7.1. PROBLEMS AND SOLUTIONS 91
Thus, we have the potential
V =e
4πǫ0r
[a
r
(
cos(wt− α) + cos(wt+ α))
(7.67)
+(a
r
)2(
3 (cos2(wt− α) + cos2(wt+ α))− 2
2
)]
=1
4πǫ0r
[2ae coswt cosα
r+a2e
r
(
3 cos 2wt cos 2α+ 1)]
=p′
4πǫ0r2+
Q′33
16πǫ0r3
Finally, we find the modified dipole and quadrupole moments which include phase andalso time-dependency.
p′ = 2ae coswt cosα (7.68)
Q′33 = 4a2e
(
3 cos 2wt cos 2α+ 1)
7.1.14 A water molucule
A water molecule is neutral but has a dipole moment of 1.8 Debye (1 D = 3.3× 10−30 C m).
(a) Using the information that the HO length is approximately 1 A and the HOH angle is105, find the partial charges on the H and O atoms.
(Solution)Due to symmetry, the total dipole moment of an H2O moleculeis twice that of single H−O system:
p = (2δq) · l = 1.8× 3.3× 10−30C m ,
where l = (1A) · cos(
105
2
)(see the figure on the right).
From the relation above, we find the partial charges on H andO atoms:
δq =5.97× 10−30C ·m
2l≈ 4.88× 10−20C ≈ 0.31|e|
(b) Suppose that a water molecule is placed in a region of uniform E field of E = 1 kV/mat an angle 90 to the field. How much work is required to rotate the dipole 180 aboutan axis perpendicular to the dipole moment vector ~p?
(Solution)
92 CHAPTER 7. INTERPRETATION IN MULTIPOLES
Work to be done = U(final) - U(initial)
U = −~p · ~E = −pE cos θ
Where, θinitial = 90 and θfinal = 270. Therefore,
W = U(final)− U(initial) = 0 .
(c) The phenomenon of hydration can be modelled by the force of attraction between awater molecule as an electric dipole and an ion in a solution as a point charge. Esti-mate the energy required to separate an ion carrying a single charge |e| from a watermolecule. Assume that initially the ion is located 2 A from the effective center of thewater molecule along the axis of the dipole moment.
(Solution)The electric potential due to dipole moment is
V =1
4πǫ0
p cos θ
r2
And the potential energy (Ep) is defined as following
Ep = |e|V =−1
4πǫ0
5.94× 10−30
(2A)2eV ≈ 1.34 eV
In the limit when b → ∞ we may remove (a/b) and (r/b) terms. Removingthe latter corresponds to having only inverse powers of r surviving, which is theexpected case for an exterior solution. The result is
Φ(r, θ)→ V
2+V
2
[32
(ar
)2
P1(cos θ)− 78
(ar
)4
P3(cos θ) + · · ·]
which agrees with the exterior solution for a sphere with oppositely charged hemi-spheres (except that here we have the average potential V/2 and that the potentialdifference between northern and southern hemispheres is only half as large).
Similarly, when a → 0 we remove (a/b). But this time we get rid of the inversepowers (a/r) instead. The result is the interior solution
Φ(r, θ)→ V
2− V
2
[32
(rb
)P1(cos θ)− 7
8
(rb
)3
P3(cos θ) + · · ·]
which is again a reasonable result (this time with the hemispheres oppositelycharged from the previous case).
3.2 A spherical surface of radius R has charge uniformly distributed over its surface witha density Q/4πR2, except for a spherical cap at the north pole, defined by the coneθ = α.
a) Show that the potential inside the spherical surface can be expressed as
Φ =Q
8πε0
∞∑l=0
12l + 1
[Pl+1(cosα)− Pl−1(cosα)]rl
Rl+1Pl(cos θ)
where, for l = 0, Pl−1(cosα) = −1. What is the potential outside?
Note that this problem specifies a spherical surface of charge, not a sphericalconductor
α
RQ
2Rπ4=σ
=0σ
We are thus interested in obtaining the potential Φ(r, θ) given a charge distri-bution. This may be done using Coulomb’s law (or, equivalently, integratingthe Green’s function with the charge density). Alternatively, in this problem,
the surface charge density specifies an appropriate jump condition on the normalcomponent of the electric field
Er out
∣∣∣r=R
= Er in
∣∣∣r=R
+1ε0σ (4)
This condition allows us to solve for the electrostatic potential Φ(r, θ).
In particular, because of the azimuthal symmetry of this problem, we may performan expansion in Legendre polynomials
Φin =∞∑l=0
αl
( rR
)lPl(cos θ)
Φout =∞∑l=0
αl
(R
r
)l+1
Pl(cos θ)
(5)
Note that the expansion coefficients αl are identical for the inside and outsideexpansion. This holds because we demand that Φ is continuous at r = R. In thiscase, the radial components of the interior and exterior electric fields are givenby
Er = − ∂
∂rΦ ⇒
Er in = −
∞∑l=1
lαlR
( rR
)l−1
Pl(cos θ)
Er out =∞∑l=0
(l + 1)αlR
(R
r
)l+2
Pl(cos θ)(6)
Substituting this into (4) gives
σ(cos θ) = ε0[Er out − Er in
]r=R
=∞∑l=0
(2l + 1)ε0αlR
Pl(cos θ)
Since this is a Legendre polynomial expansion, the coefficients of the expansionare given by the relation
(2l + 1)ε0αlR
=2l + 1
2
∫ 1
−1
σ(cos θ)Pl(cos θ)d(cos θ)
or
αl =R
2ε0
∫ 1
−1
σ(cos θ)Pl(cos θ)d(cos θ)
Using
σ(cos θ) =Q
4πR2× 0 cos θ > cosα
1 cos θ < cosα
gives
αl =Q
8πε0R
∫ cosα
−1
Pl(cos θ)d(cos θ)
This may be integrated by using the Legendre polynomial relation
Pl(x) =1
2l + 1(P ′l+1(x)− P ′l−1(x)) (7)
where P−1(x) is formally defined to be a constant, so that P ′−1(x) = 0. In thiscase, we obtain
αl =Q
8πε0R1
2l + 1[Pl+1(cos θ)− Pl−1(cos θ)
]cosα−1
Noting that Pl(−1) = (−1)l then yields
αl =Q
8πε0R1
2l + 1[Pl+1(cosα)− Pl−1(cosα)]
so long as we define P−1(x) = −1 (so that P1(−1) = P−1(−1) is true). Substi-tuting this into (5) then gives the desired potential (both inside and outside thespherical shell).
Note that, by defining
r< = min(r,R), r> = max(r,R)
the inside and outside expressions (5) may be combined into a compact form
Φ =∞∑l=0
αlRrl<rl+1>
Pl(cos θ)
=Q
8πε0
∞∑l=0
12l + 1
[Pl+1(cosα)− Pl−1(cosα)]rl<rl+1>
Pl(cos θ)
(8)
valid both inside and outside the shell.
b) Find the magnitude and the direction of the electric field at the origin.
By symmetry, the electric field at the origin must point along the z axis (eitherθ = 0 or θ = π). As a result, the radial component Er given by (6) completelyspecifies the electric field at the origin. Noting that Er in ∼ rl−1, we see that onlythe l = 1 component survives at the origin. As a result
Er(r = 0, θ = 0) = −α1
RP1(1)
= − Q
8πε0R2
13
[P2(cosα)− P0(cosα)]
= − Q
16πε0R2(cos2 α− 1) =
Q sin2 α
16πε0R2
In rectangular coordinates, this is equivalent to
~E =Q sin2 α
16πε0R2z (9)
Note that, had we chosen to look along the −z axis (θ = π), we would havegotten an identical result since P1(cosπ) = −1 would give an extra minus sign tocompensate for the −z direction.
c) Discuss the limiting forms of the potential (part a) and electric field (part b) asthe spherical cap becomes (1) very small, and (2) so large that the area withcharge on it becomes a very small cap at the south pole.
We first consider the case α → 0, when the spherical cap becomes very small.For small α, we use cosα ≈ 1− 1
2α2 as well as the Taylor expansion
Pl(cosα) ≈ Pl(1− 12α
2) ≈ Pl(1)− 12α
2P ′l (1) = 1− 2δl,−1 − 12α
2P ′l (1)
to write
Pl+1(cosα)− Pl−1(cosα) ≈ 2δl,0 − 12α
2[P ′l+1(1)− P ′l−1(1)]
Note that the delta functions take care of the special case concerning P−1(1) = −1instead of the usual +1. Using (7) now gives
Pl+1(cosα)− Pl−1(cosα) ≈ 2δl,0 −2l + 1
2α2Pl(1) = 2δl,0 −
2l + 12
α2
Substituting this into (8) yields
Φ ≈ Q
4πε01r>− Qα2
16πε0
∞∑l=0
rl<rl+1>
Pl(cos θ)
Recalling the Green’s function expansion
1|~r − ~r ′|
=∞∑l=0
rl<rl+1>
Pl(cos γ)
where cos γ = r · r′ finally gives
Φ ≈ Q
4πε01r>− Qα2/4
4πε01
|~r −Rz|
Physically, this expression corresponds to the limit where the spherical shell isalmost complete (Φ = Q/4πε0r> for a shell centered at the origin). By linear
superposition, the very small cap can be thought of effectively as an oppositelycharged particle located at Rz with charge given by
q = −σdA = − Q
4πR2(R2dΩ) = − Q
4π(πα2) = −Qα
2
4
The electric field at the origin is given by expanding (9) for α ≈ 0
~E(0) ≈ Qα2/44πε0
z
R2
Again, this makes sense for the electric field of a particle of charge −Qα2/4located at Rz. Note that the full spherical shell does not contribute any electricfield, since we are inside the shell.
Finally, we consider the case α→ π, when the spherical cap becomes very large.In this case, let α = π − β where β is the angle of the south polar cap. TheLegendre polynomial expansion is now
Pl(cosα) = Pl(cos(π − β)) = Pl(− cosβ) ≈ Pl(−1 + 12β
2) ≈ (−1)l + 12β
2P ′l (−1)
Note that the l = −1 special case is covered without any additions to this expres-sion. This gives us
Pl+1(cosα)− Pl−1(cosα) ≈ 12β
2[P ′l+1(−1)− P ′l−1(−1)]
=2l + 1
2β2Pl(−1) =
2l + 12
β2(−1)l
Substituting this into (8) gives
Φ ≈ Qβ2
16πε0
∞∑l=0
(−1)lrl<rl+1>
Pl(cos θ) =Qβ2
16πε0
∞∑l=0
rl<rl+1>
Pl(− cos θ)
=Qβ2/44πε0
1|~r +Rz|
This is clearly the potential due to a point charge of strength Qβ2/4 at the southpole (−Rz) of the spherical surface. For the electric field, we substitute α = π−βinto (9) to obtain
~E(0) ≈ Qβ2/44πε0
zR2
This is the electric field of a particle of charge +Qβ2/4 located at −Rz.
l.14 (0) Verify that
p1 6(p-p') = [-kl.(kp)J.(Icp,) dk
(b) Obtaio the (ollowing expansioo:
_1_ . = f 1-dk e ... · -·'I.(Icp)I.(Icp')e-"··-· ... Ix- a \ .-
(c) By appropriate limitiog procedures prove the (oUowi0B expansions:
J.(kJp+p·' -2pp'COIl <iI)= L e .... J.(Icp)I.(Icp·)
1""-.= t re-I.(Icp)
(d) From tbe last result obtaio an integra.! repr~tatioo of the Bessel function:
1J.(x)"'2m- f'·• e"-·--d</>
Compare the standard inlegnU representations.
("')~~~~ g(l)= r:~f(f))7r(-lf)df
iff) =: (;1e9~) Jfl(-lifr/~ .::iItht
flf') = r;-&. r:r-f(f);utif) tit .;;n(~t0t1~
Grcx.,x/) ~.z -?Jt=-..s.
~
46rt~)~ •
oI~ Il,,"-(i
dc~
~~ ~~~
ff') 1;0 (:-fifJfr:~nt (-if) ::T"t/0o!~"t aLu
1(//) c: (;fqo) d(f-rYJa'j
tntif' ~ cf(f-f/J r f' r:~(~f) ~cJ.fIjJ~
- '1 J. -
81 ~ ~·/~i
dcl-pf/):= ~f eAMCHJ ;M=~
~fte~~tM~
~-t.D.Jc.f:Jd.t(~)k~/;e-(~tP-pI'~J~'2-~ o(~
~ 4~(1)~ )= - : dCf-fY; cRp-p5/) d(r-~/)
d~A Il~ (c,:r:') _ ~'l: a.,..,(~/~/) =-~ dC-!-'Z:''') de:L (/ ~~~=2:/ Ad f Xk. ~ e:J:..d.e ~ ::t4t ~tVtf ~
~(i!J"t;I) ::= e-b > (~:>-r/)
e -I--6p:< ( t< "2/ )
tJik~~ ~~~~pd' ~ «'/;.;;;;. . ~ -
a;r~ (Y+e = (-I~-/1~) = -m~ r!:.. e
4-fC. :uc.. =9A ~ ~ = ~
:, if(i,-;-) -.!:..r: ~.<"""(f-I~~f);;;'" (I.f) e4: c->-%<Jt/-I..
tutti~
- 13 -
--.-:-/~.::. ~~ ~'#('-M (1-p5~ (/~)k (.I. d.Je~~>-~q~(t-)tll )0 ;:-6-'- "T. IC( ~
J!ct .;!I-C ~I~ ~ ,e.l:...f/~+t>"~~r/49<J'p
r;e-W~Jo(~J?-+r-.2!¥"Uld;J)cI~ - (~~~ eh~(~)k1(-i(,:)e-l!!~tI~
~~~rad~'t'C~
~(~ J~~~/~f' )~J4~~-~l~f):r~(~t)
-;:i:::; ~~r ~J~ ef?(<-6- t:-1) = £. t-"'7~(x)
~--p.t
..£d
- rf
<...tJ-n. z4.L .4 ~~
Cel~ ~XI
f;,t1<~
(d) -'~ ~ ~~ 'Y e -<-¥ t:hwI' ;da
hu;t ~~~I'
(~~ -(-kf~;/e~'~?tt'r' co Z. i"(~-t¥.~"'Id1cI"<:r~(,1,£)))0 T ~c-1>o Joe r 'I
l>4
-.::. Z .2.7C i""tFIl1tm(::Tm.. (~) ~t:-PO
t.tJ-n. ::tk ~-' a.LR ~~~ ~~ -??t/-?1t. ~
r::e)~I'~ eJ.'1h<1ell ""- .:21r.~~~ (~) ~ x=~t~~
r:-(f~')(~r;i)m';dlie ;oci"'J~t?f)
tYl- T (9\") -= I. ~eA~~-~""1 tielV~ I7C...cM)o 4 r
~17'l .7'~
J..c (?C)= *r:UJd(~ -f~1Jell
~~~ .:l..( ¥..q~ I%: .. ~ Ial
~ (~)IC" .mP(r1rl+±) Jo C.9O(?f490p;>~ PItt!
J;.('I'O~ ~~~l~~i-~)cI</
~~~,
If desired, the potential may be rearranged to read
Φ(~x ) =∑l
(2l + 1)V Nl
2(
1−(ab
)2l+1)[(1 + (−1)l+1
(ab
)l)(ar
)l+1
+ (−1)l(
1 + (−1)l+1(ab
)l+1)(r
b
)l]Pl(cos θ)
=V
2+ V
∞∑j=1
(−1)j+1(4j − 1)Γ(j − 12 )
4√πj!(
1−(ab
)4j−1) [(
1 +(ab
)2j−1)(a
r
)2j
−(
1 +(ab
)2j)(r
b
)2j−1]P2j−1(cos θ)
which agrees with the solution to Problem 3.1 that we have found earlier.
3.17 The Dirichlet Green function for the unbounded space between the planes at z = 0and z = L allows discussion of a point charge or a distribution of charge betweenparallel conducting planes held at zero potential.
a) Using cylindrical coordinates show that one form of the Green function is
G(~x, ~x ′)
=4L
∞∑n=1
∞∑m=−∞
eim(φ−φ′) sin(nπzL
)sin(nπz′
L
)Im
(nπLρ<
)Km
(nπLρ>
)
In cylindrical coordinates, the polar direction φ is periodic with period 2π. Thissuggests that the Green’s function could be expanded as a Fourier series in eimφ.Similarly, the boundary conditions G = 0 at z = 0 and z = L motivates the use ofa Fourier sine series sin(nπz/L) in the z coordinate. More precisely, a completeFourier expansion in φ and z would give
G(~x, ~x ′) =∑
m,n,m′,n′
g(ρ, ρ′)eimφeim′φ′
sin(nπzL
)sin(n′πz
L
)However, it turns out that m and m′ (and n and n′) do not need to be chosento be independent. This can be seen from the Green’s function equation (givenhere as a differential equation in ~x )
∇xG(~x, ~x ′) = −4πδ3(~x− ~x ′)
In cylindrical coordinates, this reads[1ρ
∂
∂ρρ∂
∂ρ+
1ρ2
∂2
∂φ2+
∂2
∂z2
]G(ρ, φ, z; ρ′, φ′, z′) = −4π
ρδ(ρ− ρ′)δ(φ− φ′)δ(z− z′)
(3)
Using the completeness relations
∞∑m=−∞
eim(φ−φ′) = 2πδ(φ− φ′) (4)
and∞∑n=1
sin(nπzL
)sin(nπz′
L
)=L
2δ(z − z′)
suggests that we take
G(~x, ~x ′) =∞∑n=1
∞∑m=−∞
g(ρ, ρ′)eim(φ−φ′) sin(nπzL
)sin(nπz′
L
)(5)
Substituting this decomposition into (3) gives[1ρ
d
dρρd
dρ− m2
ρ2−(nπL
)2]g(ρ, ρ′) = − 4
Lρδ(ρ, ρ′)
Making the substitutionx =
nπρ
L
converts (the homogeneous part of) this to a modified Bessel equation[d2
dx2+
1x
d
dx−(
1 +m2
x2
)]g(x, x′) = − 4
Lxδ(x, x′)
At this stage, the solution becomes standard. Noting that the modified Besselfunction Im(x) blows up as x→∞ and the function Km(x) blows up as x→ 0,we are left with
g(x, x′) =AIm(x) x < x′
BKm(x) x > x′
where the coefficients A and B are determined by the matching conditions
g< = g>,d
dxg< =
d
dxg> +
4Lx′
at x = x′. This system may be solved to yield
A =4Lx′
Km(x′)I ′m(x′)Km(x′)− Im(x′)K ′m(x′)
B =4Lx′
Im(x′)I ′m(x′)Km(x′)− Im(x′)K ′m(x′)
Noting that the modified Bessel functions satisfy the Wronskian formula
Iν(x)K ′ν(x)− I ′ν(x)Kν(x) = − 1x
finally gives
g(x, x′) =4L
Im(x)Km(x′) x < x′
Im(x′)Km(x) x > x′
=4LIm(x<)Km(x>)
wherex< = min(x, x′), x> = max(x, x′)
Converting x back to ρ and substituting into (5) then gives the desired DirichletGreen’s function
G(~x, ~x ′) =4L
∞∑n=1
∞∑m=−∞
eim(φ−φ′) sin(nπzL
)sin(nπz′
L
)Im
(nπρ<L
)Km
(nπρ>L
)
b) Show that an alternative form of the Green function is
G(~x, ~x ′) = 2∞∑
m=−∞
∫ ∞0
dk eim(φ−φ′)Jm(kρ)Jm(kρ′)sinh(kz<) sinh[k(L− z>)]
sinh(kL)
This alternative form of the Green’s function is derived by expanding in φ and ρinstead of φ and z. For the ρ expansion, we use the integral relation∫ ∞
0
kJν(kρ)Jν(kρ′)dk =1ρδ(ρ− ρ′)
along with the completeness relation (4) to motivate the decomposition
G(~x, ~x ′) =∞∑
m=−∞
∫ ∞0
kdk gk(z, z′)eim(φ−φ′)Jm(kρ)Jm(kρ′) (6)
Since the Bessel function Jm(kρ) satisfies the Bessel equation[d2
dρ2+
1ρ
d
dρ+(k2 − m2
ρ2
)]Jm(kρ) = 0
the substitution of (6) into the Greens’ function equation (3) gives[d2
dz2− k2
]gk(z, z′) = −2δ(z − z′)
Since gk(z, z′) vanishes at z = 0 and z = L, this is a standard one-dimensionalGreen’s function problem. Writing
gk(z, z′) =A sinh(kz) z < z′
B sinh[k(L− z)] z > z′
we find that the matching and jump conditions become
A sinh(kz′) = B sinh[k(L− z′)], A cosh(kz′) = −B cosh[k(L− z′)] +2k
This may be solved to give
A =2k
sinh[k(L− z′)]sinh(kL)
, B =2k
sinh(kz′)sinh(kL)
so thatgk(z, z′) =
2k sinh(kL)
sinh(kz<) sinh[k(L− z>)]
Substituting this into (6) then yields
G(~x, ~x ) = 2∞∑
m=−∞
∫ k
0
dk eim(φ−φ′)Jm(kρ)Jm(kρ′)sinh(kz<) sinh[k(L− z>)]
sinh(kL)
3.26 Consider the Green function appropriate for Neumann boundary conditions for thevolume V between the concentric spherical surfaces defined by r = a and r = b, a < b.To be able to use (1.46) for the potential, impose the simple constraint (1.45). Usean expansion in spherical harmonics of the form
G(~x, ~x ′) =∞∑l=0
gl(r, r′)Pl(cos γ)
where gl(r, r′) = rl</rl+1> + fl(r, r′).
a) Show that for l > 0, the radial Green function has the symmetric form
gl(r, r′) =rl<rl+1>
+
1(b2l+1 − a2l+1)
[l + 1l
(rr′)l +l
l + 1(ab)2l+1
(rr′)l+1+ a2l+1
(rl
r′l+1+
r′l
rl+1
)]
There are several approaches to this problem. However, we first consider theNeumann boundary condition (1.45)
∂G(~x~x ′)∂n′
∣∣∣∣bndy
= −4πS
R;,( )
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If we divide out by 1/r3, the approximate and exact potentials are
1.5 2 2.5 3 3.5 4
0.2
0.4
0.6
0.8
1
where the straight line is the approximation of c) and the sloped line is the exactresult. The approximation improves as r a.
4.7 A localized distribution of charge has a charge density
ρ(~r ) =1
64πr2e−r sin2 θ
a) Make a multipole expansion of the potential due to this charge density and deter-mine all the nonvanishing multipole moments. Write down the potential at largedistances as a finite expansion in Legendre polynomials.
This charge distribution is azimuthally symmetric. As a result, only m = 0moments will be nonvanishing. Furthermore, noting that
sin2 θ = 1− cos2 θ = 23 [P0(cos θ)− P2(cos θ)]
we may write down the moments
ql0 =∫ρ(r, θ)rlY ∗l0(θ, φ) r2 dr dφ d(cos θ)
= 2π
√2l + 1
4π
∫ρ(r, θ)rlPl(cos θ) r2 dr d(cos θ)
=2π64π
23
√2l + 1
4π
∫ ∞0
rl+4e−r dr
∫ 1
−1
Pl(cos θ)[P0(cos θ)− P2(cos θ)] d(cos θ)
=148
√2l + 1
4πΓ(l + 5)[2δl,0 − 2
5δl,2]
As a result, we read off the only nonvanishing multipole moments
q00 =
√1
4π, q20 = −6
√5
4π
The multipole expansion then yields the large distance potential
Φ =1
4πε0
∑l,m
4π2l + 1
qlmYlm(θ, φ)rl+1
=1
4πε0
∑l
√4π
2l + 1ql0Pl(cos θ)rl+1
=1
4πε0
[1r− 6r3P2(cos θ)
](1)
b) Determine the potential explicitly at any point in space, and show that near theorigin, correct to r2 inclusive,
Φ(~r ) ' 14πε0
[14− r2
120P2(cos θ)
]We may use a Green’s function to obtain the potential at any point in space. Ingeneral (since there are no boundaries, except at infinity)
G(~x, ~x ′) =1
|~x− ~x ′|=∑lm
4π2l + 1
rl<rl+1>
Y ∗lm(θ′, φ′)Ylm(θ, φ)
However, for azimuthal symmetry, it is sufficient to focus on the m = 0 terms inthe expansion
G(~x, ~x ′) =∑l
rl<rl+1>
Pl(cos θ)Pl(cos θ′) + (m 6= 0)
Then
Φ(~x ) =1
4πε0
∫ρ(~x ′)G(~x, ~x ′) d3x′
=1
4πε02π64π
23
∫ ∞0
r′4e−r′ rl<rl+1>
dr′
×∫ 1
−1
[P0(cos θ′)− P2(cos θ′)]Pl(cos θ′)Pl(cos θ) d(cos θ′)
=1
4πε0148
[1rl+1
∫ r
0
r′l+4e−r′dr′+ rl
∫ ∞r
r′3−le−r′dr′]
[2δl,0 − 25δl,2P2(cos θ)]
Instead of writing this out in terms of incomplete Gamma functions, it is betterjust to integrate for l = 0 and l = 2. The result is
Φ =1
4πε0124
[1r
(24− e−r(24 + 18r + 6r2 + r3)
)− 1r3P2(cos θ)
(144− e−r(144 + 144r + 72r2 + 24r3 + 6r4 + r5)
)]
Note that as r → ∞ the e−r factors are exponentially small. As a result, wesimply reproduce (1) in this limit. On the other hand, as r → 0, a Taylorexpansion yields
Φ =1
4πε0
[(14
+ · · ·)−(r2
120+ · · ·
)P2(cos θ)
](2)
Obtaining the correct l = 2 term involves the cancellation of the first five termsin the Taylor expansion. Note that the leading terms in the final expression havethe ‘correct’ powers of rlPl(cos θ) in order to satisfy Laplace’s equation.
c) If there exists at the origin a nucleus with a quadrupole moment Q = 10−28 m2,determine the magnitude of the interaction energy, assuming that the unit ofcharge in ρ(~r ) above is the electronic charge and the unit of length is the hydrogenBohr radius a0 = 4πε0h2/me2 = 0.529 × 10−10 m. Express your answer as afrequency by dividing by Planck’s constant h.
The charge density in this problem is that for the m = ±1 states of the 2p levelin hydrogen, while the quadrupole interaction is of the same order as found inmolecules.
We first note that if we put the correct units of electronic charge e and Bohrradius a0 into the charge distribution ρ, the potential near the origin (2) becomes
Φ = − e
4πε0a0
[14− 1
120
(r
a0
)2
P2(cos θ) + · · ·
]
where the overall minus sign is due to the negative charge of the electron. (Wetake e > 0). The interaction energy is then
W =∫ρNΦ d3x = − e
4πε0a0
∫ρN
[14− 1
120
(r
a0
)2
P2(cos θ) + · · ·
]d3x
where ρN is the charge density of the nucleus. Since∫ρNd
3x = Ze gives thetotal charge of the nucleus, we write
W = − e2
4πε0a0
[Z
4− 1
240a20
1e
∫ρNr
2(3 cos2 θ − 1)d3x+ · · ·]
where we have used P2(x) = 12 (3x2− 1). Using z = r cos θ, this may be rewritten
as
W = − e2
4πε0a0
[Z
4− 1
240a20
1e
∫ρN (3z2 − r2)d3x+ · · ·
]= − e2
4πε0a0
[Z
4− Q
240a20
+ · · ·]
where we have used the (classical) definition of the nuclear quadrupole moment
Q =1e
∫ρN (3z2 − r2)d3x
The first term is the electrostatic interaction energy. The quadrupole interactionenergy (expressed as a frequency) is
W/h =e2
4πε0hcQc
480πa30
=αQc
480πa30
≈ 1 MHz
where α ≈ 1/137.036 is the fine structure constant, and where we have put inthe numerical value of Q. This nuclear quadrupole interaction with the electricfield of the electron cloud typically gives rise to radio frequency resonances (inthe low megahertz range) that may be detected using the process of nuclearquadrupole resonance (NQR). Since NQR is sensitive to the electronic structure(ie chemical bonds), it has seen some application towards explosives detection.In particular, nitrogen is a common element in many explosives, and since 14Nhas a non-zero quadrupole moment, NQR can be used to detect what sorts ofnitrogen compounds may be present in a sample.
4.8 A very long, right circular, cylindrical shell of dielectric constant ε/ε0 and inner andouter radii a and b, respectively, is placed in a previously uniform electric field E0
with its axis perpendicular to the field. The medium inside and outside the cylinderhas a dielectric constant of unity.
a) Determine the potential and electric field in the three regions, neglecting endeffects.
Since the cylinder is very long, we treat this as a two-dimensional problem. Inthis case, the potential admits a general expansion
Φ =∑m
[αmρm + βmρ−m] cos(mφ− δm)
(where the m = 0 term should actually be α0 + β0 log ρ). Furthermore, byorienting the electric field along the +x direction, we may use the φ ↔ −φsymmetry of this problem to eliminate the phases δm. As a result, we are able towrite the potential as an expansion in each of the three regions
Φ =
Φ1 = A0 +∑m
Amρ−m cos(mφ)− E0ρ cosφ, ρ > b
Φ2 = B0 + C0 log ρ+∑m
(Bmρm + Cmρ−m) cos(mφ), a < ρ < b
Φ3 = D0 +∑m
Dmρm cos(mφ), ρ < a
For each value of m, there are four unknowns, Am, Bm, Cm and Dm. On the otherhand, there are also four matching conditions (D⊥ and E‖ both at a and at b).Note, however, that when m 6= 1 these matching conditions yield homogeneousequations which only admit the trivial solution
Am = Bm = Cm = Dm = 0 m 6= 1
(Although the m = 0 case has to be treated separately, it is easy to see thatC0 = 0. The remaining constants must satisfy A0 = B0 = D0, and may be takento vanish, since an overall constant added to the potential is unphysical.) Thuswe may focus on m = 1 and write
Φ =
Φ1 = (Aρ−1 − E0ρ) cosφ, ρ > bΦ2 = (Bρ+ Cρ−1) cosφ, a < ρ < bΦ3 = Dρ cosφ, ρ < a
(3)
We may obtain the electric field by taking a gradient
Eρ = −∂Φ∂ρ
=
E1ρ = (Aρ−2 + E0) cosφ, ρ > b
E2ρ = (−B + Cρ−2) cosφ, a < ρ < b
E3ρ = −D cosφ, ρ < a
Eφ = −1ρ
∂Φ∂φ
=
E1φ = (Aρ−2 − E0) sinφ, ρ > b
E2φ = (B + Cρ−2) sinφ, a < ρ < b
E3φ = D sinφ, ρ < a
(4)
The matching at ρ = a is
ε0E3ρ = εE2
ρ
∣∣∣ρ=a
, E3φ = E2
φ
∣∣∣ρ=a
or(ε0/ε)D −B + Ca−2 = 0, D −B − Ca−2 = 0
This may be solved for C and D in terms of B
C =1− ε0/ε1 + ε0/ε
Ba2, D =2
1 + ε0/εB (5)
Similarly, the matching at ρ = b is
εE2ρ = ε0E
1ρ
∣∣∣ρ=b
, E2φ = E1
φ
∣∣∣ρ=b
or(ε0/ε)Ab−2 +B − Cb−2 = −(ε0/ε)E0, Ab−2 −B − Cb−2 = E0
Eliminating C using (5) gives rise to the simultaneous equations(b−2 −1− 1−ε0/ε
1+ε0/ε
(ab
)2(ε0/ε)b−2 1− 1−ε0/ε
1+ε0/ε
(ab
)2)(
A
B
)= E0
(1
−ε0/ε
)
This yields a solution
A = E0∆−1(1− ε0/ε)(
1−(ab
)2)b2
B = −E0∆−1(2ε0/ε)
where
∆ = (1 + ε0/ε)
(1−
(1− ε0/ε1 + ε0/ε
a
b
)2)
is b2 times the determinant of the above matrix. Substituting B into (5) thengives the remaining coefficients
C = −E0∆−1 (1− ε0/ε)2ε0/ε1 + ε0/ε
a2
D = −E0∆−1 4ε0/ε1 + ε0/ε
These expressions may be simplified to read
A = E0b2 (ε2 − ε20)(b2 − a2)
(ε+ ε0)2b2 − (ε− ε0)2a2
B = −2E0ε0(ε+ ε0)b2
(ε+ ε0)2b2 − (ε− ε0)2a2
C = −2E0a2 ε0(ε− ε0)b2
(ε+ ε0)2b2 − (ε− ε0)2a2
D = −4E0εε0b
2
(ε+ ε0)2b2 − (ε− ε0)2a2
(6)
The potential and electric field are obtained by substituting these coefficients into(3) and (4). For the potential, we have
b < ρ : Φ1 = E0
[(ε2 − ε20)(b2 − a2)
(ε+ ε0)2b2 − (ε− ε0)2a2
b2
ρ− ρ]
cosφ
a < ρ < b : Φ2 = −2E0ε0b
2[(ε+ ε0)ρ+ (ε− ε0)a2/ρ](ε+ ε0)2b2 − (ε− ε0)2a2
cosφ
ρ < a : Φ3 = −4E0εε0b
2ρ
(ε+ ε0)2b2 − (ε− ε0)2a2cosφ
(7)
b) Sketch the lines of force for a typical case of b ' 2a.
For ε/ε0 = 1.5, the ‘electric field’ lines look like
Note that we have actually plotted the electric displacement field ~D, as Gauss’law in vacuum ~∇ · ~D = 0 ensures that the lines of electric displacement arecontinuous and unbroken. The electric field lines themselves are discontinuous atthe interface between dielectrics.
c) Discuss the limiting forms of your solution appropriate for a solid dielectric cylin-der in a uniform field, and a cylindrical cavity in a uniform dielectric.
A solid dielectric cylinder of radius b may be obtained by taking the limit a →0. In this case the expressions (6) and (7) simplify considerably. We give thepotential
Φ =
Φ1 = −E0x+ E0
1−ε0/ε1+ε0/ε
b2xρ2 , ρ > b
Φ2 = −E02ε0/ε
1+ε0/εx, ρ < b
(8)
where x = ρ cosφ. The potential Φ3 is irrelevant in this case. Here we see that thepotential Φ2 inside the cylinder is uniform (but corresponds to a reduced electricfield provided ε > ε0). The potential outside is that of the original uniform electricfield combined with a two-dimensional dipole.
For the opposite limit, we obtain a cylindrical cavity of radius a by taking thelimit b→∞. In this case, we end up with
Φ =
Φ2 = −E0
2ε0/ε1+ε0/ε
x− E02ε0/ε(1−ε0/ε)
(1+ε0/ε)2a2xρ2 , ρ > a
Φ3 = −E04ε0/ε
(1+ε0/ε)2x, ρ < a
At first glance, this appears to be considerably different from (8). However, notethat the physical electric field we measure as ρ→∞ is E0 = E0(2ε0/ε)/(1+ε0/ε).
In terms of E0, we have
Φ =
Φ2 = −E0x− E0
1−ε0/ε1+ε0/ε
a2xρ2 , ρ > a
Φ3 = −E02
1+ε0/εx, ρ < a
which may be rewritten as
Φ =
Φ2 = −E0x+ E0
1−ε/ε01+ε/ε0
a2xρ2 , ρ > a
Φ3 = −E02ε/ε0
1+ε/ε0x, ρ < a
This agrees with (8) after the replacement ε↔ ε0 (and a→ b), as it must.
4.9 A point charge q is located in free space a distance d from the center of a dielectricsphere of radius a (a < d) and dielectric constant ε/ε0.
a) Find the potential at all points in space as an expansion in spherical harmonics.
By symmetry, we may place the point charge on the z-axis at z = d. In thiscase, the problem is azimuthally symmetric, and we may expand the potentialin Legendre polynomials instead of spherical harmonics. For the potential insidethe dielectric sphere, we take
Φin =q
4πε
∑l
αl
( ra
)lPl(cos θ) (9)
where the q/4πε prefactor is taken for convenience (but can be absorbed into aredefinition of αl if so desired). Note that we do not need any source term, sincethere are no charges inside the sphere. On the other hand, the solution outsidethe sphere is given by
Φout =1
4πε0q
|~x− dz|+ Φ0
where Φ0 is a homogeneous solution to Laplace’s equation, ∇2Φ0 = 0. Expandingin Legendre polynomials allows us to write
Φout =q
4πε0
∑l
[rl<rl+1>
+ βl
(ar
)l+1]Pl(cos θ) (10)
Note that r< = min(r, d) and r> = max(r, d). Since we must match the parallelelectric field and perpendicular electric displacement at r = a, we may take r< = rand r> = d when using Φout in the matching equations. For the parallel electricfield, we have
Einθ = −1
r
∂Φin
∂θ
∣∣∣∣r=a
=q
4πε
∑l
αlaP ′l (cos θ) sin θ
Eoutθ = −1
r
∂Φout
∂θ
∣∣∣∣r=a
=q
4πε0
∑l
[al−1
dl+1+βla
]P ′l (cos θ) sin θ
,
4.11 The (a llowing data on thc variation oC dielectric constant with pressure are taken Cram the Smithsonian Physical Tabla, 9th ed., p. 424:
Air at 292"K
Pressure (atm)
20 1.0108 Relative density oC 40 1.0218 air as a function of 60 1.0333 pressure is given in 80 1.0439 AlP Handbook, 3rd
100 1.0548 ed.• 1972, p. 4-165.
Pentane (C.HII) at 303"K
Pressure (atm) Density (gmIan')
1 0.613 1.82 10' 0.701 1.96
4xl0' 0.796 2.12 8)( 10' 0.865 2.24
12x10' 0.907 2.33
Test the Clausiu's-Mossotti retatioo between dielectric eoostan( and density for air and pentane in the ranges tabulated. OOC$ it bold exactly? Approximately? If approximately, discuss fractional variatiON in dcosiry and (£-1). For pentane, compare the Clausius-Mossotti relation 10 the cruder relation, (£-1) oc density.
4.L1 Wlter VI
tempelllUIC Assumiog thi bility as I fUi
deduce a vall dipole mOmc
IJ ~-1 fo.t( 1J~1f"') (;W,Sj1a~~ ~::;i) 1:--1 d~~
(~.....('JI(~II)
lJ~i""1 -w
I ".4>" "-I¢?f" ;z.. /..O:J.. /. (l~~ ~.4'1t p.lttio
3 b.t-~ p. f'''1l o4>L b.Nt?)
I/. tJ,'J/J> o.~~, o.4¢J' ~ .. 1'111
S o·W (l.Ut" O.¥JP. o. 1t/-I7J
I ~ 0 .11/ ~.I'I~f' I"~/.
~ .. ; ftJ o. ~t/'
:1 0.16'1 o./3~ ,.1/0 b.~? :J..
~ 0.. It? 17. b19.J /,JIJ.r o_ ~
S" ~_o616 d.~#'6 /.46r d·.Iff
4.12 Wacer vapor is a polar gas wh~ dielectric constant exhibits an appreciable temperature dependence. The following table givC$ experimental data on this effect. Assuming thaI water vapor obeys the Ideal gas law, calculate tbe molecular polarizabillty as • function of inverse temperature and plot it. From the slope of the curve, deduoe a value for the permanent dipolc moment of ,he H,O molecule (exprcss tbe dipole moment in esu-slal-coulomlH:enlimccers).
T(OK) Pressure (em Hg) (e-l»( 10'
393 56.49 400.2 423 60.93 371.7 453 65.34 348.8 483 69.75 328.7
rm-J (tAtta,)
..l. .. ~19X 10-Z.J
.1. .. / s:J..6 J( (0-11
-0 ~.0/2..1 )( 10
/ .. f'l46 )( 10-1*
~ A~ ea.-n ~ ~. Pu ... .2 .01J' )((011 ~-~ .
- /J--/
Physics 505 Electricity and Magnetism Fall 2003Prof. G. Raithel
Problem Set 7
Maximal score: 25 Points
1. Jackson, Problem 5.1 6 Points
Consider the i-th cartesian component of the B-Field,
4π
µ0IB(x) · xi =
∮
∂S
xi ·[dl′ × (x− x′)
|x− x′|]
=∮
∂S
xi ·[dl′ ×∇x′
1|x− x′|
]
=∮
∂S
dl′ ·[∇x′
1|x− x′| × xi
]| Stokes′ law
=∫
S
da′ ·∇x′ ×
[∇x′
1|x− x′| × xi
]
=∫
S
da′ ·[∇x′
1|x− x′|
](∇x′ · xi)− xi
[∇2
x′1
|x− x′|]
+(xi · ∇x′)[∇x′
1|x− x′|
]− (
[∇x′
1|x− x′|
]· ∇x′)xi
use x 6= x′ always
=∫
S
da′ ·
0− 0 + (xi · ∇x′)[∇x′
1|x− x′|
]− 0
=∫
S
da′ ·
0− 0 + (xi · ∇x′)[∇x′
1|x− x′|
]− 0
=∫
S
da′ ·
∂
∂x′i
[∇x′
1|x− x′|
]= −
∫
S
da′ ·
∂
∂xi
[∇x′
1|x− x′|
]
= − ∂
∂xi
∫
S
da′ ·[∇x′
1|x− x′|
]| see Eqn. after 1.25 on page 33 of textbook
=∂
∂xi
∫
S
dΩ′ =∂
∂xiΩ(x)
Thus, Bi = µ0I4π
∂∂xi
Ω(x), and
B(x) =µ0I
4π∇xΩ(x) q.e.d.
1
PHY 5346Homework Set 10 Solutions – Kimel
2. 5.2 a) The system is described by
First consider a point at the axis of the solenoid at point z0. Using the results of problem 5.1,
dφm =μ0
4πNIdzΩ
From the figure,
Ω = ∫ r ⋅ dAr2 = ∫ dAcosθ
r2 = 2πz ∫0
R ρdρ
ρ2 + z2 3/2= 2π − z
R2 + z2 + 1
φm =μ0
2NI ∫
z0
∞z − 1
R2 + z2 + 1
z dz =μ0
2NI −z0 + R2 + z0
2
Br = − μ0
2NI ∂
∂z0−z0 + R2 + z0
2 =μ0
2NI
−z0 + R2 + z02
R2 + z02
In the limit z0 → 0
Br =μ0
2NI
By symmetry, thej loops to the left of z0 give the same contribution, so
B = B l + Br = μ0NI
H = NI
By symmetry, B is directed along the z axis, so
δφm = −δρ ⋅ B = 0
if δρ is directed ⊥ to the z axis. Thus for a given z, φm is independent of ρ, and consequently
H = NI
everywhere within the solenoid.
If you are on the outside of the solenoid at position z0, by symmetry the magnetic field must be inthe z direction. Thus using the above argument, φm must not depend on ρ. Set us take ρ far awayfrom the axis of the solenoid, so that we can replace the loops by elementary dipoles m directed alongthe z axis. Thus for any point z0 we will have a contributions
φm m ⋅ r1
r13 +
m ⋅ r2
r23
where m ⋅ r1 = −m ⋅ r2 and r1 = r2. Thus
H = 0
or A = Q/2π(ε + ε0). Hence
~E =Q
2π(ε + ε0)r
r2
Note that 12 (ε + ε0) may be viewed as the average permittivity in the volume
between the spheres.
b) Calculate the surface-charge distribution on the inner sphere.
The surface-charge density is given by σ = D⊥∣∣r=a
where either D⊥ = ε0E⊥ or
D⊥ = εE⊥ depending on region. This gives
σ =
ε
ε + ε0
Q
2πa2; dielectric side
ε0ε + ε0
Q
2πa2; empty side
(1)
Note that the total charge obtained by integrating σ over the surface of the innersphere gives Q as expected.
c) Calculate the polarization-charge density induced on the surface of the dielectricat r = a.
The polarization charge density is given by
ρpol = −∇ · ~P
where ~P = ε0χe~E = (ε − ε0) ~E. Since the surface of the dielectric at r = a is
against the inner sphere, we can take the polarization to be zero inside the metal(‘outside’ the dielectric). Gauss’ law in this case gives
σpol = −P⊥∣∣r=a
= −(ε− ε0)E⊥∣∣r=a
= −ε− ε0ε + ε0
Q
2πa2
Note that when this is combined with (1), the total (free and polarization) chargedensity is
σtot = σ + σpol =ε0
ε + ε0
Q
2πa2
on either half of the sphere. Since this is uniform, this is why the resulting electricfield is radially symmetric.
5.3 A right-circular solenoid of finite length L and radius a has N turns per unit lengthand carries a current I. Show that the magnetic induction on the cylinder axis in thelimit NL →∞ is
Bz =µ0NI
2(cos θ1 + cos θ2)
where the angles are defined in the figure.
21θ θ
We start by computing the magnetic field on axis for a single loop of wire carryinga current I. This may be done by an elementary application of the Biot-Savartlaw.
dl
αB
R
z
a
By symmetry, only the z component contributes
Bz =µ0I
4π
∫[d~× ~R ]z
R3=
µ0I
4π
∫d` R sinα
R3=
µ0I
4π2πa
a
R3=
µ0Ia2
2R3
Substituting in R2 = a2 + z2 yields
Bz =µ0Ia2
2(a2 + z2)3/2(2)
We now use linear superposition to obtain the field of the solenoid. Defining z1
and z2 as follows2
1θ θ 2
z− 1 z
(where z1 + z2 = L) we have
Bz =µ0Ia2
2
∫ z2
−z1
N dz
(a2 + z2)3/2
A simple trig substitution z = a tanα converts this integral to
Bz =µ0NI
2
∫ tan−1(z2/a)
− tan−1(z1/a)
cos α dα =µ0NI
2sinα
∣∣∣tan−1(z2/a)
− tan−1(z1/a)
A bit of geometry then demonstrates that this is equivalent to
Bz =µ0NI
2(cos θ1 + cos θ2)
5.6 A cylindrical conductor of radius a has a hole of radius b bored parallel to, andcentered a distance d from, the cylinder axis (d + b < a). The current density isuniform throughout the remaining metal of the cylinder and is parallel to the axis.Use Ampere’s law and principle of linear superposition to find the magnitude and thedirection of the magnetic-flux density in the hole.
Ampere’s law in integral form states∮V
~B · d~ = µ0ienc
For a cylindrically symmetric geometry this gives simply
B =µ0ienc
2πr=
µ0(jπr2)2πr
=µ0jr
2
where we have assumed a uniform current density j. The direction of the magneticinduction is given by the right hand rule. For a conductor oriented along the zaxis (so that the current is flowing in the +z direction), we may write
~B =µ0jr
2z × r =
µ0j
2z × ~r
where ~r is the vector from the center of the conductor to the position where weare measuring the field. We now use linear superposition to start with a solidcylindrical conductor and then subtract the ‘missing’ current from the hole
~B =µ0j
2z × ~x− µ0j
2z × (~x− ~d ) =
µ0j
2z × ~d
Here ~d is the vector displacement of the hole from the center of the cylinder. Thissomewhat remarkable result demonstrates that the magnetic induction is uniformin the hole, and is in a direction given by the right hand rule.
If desired, we note that the total current carried by the wire is I = j(πa2 − πb2),so we may express the magnetic induction in terms of I as
~B =µ0I
2π(a2 − b2)z × ~d
5.7 A compact circular coil of radius a, carrying a current I (perhaps N turns, each withcurrent I/N), lies in the x-y plane with its center at the origin.
a) By elementary means [Eq. (5.4)] find the magnetic induction at any point on thez axis
By appropriate integration of ~J(~x ′)/|~x−~x ′| we could find the magnetic inductionanywhere in space. However we have already computed the magnetic inductionwhen restricted to the z axis. The result is given by (2)
Bz =µ0Ia2
2(a2 + z2)3/2
4. Jackson, Problem 5.13 7 Points
There is an azimuthal surface current K(θ′) = φ′σ sin θ′aω. The corresponding three-dimensional currentdensity is
j(r′, θ′) = φ′K(θ′)δ(r′ − a) = φ′σ sin θ′aωδ(r′ − a) = φ′Jφ(r′, θ′) .
Using Eq. 2 of the previous problem and∫ −1
−1Pm
l (x)Pml′ (x)dx = 2
2l+1(l+m)!(l−m)!δl,l′ and P 1
l = − sin θ, it is
Aφ(r, θ) =µ0
4π
∑
l,m=1
1l(l + 1)
P 1l (cos θ)
∫rl<
rl+1>
P 1l (cos θ′)Jφ(r′, θ′)d3x′
=µ0
4π
∑
l,m=1
1l(l + 1)
P 1l (cos θ)
∫rl<
rl+1>
P 1l (cos θ′)σ sin θ′δ(r′ − a) aω r′2d cos θ′dφ′
= −µ0
4π
∑
l,m=1
1l(l + 1)
P 1l (cos θ)
2πσa3ωrl<
rl+1>
∫P 1
l (x)P 11 (x)dx
= −µ0σa3ω
4P 1
l (cos θ)r<
r2>
43
=µ0σa3ω
3sin θ
r<
r2>
Thus, it is outside the sphere
Aexterior(r, θ) = φµ0σa4ω
31r2
sin θ
and inside
Ainterior(r, θ) = φµ0σaω
3r sin θ
Using that for azimuthal A it is B = ∇×A = r 1r sin θ ∂θ [sin θAφ]− θ 1
r ∂r [rAφ] it is found:
Bexterior(r, θ) =µ0σa4ω
3
[r2 cos θ
r3+ θ
sin θ
r3
],
which is the field of a magnetic dipole m = z 4πσa4ω3 , and
Binterior(r, θ) =2µ0σaω
3
[r cos θ − θ sin θ
],
which is a homogeneous magnetic field in z-direction.
6
along the same direction as ~B0. The other two regions contain a dipole field inaddition a uniform component.
Since ~H = −~∇ΦM = −δx for ρ < a, the ratio of ~B on axis (ρ = 0) to ~B0 is givenby
B
B0= 4∆−1 =
4(1 + µr)(1 + µ−1
r ) + (1− µr)(1− µ−1r )(a/b)2
This may be plotted as follows
1 2 3 4 5 6
0.2
0.4
0.6
0.8
1
a / b ) 2
B / B0
log 10 µ r( a / b ) 2 = 0.1
( = 0.5
5.17 A current distribution ~J(~x ) exists in a medium of unit relative permeability adjacentto a semi-infinite slab of material having relative permeability µr and filling the half-space, z < 0.
a) Show that for z > 0 the magnetic induction can be calculated by replacing themedium of permeability µr by an image current distribution, ~J∗, with compo-nents,(
µr − 1µr + 1
)Jx(x, y,−z),
(µr − 1µr + 1
)Jy(x, y,−z), −
(µr − 1µr + 1
)Jz(x, y,−z)
We will end up solving parts a) and b) simultaneously. We start, however, bydefining the reflection (Parity) operator P : z → −z so that
P : (x, y, z) → (x, y,−z)
On the right (z > 0), we assume the magnetic induction is generated by boththe original current ~J (contained entirely on the right) and an image current ~J∗
(contained entirely on the left). Thus
~BR(~x ) =µ0
4π
∫( ~J(~x ′) + ~J∗(~x ′))× (~x− ~x ′)
|~x− ~x ′|3d3x′
By changing variables z′ → −z′ in the ~J∗ term, we may restrict this volumeintegral to z′ > 0
~BR(~x ) =µ0
4π
∫z′>0
(~J(~x ′)× (~x− ~x ′)
|~x− ~x ′|3+
~J∗(P~x ′)× (~x− P~x ′)|~x− P~x ′|3
)d3x′ (3)
On the left (z < 0), we assume the magnetic induction is generated by a currentof the same form as the original ~J , but with possibly modified strength (becauseof the change of permeability). Given a modified current λ ~J and permeability µ,we write
~BL(~x ) =µλ
4π
∫z′>0
~J(~x ′)× (~x− ~x ′)|~x− ~x ′|3
d3x′ (4)
Our aim is now to match the left and right magnetic field and magnetic induction.More precisely, at z = 0, both Hx and Hy (the parallel components) must becontinuous, and Bz (the perpendicular component) must also be continuous. Toperform this matching, we first note that the norms |~x − ~x ′| and |~x − P~x ′| areidentical at z = 0. (The are both equal to
√(x− x′)2 + (y − y′)2 + z′2.) Thus
all denominators are the same, and we deduce that the numerators of (3) and (4)must be matched as appropriate. For Bz, we have
(Jx + J∗x)(y − y′)− (Jy + J∗y )(x− x′) = µrλ(Jx(y − y′)− Jy(x− x′))
where any component of ~J∗ is understood to have argument P~x. For Hx and Hy
matching, we find
−(Jy − J∗y )z′ − (Jz + J∗z )(x− x′) = λ(−Jyz′ − Jz(x− x′))
(Jz + J∗z )(x− x′) + (Jx − J∗x)z′ = λ(Jz(x− x′) + Jxz′)
Since these equations hold for all values of (x, y), they separate into
λJy = Jy − J∗y λJz = Jz + J∗z
λJz = Jz + J∗z λJx = Jx − J∗x
µrλJx = Jx + J∗x µrλJy = Jy + J∗y
These equations may be solved to yield
J∗x = (1− λ)Jx, J∗y = (1− λ)Jy, Jz = −(1− λ)Jz
provided µrλ−1 = 1−λ, or λ = 2/(µr +1). This may be given in a more conciseform using the reflection operator
~J∗(~x ) = (1− λ)P ~J(P~x ) =µr − 1µr + 1
P ~J(P~x )
b) Show that for z < 0 the magnetic induction appears to be due to a currentdistribution [2µr/(µr + 1)] ~J in a medium of unit relative permeability.
From the expression (4) for ~BL, the magnetic induction appears to be due to acurrent λ ~J = [2/(µr + 1)] ~J in a medium of permeability µ. This is equivalent
to having a current distribution [2µr/(µr + 1)] ~J in a medium of unit relativepermeability.
5.19 A magnetically “hard” material is in the shape of a right circular cylinder of length Land radius a. The cylinder has a permanent magnetization M0, uniform throughoutits volume and parallel to its axis.
a) Determine the magnetic field ~H and magnetic induction ~B at all points on theaxis of the cylinder, both inside and outside.
We use a magnetic scalar potential and the expression
ΦM = − 14π
∫V
~∇ · ~M(~x ′)|~x− ~x ′|
d3x′ +14π
∮S
n′ · ~M(~x ′)|~x− ~x ′|
da′
Orienting the cylinder along the z axis, we take a uniform magnetization ~M =M0z. In this case the volume integral drops out, and the surface integral onlypicks up contributions on the endcaps. Thus
ΦM =M0
4π
[∫top
1|~x− ~x ′|
da′ −∫
bottom
1|~x− ~x ′|
da′]
where ‘top’ and ‘bottom’ denote z = ±L/2, and the integrals are restricted toρ < a. On axis (ρ = 0) we have simply
ΦM (z) =M0
4π
∫ (1√
ρ2 + (z − L/2)2− 1√
ρ2 + (z + L/2)2
)ρ dρ dφ
=M0
4
∫ a2
0
(1√
ρ2 + (z − L/2)2− 1√
ρ2 + (z + L/2)2
)dρ2
=M0
2
[√a2 + (z − L/2)2 −
√a2 + (z + L/2)2 − |z − L/2|+ |z + L/2|
]On axis, the field can only point in the z direction. It is given by
Hz = −∂zΦM = −M0
2
[z − L/2√
a2 + (z − L/2)2− z + L/2√
a2 + (z + L/2)2
− sgn(z − L/2) + sgn(z + L/2)]
Note that the last two terms cancel when |z| > L/2, but add up to 2 inside themagnet. Thus we may write
Hz = −M0
2
[z − L/2√
a2 + (z − L/2)2− z + L/2√
a2 + (z + L/2)2+ 2 Θ(L/2− |z|)
]
where Θ(ξ) denotes the unit step function, Θ = 1 for ξ > 0 (and 0 otherwise).The magnetic induction is obtained by rewriting the relation ~H = ~B/µ0 − ~M as~B = µ0( ~H + ~M ). Since the magnetization is only nonzero inside the magnet [ieMz = M0 Θ(L/2 − |z|)], the addition ~H + ~M simply removes the step functionterm. We find
Bz = µ0(Hz + Mz) = −µ0M0
2
[z − L/2√
a2 + (z − L/2)2− z + L/2√
a2 + (z + L/2)2
]
b) Plot the ratios ~B/µ0M0 and ~H/M0 on the axis as functions of z for L/a = 5.
The z component of the magnetic field looks like
-2 -1 1 2
-0.4
-0.2
0.2
0.4
z / L
M 0/H
while the z component of the magnetic induction looks like
-2 -1 1 2
0.2
0.4
0.6
0.8
1
/ L
/B µ 0 M
z
0
Note that Bz is continuous, while Hz jumps at the ends of the magnet. Thisjump may be thought of as arising from effective magnetic surface charge.
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