Chapter  31  –  solutions    

31.2. Solve: Using Equation 31.3 and Table 31.1, the electron current is

i = nAvd = 5.9 × 1028 m−3( )π 0.5× 10−3 m( )2

5.0 × 10−5 m/s( ) = 2.3× 1018 s−1

The time for 1 mole of electrons to pass through a cross section of the wire is

t =

NA × 1 molei

=6.02 × 1023

2.3× 1018 s−1 = 2.62 × 105 s ≈ 3.0 days

Assess: The drift speed is small, and Avogadro’s number is large. A time of the order of 3 days is reasonable.  31.12. Solve: (a) The current density is

J =IA=

IπR2 =

0.85 A

π12

0.00025 m( )⎡

⎣⎢

⎤

⎦⎥

2 = 1.73× 107 A/m2

(b) The electron current, or number of electrons per second, is

Ne

Δt=

Ie=

0.85 A1.60 × 10−19 C

=0.85 C/s

1.60 × 10−19 C= 5.3× 1018 s−1

 

31.19. Visualize:

Solve: The current-carrying cross section of the wire is

A = πr1

2 − πr22 = π 0.0010 m( )2

− 0.00050 m( )2⎡⎣⎢

⎤⎦⎥= 2.356 × 10−6 m2

The current density is

J =

10 A2.356 × 10−6 m2 = 4.2 × 106 A/m2

 

31.35. Model: Assume the battery is an ideal battery. Solve: We can find the current I from Equation 31.24 provided we know the resistance R of the gold wire. From Equation 31.23 and Table 31.2, the resistance of the wire is

R =ρLA=ρLπr 2 =

2.4 × 10−8 Ωm( ) 100 m( )π 0.050 × 10−3 m( )2 = 305.6 Ω⇒ I =

ΔVwire

R=

0.70 V305.6 Ω

= 2.3 mA

   

31.41. Solve: The density of aluminum is 2700 kg/m3, so 1.0 m3 of aluminum has a mass of 2700 kg. The conduction-electron density is the number of electrons in 1.0 m3. Because the atomic mass of aluminum is 27 u, 27 g of aluminum contains NA = 6.02 × 1023 atoms. Thus, the number of aluminum atoms in 2700 kg of aluminum is

6.02 × 1023

27 g( )2700 kg( ) = 6.0 × 1028

The number density of aluminum atoms is 6.0 × 1028 m−3 . Since each aluminum atom contributes one conduction electron to the metal, the number density of conduction electrons in aluminum is n = 6.0 × 1028 m−3 . Assess: The number density n obtained above agrees with that given in Table 31.1.    

31.42. Solve: (a) The moving electrons are a current, even though they’re not confined inside a wire. The electron current is

Ne

Δt=

Ie=

50 × 10−6 A1.60 × 10−19 C

= 3.1× 1014 s−1

This means during the time interval Δt = 1 s, 3.1 × 1014 electrons strike the screen. (b) The current density is

J =IA=

Iπr 2 =

50 × 10−6 A

π 0.00020 m( )2 = 4.0 × 102 A/m2

(c) The acceleration can be found from kinematics:

v1

2 = 4.0 × 107 m/s( )2= v0

2 + 2aΔx = 2aΔx ⇒

a =4.0 × 107 m/s( )2

2 5.0 × 10−3 m( )= 1.60 × 1017 m/s2

But the acceleration is a = F m = eE m . Consequently, the electric field must be

E =

mae=

9.11× 10−31 kg( ) 1.60 × 1017 m/s2( )1.6 × 10−19 C

= 9.1× 105 V/m

(d) When they strike the screen, each electron has a kinetic energy

K = 1

2 mv12 = 1

2 9.11× 10−31 kg( ) 4.0 × 107 m/s( ) = 7.288 × 10−16 J Power is the rate at which the screen absorbs this energy. The power of the beam is

P =

ΔEΔt

= KNe

Δt= 7.288 × 10−16 J( ) 3.1× 1014 s−1( ) = 0.23 J/s = 0.23 W

Assess: Power delivered to the screen by the electron beam is reasonable because the screen over time becomes a little warm.  

31.63. Model: Assume the battery is ideal. Visualize: The current supplied by the battery and passing through the wire is I = ΔVbat/R. A graph of current versus time has exactly the same shape as the graph of ΔVbat with an initial value of I0 = (ΔVbat)0/R = (1.5 V)/(3.0 Ω) = 0.50 A. The horizontal axis has been changed to seconds.

Solve: Current is I = dQ/dt. Thus the total charge supplied by the battery is

Q = I dt0

∞

∫ = area under the current-versus-time graph

= 12 (7200 s)(0.50 A) = 1.80 × 103 C

 

31.70. Solve: The total charge delivered by the battery is

Q = dQ∫ = I dt0 s

∞

∫ = 0.75 A( )e−t 6 hours( )

0 s

∞

∫ = 0.75 A( ) −6 hours( ) e−t 6 hours( )⎡⎣

⎤⎦0

∞

= 0.75 A( ) 6 × 3600 s( ) = 1.62 × 104 C

The number of electrons transported is

1.62 × 104 C1.6 × 10−19 C

= 1.01× 1023

       

31.72. Visualize:

The current density in the beam increases with distance from the center. We consider a thin circular shell of width dr at a distance r from the center to calculate the current density at the edge. Solve: (a) The beam current is 1.5 mA. This means the beam transports a charge of 1.5 × 10−

3 C in 1 s. The number of protons delivered in one second is

1.5× 10−3 C1.6 × 10−19 C

= 9.375× 1015 = 9.4 × 1015

(b) From J = I A , the current in the ring of width dr at a distance r from the center is

dI = JdA = Jedge

rR

⎛

⎝⎜⎞

⎠⎟2πrdr( ) = Jedge

2πr 2drR

The total current I = 1.5 mA is found by integrating this expression:

I total = dI∫ = 1.5× 10−3 A =Jedge

R2πr 2dr

0

R

∫ =2π Jedge R2

3

⇒ Jedge = 1.5× 10−3 A( ) 3

2π 2.5× 10−3 m( )2 = 115 A/m2