Download The Biot-Savart Law and Ampere`s Law

The Biot-Savart Law and
Ampere’s Law
The Biot-Savart Law
r
µ0 Ids × rˆ µ0 Ids × r µ0 ( Ids) sin θ
dB =
=
=
2
3
2
4π r
4π r
4π
r
– µ0 is the “permeability of
free space”
µ0 = 4π ×10 −7 T ⋅ m/A
• Always have to integrate
along whole wire
(current) to get actual
field
µ
B= 0
4π
Ids × rˆ
∫ r2
Field of a Loop of Current
• What’s at the center of a
loop of current?
µ 0 Ids × rˆ
B=
4π ∫ r 2
µ0 I
B=
ds
2 ∫
4π r
µ0 I
=
2πr
2
4π r
µ0 I
=
2r
I ds
r
ds
B
Line Integral of Field about a Loop
• Switch it around
• What’s the line integral
of the field going around
a line current?
B ds
 µ0

ˆ
∫ B ⋅ ds = ∫  2πr I × r  ⋅ ds
µ0
(I × rˆ ) ⋅ ds
=
∫
2πr
µ0 I
=
ds
∫
2πr
µ0 I
=
2πr
2πr
= µ0 I
r
ds
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:
B
⋅
d
s
=
µ
I
0
enclosed
∫loop
– Use it like Gauss’s Law in electrostatics
Example: An extended wire
• Apply Ampere’s Law to
a real (finite diameter)
wire
– Outside the wire (r > R)
∫ B ⋅ ds = 2πrB = µ I
0
B=
µ0 I
2πr
– Inside wire (r < R)
2
r
∫ B ⋅ ds = 2πrB = µ0  R  I
µI r
B= 0
2π R 2
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)
F1 = lI1 × B 2
F1 = lI1 B2
– Field of other wire (2)
µ
B 2 = 0 I 2 × aˆ
2πa
µI
B2 = 0 2
2πa
Force Between Two Parallel Wires
• Can now calculate the
force between two long
parallel wires
– Putting it together
µI
F1 = lI1 B2 ; B2 = 0 2
2πa
µII
F12 = 0 1 2 l
2πa
– Attractive for currents in
the same direction
– This is used to define the
amp as a unit (and the
coulomb)
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:
∫ B ⋅ ds = ∑ ∫ B ⋅ ds
π
d
B
ds = 0
B
⋅
s
=
cos
∫
∫
∫ B ⋅ ds = ∫ 0 ds = 0
∫ B ⋅ ds = Bl
∫ B ⋅ ds = Bl = µ NI
i i
2
2, 4
3
1
0
N
B = µ 0 I = µ 0 nI
l
N turns inside loop
n turns per unit length
Magnetic Flux
• Just as with the electric
field, we define the flux
of the magnetic field thru
a surface S
Φ B = ∫ B ⋅ dA
S
– dA is a vector
perpendicular to the
surface
– Units of flux: Weber (Wb)
1 Wb ≡ 1 T ⋅ m 2
Gauss’s Law in Magnetism
• Since all magnetic field lines form loops (no
magnetic charges), net flux thru any closed
surface is zero!
∫ B ⋅ dA = 0
S
Review of Magnetostatics
• If fields are constant (in time), we have
– Gauss’s Law
∫ E ⋅ dA =
S
– Gauss’s Law Magnetism
Qenc
ε0
∫ B ⋅ dA = 0
∫ E ⋅ ds = 0
∫ B ⋅ ds = µ I
S
– Conservative Electric Force
loop
– Ampere’s Law
loop
0
enclosed
Displacement Current
• Consider the magnetic
field around a capacitor
being charged
∫ B ⋅ ds = µ I
∫ B ⋅ ds = 0
S1
S2
0
– But S1 and S2 have the
same boundary!
Displacement Current
• Consider the magnetic
field around a capacitor
being charged
∫ B ⋅ ds = µ I
∫ B ⋅ ds = 0
S1
S2
0
– But S1 and S2 have the
same boundary!
– Have to add something to
make these consistent
Displacement Current
• If there is a current
flowing, the field (and
electric flux) in the
capacitor is changing!
Φ E = ∫ E ⋅ dA
S2
q
=∫
dAC + 0
Cε A
0 C
=
q
ε0
dΦ E 1 dq I
=
=
ε 0 dt ε 0
dt
Define the Displacement Current
I displacement
dΦ E
≡ ε0
dt
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
∫ B ⋅ ds = µ I
S1
0
dΦ E
∫SB2 ⋅ ds = µ0ε 0 dt = µ0 I displacement
Define the Displacement Current
I displacement
dΦ E
≡ ε0
dt
General Form of Ampere’s Law
• In general Ampere’s Law (Ampere-Maxwell’s
Law) should be
(
)
µ
B
s
⋅
d
=
I
+
I
0
enclosed
disp
∫loop
dΦ E
B
s
µ
µ
ε
⋅
d
=
I
+
0
enclosed
0
0
∫loop
dt