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Exercise 1 Exercise&1&&
Solution Exercise&1&&
II
II
aa
bb
A
current
I
=
2.5
A
is
flowing
from
left
to
right
through
a
straight
aluminum
wire
A current I = 2.5 A is flowing from left to right
through a straight aluminum wire with
with aa
1/2) mm and length L=5.0 m, as shown above
circular
cross
section
of
diameter
D=(0.3
/
π
1/2
circular cross section of diameter D=(0.3 / π ) mm and length
L=5.0 m, as shown above
-8
(not
(not drawn
drawn to
to scale!).
scale!).Aluminum
Aluminum has
has aa resistivity
resistivity of
of 2.7×10
2.7×10-8Ω
Ωm
m (at
(at room
room temperature).
temperature).
(a)
(a) Find
Find the
the voltage
voltage drop
drop between
between the
the two
two ends
ends of
of the
the wire,
wire, aa and
and b,
b,
V
=
V
−
V
.
ab
a
b
Vab = Va − Vb .
Exercise&1&&
Here,
Here,V
Vaa and
and V
Vbb are
are the
the electrical
electrical potential
potential values
values at
at the
the left
left end,
end, a,
a, and
and the
the right
right end,
end, b,
b,
respectively.
Is
V
>
0
or
V
<
0
?
[Hint:
By
convention
of
current
flow
direction,
ab
ab
respectively. Is Vab > 0 or Vab < 0 ? [Hint: By convention of current flow direction, the
the
electric
electric current
current in
in aa wire
wire or
or other
other resistor
resistor always
always flows
flows from
from high
high to
to low
low electric
electric potential.]
potential.]
I
I
(b)
in
Cross-­‐sectional rea of of
wheat
ire: energy
(b) Find
Find the
theaamount
amount
of
heat
energy generated
generated
in the
the wire
wire ifif the
the current
current flows
flows for
for 55 h.
h.
2
1/2
-­‐3
2
-­‐8 A =Find
π (D/2)
= π [ strength,
(0.3/(2 ×inside
π )) ×wire,
(10 m) ] = 2.25 × 10 m2 (c)
(c) Find the
the electric
electric field
field strength, E,
E, inside the
the wire, and
and the
the direction
direction of
of the
the E-vector.
E-vector.
Does
Resistance: Does EE point
point from
from left
left to
to right
right or
or from
from right
right to
to left?
left? [Hint:
[Hint: the
the electric
electric field
field vector
vector always
always
points
direction
of
electric
potential.]
R = ρinin
Lthe
the
/ A
= (2.7 ×decreasing
10-­‐8 Ω m
) (5.0 m) / (2.25 × 10-­‐8 m2) = 6.0 Ω points
direction
of
decreasing
electric
potential.]
a
b
Ohm’s Law: A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
|V
R I =section
(6.0 of
Ωdiameter
) (2.5 AD=(0.3
) = 15 / V
) mm and length L=5.0 m, as shown above
ab | =cross
circular
π1/2
Sign of V(not
: ab drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
Vab = Va − Vb = +15V >0 (a) Find the voltage drop between the two ends of the wire, a and b,
Reasoning: V = V − V .
ab
a
b
Current through resistor always flows from high to low potential; and I here flows Here, Va and Vb are the electrical potential values at the left end, a, and the right end, b,
Is VV
>
0
or
V
<
0
?
[Hint:
By
convention
of
current
flow
direction,
the
from a respectively.
to b, hence >
V
.
a ab b
ab
electric
current
in
a
wire
or
other
resistor
always
flows
from
high
to
low
electric
potential.]
(b) Find the amount of heat energy generated in the wire if the current flows for 5 h.
(c) Find the electric field strength, E, inside the wire, and the direction of the E-vector.
Joule heating Power = Rate of heat generation in wire: Does E point from left to right or from right to left? [Hint: the electric field vector always
P = I V
=direction
(2.5 A) 15V) = 3electric
7.5 Wpotential.]
points
inab
the
of(decreasing
Or use: P = R I2 = (6.0 Ω ) (2.5 A)2 = 37.5 W Note here that P is proportional to I2 for fixed resistance R. Heat energy generated during t=5h: H5h = P t = (37.5 W) (5 h) (3600s/h) = 675 kJ. Note here that H5h is also proportional to I2 for fixed resistance R and fixed time t. . Vab = Va − Vb .
Here, Va and Vb are the electrical potential values at the left end, a, and the right end, b,
respectively. Is Vab > 0 or Vab < 0 ? [Hint: By convention of current flow direction, the
electric current in a wire or other resistor always flows from high to low electric potential.]
(b) Find the amount of heat energy generated in the wire if the current flows for 5 h.
(c) Find the electric field strength, E, inside the wire, and the direction of the E-vector.
Does E point from left to right or from right to left? [Hint: the electric field vector always
points in the direction of decreasing Exercise&1&(cont’d.)&&
electric potential.]
E = |Vab| / L = (15V)/(5.0m) = 3.0 V/m I
I
The E-­‐vector always points in direction of decreasing potential, V. Since Va > Vb here [see part (a)], the E-­‐vector points from a to b, i.e., E-­‐vector points from left to right, a&
b&
inside the w
ire. A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
. (not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
I
(d) The charge carriers in the wire are electrons and the current, I, is due to the average
“drift” motion of these “conduction electrons” inside the wire (see/review FlipIt Physics).
Exercise&1&(cont’d.)&&
In which direction, on average, do these electrons move: left to right or right to left?
[Hint: An electron has negative charge, q = −e < 0. Consider the direction of the electric
force, F, acting on the electron, due to the electric field, E, found in part (c).]
(e) Find the change in the electric potential energy, ΔU, incurred by a conduction electron,
as it moves from one end of the wire to the other, due to the current flow. Does the
Force oelectron
n electron: gain potential energy (ΔU>0) or lose potential energy (ΔU<0)?
I
[Hint:
difference,
Vab, found in part (b); or calculate the work done on the
F =Use
qthe
E potential
= (-­‐e) E
electron by the electric force F , acting on the electron, due to the electric field E ].
a&
So, the F-­‐vector acting on each electron points in direction opposite to b&the E-­‐vector, A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
i.e., from right cross
to left. Hence the D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
circular
section
of diameter
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
electrons‘ drift motion is from right to left (d) The charge carriers in the wire are electrons and the current, I, is due to the average
“drift” motion of these “conduction electrons” inside the wire (see/review FlipIt Physics).
through he wire. Intwhich
direction, on average, do these electrons move: left to right or right to left?
[Hint: An electron has negative charge, q = −e < 0. Consider the direction of the electric
force, F, acting on the electron, due to the electric field, E, found in part (c).]
(e) Find the change in the electric potential energy, ΔU, incurred by a conduction electron,
as it moves from one end of the wire to the other, due to the current flow. Does the
electron gain potential energy (ΔU>0) or lose potential energy (ΔU<0)?
[Hint: Use the potential difference, Vab, found in part (b); or calculate the work done on the
electron by the electric force F , acting on the electron, due to the electric field E ].
Since the electron moves through the wire from right to left, i.e., from b to a, its change in potential energy is ΔU = Ua – Ub = q (Va – Vb) = q Vab = (-­‐1.6×10-­‐19 C) (+15V) = −2.4×10-­‐18 J < 0 , using q=−e=−1.6×10-­‐19C for the electron charge. The fact that ΔU<0 means that the electron loses potential energy as it drifts through the wire from b to a. . a&
b&
A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
(f) Suppose in some other galaxy, far, far away, the wire was made of anti-matter. The
charge carriers in this “anti-aluminum” wire are then anti-electrons, a.k.a. positrons,
each positron having a positive charge, q = +e > 0. Assuming the same direction and
strength of the electric current, I, the same wire length and diameter, and the same resistivity
as before, answer questions (a)-(e) for the anti-aluminum wire and, respectively, for the
“conduction positrons” flowing through the wire.
Note: You don’t have to re-do any of the numerical calculations. Instead just state for each
part, (a)-(e), whether, or not, the answer for the anti-aluminum wire and its positron charge
carriers is the same as for the aluminum wire with electron charge carriers; and, if not, how
the answer changes. In particular, how do the answers change, or not, for parts (d) and (e)?
Since the strength and direction of the current flow in the anti-­‐aluminum wire, the resistivity and the wire size is the same as before, the wire resistance, R, and the potential difference Vab = +15V > 0 is the same as in part (a) as before. For the same reasons, the heating power, P, and the amount of heat generated during 5h of current flow, H5h = 675 kJ is the same as in part (b); and the electric field E = 3.0 V/m, with E pointing from left to right, is the same as in part (c). However, because the positrons have positive charge, q=+e>0, the electric force acting on the positrons F = q E = (+e) E is in the same direction as the E-­‐vector. So, the F-­‐vector acting on the positrons points from left to right. Hence the positrons‘ drift motion is from left to right, opposite to electrons’ drift motion part (d) Since the positron moves through the wire from left to right, i.e., from a to b, its change in potential energy is ΔU = Ub – Ua = q (Vb – Va) = q (−Vab )= (+1.6×10-­‐19 C) (−15V) = −2.4×10-­‐18 J < 0 , which is the same as for the electron, moving from right to left, in part (e), using q=+e=+1.6×10-­‐19C for the positron charge. The fact that ΔU<0 means that the positron loses potential energy, again the same as for the electron in part (e), as the positron drifts through the wire from a to b. Notice: The positron charge, q=+e, is opposite in sign to the electron charge. However, the positron also drifts through the wire in the opposite direction as the electron. Hence, the sign of the electric potential difference traversed by the positron, Vb – Va = −Vab = −15V, is opposite in sign to the potential difference traversed by the electron, Va – Vb = +Vab = +15V. Hence the potential energy change , ΔU, i.e., the product of carrier charge, q, and traversed electric potential difference, ΔV, is the same, in sign and magnitude, for electrons and positrons. a&
b&
A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
. Returning to our own galaxy and planet earth…
Exercise&1&(cont’d.)&&
(g) Calculate the total number of electrons, N5h , flowing through the aluminum wire
during 5h. [Hint review the definition of electric current, I, and recall how much
charge, q, each electron carries.]
I
I
Recall (h)
that the urrent I is the rate at which charge Q flows though the wire, i.e., From
Nc
5h found in part (g) and ΔU found in part (e), calculate the combined total
I = or
dQ/dt gain
loss of electric potential energy, ΔUtot , from all electrons flowing through the
during
5h.tCompare
ΔUtot toothe
result from
part
(b):passed the totaltheat
energy
generated
where wire
Q(t) is the otal amount f charge that has hrough the wire up to time in
5h
by
the
current
flow.
t. So, for a constant, i.e., time-­‐independent, current, I, the total charge, Q
a&
b&5h, passing through Atcurrent
he wire d
uring t
=5h i
s: I = 2.5 A is flowing from left to right through a straight aluminum wire with a
Qcircular
× t =
(2.5A) (5h) (3600s/h) 45,000 C. L=5.0 m, as shown above
5h = Icross
section
of diameter
D=(0.3 / π1/2)=mm
and length
-8 Ω m (at room temperature).
(not drawn
to scale!). Aluminum
has a resistivity
of 2.7×10
The carriers transporting this charge are electrons, each electron carrying an -­‐19
absolute charge of |q|= e = 1.6×10 C. Hence, the total number of carriers that have Returning to our own galaxy and planet earth…
passed through the wire during t=5h is therefore: +3C) / (N
-­‐19 through
NCalculate
/ total
q = number
(45 ×of
10electrons,
1.6×10
= 2the
.8125 ×1023
electrons (g)
aluminum
wire
5h = Q5hthe
5h , flowing C) during 5h. [Hint review the definition of electric current, I, and recall how much
. charge, q, each electron carries.]
(h) From N5h found in part (g) and ΔU found in part (e), calculate the combined total
gain or loss of electric potential energy, ΔUtot , from all electrons flowing through the
wire during 5h. Compare ΔUtot to the result from part (b): the total heat energy generated
in 5h by the current flow.
ΔUtot = N5h × ΔU = (2.8125 ×1023) (−2.4×10-­‐18 J) = −675 kJ . As expected, the total potential energy lost by all electrons, −ΔUtot, passing through the wire during 5h, is exactly the same as the total amount of heat energy generated in the wire during those 5h. . a&
b&
A current I = 2.5 A is flowing from left to right through a straight aluminum wire with a
circular cross section of diameter D=(0.3 / π1/2) mm and length L=5.0 m, as shown above
(not drawn to scale!). Aluminum has a resistivity of 2.7×10-8 Ω m (at room temperature).
(i) Calculate the conduction electron number density, ne , in aluminum, that is, the number
of conduction electrons per volume.Exercise&1&(cont’d.)&&
Use this:
Since aluminum is trivalent (in case you already know any chemistry), each aluminum
provides 3 conduction electrons. The molar mass of aluminum is 27 g per mole of
Iatom
aluminum atoms. The number of aluminum atoms in one mole is given by Avogadro’s
number, NA =6.02×1023 atoms per mole. The mass density of aluminum is 2.70 g/cm3.
I
. (j) Calculate the electron drift speed, that is, the mean velocity of the electrons’ motion
3: in part (i), the current I, the wire cross section
to the current
[Hint:Vuse
ne found
Mass of due
aluminum in flow.
volume =1cm
a&
b&
and=the
charge, q = 3-e.]
M ρmelectron’s
V = (2.70g/cm
) (1cm3) = 2.70 g A current I = 2.5 A is flowing from left to right through
a
straight
aluminum
wire
with a
3: Number circular
of moles f aluminum in vD=(0.3
olume =1cm
crossosection
of diameter
/ πV1/2
) mm and
length L=5.0 m, as shown above
-8 Ω
N(not
M / to
Mscale!).
2.70g) /has
(27 g/mole) 0.100 mmole m =drawn
mol = (Aluminum
a resistivity
of =2.7×10
(at room temperature).
Number of aluminum atoms in volume V=1cm3: (i) Calculate the conduction electron number density,23ne , in aluminum, that is, the number
N
= Nm×N
mole)(6.02×10
atoms/mole)=6.02 × 1022atoms A= (0.100 ofatoms
conduction
electrons
per volume.
Use this:
Number of conduction electrons in volume V=1cm3: 22 electrons N
= trivalent
3 × (6.02 × 1you
022already
) = 18.06 10chemistry),
Since
aluminum
(in case
know×any
each aluminum
e=3×N
atoms is
atom
3 conduction
electrons.
molar mass of aluminum is 27 g per mole of
Number of cprovides
onduction electrons per vThe
olume: aluminum
atoms. The number of22aluminum atoms
in one mole is22 given -­‐3 by Avogadro’s
3
28 -­‐3 n
e = Ne/V = (18.02 × 10 )/(1cm ) = 18.06 × 10 cm = 18.06 × 10 m . number, NA =6.02×1023 atoms per mole. The mass density of aluminum is 2.70 g/cm3.
. (j) Calculate the electron drift speed, that is, the mean velocity of the electrons’ motion
due to the current flow. [Hint: use ne found in part (i), the current I, the wire cross section
and the electron’s charge, q = -e.]
. Current density in the wire: J = I / A = (2.5 A) / (2.25 × 10-­‐8 m2) = 1.111 × 108 A/m2 using wire cross-­‐section area, A=2.25 × 10-­‐8 m2, from part (a). Current density, J, is given in terms of drift speed, vDrift, carrier number density, ne, and carrier charge, |q| , see FlipItPhysics or textbook, by: J = |q| ne vDrift  vDrift = J / ( |q| ne ) = (1.111 × 108 A/m2) / [ (1.6 × 10-­‐19 C) (18.06 × 1028 m-­‐3 ) ] = 0.00384 m/s = 3.84 mm/s .