Hw7

1

Assignment 7 — Solutions 3/7/04 Problem 5.2 (a)

For a single loop of current I on the x-y plane centered at the origin and observation point at f = 0: ÷” ÷”£ ÷” m0 I ÷ ” R ÿ a B1 = ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ “ W , W = ® ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 4p R3 S ` ÷ ” a£ = v£ v£ f£ z£ ` £ , ÷R” = x” £ - x” ` + z z` , x” £ = v£ v x” = v v R2 = †x” £ §2 - †x” §2 - 2 x” £ ÿ x” = v£2 + v2 + z2 - 2 v£ v cos f£

JK R S x

K x

z JK da′

JK x′

y

For z ∫ 0: W = ‡

2p

0

f£ ‡

2p ∑W ÅÅÅÅÅÅÅÅÅÅÅÅ = ‡ f£ ∑z 0

= ‡

0

2p

f£

2p ∑W ÅÅÅÅÅÅÅÅÅÅÅÅ = ‡ f£ ∑v 0

-z v£ v£ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 0 Hv£2 + v2 + z2 - 2 v£ v cos f£ L3ê2 b 3 z v£ v£ y £ jij- ÅÅÅÅ v Å ÅÅÅ Å Å + ÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 2 zzz ‡ 3 5 R 2 R k { 0 £ £2 2 2 b v - 2 z - 2 v£ v cos f£ L £ -v Hv +ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ‡ v ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 0 Hv£2 + v2 + z2 - 2 v£ v cos f£ L5ê2 b 3 z v£ H2 v - 2 v£ cos f£ L £ ÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ v ‡ 2 Hv£2 + v2 + z2 - 2 v£ v cos f£ L5ê2 0 b

∑W ÅÅÅÅÅÅÅÅÅÅÅÅ = 0 ∑f We can imagine the solenoid to be made up of an infinite stack of circular current loops each carrying current I . Since the system is azimuthally symmetric, fix the observation point x” at f = 0, and since the solenoid is infinitely long fix z = 0. The total magnetic field is the superposition of fields due to each loop, which we have from above as a function of relative position of the observation point from a given loop. Thus ¶ ¶ m0 I N ÷” ÷” BHvL = ‡ z N B1 Hv, zL = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ‡ z 4p -¶ -¶

ij ` ∑W ` ∑W y jz ÅÅÅÅÅÅÅÅÅÅÅÅ + v ÅÅÅÅÅÅÅÅÅÅÅÅ zz ∑v { k ∑z

∑W ê ∑v is odd in z in the range -¶ < z < +¶, so that integral vanishes. ∑W ê ∑ z is even in the same range, so for some infinitesimal e > 0:

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2

m0 I N ÷” BHvL = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 4p m0 I N = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 4p

-e e ¶ i ∑W y z N jjz` ÅÅÅÅÅÅÅÅÅÅÅÅ zz ì ‡ z = ‡ z = - ‡ z k ∑z { -¶ -e e -¶ ¶ e ÄÅ ÉÑ e I N ∑W m Å Ñ 0 ` Å Ñ` ‡ z z ÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ lim+ W - lim- WÑÑÑÑ z zØ0 4 p ∑ z Ç Ö zØ0 -e

J‡

-e

z + ‡

e

z + ‡

¶

The splitting of ranges was necessary because W is discontinuous at z = 0, so ∑W ê ∑ z is proportional to a d-function. For v > a (observation point outside solenoid) limzØ0≤ W = 0 since no angle is subtended by a surface in the same plane but not inclusive of the observation point. For v < a, the area subtended approaches the entire hemisphere:

JK da′

JK da′

JK R

ε lim W = 2 p

zØ0+

JK R

−ε

lim W = -2 p

zØ0-

ï

lim W - lim- W = 4 p

zØ0+

÷” B m0 I N z` , v < a I N z` , v < a ÷” ÷÷÷” BHvL = : ï H HvL ª ÅÅÅÅÅÅÅÅÅ = : 0 , v>a 0 , v>a m0

zØ0

(b) Let's assume that a "realistic solenoid" is an infinite helix of outer radius a and with N turns per unit length, its axis being the z-axis. For a circular Amperian loop concentric with the solenoid and with radius v > a, the plane of the loop is pierced exactly once by the wire (geometrical property of helical curves), so 2p ÷” ” a f Bf Hv > a, f, zL = m0 I ‡ B ÿ l = ‡

÷” For any B -field related quantity f Hv, f, zL, define the average of f over a segment of length L of the solenoid as X f Hv, f, zL\ ª AŸzz+L z£ f Hv, f, z£ LE ë L. We can argue that X f \ is only a function of v in the limit L Ø ¶: 0

X f \ cannot depend on z because the solenoid is infinite so providing we average over enough turns X f \ should be invariant under translations in z, which can only be if it is independent of z.

X f \ cannot depend on f because when averaging over sufficiently many turns, a helix is symmetrical under rotations about its axis so the averaged quantity X f \ must also be independent of f. Thus

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3 z+L z+L 2p 2p 1 1 Xm0 I\ = ÅÅÅÅÅ ‡ z£ m0 I = ÅÅÅÅÅ ‡ z£ ‡ v f Bf Ha, f, z£ L = [‡ a f Bf _ L z L z 0 0 2p z+L 2p 1 1 ÅÅÅÅÅÅ m0 I L = ‡ v f ÅÅÅÅÅÅ ‡ z£ Bf Ha, f, z£ L = ‡ v f XBf \v>a L L z 0 0

m0 I = XBf \v>a ‡ 0 m0 I XBf \v>a = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 2pv

2p

v f = 2 p XBf \v>a

Similarly an Amperian loop of radius v < b, where b is the inner radius of the solenoid encloses no current, so XBf \v
We can show by contradiction that XBv \ = 0 everywhere. Suppose that XBv \ ∫ 0. Since XBv \ is independent of f and z, for a cylindrical Gaussian surface concentric with the solenoid ÷” ` 3 ÷ ” ÷” ‡  x “ ÿ XB\ = ® XB\ ÿ n a V S r 2p = ‡

0

+‡

v ‡

0

z0 +L z0

z

v f HXBz \z=z0 +L - XBz \z=z0 L

z ‡

2p

0

nˆ = zˆ z=z +L

r

0

z0 + L

r f XBv \v=r

nˆ = ϖˆ ϖ =r

S

z0

nˆ = − zˆ z =z

0

The integrand in the first term vanishes since XBz \ is independent of z. XBv \ is independent of f and z so ÷ ” ÷” 3 ÷ ” ÷” ‡  x “ ÿ XB\ = 2 p r L XBv \ ∫ 0 ï “ ÿ XB\ ∫ 0 somewhere V

But this is a contradiction of the fact that

z+L L 1 1 ÷ ” ÷” ÷ ” ÷” ÷ ” ÷” 0 = Y“ ÿ B] = ÅÅÅÅÅÅ ‡ z£ “ ÿ BHv, f, z£ L = ÅÅÅÅÅÅ ‡ z£ “ ÿ B Hv, f, z£ L ÷ ” ÷” L 0 L z “ ÿB = 0 ï L ÷” i 1 ÷ ” ÷” ÷” y = “ ÿ jj ÅÅÅÅÅÅ ‡ z£ BHv, f, z£ Lzz = “ ÿ XB\ kL 0 {

So it cannot be true that XBv \ ∫ 0.

For a rectangular Amperian loop G lying on a f-constant plane with the "left" edge within the solenoid and the "right" edge off at infinity:

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4 z0 +L ÷” ” z£ HBz §v - Bz §v+l L ‡ B ÿ l = ‡ G

z0

+‡

v+L

v

v£ HBv §z0 +L - Bv §z0 L

m0 Ienc = L XBz \ + ‡ Ienc = :

z

v+L

v

ϖˆ

v£ HBv §z0 +L - Bv §z0 L

zˆ

NIL , va

Xm0 Ienc \ = XL XBz \\ + ‡

v+L

v

−ϖˆ

z0 + L

− zˆ z0

l

ϖ v£ HXBv \ §z0 +L - XBv \ §z0 L ï m0 Ienc = L XBz \

Finally we get l m0 I N z` , v < b o o ÷” o m0 I ` XB\ = m o o o ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ f , v > a n 2pv

Problem 5.3 Let z be the distance from the observation point (chosen to be at the origin) to one of the loops in the solenoid (carrying current I flowing clockwise as seen in the x-y plane), which then subtends an angle q: W = ‡

2p

f£ ‡

q

= 2 p H1 - cos qL ∑W ÷” “ W = ÅÅÅÅÅÅÅÅÅÅÅÅ z` ∑z 0

0

q£ sin q£ = -2 p ‡

1

cos q

Hcos q£ L

z

a

θ1

L1

Integrate over all loops using the formula in problem 5.1 to get the magnetic field:

L1 L1 m0 I ÷ ” m0 N I ∑W m0 N I ÷” z N ÅÅÅÅÅÅÅÅÅÅÅÅÅ “ W = ‡ z ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅ z` = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ HW §z=-L2 - W §z=L1 L B = ‡ 4p 4p ∑z 4p -L2 -L2 m0 N I m0 N I = ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅ @2 p - cosHp - q2 L - 2 p + cos q1 D = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Hcos q1 + cos q2 L 2 2

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0

θ2

L2

5

Problem 5.6

÷” For any cylinder of radius R with constant current density J = J z` , Ampere's law for a circular loop G of radius ` ÷” v < R concentric with the sylinder and on a constant-z plane gives [with B = B HvL f from azimuthal symmetry]: ÷” ” ÷” ÷” = m0 J p v2 BHvL 2 p v = ® B ÿ l = m0 ‡ J ÿ a 1 1 ` ÷” B = ÅÅÅÅÅ m0 J v f = ÅÅÅÅÅ m0 J H- y x` + x y` L 2 2 G

S

Displacing the cylinder by x0 along the x-axis is accomplished by substituting x Ø x - x0 . By superposition

y

y

JK J = J zˆ d

a

x

=

y

JJK J ′ = −J zˆ d

x

JJK J ′′ = J zˆ

+

x

b 1 1 1 ÷” ÷”£ ÷”≥ B = B + B = - ÅÅÅÅÅ m0 J @- y x` + Hx - dL y` D + ÅÅÅÅÅ m0 J H- y x` + x y` L = ÅÅÅÅÅ m0 J d y` , v < b 2 2 2

Problem 5.14

÷” Neglecting end effects we have a 2D problem, i.e. the z = 0 slice taking the cylinder axis as along the z-axis. J = 0 ÷” ÷” everywhere means we can write B = -“ F where the magnetic scalar potential F satisfies Laplace's equation and can be expanded in Legendre polynomials: F = ‚ IAl vl + Bl v-Hl+1L M Pl Hcos fL

l FI , v § a o o o o II = m F , a§v§b o o o III o , b§v nF ` ÷” lim F = B0 x = B0 v cos f lim B = B0 x ï vض l=0

JK B0 = B0 xˆ

y

¶

ΦΙΙΙ

vض

The asymptotic form of F implies only the l = 1 term is nonzero in FIII :

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ΦΙΙ

ΦΙ a

µ0

µ

b

x

6

i BIII y FIII = jjjjB0 v + ÅÅÅÅÅÅÅÅÅÅÅÅÅ zzzz cos f v { k Continuity at the II-III and I-II boundaries similarly restricts the expansions to the l = 1 terms. Also in region I finiteness at v = 0 forces us to drop the 1 ê v term: i BII y FII = jjjj AII v + ÅÅÅÅÅÅÅÅÅÅ zzzz cos f ï FI = AI v cos f v { k

The boundary conditions are that B¦ = -∑F ê ∑v and H˛ = -@1 ê m HvL vD ∑F ê ∑f be continuous: -B¦

Region I

v§a

II

a§v§b

III

b§v

AIII cos f

BII zy jij AII - ÅÅÅÅÅÅÅÅ ÅÅ zz cos f jj 2z v k { III ij B y jjB0 - ÅÅÅÅÅÅÅÅÅÅÅÅÅ zzz cos f z j v2 { k

-H˛ 1 - ÅÅÅÅÅÅÅÅÅ AIII sin f m0

BII zy jij AII + ÅÅÅÅÅÅÅÅ ÅÅ zz sin f jj 2z v k { III i 1 B y - ÅÅÅÅÅÅÅÅÅ jjjjB0 + ÅÅÅÅÅÅÅÅÅÅÅÅÅ zzzz sin f m0 k v2 { 1 - ÅÅÅÅÅ m

BII v = a : AII - ÅÅÅÅÅÅÅÅÅÅ = AIII a2 m0 m0 i BII y BII i m - m0 y AII + ÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅ AIII = ÅÅÅÅÅÅÅÅÅ jjjj AII - ÅÅÅÅÅÅÅÅÅÅ zzzz ï BII = a2 jj ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ zz AIII 2 2 m m k a { a k 2 m0 { BIII BII v = b : B0 - ÅÅÅÅÅÅÅÅÅÅÅÅÅ = AII - ÅÅÅÅÅÅÅÅÅÅ b2 b2 III B m0 i BII y B0 + ÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅ jjjj AII + ÅÅÅÅÅÅÅÅÅÅ zzzz b2 m k b2 {

ï

m0 + m m0 - m BII B0 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ AII + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ 2m 2m b2 m0 + m i m0 - m y BIII = b2 jj ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ zz AII + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ BII 2m k 2m {

4 m0 m b2 B0 AI = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = 4 m0 m b B0 b2 Hm0 + mL2 - a2 Hm - m0 L2

2 m Hm + m0 L b2 B0 AII = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = 2 m Hm + m0 L b B0 b2 Hm0 + mL2 - a2 Hm - m0 L2

2 m Hm + m0 L b2 a2 B0 BII = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = 2 m Hm - m0 L a2 b B0 b2 Hm0 + mL2 - a2 Hm - m0 L2

-b2 Hm2 + m20 L Hb2 - a2 L B0 BIII = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = -Im2 + m20 M Ib2 - a2 M b B0 b2 Hm0 + mL2 - a2 Hm - m0 L2

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1 b ª ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ b2 Hm0 + mL2 - a2 Hm - m0 L2

7

FI = 4 m0 m b B0 cos f v i a2 y FII = 2 m b B0 jjjjHm + m0 L v + Hm - m0 L ÅÅÅÅÅÅÅÅÅ zzzz cos f v{ k by i FIII = B0 jjv - Im2 + m20 M Ib2 - a2 M ÅÅÅÅÅÅÅÅ zz cos f k v{ Region I

1 ∑F Hv = - ÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅ m ∑v -4 m b B0 cos f

i a2 y -2 b B0 jjjjHm + m0 L - Hm - m0 L ÅÅÅÅÅÅÅÅÅÅ zzzz cos f v2 { k b y B0 i - ÅÅÅÅÅÅÅÅÅ jj1 - Im2 + m20 M Ib2 - a2 M ÅÅÅÅÅÅÅÅÅÅ zz cos f m0 k v2 {

II III

1 ∑F Hf = - ÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ m v ∑f 4 m b B0 sin f

i a2 y 2 b B0 jjjjHm + m0 L + Hm - m0 L ÅÅÅÅÅÅÅÅÅÅ zzzz sin f v2 { k B0 i b y ÅÅÅÅÅÅÅÅÅ jj1 - Im2 + m20 M Ib2 - a2 M ÅÅÅÅÅÅÅÅÅÅ zz sin f m0 k v2 {

÷÷÷” ÷” ÷” Along the axis †H §v=0 = 4 m b B0 = †B§v=0 ë m so †B§v=0 = 4 m2 b B0 , i.e. ÷” †B§v=0 4 m2 b2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ B0 b2 Hm0 + mL2 - a2 Hm - m0 L2 4 m2r = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Hmr + 1L2 - Ha ê bL2 Hmr - 1L2

÷” †B§v=0 log10 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = log10 H4L + 2 log10 Hmr L B0 -log10 AHmr + 1L2 - Ha ê bL2 Hmr - 1L2 E

÷” †B§v=0 log10 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ B0 4 log10 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 - Ha ê bL2

Problem 5.17 (a) and (b)

mr ª m ê m0 ÷” J Hx, y, zL = Jx Hx, y, zL x` + J y Hx, y, zL y` + Jz Hx, y, zL z` mr - 1 ÷”* J Hx, y, zL ª ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ @Jx Hx, y, -zL x` + J y Hx, y, -zL y` + Jz Hx, y, -zL z` D mr + 1

We need to show that the field

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a2 êb2 =0.5 a2 êb2 =0.1

log10 Hmr

8

l m0 x” - x” £ ÷”> ÷” ” £ ÷”* ” £ o 3 £ o ÅÅÅÅÅÅÅÅÅ , z ¥ 0 B ª ÅÅÅÅÅÅÅÅÅÅ ‡  x AJ Hx L + J Hx LEä ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ o o o 4p o ÷” †x” - x” £ §3 BHx, y, zL ª m o o m0 2 mr x” - x” £ ÷”< o 3 £ ÷” ” £ o o ÅÅÅÅÅÅÅÅÅ , z§0 o B ª ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ‡  x J Hx Lä ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 4 p mr + 1 †x” - x” £ §3 n

(1)

satisfies

÷” l ÷ ” ÷” o m0 J , z > 0 “ äB = m o0 , z<0 n

(2)

÷ ” ÷” “ ÿB = 0

(3)

÷”> ÷”< < IB - B M ÿ z` •z=0 = 0 ï B> z §z=0 = Bz §z=0

(4)

< mr B> x §z=0 = Bx §z=0 ÷÷÷”> ÷÷÷”< ` `z äIH - H M ÿ z•z=0 = 0 ï mr B>y • = B
(5)

÷” (3) is automatic since we've written B in terms of currents. Since z=0

z=0

x” - x” £ ÷ ” m0 ÷è” ÷è” ” £ “ ä ÅÅÅÅÅÅÅÅÅÅ ‡ 3 x£ J Hx” £ Lä ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ Å ÅÅÅÅ = m J 0 Hx L 4p †x” - x” £ §3

÷è” for any J , we have

÷” ÷”* ÷” l m0 J + m0 J = m0 J , z > 0 o o o ÷ ” ÷” o “ äB = m 2 mr ÷” o o , z<0 o o ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ m0 J = 0 n mr + 1

÷”* ÷” since J §z>0 = 0 and J §z<0 = 0. Thus (2) is satisfied.

ƒƒ x` y` z` ƒƒƒƒ ƒƒ ƒ ƒ ÷” ƒ Jy Jz ƒ§ J ä Hx” - x” £ L = †ƒ Jx ƒƒ ƒƒ ƒƒƒ x - x£ y - y£ z - z£ ƒƒƒƒ = x` @Hz - z£ L J y - Hy - y£ L Jz D + y` @Hx - x£ L Jz - Hz - z£ L Jx D + z` @Hy - y£ L Jx - Hx - x£ L J y D

For conciseness in the following integrals the arguments to a function f Hx£ , y£ , z£ L being integrated over will be suppressed when there is no ambiguity. The image function f Hx£ , y£ , -z£ L will be written as f H-z£ L. "################################ ################## Hx - x£ L2 + Hy - y£ L2 + z£2 m0 mr - 1 mr - 1 3 £ £ ij £ £ yz £ ij £ £ yz B> z §z=0 = ÅÅÅÅÅÅÅÅÅÅ ‡  x Hy - y L j Jx Hz L + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Jx H-z Lz - Hx - x L j J y Hz L + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ J y H-z Lz mr + 1 mr + 1 4p k { k { r0 ª

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1 ÅÅÅÅÅÅÅÅ r30

9

A change of variables -z£ Ø z£ takes Ÿ x£ f Hx£ , y£ , -z£ L gHr0 L = Ÿ x£ f Hx£ , y£ , z£ L gHr0 L for any function f and function g of r0 only, since the range is over all space and r0 is invariant under -z£ Ø z£ . Performing this on the integrals involving Ji H-z£ L: m0 B> z §z=0 = ÅÅÅÅÅÅÅÅÅÅ 4p

mr - 1 zy jij1 + ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅ z ‡ 3 x£ @Hy - y£ L Jx - Hx - x£ L J y D m + 1¨¨¨¨¨{Æ k ¨¨¨¨¨¨¨≠r¨¨¨¨¨¨¨¨ ´¨¨¨¨¨¨¨¨ 2 mr êH1+mr L

1 ÅÅÅÅÅÅÅÅ r30

Evaluating (1) at z = 0 gives m0 2 mr 3 £ £ £ B< z §z=0 = ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ‡  x @Hy - y L Jx - Hx - x L J y D 4 p 1 + mr

1 ÅÅÅÅÅÅÅÅ r30

which verifies (4). B> x §z=0 =

É ÄÅ m0 mr - 1 mr - 1 ÅÅ £ ij yz ij yzÑÑÑÑ 3 £ £ £ £ £ £ Å ÅÅÅÅÅÅÅÅÅÅ ‡  x ÅÅ-z jJ y Hz L + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ J y H-z Lz - Hy - y L jJz Hz L - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Jz H-z LzÑÑ ÅÅÇ mr + 1 mr + 1 4p { k {ÑÑÖ k

1 ÅÅÅÅÅÅÅÅ r30

Again we change variables -z£ Ø z£ in the second and last integrals, under which the J y H-z£ L term picks up a minus sign and the rest are unchanged: m0 B> x §z=0 = ÅÅÅÅÅÅÅÅÅÅ 4p

mr - 1 yz ij j1 - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅ z ‡ 3 x£ @-z£ J y - Hy - y£ L Jz D + 1¨¨¨¨¨{Æ k´¨¨¨¨¨¨¨¨¨¨¨¨¨¨m¨≠r¨¨¨¨¨¨¨¨ 2êH1+mr L

1 ÅÅÅÅÅÅÅÅ r30

Comparing this to (1) at z = 0: m0 2 mr 3 £ £ £ B< x §z=0 = ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ‡  x @-z J y - Hy - y L Jz D 4 p 1 + mr

1 ÅÅÅÅÅÅÅÅ = mr B> x §z=0 r30

which verifies the part of (5) involving Bx . An identical computation with x ¨ y (there is only an overall sign switch) shows that the part involving B y is satisfied. Thus we've shown that this choice of image currents defines ÷” the correct B field for the boundary conditions (and other conditions that must be satisfied by any magnetic field). Incidentally (because we've "cheated" and used the answer) the z < 0 part of (1) is manifestly the field due to a ÷” current H2 mr ê mr + 1L J in m0 (unit relative permeability) space, thus solving (b).

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