The Biot-Savart Law and Ampere’s Law

The Biot-Savart Law

– µ0 is the “permeability of free space”

• Always have to integrate along whole wire (current) to get actual field

2

0

3

0

2

0sin)(

44

ˆ

4 r

Id

r

Id

r

Idd

θπ

µπ

µπ

µ srsrsB =×=×=

r

m/AT104 70 ⋅×= −πµ

∫×=2

0 ˆ

4 r

Id rsB

πµ

Field of a Loop of Current

• What’s at the center of a loop of current?

r

I

rr

I

dsr

IB

r

Id

2

24

4

ˆ

4

0

20

20

20

µ

ππ

µπ

µπ

µ

=

=

=

×=

∫

∫rs

Br

I sd

sd B

Line Integral of Field about a Loop

• Switch it around• What’s the line integral

of the field going around a line current?

( )

I

rr

I

dsr

I

dr

dr

d

0

0

0

0

0

22

2

ˆ2

ˆ2

µ

ππ

µπ

µπµ

πµ

=

=

=

⋅×=

⋅

×=⋅

∫

∫

∫∫

srI

srIsBr

B sd

sd I

Ampere’s Law

• This is true actually for any loop– Radial segments don’t contribute– Tangential segments contribute by angle

• Could add another current– Superposition works for magnetic field too!

• General result is Ampere’s Law:

– Use it like Gauss’s Law in electrostatics

nclosedId

e0loopµ=⋅∫ sB

Example: An extended wire

• Apply Ampere’s Law to a real (finite diameter) wire– Outside the wire (r > R)

– Inside wire (r < R)

r

IB

IrBd

πµ

µπ

2

2

0

0

=

==⋅∫ sB

20

2

0

2

2

R

rIB

IR

rrBd

πµ

µπ

=

==⋅∫ sB

Force Between Two Parallel Wires

• Can now calculate the force between two long parallel wires– Magnetic force on one

wire (1) due to field of the other (2)

– Field of other wire (2)211

211

BlI

l

=×=

F

BIF

a

Ia

πµπµ

2

ˆ2

202

20

2

=

×=

B

aIB

Force Between Two Parallel Wires

• Can now calculate the force between two long parallel wires– Putting it together

– Attractive for currents in the same direction

– This is used to define the amp as a unit (and the coulomb)

la

IIF

a

IBBlIF

πµ

πµ

2

2 ;

21012

202211

=

==

Example: Field Inside a Solenoid

• A solenoid is a wire wrapped in a helix

Example: Field Inside a Solenoid

• A solenoid is a wire wrapped in a helix

• Ideally, the coil spacing is tight…

Example: Field Inside a Solenoid

• A solenoid is a wire wrapped in a helix

• Ideally, the coil spacing is tight…

• …and it goes on forever• Field outside is small

– Using Ampere’s Law: no net current thru loop

Example: Field Inside a Solenoid

• Inside:

nIl

NIB

NIBld

Bld

dsd

dsBd

ddi i

00

0

1

3

24,2

0 0

0cos

µµ

µ

π

==

==⋅

=⋅

==⋅

==⋅

⋅=⋅

∫

∫

∫∫

∫∫

∑ ∫∫

sB

sB

sB

sB

sBsB

N turns inside loopn turns per unit length

Magnetic Flux

• Just as with the electric field, we define the flux of the magnetic field thru a surface S

– dA is a vector perpendicular to the surface

– Units of flux: Weber (Wb)

∫ ⋅=ΦSB dAB

2mT 1 Wb1 ⋅≡

Gauss’s Law in Magnetism

• Since all magnetic field lines form loops (no magnetic charges), net flux thru any closed surface is zero!

0=⋅∫S dAB

Review of Magnetostatics

• If fields are constant (in time), we have

– Gauss’s Law

– Gauss’s Law Magnetism

– Conservative Electric Force

– Ampere’s Lawenclosed0loop

loop

0

enc

0

0

Id

d

d

Qd

S

S

µ

ε

=⋅

=⋅

=⋅

=⋅

∫

∫

∫

∫

sB

sE

AB

AE

Displacement Current

• Consider the magnetic field around a capacitor being charged

– But S1 and S2 have the same boundary!

02

10

=⋅

=⋅

∫

∫

S

S

d

Id

sB

sB µ

Displacement Current

• Consider the magnetic field around a capacitor being charged

– But S1 and S2 have the same boundary!

– Have to add something to make these consistent

02

10

=⋅

=⋅

∫

∫

S

S

d

Id

sB

sB µ

Displacement Current

• If there is a current flowing, the field (and electric flux) in the capacitor is changing!

00

0

0

1

0

2

εε

ε

ε

I

dt

dq

dt

d

q

dAA

q

d

E

C CC

SE

==Φ

=

+=

⋅=Φ

∫

∫ AE

dt

dI E

ntdisplaceme

Φ≡0

ε

Define the Displacement Current

Displacement Current

• If there is a current flowing, the field (and electric flux) in the capacitor is changing!

• Use displacement current to make the two integrals consistent

ntdisplaceme

E

S

S

Idt

dd

Id

0002

01

µεµ

µ

=Φ=⋅

=⋅

∫

∫

sB

sB

dt

dI E

ntdisplaceme

Φ≡0

ε

Define the Displacement Current

General Form of Ampere’s Law

• In general Ampere’s Law (Ampere-Maxwell’s Law) should be

( )

dt

dId

IId

EΦ+=⋅

+=⋅

∫

∫

00enclosed0loop

dispenclosed0loop

εµµ

µ

sB

sB