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4). Ampere’s Law and Applications
• 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
• The analogue of Coulomb’s Law is
the Biot-Savart Law
• Consider a current loop (I)
• 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
O
r
r’
r-r’
dB(r)
I
dℓ’
3
o )
x(
'
d
4
)
(
d
r'
r
r'
r
r
B


I


p

 

I



3
o )
x(
'
d
4
)
(
r'
r
r'
r
r
B
p

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:
r
I
p

2
B o

r - r’
dB
I
dℓ
B
r’
O
r
q
a
q  p/2 + a
sin q = cos a 
 1/2
2
2
z
r
r
+
z
B-S Law examples
(2) “on-axis” field of circular loop
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.
Magnitude of element dB
Integrating around loop, only z-components of dB survive
The on-axis field is “axial”
 1/2
2
2
2
o
z
2
o
z
a
a
'
-
a
cos
cos
'
-
d
4
dB
'
-
d
4
dB
+





r
r
r
r
r
r
q
q
p

p
 I
I 

r - r’ dB
I
dℓ
z
q
dBz
r’
r
a
On-axis field of circular loop
Introduce axial distance z,
where |r-r’|2 = a2 + z2
2 limiting cases:
  3
2
o
2
o
2
o
z
axis
on
'
-
2
a
a
2
cos
'
-
4
d
cos
'
-
4
dB
B
r
r
r
r
r
r
I
I
I

p
q
p

q
p




 

 
 2
3
2
2
2
o
axis
on
z
a
2
a
B
+


I

3
2
o
a
z
axis
on
o
0
z
axis
on
2z
a
B
and
2a
B
I
I 


 



r - r’ dB
I
dℓ
z
q
dBz
r’
r
a
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
m “current times area” vs p “charge times distance”
 
 
loop
current
by
enclosed
area
z
a
or
a
m
where
sin
cos
2
r
4
m
sin
cos
2
r
4
p
2
2
3
o
3
o
a
p
a
p
q
q
q
p

q
q
q
p
ˆ
I
I
I
ˆ
ˆ
ˆ
ˆ



+


+

m
r
B
r
E
r
q
m
B field of large current loop
• 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
 
 
2
z
a
dz
a
circle)
(semi
0
z
a
dz
a
2
.d
3/2
2
2
2
o
3/2
2
2
2
o

+


+
+










I
I


C

B 



.d
B

oI/2
oI
z→+∞
z→-∞
I (enclosed by C)
C
a
Electrostatic potential of dipole sheet
• 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




.d
E

Np/2o
-Np/2o
0



+ 



 C

 .d
circle)
(semi
0
.d E
E
Electric magnetic
Field reverses no reversal
Differential form of Ampere’s Law

 

S
S
j
B .d
.d o
encl
o 
 I

S
j
B
dℓ
S
j.d
d 
I
Obtain enclosed current as integral of current density
Apply Stokes’ theorem
Integration surface is arbitrary
Must be true point wise
  

 



S
S
S
j
S
B
B .d
d
.d o

.

j
B o




Ampere’s Law examples
(1) Infinitely long, thin conductor
B is azimuthal, constant on circle of radius r
Exercise: find radial profile of B inside and outside conductor
of radius R
r
2
B
r
2
B
.d o
o
encl
o
I
I
I
p


p
 




 
B
B
r
2
B
R
2
r
B
o
R
r
2
o
R
r
p

p

I
I




B
r
R
Solenoid
Distributed-coiled conductor
Key parameter: n loops/metre
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
(use label Bvac to distinguish from core-filled solenoids)
solenoid is to magnetostatics what capacitor is to electrostatics
  I
I
I n
B
nL
L
B
.d o
vac
o
vac
encl
o
vac 

 




 
B
I
L
B
I
Relative permeability
Recall how field in vacuum capacitor is reduced when
dielectric medium is inserted; always reduction, whether
medium is polar or non-polar:
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
vac
r
r
vac
B
B
E
E 




Magnetic vector potential
0
x
x
-
.d











0
E
E
E 
For an electrostatic field
We cannot therefore represent B by e.g. the gradient of a scalar
since
Magnetostatic field, try
B is unchanged by
)
o
o



.
(
always
0
.
also
zero)
not
(
x






E
B
j
B rhs
 
  later)
(see
x
x
x
0
x
x
A
B
A
B
A
B
.
.











  0
x
x
'
x +



+




+

A
A
A
A
A'



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Amperes_Law.ppt

  • 1. 4). Ampere’s Law and Applications • 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”
  • 2. Biot-Savart Law • The analogue of Coulomb’s Law is the Biot-Savart Law • Consider a current loop (I) • 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 O r r’ r-r’ dB(r) I dℓ’ 3 o ) x( ' d 4 ) ( d r' r r' r r B   I   p     I    3 o ) x( ' d 4 ) ( r' r r' r r B p 
  • 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: r I p  2 B o  r - r’ dB I dℓ B r’ O r q a q  p/2 + a sin q = cos a   1/2 2 2 z r r + z
  • 4. B-S Law examples (2) “on-axis” field of circular loop 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. Magnitude of element dB Integrating around loop, only z-components of dB survive The on-axis field is “axial”  1/2 2 2 2 o z 2 o z a a ' - a cos cos ' - d 4 dB ' - d 4 dB +      r r r r r r q q p  p  I I   r - r’ dB I dℓ z q dBz r’ r a
  • 5. On-axis field of circular loop Introduce axial distance z, where |r-r’|2 = a2 + z2 2 limiting cases:   3 2 o 2 o 2 o z axis on ' - 2 a a 2 cos ' - 4 d cos ' - 4 dB B r r r r r r I I I  p q p  q p           2 3 2 2 2 o axis on z a 2 a B +   I  3 2 o a z axis on o 0 z axis on 2z a B and 2a B I I         r - r’ dB I dℓ z q dBz r’ r a
  • 6. 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 m “current times area” vs p “charge times distance”     loop current by enclosed area z a or a m where sin cos 2 r 4 m sin cos 2 r 4 p 2 2 3 o 3 o a p a p q q q p  q q q p ˆ I I I ˆ ˆ ˆ ˆ    +   +  m r B r E r q m
  • 7. B field of large current loop • 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
  • 8. 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     2 z a dz a circle) (semi 0 z a dz a 2 .d 3/2 2 2 2 o 3/2 2 2 2 o  +   + +           I I   C  B     .d B  oI/2 oI z→+∞ z→-∞ I (enclosed by C) C a
  • 9. Electrostatic potential of dipole sheet • 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     .d E  Np/2o -Np/2o 0    +      C   .d circle) (semi 0 .d E E Electric magnetic Field reverses no reversal
  • 10. Differential form of Ampere’s Law     S S j B .d .d o encl o   I  S j B dℓ S j.d d  I Obtain enclosed current as integral of current density Apply Stokes’ theorem Integration surface is arbitrary Must be true point wise          S S S j S B B .d d .d o  .  j B o    
  • 11. Ampere’s Law examples (1) Infinitely long, thin conductor B is azimuthal, constant on circle of radius r Exercise: find radial profile of B inside and outside conductor of radius R r 2 B r 2 B .d o o encl o I I I p   p         B B r 2 B R 2 r B o R r 2 o R r p  p  I I     B r R
  • 12. Solenoid Distributed-coiled conductor Key parameter: n loops/metre 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 (use label Bvac to distinguish from core-filled solenoids) solenoid is to magnetostatics what capacitor is to electrostatics   I I I n B nL L B .d o vac o vac encl o vac           B I L B I
  • 13. Relative permeability Recall how field in vacuum capacitor is reduced when dielectric medium is inserted; always reduction, whether medium is polar or non-polar: 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 vac r r vac B B E E     
  • 14. Magnetic vector potential 0 x x - .d            0 E E E  For an electrostatic field We cannot therefore represent B by e.g. the gradient of a scalar since Magnetostatic field, try B is unchanged by ) o o    . ( always 0 . also zero) not ( x       E B j B rhs     later) (see x x x 0 x x A B A B A B . .              0 x x ' x +    +     +  A A A A A'  