150p0)e150P01 A (a) hoop/110345 A/m, —100pa¢ Wb/mz (c) 2—p@[1 — (1 + 150p)e~15091 A/m
(406010121Z A/m2 (b) 4cm A (6)4017 A
(a) —259 A (b) —259 A (a) 21:: + 2y>/z3ax + We A/m
(b) same as part (a) (c) 1/8 A
£33") (@159 x 107aZ A/m2 (b) 7.96 x 106pa¢ A/m,
10p2¢ Wh/m2 (c) as expected (d) 1/(er)a¢ A/m,
uo/(rrp)a¢ Wh/rn2 (e) as expected 1 (a) 0.392 ,u.Wb_ (b) 1.49 awn (c) 27 “W13 @ (b) _§§az A/m2 (c) —500 MA (d) —500 MA it" (a) (90, 0, —.135) (b) 3 x 105a; * 9 x 10432 111/5 13 r (a) 77.6yaZ kA/m (b) 5.15 x 106 H/m (c) 1.5 x 10‘5 I
12‘" (a) .70ax + .703}, ~ .1232 (b) 7.25 f] (c) 4.1 (d) 241 yaz kA/m (e) 77.63,; kA/m2
(f) 2413.. 19:1ij (g) 318ax kA/m2 I‘V’ .a. Find H in cartesian components at P (2, 3, 4) if therei‘s’a dﬁiéht‘tiéﬁéht on the z axis carrying 8 mA in the a: direction: I
l Applying the ‘Biot—Savart Law, we obtain _ , _ muting” °°Iczzaz><[23a+3ay+(4—z)az1—/m ———Idzl2ay—Baa]
H“ ’ Loo W ‘ 00 _ 00 477022 — 812+29W2 Using integral tables, this evaluates as I 2(22 — sway 43am) °° I ‘
= —— ————.———~——— = —— 2 — 3
Ha 4w [ 52(22 — 82 + 29)”2 _ 26w< a‘y am) Then with I = 8 mA,we ﬁnally obtain Ha = ~—29'/la,I + 196ay uA/m b. Repeat if the ﬁlament is located ata: = —1, y = 2: In this case the .Biot—Savart integral
b’ecomes  ‘ H _? . °° Idzaz >< [(2+1)ax+(3 2)ay+(4—z)azl =
b _ _oo 477(22—82+26)3/2 I °°‘ I dz[3ay — am]
hm 47r(z2 — 82 + 26)“? i . Evaluating as before, we obtain with I = 8 i __I 2(22—3)(3ay“aw)m 1. _ =_127’ ’382 A—m 40(z2  82 + 26)1/2' _ c. Find E if both ﬁlaments are present: This will be just the sum, of the results of parts a and b, or y .
HT = Ha + Hz, = ~421am _+ 57831, [JA/m This problem can also be done (semewhat more simply) by using the known result for H from I ‘ The BiotSavart method was used here for thesalre of illustration. \ Two semiinﬁnite ﬁlaments on the z axis lie in the regions —oo <
error in problem statement) and a < z < on. Each carries a. current I in the az direction. ,
a) Calculate H as a function of p and 915 at z = 0: One way to do this is tense the ﬁeld from'
an inﬁnite line and subtract from it that portion of the ﬁeld that would arise onm the
current segment at —a < z < a, foundfrom the BiotSavart law. Thus, ~ ’ H___;I_a __/f‘ Idz_az><[pap—zaz]
i _ 2,173 e _a 4ﬂp2 +z213/2 ..:r 4» x1”; ' ‘ The integral part simpliﬁes and is evaluated: a. _ Idzpaqg _Ipa 2 “ _ Id _.__a.
I ' 4 _a 47T[,02+2.2l‘°"/2 —.47V ¢p2 p2+z2 a_2wpx/p2+<12 45
I Finally,
I a
H=—_ 1_——————— a Am
21rp‘: /p2+al2] / I b) What valueof a will cause the magnitude of H at p = 1, z = O, to be one~half the value.
. obtained for an inﬁnite ﬁlament? We require ' 11_____.___a :1 =>7 'a 2:l : a=1/\/§
\/'p2+a.2 pzl 2 1+1; 2 —— an inﬁnitelylong wire in cylindrical components, and transforming to cartesian components. «1.?
!‘, “Hf, .'_._._.'._‘,.\. __ ,' ‘.
J. < —a (note typographical 1" " "i ' ' A current sheet K = 8a; A/m ﬂows in theregion —2 < y < 2 in the plane 2 = 0. Calculate I H at PG), 0,, Using the RiotSavart law, we write K x aRdz cly 2 °° 8a,, >< (~93am —ya +33”)
H = ————————— = ———————————————' y a ‘ d d
P (“R2 [2100 47T(232+y2+9)o/2 a: y
Taking the cross product gives: 2 co __ __
HP=/ / 8( ya; Sawdrdy
 2 m 41r(2:2+y2+9)°/2 We note that the 2 component is antisymmetric in 3; about the origin (odd parity). Since the. ~ l limits are symmetric, the integral of the 2 component over 3,: is Zero. We are left ’with H _/2 /” ~24aydxdy _ 6 f2 9,—  i re d
P __2. __oo 4W($2+y2+9)3/2 Way _2 (y2+9) r—————$2+y2+9 ice y  I
' 6 2 2 4 12 1 _, y 2 4' ' _. —;r—ay [2 y2 +9 dy — ——?ay gtan _2 —I—;1_(2)(O.59)ay — —l.50ay :1 ® AsSume that there is a region with cylindrical symmetry in which the conductivity is given by
r 0' = 1551509 kS/m. An electric ﬁeld of'3_0 az V/m is present. 
3.) Find 3: Use ‘  2
J = «7E = 4562—150" 3., kA/m b) Find the total current crossing the surface p < pg, z = 0, all qﬁ: 27" P0 _ '_150 PO
. __ —150p _. P ....]_ . kA
I=ffJdS—/o f0 45e‘ pdpfict (150)26 [ i p In  =12.6[1(1+150pg)e15DP°] A 1 c) Make use of Ampere’s circuital law to ﬁnd H: Symmetry suggests that H will be ct—
directed only, and so we consider a circular path of integration, centered on and perpen—
dicular to the z sods. Ampere’s law becomes: 271'qu5 = 1mg, where I and IS the current
found in part b, except with pa replaced by the variable, ,0. We obtain H Given the ﬁeld E = ZOPZ aé A/ugciz ' Ar_ __; ,. V , , a) Determine the current density J: This is found through the curl of 11,. which simplifies
to a single term7 since H varies only with p and has only a 45 component: '
1 d(pH¢) _ 1 d ~3=VxH=~ az___ 203 az=60 az'Amz
p dp de< P) __P___L_ i q ' b) Integrate I over the circular surface p = 1, O < qﬁ < 271', z = 0, to determine the total current passing through that surface in the 21; direction: The integral is: 1 “Bar 1 V
I=//JdS=/ fﬁﬂpazpdpdqbaz=407rA
o o . c) Find the total current once more, this time by a line integral around the circular path
p=1,0<¢<27r,z=0: ‘  2w 21r
I = fﬁ  dL =f 20,32 a¢IP=1(I)d¢a¢, =/ 20dq§ = 4071'. A
o o I When 1;, y, and z are positive and less than 5, a certain magnetic ﬁeld intensity may be
expressed as H = [m2yz/(y + + 3392231, — [ryzz/(y—l— 1)]az. Find the total current in the _ ‘ 343 direction that crosses the strip, :2: = 2, 1 _<_ y S 4, 3 5 2 S 4, by a method utilizing: a) a surface integral: We need to ﬁnd the current density by taking the curl of the given H.
Actually, since the strip lies parallel to the yz plane, we need only ﬁnd the a: component
of the current density, as only this component Will contribute to the requested current. This is _
8H; dilly) __ _‘( $22 3;, " 6:: (y + 1).2 Jm=(V><H)m=< +6mzz) The current through the strip is then 4 4 ' 2Z2 . 74 _222 24 ’4 d
I'=/Jada=—f/ +242)dclz=— C + 2y) z
s z " 3 1 (y+1)2 y _/3 (y‘i'l) 1 4 ‘4 3 r, 1
= —/ (—z‘ + 722') dz = — (—23 + 3622) = —259
3 5 5 ' 3 7 ‘_“~‘r“‘—.._._ ,.__ _.__‘.__._._‘......~_.,..h .,. VV.\_‘_\ ~‘n b.) a. closed line integral: We integrate counter—clockwise around the strip boundary (using the
righthand convention), Where the path normal is positive am. The current is then 4 4 . 1v 3 r '2
‘ I=][H.dL=/l 3(2)2(3)2dyf/3 {ﬁg dzl—L 3(2)2§4)2dy_+/ 72:221.) dz s
= 108(3)  15(43 — 33) + 192(1 — 4) — $33 — 43) = ~259 ,.~._.,____v.... . i ‘_<~_,_§‘ _ s The magnetic ﬁ€1d iﬂtensitr is given in a certain region of space as _ ‘ ' _ 
‘i , ' a: + 2 ' 2
z" z ‘ a) Find V x E: For this ﬁeld, the general curl expression in rectangular coordinates simpliﬁes
. to . t
8H,, 3H,, __ 2(3: + 23/) 1 .__._. ._L._.
amI 81: az 23 aw .. zzaz A/m V H:
E X dz b) Find 1: This will be the answer of part 0,, since V X H = J. c) Use J to ﬁnd the total current passing through the surface z = 4, l < a: < 2, 3 < y < 5, in the az direction: This will be 5 2 i ll . I=//le=4'azdmdy=f / Edmdy=l/8A i  , a 1 __. —.._.._.._n_...y.__..,_,_. ,... . _ 1 Show that the same result is obtainedamusing the other side of Stokes’ theorem: We take l 39 H  dL over the square path at z = 4 as deﬁned in part c. This involves two integrals of the I y component of H over the range 3 < y < 5. Integrals over as, to complete the loop, do not exist Since there is no :1: component of H. We have i , I = frag,  dL f5”
3 16 M .1, A long straight nonmagnetic conduotor of 0.2 mm radius carries a. uniformlydistributed ‘ current of 2 A do.  _ , v i r
a) Find J within the conductor: Assuming the current is +2 directed, E 2 J:——————————— ,=1.59><107 ,Am2
7,—(02 X 10—3)2 a“ #3:; b) Use Ampere’s circuital law to ﬁnd H and B Within the conductor: Inside, at radius ,9, we have
pJ 3—
Then B = solI = (477 x 10—7)(7.96 X 105)pa¢ F 10pa¢ Wb/mg. I 27rpH¢J = 7rsz => H: age. = 7.96 x 105p aqs A/m c) Show that V x H = J Within the conductor: Using the result of part b, we ﬁnd, 1 d 1 d 1.59 x107p2 7 2
=—— =—— ——————— =1.59X10 2A =3
V XII pdp(,ofi’¢)az pdp( 2 )a2 a /m d) ’Find H and B outside the conductor (note typo in boolc): Outside, the entire current is
I enclosed by a closed path at radius p, and so ' I 1
H—ﬁa¢—;—f;a¢ A/m New B = MOII = [Jo/(WP) a4, Win/mg. e) Show that'V X H = J outside the conductor: Here we use H outside the conductor and write: . 1d 1d 1 V X H = ——(PH¢)'az = —' (9*) at: = Q (ﬁtS expected) ,, _ Pdp . .de 7”” The cylindrical shell deﬁned by 1 cm < p < 1.4 cm consists of a nonmagnetic conducting
material and carries a total current of 50 A in the a: direction. Find the total magnetic ﬂux crossing the plane 9b = O, O < z < 1:
a) 0 < p < 1.2c1n: We ﬁrst need to ﬁnd J, H, and B: The current density Will be: 50 5 I a
= m 7 = . 7 A‘ 4
J 7r[(1.4 x 10“?)2 .— (1,0 x 10—2)2] al— 1 56 X 10 ad /m Next we ﬁnd H¢ at radius p between 1.0 and 1.4 cm, by applying Ampere’s circuital law,
and noting that the current density is Zero at radii less than 1 cm: 27:" p ' ’
27rpH¢ = and = f f 1.66 X 10°pl'dp' dgb
o 102 A (P2 ‘ 10—4) 2 2
=;~» H45 = 8.30 x 10‘—7— A/m (10' m < p < 1.4 x 10— m) Then B = 11011, or ‘
p2 — 10‘4) B = 0.104 p a4; Wis/m2
New,
 1 1.2><1o—2 104
@a=//BdS=// O.104[———]d’pdz
' 0 102 P 2 —2 2 _ “'4 I I
w .. 10—4 m = 3.92 x 10—7Wb = 0.392 qu = 0104 l 2 1.0 b) 1.0 cm < p < 1.4 cm (note typo in book): This is part a over again, except we change the
upper limit of the radial integration: —2
1.4x10 10—4 1
@b=//BdS=// ]dde
o 102 P _2 2 __ —4 v
W9... _ 10—413 = 1.49 x lO‘GWb _== 149 PWb 0.104 [p —— = 0.104 [ c) 1.4 cm < p < 20 cm: This is entirely outside the current distribution, so We need B there: We modify the Ampere’s circuital law result of part a to ﬁnd:  1.4. —2 2 — 4 10’5 ‘
Bout = 0.104K—X—10—l—10—l a¢ = 21¢ Wb/m2
P P
We now ﬁnd
1 20x10“2 10—5 20
@czf / dpdz=1051n(——)=2.7><10“5Wb=27qu
o 1.4x10—2 P 14 @ nwu Assume that A = 50,028.: Wb/In in a certain region of ﬁne space, I i ' 2.) Find H and B: Use ~ , B = \7 x A =‘—%a¢ = ~100pa¢ W‘b/mL’ I
Then H = B/ﬂo = “mop/#0 345 A/m . I l .
i‘ .13) Find J: Use II 1 a 1 3 4100,92) ' 200  2 J=VxH=— H ,=— a=———azAm _ r
pap”)  ‘03” papl #0  z #0 / a /
 c) Use J to ﬁnd the total current crossing the surface 0 S p S l, 0 S <35 <I27r, z :0: The
2‘ current is ~ ' 21: 1__ —200 '
; I=//Jds;/ / 200a,.azpdpdq5: WA=—500MA
‘0 ’0 #0 _ .U'O . d) Use the value of H45 at p = 1 to calculate 3513 03L for p = 1, z = 0: Have = 400“ A = —500 MA #0 . . {I ' A point charge, Q = —0.3 MG and m = 3 X 10*15 kg, is moving through the ﬁeld E = 30 a,  v Use Eq. (1) and Newton’s laws to developthe appropriate differential equations and solve them,
i subject to the initial conditions at t = 0: v = 3 X 105 a.” m/s at the origin. At t = 3us,.ﬁnd: a) the position P(a:, y, z) of the charge: The force on the charge is given by F = qE, and Newton’s
l
i
l
I
l dgz
dig :
3 second law becomes: I
i
F = ma = m i '= qE = (—0.3 x 10—6)(‘30 a2) describing motion of the charge in the z direction. The initial velocity in z is constant, and
so no force is applied in that direction. We integrate once:  dz qE . ' at — “z  F? t 01
The initial velocity along 2, 212(0) is zero, and so 01 = 0. Integrating a second time yields the
z coordinate: ' . The charge lies at the origin at t = O, and so 02 = 0. Introducing the given values, we ﬁnd i  = (—0.3 x 105)(30) 2 X 3 X 10% = —1.5 x 10101:2 m Z At t = 3 ,u.s, z = ——(1.5 X 101°)(3 x 10‘6)2 = .—.l35 cm. NOW, considering the initial constant
velocity in w, the charge in 3 us attains an a; coordinate of a: = ct = (3X 105)(3 X 10‘6) = .90 m.
In summary, att = 3 ps we have P(a:, y, z) = (90,0, —.135). '«. ~ ‘ b) the velocity, v: After the ﬁrst integration in part 0., We ﬁnd 1;, = 37% = —(3 x 101°)(3 x 10—6) = «9 x 104 m/s including the intial mdirected velocity, we ﬁnally obtain v = 3 x 105 a3 — 9 ix 104212 m/s. I . ' = mlv!2'= .—(‘3 >< 104631113 x100)“ = 1.5 x 10‘“ J . .r_§____> .3 r  A point charge for which Q = 2 X 1016 G and m = 5 x 10‘26 kg is moving in the combined ﬁelds E = lOOaw  200a? + 300az V/m and B = —3a.s + 22y — az mT. H the charge velocity at t = 0 is 3 . V(O) = (2a.:  33y — 421:) x 105 111/5: . ‘ a r a.) give the unit vector showing the direction in Which the charge is accelerating at t = 0: Use 1 F t = O) = q[E + (v(0) x 13)], where d ' ‘ 1 1 v ’ v(0) X B =1 (2a.; — Say —.4az)105' >< + 23? — 3.2)10"3 = 11.0051m + 14003.5. 4 500.33
. 1 So the force in newtons becomee  g ‘
F(o)' = (2 x 1015)[(10b41106)am+§1400—200)ay+(3oo—500)az] = 42 10—14[6aq +5211;— az] The unit vector thatgives the accelerationdirection 1e found tom the force to be ( '63a:+6‘3»g—az 8.1: '= = freem + .7031,g — .12az $723, . h) kdnetic energy of the charge at t = O:‘ / v. ammo)? =, %(5 x 10—26. kgj(5.'39 x 105'm/s)2 = 7.25 21015 J = 7.255 ' t I; h . . ... _ ,...._,r. engage”, 32:143.}    1’) t=<1+31JHO=W
iv 1“ C) Hr_=(l+3.1)'=4_._l_, I d) M; me = (3.1)(77tyay) =. 241yaz kA/m ’
e) 3 = v x H =,(de)'/(dy) ‘am '=,77.63¢ ILA/1:12; f) Jb; v x M = (mg/(em am = 241%. kA/mz; 'j , g g) 3T e V x B/Po .= 318% kA/mg  g? " ——..;‘_._. ' ILet #51 = 2 in region 1, deﬁne'cigby QwESZI—dz nhﬂe 141.3 ¥ 5 inregion 2 wherev25n+3y—4z < 1. .—
In region 1, H1 = 5an — soay‘+;20az'A/m. Find: . ‘  .
e) HNl (normal component of H1 at the boundary): We ﬁrst need a Limit Vector'normal to the
surface, found through ‘ ’ ' ' ‘ V(2:r+.3y—4z) 'Zar+3ay_4a . . "
a = ——_* = _________.z = ._.I
N [V'(2a: + 3y — 42M vie? ‘ “9’73” + 563? ‘ ﬂat Since this vector is fonnd through the gradient, it will point in the direction” of increasing
values of 2a: + 3y — 4z, and so will be directed into regionl. Thus We write em = aygl. The
normal component of H1 will now be: ' ' ' HM ’= (H1 ' 3N21)a2\r21 = [(50.212 — 30a? + 20az) . (37% + 5day — .74a,)] (.3733 + .seay — ﬂag)
= ~4.83am — 7.24211, '+ 9.66512 A/m 41. b) ET {ta—.2321; 2::sone11t ole at the boundary): ‘ ._ v.   §
Em = E1  HNI‘. . . = (50am — '3an + 20%) —_ («4.83:1cc — 7.24217}, + 9.66az) = 54.83.21m — 22.76ay‘+ 10.34212, A/m :_— ‘Vtengential Vcomponeht of H2 at the houndarjr): Since tangential components of H are 0011. ::;ous aeross a. boundary between two media of diﬁ'erent permeabihties, we have HT; ; Em = 54.83am — 22.76% + 10.342: A/m d) Hm (nofmal comﬁonent of Hg. at the boundery): Since normal components of B are contin— uous across a boundary between media. of different permeabﬂities, We write WE N1 = #313372
01' ' Mrl
#122 Hm = HNl = 125—(—4.83am — 7.24% + 9.662%) = ~1.93am — 2.90% + 3.8635 A/m e) 491, the angle between E1 and aﬁm: This will be __ E1 50am —— 30%, + 202.3 I ' ~ '
COS 61 ' aNgl = ' + — .7432 — ' Therefore 61 = cos‘1(—.2l) = 102°.
‘f) 62,. the angle between 33 and aNgll First, ' ' ‘ ' ' _’ Hg = Hm + Hm = (54.33.21,: — 22.76% +.1o.34az) + (—1.93% — 2,9021%, + 3.86az) ’
= 52.90% — 25.56% + 14.20a5'A/m‘ ‘ New
H3 , 52.90343, — 25.66% + 14,205., ' ' ' '
9 =  = ——————————————————  . .56 — .74 = —U.09 005 2 ‘H2‘ aNzl [ 6049 (37am+ az) _ Therefore 62 = cos—1(——.DQ) = 9__5_°_.  .’ . ' ﬂied—ref'f/fl I ...
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This note was uploaded on 08/12/2010 for the course ECE ECE 302 taught by Professor Ferguson during the Spring '10 term at Cal Poly Pomona.
 Spring '10
 Ferguson
 Electromagnet

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