Download Chapter 7 Electrodynamics 7.1 Electromotive Force

Chapter 7 Electrodynamics
7.1 Electromotive Force
7.1.1 Ohm’s Law
Current Density:
∆Q (n∆V )q (nA∆l )q
I
=
=
= Aqnv Æ J = qnv =
∆t
∆t
∆t
A
r
r r
r
For electron: J = n(− e )v and I = ∫ J ⋅ da
I=
For most substances, the current density is proportional to the force per unit charge:
r
r
r r r
⎛F⎞
J = σ ⎜⎜ ⎟⎟ = σ E + v × B .
⎝q⎠
(
)
σ : conductivity, perfect conductor: σ → ∞
1
≡ ρ : resistivity
σ
If the velocity of the charges is sufficiently small, the force due to magnetic field can
be ignored.
r
r
Ohm’s law: J = σE
Simple Understanding:
From experimental results, we have V = IR . Considering uniform electric field in a
conductor, we may have V = El = IR . We can intuitively believe that R depends on
the conductor’s length and cross section as R = ρ
l
l
, so we have El = I ρ .
A
A
Finally we obtain the other form of Ohm’s law as E =
I
ρ , that is E = ρJ . Here
A
ρ is resistivity. The conductivity σ is the inverse of resistivity, i. e. σ =
r
r
We usually write the Ohm’s law as J = σE .
1
ρ
.
Simple Classical Model:
Define the mean free time τ as the elapsed time after one collision and before a next
collision event.
The average speed is vavg = aτ =
r
r
r
ne 2τ r
J = n(− e )vavg =
E = σE .
m
Material
− eEτ
. The current density is
m
σ=
ne 2τ
m
Resistivity (µΩ-cm) Material
Conductors
Resistivity (Ω-m)
Semiconductors
Silver
1.59
Germanium 4.4 X 10-2
Copper
1.68
Diamond
2.7
Gold
2.21
Silicon
2.0 X 103
Aluminum 2.65
(All values are for 1 atm, 20o C)
r
r
E = 0 inside a conductor for stationary charges J = 0 .
r
r J
For perfect conductors E = = 0 even if current is flowing.
σ
Usually we treat the connecting wires in electric circuit as equipotential since metals
are good conductors.
Example: A cylindrical resistor of cross-sectional area A and length L is made from
material with conductivity σ . If the potential is constant over each end, and the
potential difference is V, what current flows?
I = ? , I = JA , Ohm’s Law: J = σE = σ
Æ R=
A
V
Æ I =σ V #
L
L
V 1 L
L
=
=ρ
I σ A
A
Example: Two long cylinder (radii a and b) are separated by material of conductivity
σ . If they are maintained at a potential difference V, what current flows from one to
the other, in a length L?
r
λL
rˆ
Use Gauss’ Law: E =
2πε 0 rL
σλ
σλ
λ
λ
⎛b⎞
Æ I = J 2πrL =
L
ln⎜ ⎟ , J = σE =
dr =
2
2
πε
r
πε
a
2
πε
r
ε
⎝
⎠
0
0
0
0
a
b
V =∫
I=
σ 2πVL
1 ln (b / a )
Example: The field is uniform in a cylindrical resistor with a constant potential
between two ends. Prove it.
r
r
r
r
On the cylindrical surface J ⋅ nˆ = 0 and J = σE , therefore E ⋅ nˆ = 0 and hence
∂V
V
= 0 . By using uniqueness theorem, as we find a solution V ( z ) = 0 z we can
∂n
L
claim that this is the only solution.
Example: In vacuum-tube diodes, electrons are emitted from a hot cathode at zero
potential and collected by an anode maintained at a potential V0, resulting in a
convection current flow. Assuming that the cathode and the anode are parallel
conducting plates and that the electrons leave the cathode with a zero initial velocity,
find the relation between the current density J and V0.
dv
dV
V0
F = −eE = ma = m , E ( y ) = −
J, E
dt
dy
e
dv
dv
dV
dV
dV
=m , e
v=e
= m v Æ edV = mvdv
dt
dt
dy
dt
dy
1 2
2eV
mv = eV , v =
(energy conservation)
2
m
r
ρ
∇ ⋅ E = −∇ 2V = , J = n(− e )v = ρv Æ J is independent on y.
ε0
1/ 2
−
d 2V
J
J ⎛m⎞
=
= ⎜ ⎟ V −1 / 2
2
ε 0 v ε 0 ⎝ 2e ⎠
dy
−
1/ 2
2
d 2V dV
d ⎛ (dV / dy ) ⎞ J ⎛ m ⎞
−1 / 2 dV
⎟
⎜
=
Æ
=
−
⎜ ⎟ V
2
⎟
⎜
2
ε
2
dy
dy dy
dy ⎝
e
⎝
⎠
0
⎠
⎛ (dV / dy )2 ⎞ J ⎛ m ⎞1 / 2 −1 / 2
1 ⎛ dV
⎟ = ⎜ ⎟ V dV Æ ⎜⎜
− d ⎜⎜
⎟
2 ⎝ dy
2
⎝
⎠ ε 0 ⎝ 2e ⎠
1/ 2
⎛ J ⎞ ⎛ mV ⎞
dV
= 2⎜⎜ ⎟⎟ ⎜
⎟
dy
⎝ ε 0 ⎠ ⎝ 2e ⎠
1/ 2
1/ 4
1/ 2
Æ V
⎛J ⎞ ⎛m⎞
4 3/ 4
V = 2⎜⎜ ⎟⎟ ⎜ ⎟
3
⎝ ε 0 ⎠ ⎝ 2e ⎠
1/ 4
−1 / 4
1/ 2
⎛J ⎞ ⎛m⎞
dV = 2⎜⎜ ⎟⎟ ⎜ ⎟
⎝ ε 0 ⎠ ⎝ 2e ⎠
1/ 2
1/ 4
dy
⎛J ⎞ ⎛m⎞
4 3/ 4
y Æ V0 = 2⎜⎜ ⎟⎟ ⎜ ⎟
3
⎝ ε 0 ⎠ ⎝ 2e ⎠
4ε ⎛ 2e ⎞
3/ 2
-- Child-Langmuir law
J = 02 ⎜ ⎟ V0
9d ⎝ m ⎠
1/ 2
2
⎞
J ⎛m⎞
⎟⎟ = ⎜ ⎟
⎠ ε 0 ⎝ 2e ⎠
1/ 4
d
2V 1 / 2
Conductance: symbol -> G, unit -> Simens (S)
Resistance: symbol -> R, unit -> Ohm (Ω)
Conductivity: symbol -> σ , unit -> Ω-cm or Ω-m
Resistivity: symbol -> ρ , unit -> S / cm or S / m
Resistance connected in series: R = R1 + R2 + ...
1 1
1
= +
+ ...
Resistance connected in parallel:
R R1 R2
Mobility: Since the average drift velocity is directly proportional to the electric field,
r
r
r
r
we write u = − µe E and u = µ h E for electrons and holes, respectively.
r
r
r
r
J = ρv = (− ρ e µe + ρ h µ h )E = σE Æ σ = − ρ e µe + ρ h µ h
Conductivity will be related to the carrier concentration and the carrier’s moving
ability.
Joule Heating Law: P = VI = I 2 R
r r
∆W r r
dP
P=
= F ⋅ u , ∆P = (qE )u∆N = (qE )un∆v Æ
= qnuE = E ⋅ J
∆t
dv
r r P
is power density, power per unit volume
E⋅J =
V
r r
Total power P = ∫ E ⋅ Jdv = ∫ Edl ∫ Jda = VI
Boundary Conditions for Current Density:
r
r
J = σE in the conductor
r
r ρ
r
r
⎛J⎞
for steady current: ∇ ⋅ E =
= 0 , ∇ × E = 0 Æ ∇ ⋅ J = 0 , ∇ × ⎜⎜ ⎟⎟ = 0
ε0
⎝σ ⎠
Crossing the boundary: J ⊥, above = J ⊥,below ,
J //, above
σ above
=
J //, below
σ below
Example: Two conducting media with conductivity σ 1 and σ 2
are separated by an interface.
The steady current density in
medium 1 has a magnitude J1 and makes an angle α1 with the
normal. Determine the magnitude and direction of the current
density in medium 2.
J1 cos α1 = J 2 cos α 2 ,
J1 sin α1
σ1
=
J 2 sin α 2
σ2
For a homogeneous conducting medium (not crossing the boundary) the differential
r
form simplifies to ∇ × J = 0 Æ a curl free vector can be expressed as the gradient of
r
a scalar potential J = −∇ψ
r
∇ ⋅ J = 0 Æ ∇ 2ψ = 0 , a Laplace’s equation
Resistance Calculations:
Example: A conducting material of uniform thickness h and conductivity σ has the
shape of a quarter of a flat circular washer, with inner radius a and outer radius b.
Determine the resistance between the end faces.
r
r
r
V=V0
∇ × J = 0 -> J = −∇ψ , ∇ ⋅ J = 0 Æ ∇ 2ψ = 0
1 ⎡ ∂ ⎛ r ∂ψ ⎞ ∂ ⎛ 1 ∂ψ
⎜
⎟+
⎢ ⎜
r ⎣ ∂r ⎝ 1 ∂r ⎠ ∂φ ⎜⎝ r ∂φ
⎞ ∂ ⎛ r ∂ψ
⎟⎟ + ⎜
⎠ ∂z ⎝ 1 ∂z
⎞⎤
⎟⎥ = 0
⎠⎦
V=0
V = 0 at φ = 0
V = V0 at φ = π / 2
2V
d 2V
=0 Æ V = 0φ
2
dφ
π
b
r
r
1 ∂V
2σV0
2σV0 dr 2σV0 h ⎛ b ⎞
ˆ
ˆ
I
h
=
σ
=
−
φ
σ
=
−
φ
,
ln⎜ ⎟
=
=
J
E
r ∂φ
πr
π ∫a r
π
⎝a⎠
R=
π
2σh ln (b / a )
7.1.2 Electromotive Force
r
∇× E = 0
r r
E
∫ ⋅ dl = 0 Æ
r
J ≠0
What happens?
r r r
f = fs + E
r
J
∫σ
r
⋅ dl = 0
heater wire (NiCr alloy)
If σ is finite, J must be zero.
heating, hot
r
f s : a source, ordinarily confined to one portion of the loop (a battery)
r
E : an electrostatic force, to smooth out the charge flow and communicate the
influence of the source to distant parts
b r
b r
r
r
r r
r r
V = − ∫ E ⋅ dl = ε = ∫ f s ⋅ dl = ∫ f s ⋅ dl = ∫ f ⋅ dl , where ε is named electromotive
a
a
E
force (emf)
+
E
-
ε
Example: Estimate the relaxation time of volume change in a good conductor.
r ∂ρ
r ρ
r
r
J = σE , ∇ ⋅ J +
= 0, ∇⋅E =
∂t
ε0
σ
t
− t
r ∂ρ
r ∂ρ
−
∂ρ
ρ
ε0
τ
= ρ 0e
Æ
+∇⋅J =
+ σ∇ ⋅ E =
+σ
= 0 Æ ρ = ρ 0e
∂t
∂t
∂t
ε0
Relaxation time: τ =
ε0
. For copper τ ~ 1.52 × 10−19 s.
σ
7.1.3 Motional emf
Mention Phenomena of Induced Currents at First.
h
u
v
fpull
E
v
u
w
h
θ
h/cosθ
You pull the wire.
You found that the wire is moving at a constant velocity, v.
There must be one opposite force cancel the pulling force.
The force generated by the magnetic field and doing on the charge may be Lorentz
F
force. v is from pulling and u is from balancing: E = = f pull = uB
q
r r
⎛ h ⎞
5. The emf is: ∫ E ⋅ dl = (uB )⎜
⎟ sin θ = u tan θBh = vBh
⎝ cosθ ⎠
Magnetic force do not work, who is supplying the energy that heats the resistor?
The person who is pulling on the loop is supplying the energy.
1.
2.
3.
4.
Expressing the emf by the magnetic flux:
r
dΦ B
dx
Φ B = ∫ B ⋅ nˆ da = Bhx , ε = Bhv = −
= Bh
dt
dt
The flux rule for motional emf: ε = −
Proof of the flux rule:
dΦ B
dt
r r
r
r r r
ˆ
B
⋅
n
da
,
d
a
=
v
×
d
l
dt
(
area
=
A
×B)
∫
dΦ = Φ(t + dt ) − Φ (t ) =
changed _ area
r
dl is along the loop that bounds the area
r
r
r
find the w which is the component of v and is orthogonal to dl
r
r
r r r
r r r
dΦ
dΦ B
= ∫ B ⋅ w × dl = − ∫ w × B ⋅ dl = − ∫ f mag ⋅ dl Æ ε = −
dt
dt
(
)
(
)
Example: A metal disk of radius a rotates with angular velocity ω about a vertical
axis, through a uniform field B. A circuit is made by connecting one end of a
resistor R to the axle and the other end to a sliding contact, which touches the outer
edge of the disk. Find the current in the resistor.
B
a 2π
Φ = Bπa 2 = B ∫ ∫ rdθ dr
0 0
ε =−
I=
a
a
r
r
dΦ
ωBa 2
= B ∫ v⊥ dr = B ∫ ωrdr =
or ε = ∫ f mag ⋅ dl =
dt
2
0
0
ωBa 2
2R
Exercise: 7.1, 7.4, 7.7, 7.10