Φ21 Fall 2006
1
HW5 Solutions
Problem 28.4
1.0 × 1016
electrons ow through a cross section of silver wire in
320 µs
with a drift speed of
8.0 × 10−4 m/s.
What is the diameter of the wire?
Solution: The electron current is:
i=
In silver, there are
5.8 × 1028
Ne
1.0 × 1016
=
= 3.125 × 1019 electrons/s
∆t
320 × 10−6
conduction electrons per cubic meter. The electron current is
i = ne vA,
so the
area is:
A=
i
ne v
=
3.125 × 1019 s−1
= 6.735 × 10−7 m2
(5.8 × 1028 m−3 ) (8.0 × 10−4 m/s)
and the diameter is easily found to be:
d = 2r = 2
√
A/pi = 9.26 × 10−4 m
(Note MasteringPhysics asks for the answer in mm.)
2
Problem 28.12
The current in an electric hair dryer is 10.0 A. How much charge ows through the hair dryer in 5.0 min?
How many electrons ow through the hair dryer in 5.0 min?
Solution: Since the current is charge per unit time, the charge is:
Q = I ∆t = (10.0 C/s) (5.0 min) (60 s/min) = 3000 C
That charge is carried in the form of electrons:
(
)
Ne = Q/e = (3000 C) / 1.6 × 10−19 C = 1.88 × 1022 electrons
1
3
Current and Current Density at a Junction
Consider the junction of three wires as shown in Figure 1. The magnitudes of the current density and the diameters for wires 1 and 2 are given
in the table. The current directions are indicated by the arrows.
Part A. Find the current
the current density
J3
I3
in wire 3. Part B. Find the magnitude of
in wire 3. The diameter of wire 3 is 1.5 mm.
Solution: The total current entering the junction must equal the total
current exiting it.
∑
Iin =
∑
Iout
So rst the currents in wires 1 and 2 must be calculated.
(
2
)
(
I1
= J1 A1 = 3.0 A/mm
I2
(
)
= J2 A2 = 5.0 A/mm2 π
Assume that
I3
π
(
2.0 mm
2
3.0 mm
2
)2
Figure 1:
= 9.42 A
)2
= 35.34 A
is exiting the junction, then the junction equation is:
The negative sign indicates that
4
Current
Wire
Density
2
(A/mm )
I1 = I2 + I3
9.42 A = 35.34 A + I3
I3 = −25.92 A
is:
Current and Current
Density at a Junction
I3
Diameter
(mm)
1
3.0
2.0
2
5.0
3.0
Table 1: Parameters for Current
Density at a Junction
is actually entering the junction. The magnitude of the current density
( (
)2 )
1.5 mm
= 14.7 A/mm2
J3 = I3 /A3 = 25.92 A/ π
2
Problem 28.36
The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and
calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane
known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective
for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion
+
(K ) channel carries a current of 1.8 pA.
Part A. How many potassium ions pass through if the ion channel opens for 1.0 ms?
Part B. What is the current density in the ion channel?
Solution: In 1.0 ms, the charge that passes through is
(
)(
)
Q = I ∆t = 1.8 × 10−12 A 1.0 × 10−3 s = 1.8 × 10−15 C
Since each ion has a +1 charge (measured in electron charges), this represents
NK+
(
)
1.8 × 10−15 C
Q
=
=
= 11250
e
(1.60 × 10−19 C)
The current density is calculated from the current and the size of the channel.
(
)
1.8 × 10−12 A
I
(
) = 2.55 × 107 A/m2
J=
=
2
A
π (0.30 × 10−9 m/2)
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