Biot-Savart Law
I
dB(r)
•
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
r
O
r-r’
r’
dℓ
dB(r ) 
o Ι d x (r  r' )
3
4p
r  r'
dB perpendicular to both dℓ and r - r’
Inverse square dependence on distance
o/4p = 10-7 Hm-1
Integrate to get Biot-Savart Law
o Ι d x (r  r' )
B(r ) 
4p  r  r' 3

Biot-Savart Law examples
(1) Infinite straight conductor
dℓ and r, r’ in the page
dB is into the page
B forms concentric circles about
the conductor
o I  d x (r  r' )
B(r ) 
4p  r  r' 3
d x (r  r' )  d r  r' sinq nˆ
2
r  r'  r 2  z 2
q  p / 2 a
I
d x (r  r' )
dℓ q
r  r'
r’ z
O
r - r’
r
a dB
B
3
sinq  cosa 

d r  r' sinq
r  r'
3
r
nˆ 
r
2
r
o I  r dz
B
nˆ
3/2

2
2
4p  r  z 

 r

r dz
2
 z2

3/2

2
r
B
o I ˆ
n
2pr
 z2

1/2
r dz
2
z

2 3/2
nˆ
Biot-Savart Law examples
(2) Axial field of circular loop
dℓ
Loop perpendicular to page, radius a
q
dℓ out of page at top and r, r’ in the page
On-axis element dB is in the page,
perpendicular to r - r’, at q to axis.
r’
I
a
r - r’
r
z
dB || n
q
dBz
Magnitude of element dB
dB 
o I d x r - r ' o I d ˆ
 o I d

n

dB

cosq
z
3
2
2
4p
4p r - r '
4p r - r '
r - r'
a
a
cosq 

r - r'
a2  z2


1/2
Integrating around loop, only z-component of dB contributes net result
On-axis field of circular loop
dℓ
o I
Bon axis   dB z 

o I
4p r - r '
4p r - r '
2
cosq 2p a  
2
Introduce axial distance z,
where |r-r’|2 = a2 + z2
r - r’
cosq  d
o Ia
2 r - r'
Bonaxis 
I
2

a
3
o I a 2
2a z
2
2 limiting cases
B
z 0
on axis

o I
2a
and B
z a
on axis

r’
o Ia2
2z3
2

3
2
r
z
dB
q
dBz
Magnetic dipole moment
The off-axis field of circular loop is
much more complex. For z >> a (only) it is
identical to that of the electric point dipole
E
p
4p or
3
2 cosq rˆ  sinq qˆ

p
q+
q-

 om
2 cosq rˆ  sinq qˆ
3
4p r
where m  p a 2 I or m  p a 2 I zˆ
B
p a 2 area enclosed by current loop
m = current times area vs p = charge times distance
m = magnetic dipole moment
m
q r
I
Current I and current density j
Current density
j(r ) =  (r ) v(r )
Suppose current is composed of point charges
j(r)   q i (r - ri ) v i
i
Integrate current over some volume, V, say A d𝓁
where A is cross-section area of wire d𝓁 is element of length of wire
 j(r).nˆ dr    q  (r - r ) v .nˆ dr   q
i
V
i
V
i
i
i
v i . nˆ  Id
i in V
n is a unit vector parallel to the current direction
Current is commonly treated as continuous (j(r)), but is actually composed
of point particles
Continuity Equation
Rate of charge entering xz face at y = 0: jy=0DxDz
Cm-2s-1 m2 = Cs-1
Rate of charge leaving xz face at y = Dy: jy=DyDxDz = (jy=0 + ∂jy/∂y Dy) DxDz
Net rate of charge entering cube via xz faces: (jy=0 - jy=Dy ) DxDz = -∂jy/∂y DxDyDz
Rate of charge entering cube via all faces:
z
jy=0
Dz
Dy
Dx
x
-(∂jx/∂x + ∂jy/∂y + ∂jz/∂z) DxDyDz = dQencl/dt
 = lim (DxDyDz)→0 Qencl /(DxDyDz)
jy=Dy
y
. j + d /dt = 0
For steady currents d /dt = 0 (Magnetostatics) and . j = 0
Applies to other j’s (heat, fluid, etc) Conservation of charge, energy, mass, ..
Ampere’s Law
I
dB(r)
Replace I d𝓁 by j(r’) dr’ in Biot-Savart law
 j(r' ) x (r  r' )
dB(r )  o
dr'
3
4p
r  r'
r
1
r  r'
-

r  r' r  r' 3
dB(r )  
O
o
j(r' )
x
dr'
4p
r  r'
o
j(r' )
B(r ) 
 x
dr'
4p
r  r'

r’
dℓ
o
1
j(r' ) x 
dr'
4p
r  r'
o
B(r ) 
4p
r-r’
j(r' ) x (r  r' )
r  r'
3
See Homework Problems II
for intermediate steps
E(r )  
dr'
E(r ) 
1
 (r' )
4po

1
 (r' ) (r  r' )
4po

r  r'
r  r'
3
dr'  - (r )
dr'
Ampere’s Law
Evaluate Div and Curl of B(r)
NB ∇ acts on functions of r only, ∇’ acts on functions of r’
o
j(r' )
B(r ) 
 x
dr'
4p
r  r'

o 
j(r' )
.B(r ) 
.  x 
dr'   . x f(r )  0

4p 
r  r'



o
j(r' )

 x B(r ) 
x  x 
dr'   o j(r )  term containing . j(r' )


4p
r  r'


 .B  0
Absence of magnetic monopoles (generally valid)
 x B  o j Ampere’s Law (limited to magnetostatics (∇.j = 0))
Ampere’s Law
Can B(r) be expressed in terms of a potential? Yes!
B(r ) 
o
j(r' )
 x
dr'
4p
r  r'
o
 x A(r )
4p

j(r' )
A(r )  o 
dr'
4p r  r'
B(r ) 
A is the vector potential
Differential form of Ampere’s Law
Integral form of law: enclosed current is integral dS of current density j
 B.d   I
o encl
 o  j.dS
S
B
Apply Stokes’ theorem
 B.d     B.dS    j.dS
o
S
j
dI  j.d S
S
   B -  j.dS  0
o
dℓ
S
Integration surface is arbitrary
  B  o j
Must be true point wise
S
B.dℓ for current loop
• Consider line integral B.dℓ from current loop of radius a
• Contour C is closed by large semi-circle, contributes zero to line integral
I (enclosed by C)
a
z→-∞
C
z→+∞

 B.d 
C

 a

o I
2
a 2dz
2
z

2 3/2

 a

a 2dz
2
2
z

2 3/2
 0 (semi  circle)  o I
 B.d 

oI/2
 oI

E.dℓ for electric dipole
• Consider line integral E.dℓ for electric dipole with charges ±q at ±a/2
• Contour C is closed by large semi-circle, contributes zero to line integral
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q  1
1 
 (z) 


4po  z  a/2 z  a/2 
d (z)
q  sign(z  a/2) sign(z  a/2) 
E z (z)  


2
2


dz
4po  z  a/2
z  a/2

4
2
2
2
2
4

 E (z)d   0 (semi  circle)   E.d  0
z

C
Electric
Magnetic
 E.d vanishes for electrosta tic field
C
 B.d does not vanish when a current
C
is enclosed by C
Field reverses No reversal
4
Ampere’s Law examples
(1) Infinitely long, thin conductor
B is constant on circle of radius r
B
 B.d  o Iencl  B 2p r  o I  B 
o I
2p r
Exercise: Find radial profile of B inside conductor of radius R
o Ir
2p R 2
I
 o
2p r
B r R 
B
B r R
R
r
Ampere’s Law examples
(2) Solenoid with N loops/metre
B constant and axial inside, zero outside
Rectangular path, axial length L
 B.d  o Iencl  BL  o NL I  B  oNI
B
L
Exercise: Find B inside toroidal solenoid, i.e. one which forms a doughnut
solenoid is to magnetostatics what capacitor is to electrostatics
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