Wed. Fri.
6.1 Magnetization 6.2 Field of a Magnetized Object
HW10
Mon., 6.3, 6.4 Auxiliary Field & Linear Media HW11
Wed. 12 noon Exam 3 (Ch 7, 10, 4, 6)
Linear Dielectrics Chunk of induced dipoles
Ep
Everyone’s field but its own
Point along field Linearly proportional
d
pdP
Polarization = Dipole density
so E
d
d
d
EdP
Define “electric susceptibility” to be the proportionality constant (and provide convenient factor of eo.)
P e0eE
e
d
d
o
e
1
PED o
e
Always linear dielectric
EE eoo
ee Eeo
e 1
ED
e
e e0 1 e Permittivity (of not-so-free space)
Dielectric Constant
er oe
ee 1
ED ro
ee
Or, in terms of polarization
PDe
11
or PD
e
e
1
for individual induced dipole
For chunk of induced dipoles
Linear Dielectrics
o
Example: Alternate / iterative perspective on field in dielectric. Consider again a simple capacitor with dielectric. We’ll find the electric field in terms of what it would have been without the dielectric. We’ll do this iteratively and build a series solutions.
sfree
-sfree
o
-sb.o
0. Say we start with no dielectric. Initially there’s the field simply due to the free charge; Eo.
oEE
and the associated surface charges contribute a field of their own, where so in the opposite direction.
E1 -P0 e0 -eE0
oeoo EP
e
zEo
ob ˆ.1
e
s
nPobˆ
0.
s
oe E
-
1
1. This field induces a little counter polarization,
P1 e0eE1 -e0e
2E0
E2 -P1 e0 -e 2E0
oe E
2-
2. See a pattern?
...
1
2
o
n
n
e EE
-0
As long as e< 1, this converges to
Einside 1
1 e
E0
1
er
E0
Same result as we got previously
We insert the dielectric and that field induces a polarization,
Which is means a surface charge and resulting field contribution of its own
Linear Dielectrics
o
Exercise: Try it for your self. A sphere made of linear dielectric
material is placed in an otherwise uniform electric field
o. Find the electric field inside the sphere in terms of the material’s dielectric constant, er.
sfree
-sfree oEE
You can take it as a given that a sphere of uniform polarization contributes field E -P 3e0
Example: A coaxial cable consists of a copper wire of radius a surrounded by a concentric copper tube of inner radius c. The space between is partially filled (from b to c) with material of dielectric constant er as shown below. Find the capacitance per length of the cable.
a b
c
V
QC
L
For the sake of reasoning this out, say there’s charge Q uniformly distributed along the surface of the central wire.
er
enclfQadD .
-
c
a
ldEV
QsLD 2
Gaussian cylinder of some radius a<s<c.
sL
QD
2
csbssL
Q
bsassL
Q
E
ro
o
ˆ2
ˆ2
ee
e
-
c
a
ldEV
--
c
b ro
b
a o
dssL
Qds
sL
QV
eee 22
bc
ab
orL
QV lnln
21e
e-
csbD
bsaDE
ro
o
1
1
ee
e
--
c
b
b
a
ldEldE
bc
ab
o
rL
C
lnln
21e
e
Exercise: There are two metal spherical shells with radii R and 3R. There is material with a dielectric constant er = 3/2 between radii R and 2R. What is the capacitance? R
2R
3R
er
Recall Multi-pole Expansion of Vector Potential
1
r1
r
r
r
n0
n
Pn cos
cosnPObservation location
nth Legendre polynomial
P0 1
P1 u u
2
13 2
2
-
uuP
r
vdqrA o
4)(
Continuous current distribution
014
cos1
)(n
n
n
ndrJPr
rrA o
drJP
r
r
rrA
n
n
n
o
0
4cos
1)(
Re-ordering sums
r
r
r
-
...2
1cos3cos
3
22
2 r
drJr
r
drJr
r
drJ
Monopole Dipole term term
Observation location
r
vdqrA o
4)(
Continuous current distribution
r
r
r
...cos
24r
drJro
Dipole term Dipole’s integral
drJr
cos drJrr
ˆ lIdrr
ˆ
Steady current
ldrrI
ˆ
to mathland and back
- adrI
ˆ - adrI
ˆ arI- ˆ raI ˆ
Recall Multi-pole Expansion of Vector Potential
r
vdqrA o
4)(
...
ˆ24
r
raIo
Magnetic Dipole Moment
aIm
...
ˆ)(
24r
rmrA o
Dipole term for a loop Observation location
r
Magnetic Dipole Moment
aIm
...
ˆ)(
24r
rmrA o
r
m
I
zRIm ˆ2
...
ˆˆ)(
2
2
4r
rzRIrA o
...ˆsin
2
2
4
r
RIo
Same direction as current AB
3
ˆˆ3
4)(
r
rr m-mrB o
dip
If m at origin and pointing up
ˆsinˆcos24
)(3
rr
mrB o
dip
(yes, same form as E for p)
Magnetic Dipoles
...
ˆ)(
24r
rmrA o
aIm
ˆsinˆcos24
)(3
rr
mrB o
dip
From a distance much greater than the current distribution’s size, the dipole term dominates
Recall Torque for real current loop
extB
RightF
LeftF
a
I
extBaIN
sinwILBN ext
extBmN
Like for electric dipoles
extEpN
Recall Force for real current loop
extB
RightF
leftF
a
I
xILBILBFFF LeftRightleftRightnetˆ-
Like for electric dipoles
extEpF
extBmF
If m is constant, the same as
extBmF
In terms of dipole moment In terms of dipole moment
Change in energy
EpU
-
Like for electric dipoles
-- LLRR ldFldFU
- rightright ldF
2
-
dBIb a2
)(2
- dBIb a2
sin)(2
f
i
BIba
cos)(-
f
i
mBU
cos-
BmU
-
In terms of dipole moment
Magnetic Dipoles
...
ˆ)(
24r
rmrA o
aIm
ˆsinˆcos24
)(3
rr
mrB o
dip
From a distance much greater than the current distribution’s size, the dipole term dominates
extBmN
extBmF
If m is constant, the same as
extBmF
BmU
-
Exercise: you have two magnetic dipoles; find the torque m1 applies on m2.
m1
m2
r
y x
z
Dipole B
Dipole A
Magnetic Dipoles
...
ˆ)(
24r
rmrA o
aIm
ˆsinˆcos24
)(3
rr
mrB o
dip
From a distance much greater than the current distribution’s size, the dipole term dominates
extBmN
extBmF
If m is constant, the same as
extBmF
BmU
-
Exercise: find the force on dipoles A, B, C in and near a slab of uniform current
Dipole C
a
-a
2a
zJJ ˆ
infront ˆ
inside ˆ
yJa
yJxB
o
o
B
Dipole B
Dipole A
Magnetic Dipoles
...
ˆ)(
24r
rmrA o
aIm
ˆsinˆcos24
)(3
rr
mrB o
dip
From a distance much greater than the current distribution’s size, the dipole term dominates
extBmN
extBmF
If m is constant, the same as
extBmF
BmU
-
Exercise: Force between two dipoles (read, “bar magnets”). What’s the force A exerts on B?
r = z
Considering ‘real’ dipoles (with real radii), roughly sketch the field and resulting forces on the current loops
Effect of Magnetic Field on Dipoles
Para-magnetic: Rotate the loop (torque)
Dia-magnetic: Stretch the loop; changing field(s) – Faraday-effect: impede the current
Wed. Fri.
6.1 Magnetization 6.2 Field of a Magnetized Object
HW10
Mon., 6.3, 6.4 Auxiliary Field & Linear Media HW11
Wed. 12 noon Exam 3 (Ch 7, 10, 4, 6)