the biot-savart law and ampere’s lawweird101.weebly.com/uploads/1/2/1/9/12199675/bio_s... · the...
TRANSCRIPT
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