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