Transcript 4). Ampere’s Law and Applications

4). Ampere’s Law and Applications
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As far as possible, by analogy with Electrostatics
•
B is “magnetic flux density” or “magnetic induction”
•
Units: weber per square metre (Wbm-2) or tesla (T)
•
Magnetostatics in vacuum, then magnetic media
based on “magnetic dipole moment”
Biot-Savart Law
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•
I
The analogue of Coulomb’s Law is
the Biot-Savart Law
dB(r)
r
r-r’
Consider a current loop (I)
O
r’
dℓ’
•
For element dℓ there is an
associated element field dB
dB perpendicular to both dℓ’ and r-r’
same 1/(4pr2) dependence
o is “permeability of free space”
defined as 4p x 10-7 Wb A-1 m-1
Integrate to get B-S Law
oI d' x(r  r')
dB(r ) 
4p r  r' 3
B(r ) 
o I d' x(r  r' )
4p  r  r' 3

B-S Law examples
(1) Infinitely long straight conductor
dℓ and r, r’ in the page
dB is out of the page
B forms circles
centred on the conductor
Apply B-S Law to get:
I
dℓ q
r’ z
O
r - r’
r
a dB
q  p/2 + a
sin q = cos a 
o I
B
2p r
B
r
r
2
+z

2 1/2
B-S Law examples
(2) “on-axis” field of circular loop
dℓ
Loop perpendicular to page, radius a
dℓ out of page and r, r’ in the page
On-axis element dB is in the page,
perpendicular to r - r’, at q to axis.
r - r’
I
r’
a
r
z
dB
q
dBz
Magnitude of element dB
o I d
o I d
a
a
dB 
 dBz 
cosq cosq 

2
2
4p r - r'
4p r - r'
r - r' a2 + z2 1/2
Integrating around loop, only z-components of dB survive
The on-axis field is “axial”
On-axis field of circular loop
dℓ
Bon axis   dB z 

o I
4p r - r '
o I
4p r - r '
r - r’
cosq  d
2
cosq 2p a  
2
I
o Ia2
2 r - r'
a
Bon axis 
2 limiting cases:
z 0
Bon
axis 
2a
z  a
and Bon
axis 
q
r
z
dBz
3
Introduce axial distance z,
where |r-r’|2 = a2 + z2
o I
r’
dB
o Ia2
2z3

o I a 2
2a +z
2
2

3
2
Magnetic dipole moment
The off-axis field of circular loop is
much more complex. For z >> a it is
identical to that of the electric dipole
E
p
4p or 3

2 cosq rˆ + sinq qˆ

 om
2 cosq rˆ + sinq qˆ
3
4p r
w herem  p a 2 I  a I or m  p a 2 I zˆ
a area enclosed by current loop
B 
m
m “current times area” vs p “charge times distance”
q r
B field of large current loop
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Electrostatics – began with sheet of electric monopoles
Magnetostatics – begin sheet of magnetic dipoles
Sheet of magnetic dipoles equivalent to current loop
Magnetic moment
for one dipole m = I a
area a
for loop M = I A
area A
• Magnetic dipoles
one current loop
• Evaluate B field along axis passing through loop
B field of large current loop
• Consider line integral B.dℓ from loop
• Contour C is closed by large semi-circle which contributes
zero to line integral
I (enclosed by C)
a
z→-∞
 B.d 
C

 a

o I
2
a2dz
2
+z

2 3/2
C

 a

a2dz
2
2
+z

2 3/2
z→+∞
+ 0 (semi  circle)  o I

 B.d

oI/2
 oI

Electrostatic potential of dipole sheet
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•
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Now consider line integral E.dℓ from sheet of electric dipoles
m = I a I = m/a (density of magnetic moments)
Replace I by Np (dipole moment density) and o by 1/o
Contour C is again closed by large semi-circle which
contributes zero to line integral


 E.d + 0 (semi  circle)  E.d  0

Electric
magnetic
Np/2o
 E.d

C

-Np/2o
Field reverses no reversal
Differential form of Ampere’s Law
Obtain enclosed current as integral of current density
 B.d   I
o encl
  o  j.dS
S
B
Apply Stokes’ theorem
 B.d     B.dS    j.dS
j
dI  j.dS
o
S
S
dℓ
Integration surface is arbitrary
  B  o j
Must be true point wise
S
Ampere’s Law examples
(1) Infinitely long, thin conductor
B is azimuthal, constant on circle of radius r
 B.d  o Iencl  B 2p r  o I  B 
B
o I
2p r
Exercise: find radial profile of B inside and outside conductor
of radius R
 o Ir
2p R 2
I
 o
2p r
B r R 
B
B r R
R
r
Solenoid
Distributed-coiled conductor
Key parameter: n loops/metre
B
I
If finite length, sum individual loops via B-S Law
If infinite length, apply Ampere’s Law
B constant and axial inside, zero outside
Rectangular path, axial length L
B
vac
I
.d  o I encl  B vac L  o nL I  B vac  onI
L
(use label Bvac to distinguish from core-filled solenoids)
solenoid is to magnetostatics what capacitor is to electrostatics
Relative permeability
Recall how field in vacuum capacitor is reduced when
dielectric medium is inserted; always reduction, whether
medium is polar or non-polar:
E
E vac
r
 B  rBvac
is the analogous expression
when magnetic medium is inserted in the vacuum solenoid.
Complication: the B field can be reduced or increased,
depending on the type of magnetic medium
Magnetic vector potential
For an electrostatic field
 E.d  0
E  -
 x E   x   0
We cannot therefore represent B by e.g. the gradient of a scalar
since
 x B  o j (rhs not zero)
Magnetostatic field, try
B is unchanged by

also .B  0 alw ays(.E  )
o
BxA
.B  . x A   0
 x B   x  x A  (see later)
A'  A + 
 x A'   x A +     x A + 0