Problemas de Eletrodinâmica Clássica de Jackson - Capítulo 5

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Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
April 20, 2001
Chapter 5: Problems 19-27
Problem 5.19
A magnetically “hard” material is in the shape of a right circular cylinder of length L
and radius a. The cylinder has a permanent magnetization M0, uniform throughout
its volume and parallel to its axis.
(a) Determing the magnetic field H and magnetic induction B at all points on the
axis of the cylinder, both inside and outside.
(b) Plot the ratios B/µ0M0 and H/M0 at all points on the axis of the cylinder,
both inside and outside.
There is no free current in this problem, so H(ρ, z) may be derived from a
scalar potential Φm(ρ, z) satisfying the Laplace equation. Dividing space into
three regions
Φm =

















∫
∞
0
dk A(k)e−kzJ0(kρ), z > L/2
∫
∞
0
dk
[
B(k)ekz + C(k)e−kz
]
J0(kρ), − L/2 < z < L/2
∫
∞
0
dk D(k)ekzJ0(kρ), z < −L/2.
1
Homer Reid’s Solutions to Jackson Problems: Chapter 5 2
The tangential boundary condition at z = +L/2 is
∂Φm
∂ρ
∣
∣
∣
∣
z= L
2
+
=
∂Φm
∂ρ
∣
∣
∣
∣
z= L
2
−
⇒
∫
∞
0
dk kA(k)e−kL/2J1(kρ) =
∫
∞
0
dk k
[
B(k)ekL/2 + C(k)e−kL/2
]
J1(kρ)
(1)
This must hold for all ρ. Multiplying both sides by ρJ1(k
′ρ), integrating from
ρ = 0 to ρ = ∞, and using the identity
∫
∞
0
dρ ρJn(kρ)Jn(k
′ρ) =
1
k
δ(k − k′) (2)
we obtain from (1) the relation
A(k) = B(k)ekL + C(k). (3)
The perpendicular boundary condition at z = +L/2 is
Bz(z = L/2+) = Bz(L/2−)
or
µ0Hz(z = L/2+) = µ0
[
Hz(z = L/2−) + Mz(z = L/2−)
]
∂Φm
∂z
∣
∣
∣
∣
z= L
2
+
=
∂Φm
∂z
∣
∣
∣
∣
z= L
2
−
+ M(ρ)
⇒
∫
∞
0
dk kA(k)e−kL/2J0(kρ) =
∫
∞
0
dk k
[
−B(k)ekL/2 + C(k)e−kL/2
]
J0(kρ) + M(ρ)
(4)
where
M(ρ) =
{
M1, ρ < a
0, ρ > a.
Now we multiply both sides of (4) by ρJ0(k
′ρ) and integrate from ρ = 0 to
ρ = ∞ to obtain
A(k) = −B(k)ekL + C(k) + M1ekL/2
∫ a
0
ρJ0(kρ)dρ
= −B(k)ekL + C(k) + γ(k) (5)
where we defined
γ(k) = M1e
kL/2
∫ a
0
ρJ0(kρ)dρ =
aM1
k
ekL/2J1(ka).
Homer Reid’s Solutions to Jackson Problems: Chapter 5 3
The solution of eqs. (3) and (5) is
B(k) =
1
2
e−kLγ(k) C(k) = A(k) −
1
2
γ(k). (6)
From the boundary conditions at z = −L/2 we may similarly obtain the rela-
tions
B(k) + C(k)ekL = D(k)
B(k) − C(k)ekL = D(k) − γ(k)
which may be solved to yield
B(k) = D(k) −
1
2
γ(k) C(k) =
1
2
e−kLγ(k). (7)
Comparing (6) and (7) we find
A(k) = D(k) =
M1a
k
cosh
kL
2
J1(ka)
B(k) = C(k) =
M1a
2k
e−kL/2J1(ka).
Then the components of the H field are
Hρ =

















M1a
∫
∞
0
dk cosh
kL
2
e−kzJ1(ka)J1(kρ), z > L/2
M1a
∫
∞
0
dk e−kL/2 cosh(kz)J1(ka)J1(kρ), − L/2 < z < L/2
M1a
∫
∞
0
dk cosh
kL
2
ekzJ1(ka)J1(kρ), z < −L/2
Hz =

















M1a
∫
∞
0
dk cosh
kL
2
e−kzJ1(ka)J0(kρ), z > L/2
−M1a
∫
∞
0
dk e−kL/2 sinh(kz)J1(ka)J0(kρ), − L/2 < z < L/2
−M1a
∫
∞
0
dk cosh
kL
2
ekzJ1(ka)J0(kρ), z < −L/2.
Homer Reid’s Solutions to Jackson Problems: Chapter 5 4
Problem 5.23
A right circular cylinder of length L and radius a has a uniform lengthwise magne-
tization M .
(a) Show that, when it is placed with its flat end against an infinitely permeable
plane surface, it adheres with a force
F = 2µ0aLM
2
[
K(k) − E(k)
k
−
K(k1) − E(k1)
k1
]
where
k =
2a
√
4a2 + L2
, k1 =
a
√
a2 + L2
.
(b) Find the limiting form of the force if L � a.
We’ll define our coordinate system so that the z axis is the cylinder axis,
and we’ll take the surface of the permeable medium at z = 0.
Our general strategy for this problem will be as follows. First, we’ll find
the magnetic field H0 that exists in all space when the cylinder is pressed up
flat against the infinitely permeable medium. Then we’ll calculate the shift dE
in the energy of the magnetic field incurred by moving the cylinder up a small
distance dz off the surface of the medium. The force on the cylinder is then
readily calculated as F = −dE/dz.
To calculate the energy shift incurred by moving the cylinder a distance dz
away from the permeable medium, we won’t have to go through and completely
recalculate the fields and their energy in the new configuration. Instead, we
can use the following little trick. When we move the cylinder up a distance dz,
two things happen. First a gap of height dz opens between the surface and the
face of the cylinder, where previously there had been a fixed magnetization M,
but now there is just free space. Second, between L and L + dz there is now a
fixed magnetization M where previously there was none. Moving the cylinder of
fixed M up a distance dz is thus formally equivalent to keeping the cylinder put
and instead introducing a cylinder of the opposite magnetization −M between 0
and dz, while also introducing a cylinder of magnetization +M between L and
L + dz. The increase in field energy in this latter case is fairly easily calculated
by taking the integral of µ0Mċ
H0 over the regions in which the fixed magnetization changes.
So the first task is to find the field that exists when the cylinder is pressed
flat against the surface. Since there are no free currents in the problem, we
may derive H from a scalar potential satisfying the Laplace equation. To be-
gin we write down the general solutions of the Laplace equation in cylindrical
coordinates, observing first that by symmetry we can only keep terms with no
Homer Reid’s Solutions to Jackson Problems: Chapter 5 5
azimuthal angle dependence:
Φ(m) =

















∫
∞
0
dk A(k)e−kzJ0(kρ), z > L
∫
∞
0
dk [B(k)ekz + C(k)e−kz ]J0(kρ), 0 < z < L
∫
∞
0
dk D(k)e+kzJ0(kρ), z < 0.
(8)
The boundary conditions at z = 0 are that Hρ and Bz be continuous. Assum-
ing first of all that the medium existing in the region below z = 0 has finite
permeability µ, the tangential boundary condition is
∂Φm
∂ρ
∣
∣
∣
∣
z=0−
=
∂Φm
∂ρ
∣
∣
∣
∣
z=0+
∫
∞
0
dk k D(k)J1(kρ) =
∫
∞
0
dk k [B(k) + C(k)]J1(kρ). (9)
Multiplying (9) by ρJ1(k
′ρ), integrating from 0 to ∞, and using the identity
(2), we find
D(k) = B(k) + C(k). (10)
The normal boundary condition at z = 0 is of a mixed type. Below the line
we have simply Bz = µHz. Above the line we may write Bz = µ0[Hz + M(ρ)],
where M(ρ) represents the fixed magnetic polarization of the cylinder:
M(ρ) =
{
M, ρ < a
0, ρ > a.
(11)
The normal boundary condition at z = 0 is then
−µ
∂
∂z
Φm
∣
∣
∣
∣
z=0−
= −µ0
∂
∂z
Φm
∣
∣
∣
∣
z=0+
+ µ0M(ρ)
−
µ
µ0
∫
∞
0
dk k D(k)J0(kρ) = −
∫
∞
0
dk k [B(k) − C(k)]J0(kρ) + M(ρ)
Now multiplying by ρJ0(k
′ρ), integrating from ρ = 0 to ∞, and using (2) yields
µ
µ0
D(k) = −B(k) + C(k) −
∫
∞
0
ρ M(ρ)J0(kρ) dρ. (12)
Using (11), the integral on the RHS is
M
∫ a
0
ρJ0(kρ) dρ =
Ma
k
J1(ka) ≡ γ(k)
where we defined a convenient shorthand. Then (12) is
µ
µ0
D(k) = −B(k) + C(k) − γ(k).
Homer Reid’s Solutions to Jackson Problems: Chapter 5 6
Now taking µ → ∞, we see that, to keep the B and C coefficients from blowing
up, we must have D → 0. Then equation (??) tells us that B(k) = −C(k), so
the middle entry in (8) may be rewritten:
Φm(z, ρ) =
∫
∞
0
dk β(k) sinh(kz)J0(kρ), (0 < z < L).
The boundary conditions at z = L are
∂Φm
∂ρ
∣
∣
∣
∣
z=L+
=
∂Φm
∂ρ
∣
∣
∣
∣
z=L−
−
∂Φm
∂z
∣
∣
∣
∣
z=L+
= −
∂Φm
∂z
∣
∣
∣
∣
z=L−
+ M(ρ)
with M(ρ) defined as above. Working through the same procedure as above
yields the conditions
A(k)e−kL = β(k) sinh(kL)
A(k)e−kL = β(k) cosh(kL) + γ(k)
with γ(k) defined as above. The solution is
β(k) = −γ(k)e+kL
A(k) = γ(k) sinh(kL).
Plugging these back into (8) and differentiating, we find for the z component of
the H field
Hz(ρ, z) =







Ma
∫
∞
0
dk e−kz cosh(kL)J0(kρ)J1(ka), z > L
−Ma
∫
∞
0
dk e−kL cosh(kz)J0(kρ)J1(ka), 0 < z < L.
(13)
Now that we know the field, we want to find the change in energy density
incurred by putting into this field a short cylinder (radius a, heightdz) of
magnetization −M k̂ between z = 0 and z = dz, and another cylinder of the
same size but with magnetization +M k̂ between z = L and z = L + dz. The
change in field energy is just the integral of µ0M · H over the volume in which
the magnetization density has changed:
dU = −2πµ0M
∫ dz
0
∫ a
0
Hz(z, ρ)ρ dρ dz + 2πµ0M
∫ L+dz
L
∫ a
0
Hz(z, ρ)ρ dρ dz
= 2πµ0Mdz
(
∫ a
0
Hz(L, ρ)ρ dρ −
∫ a
0
Hz(0, ρ)ρ dρ
)
(14)
where in the last step we assumed that Hz remains essentially constant over
a distance dz in the z direction, and may thus be taken out of the integral.
Homer Reid’s Solutions to Jackson Problems: Chapter 5 7
Inserting (13) into (), and exchanging the order of integration, we first do the ρ
integral:
∫ a
0
J0(kρ)ρdρ =
a
k
J1(ka).
Then () becomes