Download Lecture 3. Magnetostatics with magnetics 1 Magnetization December 28, 2006

Lecture 3. Magnetostatics with magnetics
December 28, 2006
1
Magnetization
A magnetic moment m produces the vector potential
A(r) =
m × (r − r 0 )
|r − r 0 |3
1
= −m × ∇r
|r − r 0 |
1
= rot m
|r − r 0 |
(1)
(2)
(3)
The effective current density should be obtained from
4bmA = −(4π/c)j
(4)
which gives
jef f
1
c X
rot m
=− 4
4π
|r − r 0 |
X
0
= c rot
mδ(r − r )
= c rothnmi
= c rot M
(5)
(6)
(7)
(8)
where M is the magnetization. Substituting into the corresponding Maxwell equation
one finds
4π
jext + 4π rot M ,
(9)
rot B =
c
4π
rot(B − 4πM ) =
j,
(10)
c
4π
j, B = H + 4πM
(11)
rot H =
c
1
Electrodynamics 2
2
Lecture 3
Diamagnetism
If atoms/molecules do not have an intrinsic magnetic moment, the medium is diamagnetic. The magnetic moment should be induced, and according to the Lenz law, the
direction of the induced magnetic moment is opposite to the direction of the external
magnetic field. Denoting M = αB, where α < 0, one has
B = H(1 − 4πα)−1
B = µH, µ = (1 − 4πα)−1 < 1
(12)
(13)
There is only diamagnetism in classical physics since all magnetic moments are due to
currents and can be produced only due to the electromagnetic induction.
3
Paramagnetism
If atoms/molecules possess an intrinsic magnetic moment (due to electron spin) then α
can be positive. Let us consider a model of an ensemble of magnetic moments which
cannot move but can rotate. If we apply an external magnetic field B the energy of a
magnetic moment is H = −m · B, so that
Z 1
Z 1
aµ
eaµ dµ)−1 ,
(14)
µe dµ)(
Mk = nm(
−1
−1
T
mB
−
(15)
= nm coth
T
mB
For weak fields mB/T 1 one finds
M=
nm2
B
3T
(16)
Therefore
H=
nm2
1−
3T
B,
B = µH,
4
µ=
nm2
1−
3T
(17)
−1
>1
(18)
Ferromagnetism
Ferromagnets a) have large µ and b) can be magnetized even in the absence of the
external (applied) magnetic field.
2
Electrodynamics 2
5
Lecture 3
Energy
Magnetic field itself does not produce work. However, if the magnetic field changes,
and induction electric field appears which is responsible for the work. The infinitesimal
fork produced by the electric field on a system of currents is
Z
δW = E · jdV δt
(19)
Z
= (c/4π) E rot HdV δt
(20)
Z
Z
= (c/4π) div(E × H)dV δV − (c/4π) H rot EdV δt
(21)
Z
∂
BδtdV
(22)
= (1/4π) H
∂dt
Z
= (1/4π) H · δBdV
(23)
For the linear relation B = µH one gets
Z
U=
Z
=
Z
=
B2
dV
8πµ
µH 2
dV
8π
H ·B
dV
8π
On the other hand, from the same expression it is easy to derive
Z
δU = (1/c) j · δAdV,
Z
U = (1/2c) A · jdV
6
(24)
(25)
(26)
(27)
(28)
Energy of current system
Starting with
Z
U = (1/2c)
A · jdV
(29)
let us assume that the currents are closed loops of linear currents, that is, jdV = Idl
and
I
X
U=
(1/2c) Ia A · dl
(30)
a
a
3
Electrodynamics 2
Lecture 3
=
X
Z
X
(31)
a
a
=
B · dS
(1/2c)Ia
Ia Φa /2c
(32)
a
We can proceed further in a uniform medium with µ = const where the vector potential
satisfies the equation
4A = −
4πµ
j
c
(33)
with the solution
Z
µ
j(r 0 )dV 0
A(r) =
c
|r − r 0 |
X µIb I
dl0
=
c b |r − r 0 |
b
(34)
(35)
Substituting one has
U=
X Ia Ib
I I
Lab =
a
Lab ,
(36)
dla dlb
|ra − rb |
(37)
2c
ab
b
where for calculation of Laa we have to take into account the finite thickness of the
currents.
7
Volume forces in weakly dielectric and weakly magnetic fluid
Let us assume that ∼ 1, µ ∼ 1 and both are constant. In such fluid the main force is
1
fi = ρEi + ilm jl Bm
c
1
∂
1
∂
=
Ei
Dl +
ilm lab Bm
Hb
4π ∂xl
4π
∂xa
1 ∂
1
∂
=
(Ei Dl ) −
Dl
Ei
4π ∂xl
4π ∂xl
1
∂
1
∂
+
Bl
Hi −
Bl
Hl
4π ∂xl
4π ∂xi
4
(38)
(39)
(40)
Electrodynamics 2
Lecture 3
1
∂
1 ∂
(Ei Dl ) −
Dl
El
4π ∂xl
4π ∂xi
1 ∂
1
∂
+
(Hi Bl ) −
Bl
Hl
4π ∂xl
4π ∂xi
∂ Ei El E 2 δil
−
=
∂xl
4π
8π
2
µHi Hl µH δil
+
−
4π
8π
=
8
(41)
(42)
Total force on embedded body
If a body in inserted into a magnetic, the total force on this body can be obtained from
the Maxwellian stress-tensor in the ambient medium as follows:
Z
Fi = − Tij nj dS
(43)
where the normal n̂ is from inside the body outwards. This gives
Z
µ
F =−
[2H(H · n̂) − H 2 n̂]dS
8π
9
(44)
Boundary conditions
It is clear that the boundary conditions read
[B · n̂] = 0,
4π
I
[H × n̂] =
c
10
(45)
(46)
Magnetics without currents
In the regions where j = 0 the equation
rot H = 0
(47)
H = grad ψ
(48)
means that
Since B = µH the equation to solve reduces to
div µ grad ψ = 0
5
(49)
Electrodynamics 2
Lecture 3
very similar to the equation of the electrostatics. Respectively, the boundary conditions
read
4π
I,
c
[µ grad ψ · n̂] = 0
[grad ψ × n̂] =
11
(50)
(51)
Permanent magnets: Independent magnetization
If j = 0 then H = grad ψ and
div B = div(H + 4πM ) = 0,
4ψ = −4π div M
12
(52)
(53)
Effective surface currents
Effective surface current is obtained by integration of
j = c rot M
(54)
over an infinitely thin layer around the discontinuity, which gives
I = c[n̂ × M ]
= (c/4π)[n̂ × (B − H)]
= (c/4π)[n̂ × B]
= (c/4π)[n̂ × µH]
= (c/4π)(n̂ × H)[µ]
13
(55)
(56)
(57)
(58)
(59)
Problem: Magnetic spherical shell in external magnetic field
Into initially uniform B0 a spherical shell with the magnetic constant µ is inserted. The
magnetic constant of the ambient medium is µ = 1. Outer and inner radii are a, and b,
a > b. Since there no currents we have H = grad ψ and ψ satisfies the Laplace equation
in each region r < a (µ = 1), a < r < b (µ = µ), and b < r (µ = 1). The boundary
conditions would read [ψ] = 0, [(∂ψ/∂θ)] = 0, [(∂ψ/∂r)µ] = 0 at both surfaces r = a
and r = b. From our previous experience we shall seek for a solution in the form
ψ = (A/r2 − H0 r)P1 ,
r > b,
6
H0 = B0 /µ2 ,
(60)
Electrodynamics 2
Lecture 3
= (Cr + D/r2 )P1 ,
= ErP1 ,
r<a
a < r < b,
(61)
(62)
The boundary conditions give
A/b2 − H0 b = Cb + D/b2 ,
(2A/b3 + H0 ) = (2D/b3 − C)µ,
Ca + D/a2 = Ea,
(C − 2D/a3 )µ = E
(63)
(64)
(65)
(66)
Eventually we have the magnetic field inside the void r < a
9µB0 b3
B = H = −E = −
2(1 − µ2 )a3 − (2µ + 1)(2 + µ)b3
(67)
When µ 1 the field is
9B0 b3
B0
B≈
2µ(a3 + b3 )
(68)
which is nothing but the magnetic shielding.
14
Problem: field of infinite uniformly magnetized cylinder
Let an infinite cylinder me uniformly magnetized (M ) in the direction perpendicular to
the cylinder axis. We seek for a solution in the form H = grad ψ
ψ = A cos ϕ/r,
= Br cos ϕ,
r>a
r<a
(69)
(70)
Boundary conditions give
[ψ] = 0 ⇒ A = Ba2 ,
∂ψ
∂ψ
(r → a+) =
(r → a−) + 4πMr ⇒
∂r
∂r
− A/a2 = B + 4πM
(71)
(72)
(73)
from which
H(r < a) = −2πM x̂,
2πM a2
H(r > a) =
(cos ϕr̂ + sin ϕϕ̂)
r2
The magnetic field inside is Bx = Hx + 4πMx = 2πMx .
7
(74)
(75)
Electrodynamics 2
15
Lecture 3
Problem: magnetization of a long cylinder
Let a long cylinder (a, l, µ) be inserted into an external magnetic field B0 . What is the
magnetization and magnetic moment of the cylinder ? It is sufficient to consider the
case where B0 is parallel to the axis, and when they are mutually perpendicular, since
any other case can be solved by superposition.
If the magnetic field is parallel to the axis, the continuity of the tangential component
of H simply gives Hin = B0 , so that M = (µ − 1)H/4π = (µ − 1)B0 /4π and the
magnetic moment m = M V = πa2 lM .
If the magnetic field is perpendicular to the axis the problem becomes a two-dimensional
problem for the magnetic potential which in the cylindrical coordinates will take the
form
ψ = (B0 r + A/r) cos ϕ,
= Cr cos ϕ,
r<a
r>a
(76)
(77)
The boundary conditions give
B0 a2 + A = Ca2 ,
(B0 a2 − A) = µCa2
(78)
(79)
The magnetic field inside
H(r < a) =
2B0
µ+1
(80)
Again
M=
µ−1
(µ − 1)B0
H=
4π
2π(µ + 1)
8
(81)