Chapter 27
Current and Resistance. Solutions of
Selected Problems
27.1
Problem 27.10 (In the text book)
A Van de Graaff generator produces a beam of 2.00-M eV deuterons, which are heavy hydrogen nuclei containing a proton and a neutron.
(a) If the beam current is 10.0 A, how far apart are the deuterons?
(b) Is the electric force of repulsion among them a significant factor in beam stability?
Explain.
Solution
(a) The speed of each deuteron is given by:
1
K = mv 2
2
r
or
v=
2K
m
where K is the kinetic energy of the deuteron and m is its mass. The mass of the
deuteron is double the mass of a proton. We then get:
s
2 × 2.00 × 106 (eV ) × 1.60 × 10−19 (J/eV )
v=
= 1.38 × 107 m/s
2 × 1.67 × 10−27 kg
2
CHAPTER 27. CURRENT AND RESISTANCE. SOLUTIONS OF SELECTED
PROBLEMS
The deuteron current is due the passage of certain number of deuterons per second at
a stationary point in their path. If the charge on each deuteron is q = 1.60 × 10−19 C,
then the time between individual deuterons t is given by:
I=
q
t
or
t=
q
1.60 × 10−19
=
= 1.6 × 10−14 s
−6
I
10.0 × 10
So, the distance between deutrons is d = vt, i.e.
d = vt = 1.38 × 107 × 1.6 × 10−14 = 2.21 × 10−7 m
(b) One nucleus with a charge q puts its nearest neighbor at a potential V of:
V =
8.99 × 109 × 1.60 × 10−19
ke q
= 6.51 × 10−3 V
=
r
2.21 × 10−7
Th is very small voltage compared to the voltage of 2 M V that is used to accelerate
the deuterons to a 2.00 M eV energy. So Coulomb repulsion between the deuterons is a
small effect.
Physics 111:Introductory Physics II, Chapter 27
Winter 2005
Ahmed H. Hussein
27.2. PROBLEM 27.20 (IN THE TEXT BOOK)
27.2
3
Problem 27.20 (In the text book)
A solid cube of silver (density = 10.5 g/cm3 ) has a mass of 90.0 g.
(a) What is the resistance between opposite faces of the cube?
(b) Assume each silver atom contributes one conduction electron. Find the average drift
speed of electrons when a potential difference of 1.00 × 10−5 V is applied to opposite
faces. The atomic number of silver is 47, and its molar mass is 107.87 g/mol.
Solution
The distance ` between opposite faces of the cube is:
1/3
` = (V )
=
m 1/3
d
=
90.0
10.5
1/3
= 2.05 cm
where d is the density of silver.
(a) The resistance between opposite faces is then:
R=
ρ`
1.59 × 10−8 Ω · m
ρ`
ρ
= 2 = =
= 7.76 × 10−7 Ω
A
`
`
2.05 × 10−2 m
(b) When a potential difference of 1.00 × 10−5 is applied to opposite faces a current I passes
through,
∆V
1.00 × 10−5
=
I=
= 12.9 A
R
7.76 × 10−7
In any solid there is Avogadro’s number NA = 6.02 × 1023 of atoms in every molar mass.
The number of silver atoms in one gram N of silver is then:
N=
NA
107.87
To get the number of atoms per unite volume n we multiply N by the density of silver
d in g/m3 , or:
n = Nd =
6.02 × 1023
10.5(g/cm3 )
= 5.86 × 1028 atoms/m3
×
107.87
1.00 × 10−6 (m3 /cm3 )
Physics 111:Introductory Physics II, Chapter 27
Winter 2005
Ahmed H. Hussein
CHAPTER 27. CURRENT AND RESISTANCE. SOLUTIONS OF SELECTED
PROBLEMS
4
Since every silver atoms contributes one conduction electron, then number of conduction
electron per unit volume is also n = 5.86 × 1028 electrons/m3 . The current through the
silver cube can also be expressed as:
I = nqvd A
and
v=
I
I
12.9
=
=
= 3.27 × 10−6 m/s
2
28
nqA
nq`
5.86 × 10 × 1.60 × 10−19 × (2.05 × 10−2 )2
Physics 111:Introductory Physics II, Chapter 27
Winter 2005
Ahmed H. Hussein
27.3. PROBLEM 27.31 (IN THE TEXT BOOK)
27.3
5
Problem 27.31 (In the text book)
An aluminum wire with a diameter of 0.100 mm has a uniform electric field of 0.200 V /m
imposed along its entire length. The temperature of the wire is 50.0◦ C. Assume one free
electron per atom.
(a) Use the information in Table 27.1 and determine the resistivity.
(b) What is the current density in the wire?
(c) What is the total current in the wire?
(d) What is the drift speed of the conduction electrons?
(e) What potential difference must exist between the ends of a 2.00-m length of the wire to
produce the stated electric field?
Solution
(a) Table 27.1 gives the resistivity of aluminum ρ◦ = 2.82 × 10−8 Ω · m at T◦ = 20◦ C, and
the temperature coefficient α = 3.90 × 10−3 , so the resistivity of aluminum at 50◦ C is:
ρ = ρ◦ [1 + α(T − T◦ )] = 2.82 × 10−8 [1 + 3.90 × 10−3 × (50.0 − 20.0)] = 3.15 × 10−8 Ω · m
(b) The current density is given by;
E
0.200
= 6.35 × 106 A/m2
J=
=
ρ
3.15 × 10−8
(c) the total current in the wire is:
2
πd
π × (1.00 × 10−4 )2
= 6.35 × 106 ×
= 4.99 × 10−2 A = 49.9 mA
I = JA = J
4
4
(d) We can calculate the number of electron per unit volume n as we did in the previous
problem. The molar mass of aluminum is 26.98 g/mol and density of 2.7 × 106 g/m3 , n
is:
6.02 × 1023 × 2.7 × 106
n=
= 6.02 × 1028 electrons/m3
26.98
the drift velocity of the electrons is then:
J
6.35 × 106
vd =
=
= 6.59 × 10−4 m/s
ne
6.02 × 1028 × 1.60 × 10−19
(e) The potential difference is:
∆V = E` = 0.200 × 2.00 = 0.4 V
Physics 111:Introductory Physics II, Chapter 27
Winter 2005
Ahmed H. Hussein
CHAPTER 27. CURRENT AND RESISTANCE. SOLUTIONS OF SELECTED
PROBLEMS
6
27.4
Problem 27.42 (In the text book)
A 500-W heating coil designed to operate from 110 V is made of Nichrome wire 0.500 mm
in diameter.
(a) Assuming that the resistivity of the Nichrome remains constant at its 20.0 ◦ C value, find
the length of wire used.
(b) What If? Now consider the variation of resistivity with temperature. What power will
the coil of part (a) actually deliver when it is heated to 1200 ◦ C?
Solution
(a) The power delivered by the wire is;
(∆V )2
P = I(∆V ) =
R
and
(∆V )2
(110)2
R=
=
= 24.2 Ω
P
500
using the resistance we can get the length of the wire;
R=
ρ`
A
and
`=
RA
Rπr2
24.2 × π × (2.5 × 10−4 )2
= 3.17 m
=
=
ρ
ρ
1.50 × 10−6
(b) The resistance of the wire at 1200◦ C is:
R = R◦ [1 + α(T − T◦ )] = 24.2[1 + 0.4 × 10−3 (1200 − 20)] = 35.6 Ω
The power delivered at 1200◦ C is:
P =
Physics 111:Introductory Physics II, Chapter 27
(∆V )2
(110)2
=
= 340 W
R
35.6
Winter 2005
Ahmed H. Hussein
27.5. PROBLEM 27.68 (IN THE TEXT BOOK)
27.5
7
Problem 27.68 (In the text book)
An oceanographer is studying how the ion concentration in sea water depends on depth. She
does this by lowering into the water a pair of concentric metallic cylinders (Figure (27.68))
at the end of a cable and taking data to determine the resistance between these electrodes
as a function of depth. The water between the two cylinders forms a cylindrical shell of
inner radius ra , outer radius rb , and length L much larger than rb . The scientist applies a
potential difference ∆V between the inner and outer surfaces, producing an outward radial
current I. Let ρ represent the resistivity of the water.
(a) Find the resistance of the water between the cylinders in terms of L, ρ, ra , and rb .
(b) Express the resistivity of the water in terms of the measured quantities L, ra , rb , ∆V ,
and I.
Figure 27.68:
Solution
Consider a thin cylindrical shell between the two cylindrical conductors with radius r, thickness dr, and length L. The resistance dR of this shell is given by:
ρ dr
ρdr
ρdr
dR =
=
=
A
2πrL
2πL r
Physics 111:Introductory Physics II, Chapter 27
Winter 2005
Ahmed H. Hussein
8
CHAPTER 27. CURRENT AND RESISTANCE. SOLUTIONS OF SELECTED
PROBLEMS
The resistance of the whole region between the two cylindrical conductors is then:
Z rb
ρ
dr
ρ
rb
R=
=
ln
2πL ra r
2πL
ra
Since:
∆V
ρ
=
ln
R=
I
2πL
Physics 111:Introductory Physics II, Chapter 27
rb
ra
and
Winter 2005
ρ=
2πL∆V
I ln(rb /ra )
Ahmed H. Hussein