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6.2 The Field of a Magnetized Object
6.2.1 Bound Currents
Use current model to understand magnetic field of a magnetized object.
r r
µI
A(r ') = 0
4π
∫
r
dl '
r
r + r ' −2rr ' cos θ '
l
r
1 ⎛ r' ⎞
µ0 I
(
)
cos
'
'
P
θ
d
l
=
⎟
⎜
∑
l
θ’
4π ∫ r ⎝ r ⎠
r’
r 1
⎤
µ0 I ⎡ 1 r 1
1⎞ r
2⎛3
2
dl ' + 2 ∫ r ' cos θ ' dl ' + 3 ∫ r ' ⎜ cos θ '− ⎟dl ' + ...⎥
=
4π ⎢⎣ r ∫
r
r
2⎠
⎝2
⎦ wire
r
r
r
r r
µI
µI
magnetic dipole: Adipole (r ) = 0 2 ∫ r ' cosθ ' dl ' = 0 2 ∫ (rˆ ⋅ r ')dl '
4πr
4πr
r 1 r r
r r
r
a = ∫ r '×dl ' = ∫ nˆ ' da' , let v = c T , where c is a constant vector
2
r
r
r
r r
r
ˆ
(
)
(
)
c
T
⋅
d
l
= ∫ (∇ × (c T )) ⋅ nˆ da
v
⋅
d
l
=
∇
×
v
⋅
n
da
Æ
∫
∫
∫
2
2
area
r
r
r
r
r
∇ × (c T ) = (∇ × c )T + ∇T × c = (∇ × c )T − c × ∇T
r
r
r
r
Æ ∫ (c T ) ⋅ dl = − ∫ nˆ ⋅ (c × ∇T )da = − ∫ c ⋅ (∇T × nˆ )da
r
∫ T ⋅ dl
r
r
= − ∫ (∇T × nˆ )da , let T = rˆ ⋅ r ' Æ
r
r
r
r
' (rˆ ⋅ r ') × nˆ ')da '
∫ (rˆ ⋅ r ') ⋅ dl ' = − ∫ (∇area
⎛1 r
r
r⎞
× nˆ ')da ' = − rˆ × ∫ nˆ ' da' = −rˆ × a ' = −rˆ × ⎜ ∫ r '×dl ' ⎟
∫ (rˆ ⋅ r ') ⋅ dl ' = −∫ (rˆarea
area
⎝ 2 line integral
⎠
r
r
µI
Adipole (r ) = 0 2
4πr
r
r
r
r
r
µI
µ r
∫ (rˆ ⋅ r ')dl ' = 4πr (− rˆ × a ) = 4πr (m × rˆ) , where m = Ia = I ∫ nˆda .
0
0
2
2
r r
r µ m
(r ') × (rr − rr ')
A= 0
r r3
4π
r − r'
r
In the magnetized object, each volume carries a dipole moment Mdτ ' and the vector
r
r µ M (rr ') × (rr − rr ')
dτ ' .
potential of the whole object is A = 0 ∫
r r3
4π
r − r'
r µ
We try to modify the equation A = 0
4π
r r
r r
M (r ') × (r − r ')
∫ rr − rr' 3 dτ ' to fit the current
model:
⎛ 1 ⎞
∂
∇' ⎜⎜ r r ⎟⎟ = iˆ
∂x'
⎝ r − r' ⎠
r r
µ
A(r ) = 0
4π
1
(x − x')2 + ( y − y ')2 + (z − z ')2
+ ... = −
r r
1 2(r − r ')(− 1)
=
r r
2 r − r'3
(rr − rr ')
r r3
r − r'
⎡ r r
⎛ 1 ⎞⎤
⎜ r r ⎟⎥ dτ '
(
)
×
∇
'
'
M
r
⎢
∫⎢
⎜ r − r' ⎟
⎝
⎠⎥⎦
⎣
r
r r
⎛ 1 ⎞ r r
⎛ M (rr ') ⎞
1
∇'×⎜⎜ r r ⎟⎟ = ∇' ⎜⎜ r r ⎟⎟ × M (r ') + ∇'×M (r ') r r
r − r'
⎝ r − r' ⎠
⎝ r − r' ⎠
Use the product rule:
r r
r r
⎛ 1 ⎞
1
= − M (r ') × ∇' ⎜⎜ r r ⎟⎟ + ∇'×M (r ') r r
r − r'
⎝ r − r' ⎠
(
(
r r
µ
A(r ) = 0
4π
)
r
r
⎡ ∇'×M (rr ')
⎛ M (rr ') ⎞⎤
∫ ⎢⎢ rr − rr ' − ∇'×⎜⎜ rr − rr' ⎟⎟⎥⎥ dτ '
⎝
⎠⎦
⎣
r r
r r
µ0 ∇'×M (r ')
µ
A(r ) =
r r dτ ' + 0
∫
r − r'
4π
4π
r
)
r
r r
M (r ')
µ
∫ rr − rr' × nˆ ' da' = 4π0
∫
r r
J b (r ')
µ
r r dτ ' + 0
r − r'
4π
r r
K b (r ')
∫ rr − rr' da'
r r r
v = v × c , we have
∫ ∇ ⋅ v dτ = ∫ v ⋅ nˆda , Let
∫ ∇ ⋅ (v × c )dτ = c ⋅ ∫ ∇ × v dτ = ∫ v × c ⋅ nˆda = c ⋅ ∫ (nˆda × v ) = c ⋅ (− ∫ v × nˆda )
r r
r
r
r
r r
r
r
r
r
Æ
r
∫ ∇ × v dτ = − ∫ v × nˆda
r r
r r
r r
r r
J b (r ') = ∇'×M (r ') K b (r ') = M (r ') × nˆ '
I
M
Example: Find the magnetic field of a uniformly magnetized sphere.
r
r
J 'b = ∇'×M ' = 0
r
r
K 'b = M '×nˆ ' = M ' sin θ 'φˆ'
M
⎧ µ0 M
ˆ
r r ⎪ 3 r sin θφ , _ r ≤ R
r 2µ M
A(r ) = ⎨
, B = 0 zˆ
3
3
⎪ µ0 R M sin2θ φˆ, _ r ≥ R
r
⎩ 3
r
r µ R 3 M sin θ
4πR 3 M µ0 sin θ ˆ µ0 m × rˆ
0
ˆ
r ≥ R, A=
φ=
φ=
3
r2
3
4πr 2
4πr 2
The field outside is the same as a pure dipole with a total magnetization of
r 4πR 3 r
m=
M.
3
Hint: What is the best calculation way to find the magnetic field of a uniformly
magnetized sphere?
r
Because of no externally applied current flow, ∇ × H = 0 , we can try to find a scalar
r
potential for the auxiliary field H .
r
σ b = M ⋅ rˆ = M cosθ
azimuthal symmetry and spherically symmetrical problem:
B ⎞
⎛
V (r ,θ ) = ∑ Pl (cosθ )⎜ Al r l + l +l 1 ⎟
r ⎠
⎝
l
B
Vr ≥ R = ∑ l +l 1 Pl (cosθ ) , Vr ≤ R = ∑ Al r l Pl (cosθ )
l r
l
M cosθ = −
∂Vr > R
∂V
2B ⎞
⎛
+ r<R
= ⎜ A1 + 31 ⎟ cosθ
∂r r = R
∂r r = R ⎝
R ⎠
Continuous at r = R : A1R cosθ =
B1
cosθ
R2
M
M
M 3
M R3
, B1 =
cosθ , Vr < R =
r cosθ
R Æ Vr > R =
2
3 r
3
3
3
r
r
r r
M
M
2
2 r
Vinside =
z Æ H in = − zˆ Æ Bin = µ0 H + M = Mzˆ = M
3
3
3
3
A1 =
(
)
r
r 4πR 3
r
1 m ⋅ rˆ M R 3
m = VM =
Mzˆ Æ Vout =
=
cosθ
3
4π r 2
3 r2
r
r
r
µ m
1 m
H out =
2 cosθrˆ + sin θθˆ Æ Bout = µ0 H out = 0 3 2 cosθrˆ + sin θθˆ
3
4π r
4π r
(
)
(
)
6.2.2 Physical Interpretation of Bound Currents
There is no adjacent loop to do the canceling at the edge Æ edge or bound currents
Bound currents are different from current flow of free charges. Æ No free charges
flow inside the object.
Each charge moves only in a tiny loop within a single atom.
r
r
K b = M × nˆ
When the magnetization is non-uniform, the interal currents no longer cancel.
r
r
Jb = ∇ × M
Example: Diluted Magnetic Semiconductor
Mn
GaAs or ZnO
PRB 57, R2037 (1998). Prof. H. Ohno’s group
r
r
Jb = ∇ × M
r
∇ ⋅ J b = 0 Æ no charge piled up
6.2.3 The Magnetic Field Inside Matter
The actual microscopic magnetic field inside matter fluctuates widely from point to
point and from instant to instant. The meaning of macroscopic field is: average over
regions large enough to contain many atoms.
Example: An iron rod, of length L and square cross section with
side length a, is given a uniform longitudinal magnetization M,
and then bent around into a circle with a narrow gap (width w).
Find the magnetic field at the center of the gap, assuming w << a
<< L.
r
M = Mφˆ
r
r
Inner face: K b = M × nˆ = Mzˆ
a/2
r
Apply Ampere’s law: 2πrB = µ0 2πrM , B = µ0 Mφˆ
r µ
For a square loop, B = 0
4π
I = wM
a/2
ˆ
2 2 wMµ
µ0
Iˆ × Π
wM ⋅ a / 2
'
8
=
⋅
⋅
∫ Π 2 dl 4π ∫0 x 2 + a 2 / 4 3 / 2 dxφˆ = πa 0 φˆ
(
)
r
⎛ 2 2w ⎞ ˆ
⎟φ
Superposition rule: B = µ0 M ⎜⎜1 −
πa ⎟⎠
⎝
Compare electric and magnetic fields in the matter.
The electric field produced by a uniformly polarized sphere of radius R.
V (r ,θ ) =
P
r cosθ
3ε 0
r
r
P
P
E=−
=−
zˆ
3ε 0
3ε 0
r, θ, φ
P
The magnetic field of a uniformly magnetized sphere.
r, θ, φ
r 2
r 2
B = µ0 M = µ0 Mzˆ
3
3
M
What is a demagnetizing factor? (Ref: Stephen Blundell, Appendix D in “Magnetism
in condensed matter”, Oxford University Press.)
r
When the magnetization M inside a ferromagnetic body meets the surface, it has to
r
r
suddenly stop. Hence there is a divergence of M . Using the equations ∇ ⋅ B = 0
r r
r
r
and ∇ ⋅ H + M = 0 → ∇ ⋅ H = −∇ ⋅ M , we find that there is an equal and opposite
r
divergence of H . The situation is as if magnetic monopoles have been left on the
r
surface of the ferromagnet, and these monopoles act as sources of H . The resulting
r
H field is known as a demagnetizing field.
(
)
Exercise: 6.8