wed. (c 21.6-7,.9) 1.3.4-1.3.5, 1.5.2-1.5.3, 5.3.1-.3.2
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Wed. Fri.
(C 21.6-7,.9) 1.3.4-1.3.5, 1.5.2-1.5.3, 5.3.1-.3.2 Div & Curl B (C 21.6-7,.9) 5.3.3-.3.4 Applications of Ampere’s Law
Mon. Wed. Thurs. Fri.
1.6, 5.4.1-.4.2 Magnetic Vector Potential 5.4.3 Multipole Expansion of the Vector Potential Review
HW8
q1
1r
1E
q2
2r
2E
netE
Memory Lane: Electrostatics Field of charge distributions
rr 2
ˆ1
1
41
1
qrE
o
drdqd
dqr )()(
drrEo
24
1ˆ
r
r
ech i
dqqi i
inet dq
qrE
oo
arg
41
lim1
41
ˆˆ
2i2 r
rr
r
Found Electric fields due to variety of charge distributions
-orr
or
q
Biot-Savart Law
24
ˆ
r
r
ldIrBd o
I
r
r
Source
Observation location
dL
r
2
0 ˆ4 r
r
IdrB
I
r
r
dl
r
y
x
z
x y
z
r
f f
r
l
r
r
a
aa
dr
2sin
top
bottom
o dIn
a
a
aasin
2
r
z
𝑥
𝑦
𝑧 𝜌 𝑟 ′ = 𝜌𝑜𝑐𝑜𝑠 𝜃′
r’dq’
r’sin(q’)df’ r q’
f’
dr’
Magneto-statics Fields of current distributions
Del Operator
zz
yy
xx
ˆˆˆ
Gradient – vector representing the local slope of a scalar field. Example: temperature across a sheet of metal heated at one corner.
T T
xˆ x
T
yˆ y
T
zˆ z
Divergence – scalar representing in/out flow from a point in a vector field. Example: velocities of water filling a bath tub.
v vx
x
vy
y
vz
z
Curl – vector representing circulation of a vector field. Example: velocities of water swirling down a drain.
v
ˆ x ˆ y ˆ z
x y z
vx vy vz
ˆ x vz
y
vy
z
q3
q2
Memory Lane: Electrostatics Flux from Charge Sources
q1
11 EadE
441
1
o
q
E
o
q
EadE
1
11
||daEE 11
2
2
2
1
2
0
2
41sin
f
q
qqf ddo
q
E rr
Ditto for q2, q3, …
o
enclosednetQ
net adE.
(integral form) Gauss’s Law
Vol
QE encl
VolVol o1
0
E
0lim
Vollimdiv
x
y
z
x ,y,z
Ex(x)
Ex(x x)
o
E 1
oVol
adEE
Vol
1
0limdiv
(derivative form)
0 ldE
0 E
Motivated scalar potential
Simplified symmetric problems
ldEV
EV
Pause & Put together Gauss’s Th’m
Gauss’s Theorem – explicitly putting it together
but o
enclosedQadE
oE 1
Though we were thinking specifically about electric field while we did the math that got us to this relation, it’s quite general and true for any vector field. So, as expressed in Ch. 1, for generic function F,
denclQ And we’d gone off and proven
Putting these together:
dadE
o
dEadE
dFadF
24
ˆ
r
r
ldIrBd o
I
r
r
Source
Observation location
dL
r
Magneto-Statics Flux from Current Sources
ad
ldIdadrBd o
B
24
ˆ
r
r
Not so easy; start with special case to suggest / understand general result infinite line charge
f
ˆ2s
IrB o
line
ad
IadrB o
Bline
f
ˆ2s
Surface of any radius would work
ss
sˆˆ
2dzd
Io ff
sˆ2
ff
dzIdo 0
Similarly,
zsline Bz
Bs
sBss
rB
f
f
11
dzsfds ss ˆdzdad f
01
01
2z
I
ss
sso
s
f0
24
ˆ
r
r
ldIrBd o
I
r
r
Source
dL
r
Magneto-Statics Circulation from Current Sources
ld
ldIldrBd o
24
ˆ
r
r
Not so easy; start with special case to suggest / understand general result infinite line charge
f
ˆ2s
IrB o
line
0 adrB line
fff
ˆˆ2
dI
o ss
fff
ˆˆ2
Ido
What of ?
0 rBline
s
rBCurl line
ld
IldrB o
line
f
ˆ2s
ff ˆdld s
path of any radius would work
2
2Io Io
Takes a little work to get right answer and understand.
Goes Differential: Curl
Ampere’s Law
JA
I
A
I
A
I
A
dB
A
dB
A
dB
JA
I
A
dB
z
zpiercing
y
ypiercing
x
xpiercing
zyx
I
piercing
I
0
,,,
0
00
,,,,
patchz
patchz
Acomponentz
A
dBB
patchz.
.
0.
.
limcurl
Curl=circulation density (per area encircled) Zoom in to differential scale:
Break closed path into paths around differential patches
Break area into differential area ‘patches’
(ultimately all internal path legs cancel with each other)
I
B 𝐵 ∙ 𝑑𝑙 = 𝜇𝑜𝐼𝑝𝑖𝑒𝑟𝑐𝑖𝑛𝑔
Project onto coordinate planes
...curl . componentxB ...curl . componentyB
Goes Differential: Curl
Ampere’s Law
...curl zB
Curl=circulation density (per area encircled)
I
B 𝐵 ∙ 𝑑𝑙 = 𝜇𝑜𝐼𝑝𝑖𝑒𝑟𝑐𝑖𝑛𝑔
patchx
patchxo
Apatchx
patchx
Ax
A
I
A
dBB
patchxpatchx.
.
0.
.
0 ..
limlimcurl
...curl yB
z
y
By(z+z)
By(z)
Bz(y)
Bz(y+y) Ix
zy
zyByzzBzyyByzB zyzy
zy 0lim
zy
yzByzzB
zy
zyBzyyB yyzz
zy 0lim
z
B
y
B yz
xB
curl
patchx
xo
dA
dI
.
xo J
Similarly, yozx
y Jx
B
z
BB
curl and zo
xy
z Jy
B
x
BB
curl
zyxoxyzxyz JJJ
y
B
x
B
x
B
z
B
z
B
y
BB ,,,,curl
or JB
0
Magneto-Statics Divergence and Circulation from Current Sources
f
ˆ2s
IrB o
line
0 adrB line
0 rBline
s
IldrB oline
JB
0
Derivation specific to constant, Infinite Line-Current
Gauss’s Theorem Stokes’ Theorem
Note: both follow from applying Biot-Savart, which holds only for steady currents
Now for more general (more mathematical / less intuitive) proof
Pause and put together Stoke’s:
Stoke’s Theorem – explicitly putting it together
but
Though we were thinking specifically about magnetic field while we did the math that got us to this relation, it’s quite general and true for any vector field. So, as expressed in Ch. 1, for generic function F,
daJI
And we’d gone off and proven
Putting these together:
adFldF
IldrB oline
JB
0
daJldrB oline
daBldrB line
Derive Divergence of B 𝛻𝑟 ∙ 𝐵 𝑟 = 𝛻𝑟 ∙ 𝜇𝑜
4𝜋 𝐽 ×r r2 𝑑𝜏′
observation location
𝐵 𝑟
𝑟
𝑟 ′
r = 𝑟 -𝑟 ′
source currents
𝛻𝑟 ∙ 𝐵 𝑟 =𝜇𝑜
4𝜋 𝛻𝑟 ∙ 𝐽 ×
r r2 𝑑𝜏′
Can slip del inside integral since not taking derivative with respect to integration variable
Use Product Rule (6): BAABBA
𝛻𝑟 × 𝐽 ×r r2 =
r r2 ∙ 𝛻𝑟 × 𝐽 - 𝐽 ∙ 𝛻𝑟 ×
r r2
taking derivative to see how field varies from one observation location to another, not for changes in source locations
using
dJdaldda
IldI
r r2 has no curl (can write out in Cartesian to convince, or see in spherical)
𝛻𝑟 ×r r2 =
1
𝑟
1
𝑠𝑖𝑛𝜃
𝜕
𝜕𝜑
1
r2 𝜃 - 1
𝑟
𝜕
𝜕𝜑
1
r2 𝜑 = 0
𝛻𝑟 ∙ 𝐵 𝑟 =𝜇𝑜
4𝜋 0𝑑𝜏′ = 0
So by Gauss’ Theorem
𝛻𝑟 ∙ 𝐵 𝑟 𝑑𝜏 = 𝐵 𝑟 ∙ 𝑑𝑎 = 0
Derive Ampere’s Differential form 𝛻𝑟 × 𝐵 𝑟 = 𝛻𝑟 × 𝜇𝑜
4𝜋 𝐽 ×r r2 𝑑𝜏′
observation location
𝐵 𝑟
𝑟
𝑟 ′
r = 𝑟 -𝑟 ′
source currents
𝛻𝑟 × 𝐵 𝑟 =𝜇𝑜
4𝜋 𝛻𝑟 × 𝐽 ×
r r2 𝑑𝜏′
Can slip del inside integral since not taking derivative with respect to integration variable
Use Product Rule (8): ABBABAABBA
𝛻𝑟 × 𝐽 ×r r2 =
r r2 ∙ 𝛻𝑟 𝐽 - 𝐽 ∙ 𝛻𝑟
r r2+𝐽 𝛻𝑟 ∙
r r2 -
r r2 𝛻𝑟 ∙ 𝐽
taking derivative to see how field varies from one observation location to another, not for changes in source locations
= - 𝐽 ∙ 𝛻𝑟r r2+𝐽 𝛻𝑟 ∙
r r2
using
dJdaldda
IldI
𝛻𝑟 ∙r r2=4𝜋𝛿3 𝑟
Motivation
add
22 r
r
r
r ˆˆ
by Gauss’s theorem
fqq ddsin
1 2
2r
r4
𝛻𝑟 × 𝐵 𝑟 =𝜇𝑜
4𝜋 − 𝐽 ∙ 𝛻𝑟
r
r2 𝑑𝜏′ + 𝐽 𝛻𝑟 ∙
r
r2 𝑑𝜏′
𝐽 4𝜋𝛿3 𝑟 𝑑𝜏′
4𝜋𝐽 𝑟
+ 𝐽 ∙ 𝛻𝑟′r
r2 𝑑𝜏′
Looking at just one component
𝐽 ∙ 𝛻𝑟′𝑥−𝑥′
r3 𝑥 = 𝛻𝑟′ ∙ 𝐽 𝑥−𝑥′
r3 −𝑥−𝑥′
r3 𝛻𝑟′ ∙ 𝐽 𝑥
Derive Ampere’s Differential form 𝛻𝑟 × 𝐵 𝑟 = 𝛻𝑟 × 𝜇𝑜
4𝜋 𝐽 ×r r2 𝑑𝜏′
observation location
𝐵 𝑟
𝑟
𝑟 ′
r = 𝑟 -𝑟 ′
source currents
using
dJdaldda
IldI
𝛻𝑟 × 𝐵 𝑟 =𝜇𝑜
4𝜋 𝐽 ∙ 𝛻𝑟′
r
r2 𝑑𝜏′ +4𝜋𝐽 𝑟
𝛻𝑟′ ∙ 𝐽 =-𝑑𝜌
𝑑𝑡
where
= 0
𝛻𝑟′ ∙ 𝐽 𝑥 − 𝑥′
r3𝑑𝜏′ = 𝐽
𝑥 − 𝑥′
r3∙ 𝑑𝑎 ′
𝐽 ∙ 𝛻𝑟′𝑥−𝑥′
r3 𝑥 = 𝛻𝑟′ ∙ 𝐽 𝑥−𝑥′
r3 −𝑥−𝑥′
r3 𝛻𝑟′ ∙ 𝐽 𝑥
Ditto for other two components
For now, with electro-magnetic statics
Gauss’s theorem
If area fully encloses current, then no current penetrates area
= 0
Ditto for other two components
𝛻𝑟 × 𝐵 𝑟 =𝜇𝑜
4𝜋0 +4𝜋𝐽 𝑟
𝛻𝑟 × 𝐵 𝑟 = 𝜇𝑜𝐽 𝑟
So by Stokes’ Theorem
𝐵 𝑟 ∙ 𝑑𝑙 = 𝜇𝑜 𝐽 𝑟 ∙ 𝑑𝑎 ′ = 𝜇𝑜𝐼
Limitations: holds when Biot-Savart holds - statics
Maxwell’s laws for electro-statics
𝛻𝑟 ∙ 𝐵 𝑟 = 0
𝐵 𝑟 ∙ 𝑑𝑎 = 0
𝛻𝑟 × 𝐵 𝑟 = 𝜇𝑜𝐽 𝑟
𝐵 𝑟 ∙ 𝑑𝑙 = 𝜇𝑜𝐼
𝛻𝑟 × 𝐸 𝑟 = 0
𝐸 𝑟 ∙ 𝑑𝑙 = 0
𝛻𝑟 ∙ 𝐸 𝑟 = 1𝜀0
𝜌 𝑟
𝐸 𝑟 ∙ 𝑑𝑎 = 1𝜀𝑜
𝑄
For arguably symmetric fields, useful for finding fields
Gauss’s
Ampere’s
Wed. Fri.
(C 21.6-7,.9) 1.3.4-1.3.5, 1.5.2-1.5.3, 5.3.1-.3.2 Div & Curl B (C 21.6-7,.9) 5.3.3-.3.4 Applications of Ampere’s Law
Mon. Wed. Thurs. Fri.
1.6, 5.4.1-.4.2 Magnetic Vector Potential 5.4.3 Multipole Expansion of the Vector Potential Review
HW8
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