Physics II
Current
Degree in Informatic, Mechatronic and
Cinema&Media Engineering
Politecnico di Torino
Lecturer: Alessandro Battaglia
Images in these slides are taken from the Young&Friedman textbook or
are provided by EdiSES (extracted from the reference text: Fisica II di
Mazzoldi, Nigro, Voci)
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Learning outcomes
1. The meaning of electric current, and how charges move in a
conductor.
2. What is the continuity equation
3. What is meant by the resistivity and conductivity of a
substance.
4. How to calculate the resistance of a conductor from its
dimensions and its resistivity.
5. How an electromotive force (emf) makes it possible for current
to flow in a circuit.
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Free electrons in a conductor
The number of free electrons per
unit of volume in a metallic
conductor coincides with the
number of atoms (if there is one
free e − per atom):
n=
NA ρ
A
where NA is the Avogadro
number, ρ is the density and A is
the mass number
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Free electrons in a conductor: numerical example
n=
NA ρ
A
In copper:
n=
6.022 · 1026 · 8.96 · 103
= 8.49 · 1028 electrons/m3
63.55
In silver:
n=
6.022 · 1026 · 10.5 · 103
= 5.86 · 1028 electrons/m3
107.87
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Conductor in electrostatic conditions
vm =
1 X
vi = 0
N
i
In electrostatic situations the electric field is zero everywhere within
the conductor, and there is no current (but electrons are moving
around at speed of the order of 106 m/s!).
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Electrical conduction
Two isolated conductors C1 and
C2 at potential V1 and V2 put into
contact become a conductor at
the same potential. Electrons pass
from the conductor with a lower
potential to that with a higher potential under the action of an electric field associated with the d.d.p.
EMF generator = device capable of maintaining a ∆V between two
points of the same conductor.
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Conductor with internal electric field
If an electric field E
is present, the electric
E imposes
force F = qE
a small drift (greatly
exaggerated) that takes
the electron to point P20
a distance vd ∆t from P2
in the direction of the force.
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Definition of current
The same current can be produced
by:
(a) positive charges moving in the
direction of the electric field E ;
(b) the same number of negative
charges moving at the same speed
in the direction opposite to E .
The current through the crosssectional area Σ is defined as the
net charge flowing through the area
per unit time:
i=
dq
dt
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Current and Drift Velocity
We can express the current in terms of the drift velocity of the
moving charges
i=
dq
∆q
= lim
∆t→0 ∆t
dt
cos θ dΣ is the area orthogonal to E k vd
dτ = vd ∆t dΣ cos θ
volume swept by moving charges
∆q = n+ e dτ = n+ e vd cos θ dΣ ∆t
charge flowing trough Σ
di = n+ e vd dΣcosθ
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Current, Drift Velocity and Current Density
di = n+ e vd dΣcosθ
j = n+ evvd
di = j · un dΣ
Z
j · un dΣ = ΦΣ (jj )
iΣ =
Σ
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Current Density ↔ electric current
When the surface Σ is orthogonal to j and j has the same value on
all points of Σ:
i = jΣ
j=
i
Σ
the current density is the current flowing through a unit surface whose
normal is perpendicular to the direction of the charge motion.
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Direction of current density
If, as in metals, the charge carriers are negative:
j = −n− evv−
If, on the other hand, there are
carriers of the two signs (as in
semiconductors or ionic
solutions):
j = n+ evv + − n− evv −
The two contributions are in the same direction: k E .
It is not possible to determine the sign of the charge carriers from
the direction of the current
Convention: the current density is directed along the motion of the
positive charges.
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Units
Units for current is the Ampere:
A=
C
s
Ampere is one of the seven SI base units.
Units for current density is:
[j] =
A
m2
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Multiple choice test
The electric field, E , when expressed in SI base units is measured in:
(A) kg m2 s −1 A−1
(B) kg m s −2 A−2
(C) kg m2 s −3 A−1
(D) kg m s −3 A−1
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Answer
[E ] =
[F ]
Nt
kg ms −2
kg ms −3
=
=
=
= kg m s −3 A−1
[Q]
C
C
C /s
Answer: D
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Current density versus current
Current density is a vector, current is not.
The current density describes how charges
flow at a certain point, and the vector’s direction tells you about the direction of the
flow at that point.
By contrast, the current describes how
charges flow through an extended object
such as a wire.
i has the same value at all points in the circuit on the right, but j does not: the current
density is directed downward in the left-hand
side of the loop and upward in the right-hand
side.
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Law of charge Conservation
The total charge that passes in
the unit time through the closed
surface Σ is:
I
i=
j · un dΣ
Σ
I
j · un dΣ = −
i=
Σ
∂qint
∂t
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Charge continuity equation
Z
qint =
ρdτ
τ
I
∂qint
j · un dΣ = −
⇒
∂t
Σ
I
Z
j · un dΣ = −
Σ
Z
j · un dΣ =
Σ
I
τ
∂ρ
dτ
∂t
(Divergence theorem)
∇ · j dτ
τ
Z ∂ρ
∇·j +
dτ = 0
∂t
τ
∀τ
⇒
∇·j +
∂ρ
=0
∂t
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Stationary conditions
If inside the closed surface Σ the charge does not vary, then
I
∂ρ
∂t
= 0:
j · un dΣ = 0
∇·j = 0
In stationary conditions the vector field j is solenoidal or an
incompressible vector field.
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Current carrying conductor in stationary conditions
The grey volume represents a wire with cross sections changing
with distance. There is no flux of j through the lateral surfaces.
I
Z
j · un dΣ =
Z
j1 · u1 dΣ1 +
Σ1
j2 · u2 dΣ2 = 0
Σ2
Z
Z
j2 · u2 dΣ2 = −
Σ2
j1 · u1 dΣ1
Σ1
i1 = i2
20/39
Electromotive force
If a charge goes around a complete circuit and
returns to its starting point, the potential energy must be the same at the end of the round
trip as at the beginning. Since there is always a
decrease in potential energy when charges move
through an ordinary conducting material with resistance there must be some part of the circuit
in which the potential energy increases (typically
a battery).
Steady current in a complete circuit: analogous to an ornamental
water fountain that recycles its water ⇒ a pump is needed that lifts
the water back to the top (increasing the potential energy). Without
the pump, the water would just fall to the basin at the bottom and
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stay there.
Theory of Metallic Conduction: no E field
Drude (1900) and Lorentz (1906).
• each atom is part of a crystal lattice and gives up one or more
of its outer electrons (e − );
• e − move through the lattice in a completely disordered way.
• e − undergo continuous interactions (= collisions) with ions;
between successive collisions the motion is free and the
trajectory is straight with v ≈ 106 m/s (Fermi’s theory).
• The set of trajectories is completely random and there is no
net charge flow: i = 0.
• Between collisions the e − travels an average free path ` in an
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average time τ (≈ 3 · 10−14 s for Cu at T = 300 K ):
Theory of Metallic Conduction: presence of E field
τ = `/v
v i+1 = v i −
eE
τ
me
When averaging over a large number of
electrons, N
vd =
N
N
1 X
1 X
eE
v i+1 =
vi −
τ
N
N
me
i=1
mean drift velocity
i=1
Since after each collision the mean velocity of the electrons is zero
P
i vi = 0
e τ
vd = −
E
me
(≈ mm/s)
e τ
≈ 0.5 × 10−2 C s kg −1
me
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Multiple choice test
When applying an electric field E :
(A) Each electron acquires an acceleration opposite to the E field
and moves between one collision and the next between
parabolic arcs.
(B) Each electron acquires an acceleration opposite to the E field
and moves between one collision and the next between
rectilinear arcs.
(C) Each electron acquires an acceleration concordant with the
field E and moves between one collision and the next between
rectilinear arcs.
(D) Each electron acquires an acceleration concordant with the
field E and moves between one collision and the next between
parabola arcs.
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Answer
E /m
a = F /me = −eE
The motion is parabolic as it is accelerated along E and at a
constant velocity in the orthogonal plane at E .
Answer: A
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Ohm’s law: conductivity
j = −n− evvd =
σ=
n− e 2 τ
me
n− e 2 τ
E
me
conductivity [C 2 s/(kg m3 ) = Ω−1 m−1 ]
σ ≈ 108 Ω−1 m−1 for copper at ambient temperatures
σ ≈ 10−16 ÷ 10−12 Ω−1 m−1 for insulators.
E
j = σE
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Multiple choice test
A conductor wire is connected to a generator which maintains a
potential difference ∆V between the two extremes. How does the
drift speed vd of the electrons vary if: a) the potential difference is
doubled; b) the length of the wire is doubled; c) the section of the
wire is doubled, ceteris paribus.
(A)
(B)
(C)
(D)
doubles in a), b) and c).
doubles in a), halves in b) and does not change in c).
does not change in a) and c), it is halved in c).
doubles in a) and c), halves in b).
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Answer
⇒
j = −n− evvd
E
j = σE
⇒
vd ∝ j
vd ∝ E
If I double ∆V (when keeping the wire length constant) E doubles
If I double the length of the wire (when keeping ∆V constant), E
is halved
If I double the section (when keeping ∆V constant) E remains
unchanged.
Answer: B
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Ohm’s law: resistivity
The previous equation can also be written:
E = ρjj
ρ=
1
me
=
σ
n− e 2 τ
resistivity of the material
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Ohm’s law: generic charge carriers
v+ =
e τ+
E
m+
v− = −
j = n+ evv+ − n− evv− = e
σ=
2
e τ−
E
m−
n+ τ+ n− τ−
+
m+
m−
E
n+ e 2 τ+ n− e 2 τ−
+
m+
m−
E
j = σE
30/39
Ohm’s law: electric resistance
If a d.d.p. (VA − VB ) is applied through the cylindrical metallic
conductor with constant section Σ then a current flows:
1
E
j = E = σE
ρ
in stationary regime:
E
ρ
i = jΣ = Σ ⇒ E = i
ρ
Σ
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Ohm’s law: electric resistance
Z
B
V = VA − VB =
E · dss = Eh
A
V =
V = Ri
ρh
i
Σ
with
R≡ρ
h
Σ
Resistance is measured in Ohm: Ω =
V
A
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Thermal effects
In metallic conductors, if the temperature increases, the collision time τ decreases; correspondingly σ decreases or ρ increases. In
insulators and semiconductors, the increase
in temperature produces an increase in free
carriers and therefore an increase of σ (decrease of ρ).
ρ = ρ20 ◦ C (1 + α∆T )
The α coefficient, called thermal coefficient
of resistivity is defined as:
α=
1
ρ20 ◦ C
∆ρ
∆t
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Power associated to Joule effect
E to keep the single charge in
The power spent by the force F = eE
motion with velocity at v d is:
E · vd
Pe − = F · vd = eE
The power spent per unit of volume (W /m3 ) is:
Pτ = n− Pe − = E · n− evvd = E · j
E we get:
and from the relationship j = ρ1 E = σE
Pτ = σE 2 = ρj 2
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Joule’s effect
Pτ = σE 2 = ρj 2
The total power that must be spent to circulate the electric current
i in a section of conductor with section Σ and along h is obtained
from:
P = Pτ Σh = ρ
i2
h
Σh = ρi 2
Σ2
Σ
P = R i2 = V i
35/39
Joule’s effect
The same result is also obtained by considering the work performed
by the field to move an infinitesimal charge dq between one point of
the conductor at the potential VB to another at potential VA . For
this displacement the work done is:
dW = V dq = V i dt
and the power is then calculated as:
P=
dW
= Vi
dt
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Joule’s effect: summary
The power spent to circulate the
electric current i in the circuit is:
P = Ri 2 = Vi =
V2
R
If you keep the current for a time t then the work done:
Z t
Z t
W =
Pdt =
R i 2 dt
0
0
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Multiple choice test
A washing machine and a vacuum cleaner absorb a power of 2 kW and
1.3 kW respectively. Which of the two appliances has the greatest
electrical resistance? In which one circulates the most current?
(A) The washing machine has more resistance and more current
flows in the vacuum cleaner.
(B) The washing machine has more resistance and more current
flows in the washing machine.
(C) The vacuum cleaner has more resistance and more current
flows in the vacuum cleaner.
(D) The vacuum cleaner has more resistance and more current
flows in the washing machine.
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Answer
V2
R
As the power grid sustains potential differences of 220 V then the
vacuum cleaner (which draws less power) will have a greater
resistance.
P=
P = Vi
So the washing machine (which draws more power) will have a higher
current.
Answer: D
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