PHYS 100B (Prof. Congjun Wu)
Solution to HW 5
February 21, 2011
Problem 1 (Griffiths 6.26)
At the interface between two linear magnetic materials, the magnetic field lines band. Show that tan θ2 / tan θ1 =
µ2 /µ1 , assuming there is no free current at the boundary. Compare Eq. 4.68.
∥
∥
∥
∥
Solution: At the interface, (1)B1⊥ = B2⊥ , (2) B1 = B2 + µ0 (K × n̂) ⇒ H1 = H2 + Kf × n̂ (Since µ10 B = H + M).
∥
∥
∥
∥
1
µ2 B2 .
∥
∥
No free current at the boundary ⇒ H1 = H2 ⇒ µ11 B1 =
tan θ2 / tan θ1 =
|B2 | |B1 |
∥
∥
/ ⊥ = |B2 |/|B1 | = µ2 /µ1 .
⊥
B2 B1
Problem 2 (Griffiths 6.27)
A magnetic dipole m is imbedded at the center of a sphere (radius R) of linear magnetic material with permeability
µ. Show that the magnetic field inside the sphere (0 < r ≤ R) is
{
}
µ
1
2(µ0 − µ)m
[3(m · r̂)r̂ − m] +
.
4π r3
(2µ0 + µ)R3
What is the field outside the sphere?
Solution: In view of Eq. 6.33, the volume bound current density is proportional to the free current density
Jb = ∇ × M = ∇×(χm H) = χm Jf .
There is a bound dipole at the center
mb = χm m.
⇒ Net dipole moment at the center is
mcenter = m + mb = (1 + χm )m =
µ
m.
µ0
⇒ (Eq. 5.87) Magnetic field produced by the dipole:
Bcenter
dipole
µ
=
4π
{
}
1
[3(m · r̂)r̂ − m] .
r3
Magnetic field produced by the bound surface current Kb at r = R: (No volume bound current Jb since no free
current Jf flowing through the material)
Bsurf ace = Am.
current
The magnetization
χm
χm
M = χm H =
B=
µ
4π
⇒ bound current density
{
}
1
χm
[3(m · r̂)r̂ − m] +
Am.
r3
µ
Jb = ∇ × M = 0.
1
(
}
)
(
)
1
1
χm
A
[−m
×
r̂]
+
Am
×
r̂
|
=
χ
m
−
+
sin θϕ̂.
r=R
m
r3
µ
4πR3
µ
(
)
1
This is the surface current produced by a spinning sphere: Kb = σωR sin θϕ̂, σωR = χm m A
µ − 4πR3 . ⇒
Kb = M × r̂ =
χm
4π
{
Bsurf ace =
current
and χm =
µ
µ0
2
2
µ0 σωR = µ0 χm m
3
3
(
A
1
−
µ
4πR3
)
= A.
− 1, ⇒
A=
⇒
B=
µ
4π
{
µ 2(µ0 − µ)
.
4π R3 (2µ0 + µ)
2(µ0 − µ)m
1
[3(m · r̂)r̂ − m] + 3
3
r
R (2µ0 + µ)
}
.
The field outside the sphere is that of the central dipole plus the dipole field from the surface current,
msurf ace =
current
4πR3
4πR3 3 µ 2(µ0 − µ)m
4πR3 3
µ(µ0 − µ)m
(σωR) =
Bsurf ace =
=
.
3
3
3 2µ0 current
3 2µ0 4π R (2µ0 + µ)
µ0 (2µ0 + µ)
⇒
mtot =
µm µ(µ0 − µ)m
3µm
+
=
.
µ0
µ0 (2µ0 + µ)
(2µ0 + µ)
and hence the field for r > R is
B =
=
{
}
µ
1
2(µ0 − µ)mtot
[3(mtot · r̂)r̂ − mtot ] +
4π r3
R3 (2µ0 + µ)
{
}
µ
3µ
1
2(µ0 − µ)m
[3(m · r̂)r̂ − m] + 3
.
4π (2µ0 + µ) r3
R (2µ0 + µ)
Problem 3 (Griffiths 7.60)
(a) Show that Maxwell’s equations with magnetic charge (Eq. 7.43) are invariant under the duality transformation
E′
= E cos α + cB sin α,
′
cB
cqe′
= cB cos α − E sin α,
= cqe cos α + qm sin α,
′
qm
= qm cos α − cqe sin α.
√
where c = 1/ ϵ0 µ0 and α is an arbitrary rotation angle in E/B-space. Charge and current densities transform in the
same way as qe and qm . [This means, in particular, that if you know the fields produced by a configuration of electric
charge, you can immediately (using α = 90◦ ) write down the fields produced by the corresponding arrangement of
magnetic charge.]
2
Solution:
∇ · E′
∇ · B′
∇ × E′
∇ × B′
1
= ∇ · E cos α + c∇ · B sin α = ρe cos α + cµ0 ρm sin α
ϵ0
(
)
1
1
1
=
ρe cos α + ρm sin α = ρ′e .
ϵ0
c
ϵ0
1
1 1
= ∇ · B cos α − ∇ · E sin α = µ0 ρm cos α −
ρe sin α
c
c ϵ0
= µ0 (ρm cos α − cρe sin α) = µ0 ρ′m .
(
)
(
)
∂B
∂E
= ∇ × E cos α + c∇ × B sin α = −µ0 Jm −
cos α + c µ0 Je + µ0 ϵ0
sin α
∂t
∂t
)
(
∂
1
= −µ0 (Jm cos α − cJe sin α) −
B cos α − E sin α
∂t
c
′
∂B
= −µ0 J′m −
,
∂t
(
)
(
)
1
∂E
1
∂B
= ∇ × B cos α − ∇ × E sin α = µ0 Je + µ0 ϵ0
cos α −
−µ0 Jm −
sin α
c
∂t
c
∂t
(
)
1
∂
= µ0 Je cos α + Jm sin α + µ0 ϵ0 (E cos α + cB sin α)
c
∂t
′
∂E
= µ0 J′e + µ0 ϵ0
.
∂t
(b) Show that the force Law (Prob. 7.35)
F = qe (E + v × B) + qm (B −
1
v × E),
c2
is also invariant under the duality transformation.]
Solution:
F′
1
′
(B′ − 2 v × E′ )
= qe′ (E′ + v × B′ ) + qm
c
(
)[
(
)]
1
1
=
qe cos α + qm sin α (E cos α + cB sin α) + v× B cos α − E sin α
c
c
[(
)
]
1
1
+ (qm cos α − cqe sin α) B cos α − E sin α − 2 v× (E cos α + cB sin α)
c
c
[
(
)]
1
= qe (E cos α cos α + cB sin α cos α) + v× B cos α cos α − E sin α cos α
c
[
(
)]
1
+qe − (Bc sin α cos α − E sin α sin α) + v×
E sin α cos α + B sin α sin α
c
)
(
)]
[(
1
1
1
B sin α cos α − 2 E sin α sin α
+qm E sin α cos α + B sin α sin α + v×
c
c
c
[(
)
(
)]
1
1
1
+qm B cos α cos α − E sin α cos α − v× 2 E cos α cos α + B sin α cos α
c
c
c
[
]
1
= qe (E + v × B) + qm B − v× 2 E = F.
c
3