biot-savart law
DESCRIPTION
I. d B ( r ). r-r’. r. O. r’. d ℓ. Biot-Savart Law. The analogue of Coulomb’s Law is the Biot-Savart Law Consider a current loop ( I ) For element d ℓ there is an associated element field d B d B perpendicular to both d ℓ and r - r’ Inverse square dependence on distance - PowerPoint PPT PresentationTRANSCRIPT
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’Inverse square dependence on distance o/4 = 10-7 Hm-1
Integrate to get Biot-Savart Law
O
r
r’
r-r’
dB(r)
dℓ
3o )(x d
4)(d
r'r
r'rrB
Ι
3
o )( x d
4)(
r'r
r'rrB
Ι
Biot-Savart Law examples
(1) Infinite straight conductor
dℓ and r, r’ in the page dB is into the pageB forms concentric circles about the conductor
Br - r’
dB
I
dℓ
r’
Or
z
nr2r
2
zr
dz r
nzr
dz r
4
nzr
dz rn
sind)(x d
zr
rcossin
zr
nsind)(x d
)(x d
4)(
o3/222
3/222
o
3/22233
1/222
222
3o
ˆ
ˆ
ˆˆ
2/
ˆ
B
B
r'r
r'r
r'r
r'r
r'r
r'rr'r
r'r
r'rrB
Biot-Savart Law examples(2) Axial field of circular loop
Loop perpendicular to page, radius a
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 to axis.
Magnitude of element dB
Integrating around loop, only z-component of dB contributes net result
1/222
2o
z2o
3o
za
a
'-
acos
cos'-
d
4dBn
'-
d
4'-
'- x d
4dB
rr
rrrrrr
rr
IˆII
r - r’ dB || n
I
dℓ
zdBz
r’r
a
On-axis field of circular loop
Introduce axial distance z, where |r-r’|2 = a2 + z2
2 limiting cases
3
2o
2o
2o
zaxison
'-2
aa2cos
'-4
dcos'-4
dBB
rrrr
rr
I
I
I
23
22
2o
axisonza2
aB
I
3
2oaz
axisono0z
axison 2z
aBand
2aB
I
I
r - r’ dB
I
dℓ
zdBz
r’r
a
Magnetic dipole momentThe off-axis field of circular loop ismuch more complex. For z >> a (only) it is identical to that of the electric point dipole
m = current times area vs p = charge times distance
m = magnetic dipole moment
loop currentby enclosed area a
z a oramwhere
sincos 2r4
m
sin cos 2r4
p
2
22
3o
3o
ˆI I
ˆ ˆ
ˆˆ
m
rB
rE
rm
I
q+
q-
p
Current density
Suppose current is composed of point charges
Integrate current over some volume, V, say A d𝓁where A is cross-section area of wire d𝓁 is element of length of wire
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
Current I and current density j
i
iii )-(q)( vrr rj
)( )(=)( rvr rj
Id nvr nvrrr nrj ˆ . qdˆ. )-(qd ˆ).(Vin i
iii V
iii
V
Continuity Equation
Rate of charge entering xz face at y = 0: jy=0xz Cm-2s-1 m2 = Cs-1
Rate of charge leaving xz face at y = y: jy=yxz = (jy=0 + ∂jy/∂y y) xz
Net rate of charge entering cube via xz faces: (jy=0 - jy=y ) xz = -∂jy/∂y xyz
x
z
y
z
y
x
Rate of charge entering cube via all faces:
-(∂jx/∂x + ∂jy/∂y + ∂jz/∂z) xyz = dQencl/dt
= lim (xyz)→0 Qencl /(xyz)
. j + d/dt = 0
jy=yjy=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 LawReplace I d𝓁 by j(r’) dr’ in Biot-Savart law
O
r
r’
r-r’
dB(r)
dℓ
r'r'r
r'rr'rE r'
r'r
r'rr'jrB
r r'r'r
r'rE r'
r'r
r'jrB
r'r'r
r'j
r'r'r
r'jrB
r'r
r'r
r'r
r'r'r
r'rr'jrB
d)( )(
4)(d
)(x )(
4)(
)(d)(
4)(d
)(x
4)(
d)(
x 4
dx )(4
)(d
d)(x )(
4)(d
3o
3o
o
o
o
o
3
3o
1
- 1
1 -
1-
See Homework Problems II for intermediate steps
Ampere’s LawEvaluate Div and Curl of B(r)
NB ∇ acts on functions of r only, ∇’ acts on functions of r’
Absence of magnetic monopoles (generally valid)
Ampere’s Law (limited to magnetostatics (∇.j = 0))
j B
B
r'jrj r'r'r
r'jrB
r r'r'r
r'jrB
r'r'r
r'jrB
o
oo
o
o
x
)( containing term )(d)(
xx 4
)(x
)f(x d)(
x4
)(
d)(
x4
)(
0 .
.
0 . . .
Ampere’s Law
Can B(r) be expressed in terms of a potential? Yes!
A is the vector potential
r'r'r
r'jrA
rArB
r'r'r
r'jrB
d)(
4)(
)(x 4
)(
d)(
x4
)(
o
o
o
Integral form of law: enclosed current is integral dS of current density j
Apply Stokes’ theorem
Integration surface is arbitrary
Must be true point wise
Differential form of Ampere’s Law
S
SjB .d.d oenclo I
S
j
B
dℓ
Sj.dd I
0.
.
S
SS
SjB
SjSBB
d-
.dd.d
o
o
jB o
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
2za
dza
circle)(semi 0za
dza
2.d
3/222
2
o3/222
2o
II
C
B
.dB
oI/2
oI
z→+∞z→-∞
I (enclosed by C)
Ca
4 2 2 4
4
2
2
4
Electric Magnetic
Field reverses No reversal
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
Cby enclosed is
current a whenvanish not does .d
fieldtic electrosta for vanishes .d
.dcircle)(semi 0(z)dEz
C
C
C
B
E
E 0
224
11
4
a/2z
a/2)sign(z
a/2z
a/2)sign(zq
dz
(z)d(z)E
a/2za/2z
q(z)
zo
o
Ampere’s Law examples(1) Infinitely long, thin conductor
B is constant on circle of radius r
Exercise: Find radial profile of B inside conductor of radius R
r2Br 2 B.d o
oenclo
I II
B
B
r 2B
R 2
rB
oRr
2o
Rr
I
I
B
rR
(2) Solenoid with N loops/metre
B constant and axial inside, zero outsideRectangular path, axial length L
Exercise: Find B inside toroidal solenoid, i.e. one which forms a doughnut
solenoid is to magnetostatics what capacitor is to electrostatics
Ampere’s Law examples
I II NBNLBL.d ooenclo B
B
IL