Lecture 28 Carl Bromberg - Prof. of Physics
PHY481: Electromagnetism
Magnetic dipole
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Magnetic Dipole Expansion and Vector Potential in Electromagnetism, Study notes of Physics
The magnetic multipole expansion and the calculation of the vector potential of a magnetic dipole. The text also covers legendre polynomials and their use in describing the magnetic field of a current loop. The professor, carl bromberg, is delivering a lecture on phy481: electromagnetism.
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PHY481: Electromagnetism
Magnetic dipole
Magnetic dipole
In the many attempts to find a magnetic
monopole, no one as ever seen one.
Maybe there is only one in the universe (Dirac and monopoles)
∇ ⋅ B = 0 ⇒ no magnetic monopole
A ( x ) =
0
J ( x ′) d
3
x ′
x − x ′
∫
The vector potential of an arbitrary current
distribution can tough to calculate.
However it can be described in a “multipole” expansion.
x − x ′
r
2
− 2 rr ′cos θ + r ′
2
r
r ′cos θ
r
2
r ′
2
r
3
cos
2
⎟ +^ ...
Also used for electric “multipole” expansion earlier in semester
N S
N S N S N S NS
P
1
cos θ
= cos θ
P
2
cos θ
cos
2
Legendre polynomials
Another approach
A ( x ) =
0
I
I d ′
x − x ′
∫
d ′= Rd φ ′ −
i sin φ ′+
( (^) j cos φ′)
x =
j y +
k z
x ′=
i R cos φ ′+
j R sin φ′
x − x ′ = −
i R cos φ ′+
j ( y − R sin φ ′) +
k z
x − x ′
2
= R
2
cos
2
φ ′+ ( R sin
2
φ ′+ y
2
− 2 yR sin φ ′) + z
2
x − x ′ = R
2
- y
2
- z
2
− 2 yR sin φ′
A ( x ) =
0
IR
d φ ′
i sin φ ′+
( (^) j cos φ′)
R
2
- y
2
- z
2
0 − 2 yR sin φ′
2 π
∫
μ 0
IR
4 π
d φ ′
i sin φ ′+
( (^) j cos φ′)
R
2
- r
2
(^0) − 2 rR sin θ sin φ′
2 π
∫
A
y
j @ φ ′= 0, π have equal distance to P, average to zero
if point P is placed elsewhere −
i direction becomes
φ
A
φ
μ 0
IR
4 π
sin φ ′ d φ′
R
2
- r
2
(^0) − 2 rR sin θ sin φ′
2 π
∫
Now introduce Legendre Polynomials
Compare with expansion on slide 2
r
2
+ R
2
− 2 rR cos θ
r
RP
1
(cos θ)
r
2
cos θ → sin θ sin φ′
A
φ
0
IR
4 π r
R
r
n
n = 0
∞
P
n
0
2 π
(sin θ sin φ ′)sin φ ′ d φ′
A = 0, for all even n, with first non zero term, n = 1
A
φ
0
IR
2
sin θ
4 π r
2
sin
2
φ ′ d φ ′=
0
2 π
0
I π R
2
sin θ
4 π r
2
A
φ
μ 0
m sin θ
4 π r
2
; m = I π R
2
B
θ
r
∂ r
rA φ
0
m
r
∂ r
sin θ
r
0
m
4 π r
3
sin θ
B
r
r sin θ
∂ r
sin θ A
φ
0
m
4 π r
3
sin θ
sin
2
0
m
4 π r
3
( 2 cos θ)
B ≈
0
m
4 π r
3
( 2 cos θ r ˆ + sin θ
θ)
point dipole
B
point dipole
0
3 r ˆ m ⋅ r ˆ
− m
r
3
; m = m
k
HW8 hints
Problem 8.
B ( x ) =
μ 0
4 π
J ( x ) × r ˆ
r
∫ (^2)
μ 0
I
4 π
d × r ˆ
r
∫ 2
d = rd φ
φ
d
2
B
z
dz
2
HW8 Hints
∇ ⋅ B = 0
∇ × B = μ 0
Problem 8.14 J
B
∫
⋅ d = μ 0
I
encl
= μ 0
J
∫
⋅ d A
Problem 8.
Ampere’s Law:
Problem 8.
a
z
ω = ω
ˆ k
θ
β
x = r r ˆ
a sin β
B
in
= ∇ × A
in
= ∇ ×
0
σ a ω
k × r r ˆ
B
out
= ∇ × A
out
= ∇ ×
μ 0
σ a
4
ω
k ×
r ˆ
r
2
Problem 8.34 Use result of Problem 8.
Constraints:
A
in
& A
out
given in text