Physics 417G : Problem set 5
Due : February 26, 2016
Please show all the details of your computations including intermediate steps.
1
Problem 1
~ = kρẑ parallel to
a) An infinitely long circular cylinder of radius R carries a uniform magnetization M
~
its axis, where k is a constant. Find the magnetic field (due to M ) inside and outside the cylinder.
~ = kρϕ̂, where k is a constant,
b) The same long circular cylinder of radius R carries a magnetization M
ρ is the distance from the axis, and ϕ̂ is the usual azimuthal unit vector. Find the magnetic field due
~ , for points inside and outside the cylinder.
to M
c) For a short circular cylinder of radius R and length L carries a “frozen-in” uniform magnetization
~ parallel to its axis. Find the bound current, and sketch the magnetic field of the cylinder. Make
M
three sketches: one for L R, one for L R, and one for L ≈ R.
~ and B
~ are in ẑ directions.
Sol: a) Due to symmetry, currents are in the direction ϕ̂ and the field H
We can use Gauss’s law to compute
I
~ · d~l = J~f = 0 ,
~ =0.
H
to
H
This tells us
~
~ ,
~ = B −M
H
µ0
to
~ = µ0 M
~ = µ0 kρẑ ,
B
~ = 0 outside the cylinder.
inside the cylinder, while B
There is another way to compute the magnetic field by computing the bound currents.
~ ×M
~ = −k ϕ̂ ,
J~b = ∇
~b = M
~ × n̂ = kRϕ̂ .
K
~ = 0 outside the
These are nothing but the currents of the solenoid and their superposition. Thus B
cylinder. To get the magnetic field inside, one can use the Ampere’s law. See the left figure below.
Thus we get
I
Z
~
~
~ + Kb l = µ0 [−kl(R − ρ) + kRl] = µ0 klρ.
B · dl = Bl = µ0 Ienc = µ0 J~b · dS
This gives the desired result.
~ = 0 and thus the magnetic field is
b) Similarly, the field H
~ = µ0 M
~ = µ0 kρϕ̂ ,
B
~ = 0 outside. Once can also check this results with the bound currents. They
inside the cylinder and B
are
~ ×M
~ = 1 ∂s (sks)ẑ = 2kẑ ,
J~b = ∇
ρ
~
~
Kb = M × n̂ = kRϕ̂ × ρ̂ = −kRẑ .
One can check that the bound charges add up to 0.
~ b = M ϕ̂. Thus they are nothing but a solenoid with different
c) The bound currents are J~b = 0, K
length. The magnetic fields are sketched in the right figure.
2
Problem 2
a) A current I flows down a long straight wire of radius a. If the wire is made of linear material with
a permeability µ, and the current is distributed uniformly.
~ a distance ρ from the axis?
i) What is the magnetic field B
ii) Find all the bound currents. What is the net bound current flowing down the wire?
b) A coaxial cable consists of two very long cylindrical tubes, with radii r1 and r2 , separated by
linear insulating material of magnetic susceptibility χm . A current I flows down the inner conductor
and returns along the outer one. In each case the current distributes itself uniformly over the surface
as in the figure.
~ using H
~ in the region between the tubes.
i) Find the magnetic field B
ii) Calculate the magnetization and the bound currents, and confirm that they generate the magnetic
field computed in a).
H
~
~ ·d~l = H2πρ = If,enc where If,enc = Iρ2 /a2
Sol: a) i) Using Ampere’s law, we compute the field H,
H
for ρ < a and If,enc = I for ρ > a. Thus
(
~ = µIρ2
B
2πa
~ = µ0 I
B
2πρ
~ = µH
~ =
B
~ = χm H
~ =
ii) Using M
µ−µ0 ~
µ0 H,
ρ<a
.
ρ>a
we compute
~
~
~ ×M
~ =∇
~ × (χm H)
~ = χm J~f = χm I = µ − µ0 I ,
J~b = ∇
2
πa
µ0 πa2
~
~
~b = M
~ × n̂ = χm H
~ × n̂ = −χm I = − µ − µ0 I .
K
2πa
µ0 2πa
One can check the net current flowing down the wire add up to 0.
b) IT is obvious that the magnetic fields inside the inner tube and outside of the outer tube are 0.
H
~ · d~l = H2πρ = If,enc = I. Thus
Inside these tubes, we have H
~ = I ϕ̂ ,
H
2πρ
to
~ = µ0 (1 + χm )H
~ = µ0 (1 + χm ) I ϕ̂ ,
B
2πρ
~ = χm H
~ = χm I ϕ̂. Now the bound currents are
between the tubes. Thus M
2πρ
χm I
ρ=a
2πa ẑ
~
~
~
~
~
Jb = ∇ × M = 0 ,
Kb = M × n̂ =
.
χm I
− 2πb ẑ
ρ=b
mI
Total enclosed current between the tubes are I + χ2πa
2πa = (1 + χm )I. Thus
I
~ · d~l = B2πρ = µ0 (1 + χm )I ,
~ = µ0 (1 + χm )I ϕ̂ .
B
to
B
2πρ
3
Problem 3
~ vanishes as ∇
~ ×H
~ = 0, and we can express H
~ as
a) If free current vanishes everywhere, the curl of H
~ = −∇Φ.
~
~ ·H
~ = −∇
~ ·M
~ , we can get
the gradient of a scalar potential Φ, H
According to ∇
~ 2Φ = ∇
~ ·M
~ ,
∇
~ ·M
~ as a source. Find the field inside and outside a uniformly
so Φ obeys Poisson’s equation, with ∇
magnetized sphere.
~ ·M
~ = 0 everywhere except at the surface r = R, so Φ satisfies Laplace’s equation in the
Hint: ∇
regions r < R and r > R. To do so, one can use
∞ X
Bl
l
Φ(r, θ) =
Al r + l+1 Pl (cos θ) ,
r
l=0
as we have done several times already. For this we need to figure out the proper boundary condition
on Φ.
b) A sphere of linear magnetic material with magnetic susceptibility χm is placed in an otherwise
~ 0 . Find the magnetic field inside and outside the sphere.
uniform magnetic field B
Hint : Use the same potential Φ given in a). Now the boundary conditions are different.
Sol: a) If the current vanishes everywhere we can write the potential as
P∞
Φi = l=0 Al rl Pl (cos θ)
r<R
P∞
.
l
r>R
Φo = l=0 rB
l+1 Pl (cos θ)
The boundary conditions are
Φi (R, θ) = Φo (R, θ)
~ · r̂ = M cos θ
∂r Φi (R, θ) − ∂r Φo (R, θ) = M
−→
P
l (l
+
Al R l
l+2
1)Bl /R Pl
= Bl /Rl+1
P
.
+ l lAl Rl−1 Pl = M cos θ
~ is discontinuous due to the magnetization and we take the direction
where the normal component of H
of the magnetization to be ẑ.
By combining these boundary conditions we get
A1 =
M
,
3
B1 = R3 A1 .
Thus
(
Φi =
Φo =
M
3 r cos θ
M R3
3 r 2 cos θ
−→
~
~ i = −∇Φ
~ i = − M ẑ = − M
H
3
3
3 3(r̂·M
~
~
)r̂−M
~ o = −∇Φ
~ o=R
H
3
r3
−→
~
~ = µ0 2M
B
3
~ = µ0 H
~o
B
r<R
r>R
.
The middle expression in the second line comes
from
the similar computation of the fields from a
~ = µ0 H
~ +M
~ , and M
~ = 0 outside of the sphere.
dipole. In the last expression we use B
b) Similar to a), we can use the potential Φ with a different boundary conditions. For large r, we
~ =B
~ 0 = B0 ẑ, and thus H
~ =B
~ 0 /µ0 = B0 /µ0 ẑ. This gives Φ = −B0 z/µ0 = −B0 r cos θ/µ0 .
want B
With this boundary condition included, we can write
P∞
Φi = l=0 Al rl Pl (cos θ)
r<R
P∞
.
l
Φo = −B0 r cos θ/µ0 + l=0 rB
P
(cos
θ)
r
>R
l
l+1
Now the remaining boundary conditions are
P
P
l
l+1
Φi (R, θ) = Φo (R, θ)
−→
Pl
l = −B0 r cos θ/µ0 +
l /R
l Al R PP
l BP
.
µ∂r Φi (R, θ) = µ0 ∂r Φo (R, θ)
µ0 B0 cos θ/µ0 + l (l + 1)Bl /Rl+2 Pl + µ l lAl Rl−1 Pl = 0
where second boundary condition comes from the requirement that magnetic field is continuous in its
normal component. For l 6= 1, the two conditions give Bl = R2l+1 Al and (µ0 (l + 1) + µl)Al Rl−1 = 0.
Thus Al = 0. For l = 1, they are A1 R = −B0 R/µ0 + B1 /R2 and B0 + 2µ0 B1 /R3 + µA1 = 0, so
A1 = −
3B0
,
2µ0 + µ
B1 =
B0 R3 (µ − µ0 )
.
µ0 (2µ0 + µ)
Thus inside the sphere we get
Φi = −
3B0
r cos θ ,
2µ0 + µ
~ i = −∇Φ
~ i=
H
~0
3B
,
2µ0 + µ
~
~ = µH
~ = 3µB0 = 1 + χm B
~0 .
B
2µ0 + µ
1 + χm /3
Outside of the sphere we have
B0 R3 (µ − µ0 ) 1
cos θ ,
µ0 (2µ0 + µ) r2
3
~
~
~ o = −∇Φ
~ o=B
~ 0 /µ0 + R (µ − µ0 ) 3(r̂ · B0 )r̂ − B0 ,
H
3
µ0 (2µ0 + µ)
r
Φo = −B0 r cos θ/µ0 +
4
~ = µ0 H
~o .
B
Problem 4
Let us consider the magnetic vector potential
~ x) = µ0
A(~
4π
Z
d3 x0
~ (~x0 ) × (~x − ~x0 )
M
,
|~x − ~x0 |3
~ (~x0 ) is magnetization.
where M
~b = M
~ × n̂ and J~b = ∇×
~ M
~
a) derive the expression of the vector potential in terms of bound currents, K
as we have done in class.
b) Compute the corresponding magnetic field in terms of the bound currents.
~ ×M
~.
Now we want to revisit the Ampere’s law using the J~b = ∇
c) Argue that there is another contribution for the bound current J~btot from the bound charge ρb =
~ · P~ using the continuity equation for ρb .
−∇
d) Show that the Maxwell equation
~
~ ×B
~ = µ0 0 ∂ E + µ0 (J~f + J~btot ) ,
∇
∂t
can be rewritten as
~
~ ×H
~ = ∂ D + J~f ,
∇
∂t
where
~ = 0 E
~ + P~ ,
D
~
~ = B −M
~ .
H
µ0
~ ·D
~ = ρf .
This completes the Maxwell’s equations in matter along with ∇
Sol: a) The derivation goes as
~
Z
) × (~x − ~x0 )
µ0
1
~ (~x0 ) × ∇
~0
d x
=
d3 x0 M
|~x − ~x0 |3
4π
|~x − ~x0 |
Z
~0×M
~ (~x0 ) µ0 Z
~ 0
µ0
∇
~ 0 × M (~x )
=
d3 x0
−
dS
0
4π
|~x − ~x |
4π
|~x − ~x0 |
Z
Z
~b
µ0
J~b
µ0
K
0
=
+
,
d3 x0
dS
×
4π
|~x − ~x0 | 4π
|~x − ~x0 |
~ x) = µ0
A(~
4π
Z
x
3 0 M (~
0
~0×M
~ (~x0 ), K
~b = M
~ (~x0 ) × n̂ with
where J~b = ∇
~ )×A
~ + f (∇
~ × A).
~
(∇f
b) The corresponding magnetic field is
~ =∇
~ × A(~
~ x) = µ0
B
4π
Z
R
~0 =
dS
R
~ × (f A)
~ =
dS 0 n̂. We also used the identity ∇
~ × (~x − ~x0 ) µ0 Z
~ b × (~x − ~x0 )
K
0
d x
+
dS
×
,
|~x − ~x0 |3
4π
|~x − ~x0 |3
3 0 Jb
~ acts only on ~x as ∇
~ 1 0 = − ~x−~x00 3 and A
~×B
~ = −B
~ × A.
~
where we use that ∇
|~
x−~
x|
|~
x−~
x|
c) We consider the Maxwell equations in matter. The bound charge due to polarization has the form
~ · P~ . Thus from continuity equation we have
ρb = −∇
~ · J~b = 0 ,
∂t ρb + ∇
−→
~ · −∂t P~ + J~b = 0 .
∇
~ ×M
~ + ∂t P~ .
Thus we have J~btot = ∇
d) The Ampere’s law becomes
~
~
~ ×M
~ + ∂t P~ ) ,
~ ×B
~ = µ0 0 ∂ E + µ0 (J~f + J~btot ) = µ0 0 ∂ E + µ0 (J~f + ∇
∇
∂t
∂t
which becomes
~ ×
∇
~
B
~
−M
µ0
!
=
∂ ~
0 E + P~ + J~f ,
∂t
and thus gives
~
~ ×H
~ = ∂ D + J~f .
∇
∂t