Download Re-run: Bound and free currents: π µ τ π µ uM K χ = “Bound currents”

Re-run: Bound and free currents:
 
 
  µ0
J b (r ' )
µ0 K b ( r ' ) 
A(r ) =
d
τ
'
+
da '
∫∫∫
∫∫
4π
d
4π S d
  
volume current J b = ∇ × M

 
K b = M × un surface current
The magnetic potential in all cases can be expressed as
  

A(r ) = Abound volume + Abound surface
Magnetic potential (and magnetic field) of a polarized
object can always be produced by a
  
volume current: J b = ∇ × M
“Bound currents”
and a

 
surface current: K b = M × un
Bound currents result from the contribution of all (aligned)
atomic dipoles.
Free currents involve flow of charge (source like battery).
It might flow through wires embedded in a magnetic
material or the material itself (in conductor).
Summary last session
The concept of bound currents and free currents allows
to express the total current as the sum:
 

J = J bound + J free
This allows to distinguish auxiliary field H, magnetization M
and total magnetic field B:


1 
H :=
B−M
µ0
free-
total-
bound- currrents
As any vector field, H is determined completely when the
following things are determined (Helmholtz theorem):
  
• Curl ∇ × H = J free Ampère’s law for H
   1   
 
B − ∇ ⋅ M = −∇ ⋅ M
• Divergence ∇ ⋅ H = ∇ ⋅
µ0
• Boundary conditions for H:
H
||
above
−H
||
below
 
= K f × unormal
(
⊥
⊥
⊥
⊥
H above
− H below
= − M above
− M below
)
Today’s lecture
VI. Magnetostatics in matter
• 
Dia-, Para and Ferromagnetism
Two questions
1. 
Can you envision a scenario in which you have a
free current alone (with no bound current)?
2. 
Problem 5.18
In calculating the current enclosed by an amperian loop,
one must, in general, evaluate an integral of the form
I enclosed
 
= ∫∫ J ⋅ da
S
The trouble is, there are infinitely many surfaces that
share the same boundary line. Which one are we
supposed to use?
Remember that you will have to use the above
integration when you want to use Ampere’s law in
integral form in the presence of current density that is not
homogenous.
Remembering the bar electret

 
D := ε 0 E + P
electric
displacement
total Polarization
field
Example: (4.11) Bar electret (electric analog to a magnet)
A short cylinder, radius a, length L carries a “frozen-in”
uniform polarization P parallel to its axis.

Find the bound charge
ρb = −∇P = 0
 
σ b = P ⋅ un = ± P
Sketch the electric field for
L>>a
L<<a

P
P, E and D:
L≈a

P

E

P
No free charges anywhere!

P

D
The magnetic analog to the bar electret

1  
H :=
B−M
µ0
auxiliary total Magnetization
field field
Example: (Problem 6.9) A short circular cylinder of radius
a and length L carries a “frozen-in” uniform magnetization
M parallel to its axis. Find the bound current and sketch the
   
magnetic field of the cylinder.
Jb = ∇ × M = 0
Sketch the electric field for
L>>a
L<<a
Solenoid

 

Kb = M × un = M ⋅ uφ
Physical dipole
L≈a
intermediate case
Conclusion: The external fields are the same as in the
electrical case. The internal fields (inside bar) are opposite
in direction.
No free currents anywhere!
Comparison of fields 1
No free charges anywhere!

   

 
∇D = 0; ∇ × D = ∇ × P but D = ε 0 E + P ≠ 0 !!!
No free currents anywhere!

 
  
 B

∇H = 0; ∇ × H = J free but H =
− M ≠ 0 !!!
µ0
The external fields are the same as in the electrical
case. The internal fields (inside bar) are opposite in
direction.
Comparison of fields 2
L>>a
L<<a
L≈a
No free charges anywhere!
dipole
parallel plate capacitor intermediate case
No free currents anywhere!
Long Solenoid
Physical dipole
intermediate case
Magnetism
• The magnetic susceptibility χm (and the electric
susceptibility χe) expresses the response of matter to
an external magnetic field (and an electric field).


Linear media: M = χ m H


B = µ0 (1 + χ m ) ⋅ H

= µ⋅H


P = ε0 ⋅ χe ⋅ E


D = ε 0 (1 + χ e )⋅ E

=ε ⋅E
• Electrons orbiting around the nuclei and “spinning around
their axes” are currents that cause magnetic dipole moments.
• Usually all magnetic moments cancel out.
• Application of a magnetic field causes a net alignment of
the magnetic dipole moments resulting in a magnetic field.
• The material becomes magnetized.
Dia- and Paramagnetism
Two different effects enter the magnetization of matter:
1. Distortion of electron orbit → diamagnetism
• Perturbation of the electron motion
by the magnetic field induces current
with magnetic moment opposite to field
• Effect for all atoms and material has
magnetization opposite to external field:
e−
χm < 0 : magnetization opposite to H

B
e+

m
2. Orientation of electron spin → paramagnetism
• An atom (or molecule) may have a
permanent magnetic dipole moment
• Effect for all atoms and material has
magnetization parallel to external field

mspin
χm > 0 : magnetization parallel to H
• Paramagnetism is usually absent in
  
N = m×B
materials with even number of electrons
(due to Pauli principle).
Problem
Of the following materials, which would you expect to be
paramagnetic and which diamagnetic? (Problem 6.6)
Al, Cu, C, Pb, N2, Na, NaCl, S, H2O ?
Magnetism
Diamagn.
Bismuth
Gold
Silver
Copper
Water
CO2
H2
χm
-1.6⋅10-4
-3.4⋅10-5
-2.4⋅10-5
-9.7⋅10-6
-9.0⋅10-6
-1.2⋅10-8
-2.2⋅10-9
Ne Paramagnetics χm
83 Oxygen (1atm) 2.1⋅10-6
79 Sodium
8.5⋅10-6
47 Aluminum
2.1⋅10-5
29 Tungsten
2.4⋅10-5
10
Platinum
2.8⋅10-4
22 Oxygen(-200C) 3.9⋅10-3
2
Gadolinium
4.8⋅10-1
Ne
8
11
13
74
78
8
64
• Diamagnetism is usually much weaker and is observed for
atoms where paramagnetism is absent.
Note: The number of electrons (even or odd) is not always
a good criterion. (Cu is slightly diamagnetic!)
• In both cases (dia- and paramagnetism) the magnetization
vanishes when the external magnetic field is turned off.
• Only in Ferromagnets the magnetization is retained
without an external magnetic field (“permanent magnets”).
Magnetism
Diamagn.
Bismuth
Gold
Silver
Copper
Water
CO2
H2
χm
-1.6⋅10-4
-3.4⋅10-5
-2.4⋅10-5
-9.7⋅10-6
-9.0⋅10-6
-1.2⋅10-8
-2.2⋅10-9
Ne Paramagnetics χm
83 Oxygen (1atm) 2.1⋅10-6
79 Sodium
8.5⋅10-6
47 Aluminum
2.1⋅10-5
29 Tungsten
2.4⋅10-5
10
Platinum
2.8⋅10-4
22 Oxygen(-200C) 3.9⋅10-3
2
Gadolinium
4.8⋅10-1
Dielectric constant
εr
Vacuum
1
He
1.000065
Ne
1.00013
H2
1.00052
Water vapor
1.00587
Benzene
2.28
Diamond
5.7
Water 80.1
KTaNbO3
34,000
ε r :=
ε
= 1 + χe
ε0
Ne
8
11
13
74
78
8
64
Ferromagnetism
• Magnetic field M maintained when external field is off.
• The origin of Ferromagnetism is the spin of the (unpaired)
electrons therefore all ferromagnetic materials are also
paramagnetic.
• Due to a correlation of the atoms the dipoles tend to align
automatically in larger areas called domains. (10-8-10-12m3
with 1021-1017 atoms).
• In the presence of external magnetic fields two things occur
The domains tend to grow and
The domains tend to align parallel to the magnetic field.
• Curie temperature defines phase transition:
TC=1040°C for Fe, TC=630°C for Ni
  
• Hysteresis B = F (H )
M or B
I or H
Problem
How would you go about demagnetizing a permanent
magnet at point c in the hysteresis loop? That is how could
you restore its original state with M=0 at I=0?
(Problem 6.20)