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Lecture 13

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Magnetic Dipole Moment
µ0 2πR 2 I
far from coil: Bz =
4π z 3
µ0 2 µ
Bz =
3
4
π
z
magnetic
2
µ
=
π
R
I = AI
dipole moment:
1 2p
far from dipole: E z =
4πε 0 z 3
p = sq
µ - vector in the direction of B
Twisting of a Magnetic Dipole
The magnetic dipole moment µ acts like a compass needle!
In the presence of external magnetic field a current-carrying
loop rotates to align the magnetic dipole moment µ along the
field B.
The Magnetic Field of a Bar Magnet
How does the magnetic field around a bar magnet look like?
N
S
Magnets and Matter
How do magnets interact with each other?
Magnets interact with iron or steel, nickel, cobalt.
Does it interact with charged tape?
Does it work through matter?
Does superposition principle hold?
Similarities with E-field:
• can repel or attract
• superposition
• works through matter
Differences with E-field:
• B-field only interacts with some objects
• curly pattern
• only closed field lines
Magnetic Field of Earth
The magnetic field of the earth has a pattern that looks like that
of a bar magnet
Horizontal component of
magnetic field depends on
latitude
Maine: ~1.5.10-5 T
Indiana: ~2.10-5 T
Florida: ~3.10-5 T
Can use magnetic field of Earth
as a reference to determine
unknown field.
Magnetic Monopoles
An electric dipole consists of two opposite charges – monopoles
Break magnet:
S
S N
There are no magnetic monopoles!
N
The Atomic Structure of Magnets
The magnetic field of a current loop and the magnetic field of a bar
magnet look the same.
Electrons
Batom
µ0 2 µ
=
,
3
4π z
µ = π R2 I
What is the direction?
One loop:
What is the average current I?
e
current=charge/second: I =
t
ev
2π R
I=
T=
2πR
v
1
2 ev
µ = πR
= eRv
2πR 2
N
S
Magnetic Dipole Moment
1
Magnetic dipole moment of 1 atom: µ = eRv
2
Method 1: use quantized angular momentum
Orbital angular momentum: L = Rmv
1
1 e
1 e
µ = eRv =
Rmv =
L
2
2m
2m
Quantum mechanics: L is quantized:
L = n,  = 1.05 × 10−34 J ⋅ s
1 e
L = 0.9 × 10 −23 A ⋅ m 2 per atom
If n=1: µ =
2m
Magnetic Dipole Moment
1
Magnetic dipole moment of 1 atom: µ = eRv
2
Method 2: estimate speed of electron
 
dp
= Fnet
Momentum principle:
dt

Circular motion: p = p = const

dp
v
= ω p = mv = Fnet
dt
R
mv 2
1 e2
=
R
4πε 0 R 2
1 e2
v=
≈ 1.6 × 106 m/s
4πε 0 Rm
ω – angular speed
ω =v/R
µ ≈ 1.3 × 10 −23 A ⋅ m 2 /atom
Magnetic Dipole Moment
−23
A ⋅ m2 /atom
Magnetic dipole moment of 1 atom: µ ≈ 10
Mass of a magnet: m~5g
6.1023 atoms
Assume magnet is made of iron: 1 mole – 56 g
number of atoms = 5g/56g . 6.1023 ~ 6.1022
µ magnet ≈ 6 × 10 22 ⋅10 −23 = 0.6 A ⋅ m 2
Modern Theory of Magnets
1. Orbital motion
There is no ‘motion’, but a distribution
Spherically symmetric cloud (s-orbital)
has no µ
Only non spherically symmetric orbitals (p, d, f) contribute to µ
There is more than 1 electron in an atom
Modern Theory of Magnets
Alignment of atomic dipole moments:
ferromagnetic materials:
iron, cobalt, nickel
most materials
Modern Theory of Magnets
2. Spin
Electron acts like spinning charge
- contributes to µ
Electron spin contribution to µ is of the same order as one due to
orbital momentum
Neutrons and proton in nucleus also have spin but their µ‘s are
much smaller than for electron
1 e

same angular momentum: µ ≈
2m
NMR, MRI – use nuclear µ
Nuclear Magnetic Resonance
Proton spin
Magnet
N
S
Felix Bloch
(1905 -1983)
Edward Purcell
(1912-1997)
B field
Magnetic Resonance Imaging
B
Modern Theory of Magnets
Magnetic domains
Very pure iron – no residual magnetism spontaneously disorders
Hitting or heating can also demagnetize
Why are there Multiple Domains?
Magnetic domains
Iron Inside a Coil
Multiplier effect:



Bnet = Bcoil + Biron


Bnet > Bcoil
Electromagnet:
Magnetic Field of a Solenoid
Step 1: Cut up the distribution
into pieces
Step 2: Contribution of one piece
origin: center of the solenoid
µ0
2π R 2 I
one loop: Bz =
4 π R 2 + ( d − z )2
(
)
3/2
B
Number of loops per meter: N/L
Number of loops in Δz: (N/L) Δz
µ0
2π R 2 I
Field due to Δz: ΔBz =
4 π R 2 + ( d − z )2
(
)
3/2
N
Δz
L
Magnetic Field of a Solenoid
Step 3: Add up the contribution
of all the pieces
µ0
2π R 2 I
dBz =
4 π R 2 + ( d − z )2
(
2
)
3/2
N
dz
L
L /2
µ0 2π R NI
dz
Bz =
∫
2
2
4π
L
− L /2 R + ( d − z )
(
)
3/2
B
Magnetic field of a solenoid:
µ0 2π NI ⎡⎢
Bz =
4π L ⎢
⎣
⎤
⎥
−
2
2
2
d
+
L
/
2
+
R
d
−
L
/
2
(
)
(
) + R 2 ⎥⎦
d+ L/2
d− L/2
Magnetic Field of a Solenoid
µ0 2π NI ⎡⎢
Bz =
4π L ⎢
⎣
⎤
⎥
−
( d + L / 2 )2 + R 2 ( d − L / 2 )2 + R 2 ⎥⎦
d+ L/2
d− L/2
Special case: R<<L, center of the solenoid:
µ0 2π NI ⎡⎢ L / 2
−L / 2 ⎤ µ0 2π NI
⎥ =
Bz ≈
−
[2]
2
2
4π L ⎢ ( L / 2 )
⎥ 4π L
L
/
2
(
)
⎣
⎦
Bz ≈
µ0 NI
L
in the middle of a long solenoid
Compact Muon Solenoid Detector
Particle Physics experiment at CERN
Largest superconducting magnet in
the world
B = 4 Tesla
Diameter: 15 m
Length: 21.5 m
Total weight: 12500 Tons
STAR Time Projection Chamber
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