Chapter 28 by John Smith

CURRENT AND CONDUCTIVITY

28.1. Solve: The wire's cross-sectional area is A = mz = ~ ( 1 . x0 10" m)' = 3.1415 x lo4 m2, and the electron current through this wire is 2.0 X loi9 s-l . Using Table 28.1 for the electron density of iron and Equation 28.3, the drift velocity is i

2.0~101 S-l~

=-=

(8.5 x

m")(3.1415 x 10" m')

= 7.5 x 10-5m/s= 75 pm/s

Assess: The drift speed of electrons in metals is small.

28.2. Solve: We estimate a distance of 5 ft from the wall switch to the ceiling and then a distance of 8 ft to the center of the room. This yields a total length of approximately L = 4.0 m that an electron will travel in time t. The drift speed of electrons in copper is vd = 1.O x IO4 m / s . Thus, L

t=-=

4.0 m =4.0x104 s = 11 hours 1 . 0 ~ 1m 0 /~s

Assess: Given a drift speed of 100 p d s , a time of 11 hours to cover 4.0 m is physically reasonable.

28.3. Solve: Using Equation 28.3 and Table 28.1, the electron current is i = nAv, = (5.9 x 10'' m - 3 ) ~ ( 0 . 5x lo-' m)'(S.O x 10" m / s) = 2.3 x 10" S-' The time for 1 mole of electrons to pass through a cross section of the wire is

Assess: The drift speed is small, and Avogadro's number is large. A time of the order of 3 days is reasonable.

28.4. Solve: Equation 28.2 is Ne = nAvdAt. Using Table 28.1 for the electron density, we get

4( 1.O x 10l6)

n(5.8 x 10" m-3)(8.0x lo4 m / s)(320x 10" s)

= 9.26 x lo4 m = 0.926 mm

28.5. Solve: Using Equation 28.2, N e *1.44 x 1014 = 6.0 x Av,Ar (4.0 x 10" m2)(2.0x lo4 m / s)(3.0x 10" s )

m-j

From Table 28.1, the metal is aluminum.

28.6. Solve: For L = 1.0 cm

1.0 x lo-* m ,the surface area of the wire is

A=(2m)L=nDL=n(1.0x10-3m)(1.0x1 0 ~ z m ) = ( 1 . 0 x 1 0 ~ 5 m ' ) ~

28-1

28-2

Chapter 28

The surface charge density of the wire is

(lo00 cm-' x 1 cm)1.60 x

77=x=

(1.0 x 10-~m 2 ) z

= 5.1 x lo-'' C / m2

28.7. Solve: (a) Each gold atom has one conduction electron. Using Avogadro's number and n as the number of moles, the number of atoms is

The density of gold is p = 19,300 kg/m3, the atomic mass is MA = 197 g mol-', r = 0.5 x m, L = 0.1 m, and NA = 6.02 x loz3mol-'. Substituting these values, we get N = 4.63 x 10" electrons. (b) If all the electrons are transferred a charge of (4.63 x 1021)(-1.60 X C) = -740.8 C will be delivered. To deliver a charge of -32 nC, however, the electrons within a length 1 have to be delivered. Thus, -32 x 10-~c 1= (10 cm) = 4.32 x lo-'' cm = 4.32 x m -740.8 C

28.8.

Model: Use the conduction model to relate the drift speed to the electric field strength. Solve: From Equation 28.7, the electric field is mvd - ( 9 . 1 1 ~ 1 O - ~ ' k g ) ( 2 . 0 ~m 1 0/ s~)

E=-

(1.60 x

C)(5.0 x

= 0.023 N / C

28.9. Model: We will use the model of conduction to relate the electric field strength to the mean free time between collisions. Solve: From Equation 28.8, the electric field is kg)(5.0 x 1019 8 )

mi (9. I I x E=-nez4 (8.5 x 10%m-3)(1.6x

28.10. Visualize:

C)(4.2 x

~~1-

=0.31 N / C

+ +

s)a(0.9x

c----,

Solve: (a) The charge density is not uniform along the wire. If it was, there would be no electric field inside the wire. The charge density is most positive near the positive terminal of the battery. It gradually decreases until becoming neutral halfway around the circuit. It then becomes increasingly negative, and is most negative near the negative terminal of the battery. (b) The electric field values are shown in the figure.

28.11. Model: A battery is a charge escalator. Solve: When a wire is connected to a battery, there is a sustained motion of electrons. A current of 1.5 A means that a charge of 1.5 C flows through a cross section of the wire per second. Because Q = N e e , the number of electrons transported per second is

N = -Q= = 9.4 x 10" electrons e e 1.60~10-C '~

28.12.

Solve: Equation 28.10 is Q = IAt. The amount of charge delivered is 1 min

The number of electrons that flow through the hair dryer is

Current and Conductivity

28-3

28.13. Solve: From Equation 28.10,

28.14. Visualize:

(2.0 X 1OL3)(l.60 x

I=-=-= At At

1.0~10-~

+ i-

= 0.0032 C I s = 0.0032 A = 3.2 mA

@-+-

@-+*

@-+z

The direction of the current 7 in a material is opposite to the direction of motion of the negative charges and is the same as the direction of motion of positive charges. Solve: The charge due to positive ions moving to the right per second is

q+= N+(2e) = (5.0 x IOl5)(2x 1.60 x

C) = 1.60 x lo-' C

The charge due to negative ions moving to the left per second is

q- = N ( + ) = (6.0 x ioL5s)(-I.~ox 10-l~c)= -0.96 x

Thus, the current in the solution is

. 1=--

q+ - q-

1.60 x IO-' C - ( 4 . 9 6 x

= 2.56 x

A = 2.56 mA

28.15. Solve: (a) The current density is I 2.5 A = 6.25 x lo5 A I m2 A 4 . 0 ~ 1 0 " m2 (b) Using Equation 28.13 and Table 28.1, the drift speed is J=-=

d '

=-ne

6.25 x lo5 A I m2 m-3)(1,6x

- (6.0 x IO"

28.16. Visualize:

= 6.51 x

mIs

Wire /

The current density J in a wire, as given by Equation 28.13, does not depend on the thickness of the wire. Solve: (a) The current in the wire is

Iw,, = J,,,A,+,

= (450,000 A I rn')a[+(1.5 x

m)]* = 0.795 A

Because current is continuous, I,,,, = Ililamm,. Thus, I,,, = 0.795 A. (b) The current density in the filament is Jfrlamrnl

=lRlmenl Arrlmmt

0.795 A

, = 7 . 0 3 ~ 1 0 'A l m '

lr[+(0.12 x IO-' mi]-

28-4

Chapter 28

28.17. Solve: (a) The current density is I A

I rtR2

J=-=-=

0.85 A = 1.73 x IO7 A / m2 ~[+(0.00025m)]‘

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

Ne At

-Z

0.85 A 0.85 C / s = 5.31 x 10“ 1.60 x C 1.60x C

---=

28.18. Solve:

S-‘

(a) From Equation 28.13 and Table 28.1, the current density in the gold wire is

J = nevd = (5.9 x IOz8 m-3)(1.60 x lo-’’ C)(3.0 x lo4 m / s) = 2.83 x IO6 A / m2 (b) The current is

Z = JA = (2.83 x IO6 A / m2)n(0.25 x

m)’ = 0.556 A

28.19. Solve: From Equation 28.13, the current in the wire is 1= JA = (7.50 x 10’ A / m2)(2.5 x IO6 m x 75 x 10” m) = 0.141 mA

28.20. Visualize:

Cross section of wire

Solve: The current-carrying cross section of the wire is A = mi2 - m2’ = z[(O.OOlO m)’ - (0.0005 m)’] = 2.356 x lo4 m2 The current density is

10 A = 4.24 x lo6 A / m2 2.356 x 10” m2

28.21. Model: Use the model of conduction to relate the mean time between collisions to conductivity. Solve: From Equation 28.17, Table 28.1, and Table 28.2, the mean time between collisions for aluminum is kg)(3.5 x lo7 R-‘m-’) m0.N - (9.11 x = 2.08 x 10-l4 nMe2 (6.0 x 10” m-3)(1.60 x lO-I9 C)*

%I=--

Similarly, the mean time between collisions for iron is T,, = 4.19 x lo-’’ s . Assess: The mean time between collisions in metals are of the order of 1O-I4 s.

28.22. Model: We will use the model of conduction to relate the mean time between collisions to conductivity. Solve: From Equation 28.17, Table 28.1, and Table 28.2, the mean time between collisions for silver is ‘silva

(9.1 1 x mcsilver = -nsilvere2

kg)(6.2 x lo7 Q-lrn-’)

(5.8 x IOz8 m-3)(1.60 x lo-’’ C)’

Similarly, the mean time between collisions for gold is rgold = 2.47 x Assess: Mean free times in metals are of the order of = s.

= 3.80 x 10-l4

28.23. Solve: The current,density is J = DE. Using Equation 28.18 and Table 28.2, the current in the wire is I = oEA = (3.5 x IO’ Q-’m-’)(0.012 N / C)(4 x 10” m’) = 1.68 A

28.24. Solve: From Equation 28.18 and Table 28.2, the electric field is E = - J= - = I OA

5.0 A

=0.159N/C ( 1 . 0 ~ 1 0Q~- ~ ~ - ~ ) I c ( ~ .m)’ ox~o-~

Current and Conductivity

28-5

28.25. Solve: From Equations 28.18 and 28.19, the resistivity is Em2 - (0.085 N / C)lr(1.5x I 12 A

p=-E = - = - =E- - EA J I/A I

m)*

= 5 . 0 1 ~ 1 0 -a~ m

28.26. Solve: From Equations 28.18 and 28.19, the resistivity is E E EA Em‘ (0.0075 N / C)n(0.5 x 10” m)’ --= =1.51x104 Q m p=-=-=-J IIA I I 3.9 x 10” A From Table 8.2, we see that the wire is made of nichrome.

28.27. Solve: (a) Since J = OE and J = I / A , the electric field is I E=-=-= OA

I ?D-’O

0.020 A ~ ( 0 . 2 5x

m)’(6.2 x IO7 G-Im-’)

=I.64xlO” N / C

(b) Since the current density is related to v,, by J = I / A = nev, ,the drift speed is ’d

=--

? ~ - ~ n n(0.25 e x

= 1.10 x IO-’ m / s

0.020 A m)2(5.8 x 10“ m-3)(1.60 x

Assess: The values of n and o for silver have been taken from Table 28.1 and Table 28.2. The drift velocity is typical of metals.

28.28. Visualize:

1 cm X 1 cm X 1 cmblock

Solve: The c m n t density through the cube is J = oE, and the actual current is I = AJ. Combining these equations, the conductivity is I 9.0 A O=-= = 1.8 x io7 W m - ’ AE (lo4 mZ)(5.0 x N / C) If all quantities entering the calculation are in SI units then the result for o has to be in SI units. From the value for O,we can identify the metal as being tungsten.

28.29. 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 x loz3atoms. Thus, the number of aluminum atoms in 2700 kg of aluminum is 6.02 x

(2700 kg) = 6.0 x 10” (27 g> The number density of aluminum atoms is 6.0 X loz8m-3. Since each aluminum atom contributes one conduction electron to the metal, the number density of conduction electrons in aluminum is n = 6.0 x 10” m-3. Assess: The number density n obtained above agrees with that given in Table 28.1.

28.30. Solve: Equation 28.18 will be used to relate electric field strength with the diameter. We have

Because the current is the same in the two wires,

----

. .-I .

28-6

Chapter 28

Using the values of o from Table 28.2, ENChrn,

(

lo7n'm-')( 6 . 7 ~ 1 0 'Q-'m-'

= 3S

''O m ) 2 ( 0 . 0 0 8 0 N / C) = 0.104 N/C 2.0mm

28.31. Solve: (a) Let the new current be I' such that I' = 21 where I is the original current. The current density is J = Z/A . Since the area of cross section of the wire remains the same, we have

That is, doubling the current doubles the current density. (b) Conduction-electron density is a material's characteristic value, and does not depend on external factors. Thus n ,=n. (c) The mean time between collisions is a material's characteristic value, and does not depend on external factors. Thus, 7= z: (d) From the model of conduction, Equation 28.7 is

That is, doubling the current doubles the drift speed.

28.32. Solve: (a) We will denote the new (or changed) quantities by primes. The new electric field in the wire is E'such that E'=2E. Because J = IJA = oE, I / E = oA is a constant. Therefore I -

Thus, doubling the electric field doubles the current in the wire. (b) The conduction-electron density is an intrinsic property of the material and cannot be changed by changing the external electric field. (c) The mean time between collisions is an intrinsic property of the material and cannot be changed by changing the external electric field. (d) From the model of conduction, Equation 28.7 is vd = ezE/rn. Because e, z and m do not change with electric field,

That is, doubling the electric field doubles the drift speed.

28.33. Solve:

(a) The moving electrons are a current, even though they're not confined inside a wire. The

electron current is

This means during the time interval A? = 1 s, 3.12 x 10'"electrons strike the screen. (b) The current density is

50x10"A

J=-=-=

=398A/m?

m2 ~ ( 0 . 0 0 0 2 0 m ) ~

v: =(4.Ox1O7 m l s ) ' = v i + 2 a h = 2 a h * a =

(4.0 x lo7 m / s)'

2(5.0 x

= 1 . 6 0 ~ 1 0 m/s' '~

But the acceleration is a = FJm = eEJm . Consequently, the electric field must be

ma E=-=

(9.11~10--"kg)(1.60x10i7 m / s ' ) =9.1l~lO'N/C e 1.6 x 1O-l9C (a) When they strike the screen, each electron has a kinetic energy K=+rnv12= + ( 9 . 1 1 ~ 1 0 - ~kg)(4.0x107 ' m/s)=7.288x10-I6 J

Current and Conductivity

28-7

Power is the rate at which the screen absorbs this energy. The power of the beam is

AE P = -= K S = (7.288 x

J)(3.12 x 1014 s-I) = 0.227 J / s = 0.227 W At At Assess: Power delivered to the screen by the electron beam is reasonable because the screen over time becomes a little warm.

28.34.

Visualize: Please refer to Figure P28.34. Solve: (a) The current associated with the moving film is the rate at which the charge on the film moves past a certain point. The tangential speed of the film is rev 1 min 2zrad v = wr = (90 rpm)(4.0 cm) = 90-x-xx 1.0 cm = 9.425 cm / s 1 rev nun 60s In 1.0 s the film moves a distance of 9.425 cm. This means the area of the film that moves to the right in 1.0 s is (9.425 cm)(4.0 cm) = 37.7 cm'. The amount of charge that passes to the right in 1.0 s is

Q = (37.7 cm*)(-2.0 x 10 -9 C/cm') = -75.4 x 10 -9 C

Since I = Q/ A t , we have 1-(75.4 x

c)(

= 75.4 nA

(b) Having found the current in part (a), we can once again use I = QfAt to obtain At:

28.35. Model: The current is the rate at which the charge of the ions moves through the ionic solution. Solve:

Because the atomic mass of gold is 197 g, the number of gold atoms is

A4 N = --NA =

(

0.5 g h.02 x 197 g mol-'

mol-' = 1.528 x lo2' atoms

C , in 3 hours on the statue. The

We need to deposit N = 1.528 x 1021gold ions, each with a charge of -1.60 x current is Q

I=-= At

(1.528 X 102')(1.60 X

3 x 3600 s

= 22.6 mA

28.36. Solve: (a) A current of 1.8 pA for the potassium ions means that a charge of 1.8 pC flows through the potassium ion channel per second. The number of potassium ions that pass through the ion channel per second is 1.8 x lo-'? c / s = 1.125 x 10' s-l I .6 x 1 0 - 1c ~

Since the channel opens only for 1.0 ms, the total number of potassium ions that pass through the channel is (1.125~10' ~ - ' ) ( ~ . O X ~ O - ' S ) 1.13~10'atoms. = (b) The current density in the ion channel is

28.37. Solve: (a) Current is defined as I = Q/At, so the charge delivered in time At is Q = IAt = (150 A)(0.80 s) = 120 C

(b) The drift speed is J I/A " =-=-=-ne

I m'ne

150 A ~(0.003-5 m)'(8.5 x 10" m-')(1.60 x

= 5.617 x l o 4 m / s

At this speed the electrons drift a distance d = (5.617 x loJ m/s)(0.80 s) = 4.49 x lo" m = 0.449 mm

28-8

Chapter 28

28.38. Solve: The total charge in the battery is Q = IAt = (90 A)(3600 S ) = 3.2 X lo5 C 28.39. Model: We assume that the charge carriers are uniformly distributed throughout the wire. Solve: Using Equation 28.16, we can write the current density as

For a given wire, the current is thus proportional to the area through which the current flows: Ima = R2 and

I,,., ( 5 ~ ) 'Therefore, . Oc

1 = 25% I,,,,= -

That is, 25% of the total current flows in the part of the wire with radius r 2 WZ.

28.40. Solve: Equation 28.13 defines the current density as J = I / A . This means

Assess: Fuse wires are usually thin.

28.41. Solve: The density of copper p = 8940 kg/m3= 8.94 g/cm3. The volume of the hollow copper cylinder is 62.0 g V = -M= p

8.94glcm3

= 6.935 cm3

Because cylinder's length is 10 cm, the area of cross section of the hollow cylinder is V A=-=

6.935cm3

lOcm

= 0.6935 cm2

Thus, the current I = JA = (150,000 A/m')(0.6935 x lo" m') = 10.4 A.

28.42. Visualize:

Axis

Surface at radius

r from the cylinder I

Solve: (a) Consider a cylindrical surface inside the metal at a radial distance r from the center. The current is flowing through the walls of this cylinder, which have surface area A = (2m) L. Thus I = JA = o E ( 2 m ) L Thus the electric field strength at radius r is

E=-

I 21roLr

Current and Conductivity

28-9

(b) For iron, with o= 1.0 x lo7Q-'m-',

E-

25 A 27r(0.10m)(l.ox10' W m - ' )

[

= 27r(0.25 ml(l.0 25 A x 10' Q-'m-')

28.43.

K L ) = 3.98 x 10" N / C 0.01 m

xi)

0.025 m = 1.59 x

N/C

Visualize:

Hollow sphere

Solve: (a) Consider a spherical surface inside the hollow sphere at a radial distance Y from the center. The current is flowing outward through this surface, which has surface area A = 4m2. Thus

I = JA = 0E(4m2) Thus the electric field strength at radius r is I E=4nm2

(b) For copper. with

0= 6.0 x

E,,

1 O7 Q-h-', =

25 A =3.32x104 N / C 4n(6 x lo' X ' m - ' ) (0.01 m)*

25 A Eo,,, = 4n(6x IO7 C ' m - ' ) (0.025 m)*

28.44.

=5.31x10" N / C

PCO

Solve: (a)

0 1 2 (b) Since I = A Q / A t , for infinitesimal changes

= d(4t- f?) = 4 - 2t

I=dt

I (d) The value of I at t = 2.0 s is zero. T h i s is because the charge, having reached its maximum value, has stopped entering the wire. The negative values of I mean that charge is flowing out of the wire.

__.- .. ..- - . .. ..-

Chapter 28

28-10

28.45.

Solve: (a) The value of Q (C) for selected values o f t are t (4 Q (C)

0 0

1 7.87

2 12.6

4 17.3

6 19.0

8 19.6

10 19.9

2okf 10

(b) Since I = AQ/At, for infinitesimal changes

(c) The maximum value of the current occurs at t = 0 s and is 10.0 A. (d) The values of I (A) for selected values o f t are

t (4 I (A)

28.46.

1 6.07

0 10

2 3.68

4 1.35

6 0.50

8 0.18

10 0.07

Solve: (a) The values of I (A) for selected values of t are ~~~

t (PI

0 2.0

I (A)

1.21

0.74

0.27

0.10

0.036

0.014

(b)Because I = dQ/dt ,

where we have used the condition Q = 0 C at t = 0 ps. (c) The values of Q (pC) for selected values of t are (P.4

Q (M3

1.57

2.52

3.46

3.80

3.92

3.98

Q (PC)

Current and Conductiviry

Q=-20e

28-1 1

kl.0mm-l

i--2.0 mm ___)/

Q=+20e

Both the rings contribute equally to the field strength. The radius of each ring is R = 1.5 mm. The left ring is negatively charged and the right ring is positively charged because 20 electrons have been transferred from the right ring to the left ring. Solve: From Chapter 26, the electric field of a ring of charge +Q at z = -1.0 mm on the axis where the axis of each ring- is k is (-1 .O x 1 0-3 m)(20 x 1.6 x 1O-I9 C) 1 E+ =k^ = (9x109 N m 2 / C Z ) 312 k^ 4Z&, ( z 2 + RZ)3'2 [(-1.0 x 1O-j m)' + (1.5 x

]

= -4.92 x 10-3k^N / C The left ring with charge -Q makes an equal contribution

E-

= - 4 . 9 2 ~ 1 0 - ~Ni / C

= E,, = E++ Z-= -9.84x 1 0 - ~I; N c The negative sign with E,, E , and E,, means these electric fields are in the -z direction. Using J = I / A = DE, the current is I = n(1.5 x 10 -3 m)'(3.5 x lo7 SX'm-')(9.84 x lo-, N/C) = 2.43 A Assess: This result is consistent with the value given in Table 26.1 for the electric field strength in a currentcarrying wire.

28.48.

Model: Because current is conserved, the current flowing in the 2.0-mm-diameter segment of the wire is the same as in the 1.O-mm-diameter segment. Visualize: Please refer to Figure P28.48. We will denote all quantities for the 1.0-mm-diameter wire with the subscript 1, and all quantities for the 2.0-mm-diameter wire with the subscript 2. Solve: Equation 28.13 is J = nev* This means the current densities in the two segments are J , = nev,,

J2 =nevd2

Dividing these equations, we get va = ( J 2 / J , ) v dBecause l. current is conserved, I , = I, = 2.0 A. So,

Assess: A drift velocity which is small and only

(a) of the drift velocity in the 1.O-mm-diameter wire is reasonable.

28.49.

Model: Because current is conserved, the current in the 3.0-mm-diameter end of the wire is the same as in the current in the 1 .O-mm-diameter end of the wire. Visualize: Please refer to Figure P28.49. We will denote all quantities for the 1.0-mm-diameter end of the wire with the subscript 1, and all quantities for the 3.0-mm-diameter end of the wire with the subscript 3. Solve: Equation 28.13 is J = nev,. This means the current densities at the two ends are

J , = nev,,

Dividing these equations, we obtainv, = ( J , / J , ) v , ,. Because current is conserved, I , = I,. So,

(

=1 7 1 4 vd, = v,, = ? ] v , , 111AI A3

= ( E r ( 0 . 5 x IO4 m / s) = 5.6 x lo6 m / s 3.0 rnm Assess: The smallness of the drift velocity is physically reasonable. Vd3

28-12

Chapter 28

28.50. Model: Because current is conserved, the currents in the aluminum and the nichrome segments of the wire are the same. Visualize: Please refer to Figure p28.50. Solve:

From Equations 28.13 and 28.18 we have J = Z/A = o E . This means

the above equation simplifies to where we used the conservation of current and A = 4 D 2 .For Emch, = Ealuminumr

28.51. Model: Because current is conserved, the currents in the two segments of the wire are the same. Visualize: Please refer to Figure P28.5 1. Solve: The currents in the two segments of the wire are related by Il = Z2= I. But I =AJ, so we have A J , = AJ2. The wire’s diameter is constant, so J , = J2 and olEl = qE2.The ratio of the electric fields is -----=E2 - 0 1

6 2

o,/o, 2

28.52. Model: Electric current is conserved. Visualize: Please refer to Figure P28.52. For the top, middle, and bottom segments the subscripts “top,” “mid,” and “bot” are used. =,,Z = 10 A. Solve: (a) Since current is conserved, Ilap= Irmd (b) The current density is J = I/A =I/&. Thus, J

= J

lo A =3 . 1 8 ~ lo6 A / m’ lr(0.001 m)*

Jmd

= n(0.0005 m)’

=1.27x107 A / m 2

(d) The drift speed is v, = h e . Thus

(vd)m’d

1.27 x 10’ A / m’

ne - (6.0 x 10” m-3)(1.60x J*d

= 1.33 x lo-’ m / s

(e) The mean time between collisions is r = rnvJeE. Thus, TIOP= Tbot =

Y ? Z ( V ~ ) ~ ~( 9 . 1 1 ~ l O - ~ kg)(3.31x104 ’

eElop

Tmld = m(v*)mid -

eE,,,

m/s)

C)(0.0909 N / C)

(1.60 x

= 2.07 x 10-l4 s

(9.11~10~~’kg)(1.33m ~ 1/ 0 s )- ~ (1.60 x

C)(0.364 N / C)

= 2.08 x 10-l4s

The collision timesme the same in all three segments because collisions are part of the microphysics of electrons inside the metal, independent of the macrophysics of fields and currents. (f) The electron current is i = NJAr = I k . Because I,= I? = I? = 10 A, lo A = 6.25 x 1019 s-’ C = = %)bo, = 1.60 x

(2)lw (2)m,d (

Current and Conductivity

Quantity I J E

TOP 10 A 3.18 x lo6 A/m 0.0909 NIC 3.31 x io4 dS 2.07 x 10-l4 6.25 x ioi9s-i

z i

Middle 10 A 1.27 x lo' Nm' 0.364 N/C 1.33 x m/s 2.07 x 10-l4 6.25 x IOl9 s-I

28-13

Bottom 10 A 3.18 x lo6A/m2 0.0909 N/C 3.31 x io4 d S 2.07 x 10-l4 6.25 x ioi9s-l

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

Q = jdQ =

Idt = (0.75 A)e-''(6houn) = (0.75 A)(-6 ho~rs)[e-"'~~""']~ O S

O S

= (0.75 A)(6 x 3600 S) = 1.62 x lo4 C The number of electrons transported is lo4 = 1.01 1.6 x 10-l9 c

1023

28.54. Model: Current is the rate at which the charge moves across a certain cross section. Solve: (a) For the circular motion of the electron around the proton, Coulomb's force between the electron and the proton causes the centripetal acceleration. Thus, 1 I-eIe --=--

mv'

4m0 r2

(9.0 x lo9 N m2 I C2)(1.60x

C)' = 6.56 x 1 0 ' ~HZ 4 8 ' ( 0 . 0 5 3 ~ 1 0 - ~m ) 1 ( 9 . 1 1 ~ 1 0 -kg) ~~

(b) Charge Q = e passes any point on the orbit once every period. Thus the effective current is Q =e = ef I =-

= (1.6 x

C)(6.56 x 10" Hz) = 1.05 x 10" A

28.55. Model: Assume that the conduction electrons are point particles. Solve: (a) The rms velocity of the conduction electrons at room temperature is obtained from the relationship + m v i , = + k , T . Wehave = 1.15 x lo5 m / s

(b) The mean free path of an electron depends on the number of atoms of copper per volume. The electron will collide with an atom if it comes within a distance r from the atom, where r is the radius of a copper atom. From Chapter 18, the mean free path is L L 1 -

A=---

Ncollis,o" (NlV)m'L

- (NIV)m2

For atoms, r = 0.5 x lo-'' m. Using N / V = 8.5 x IOz8 m-3 from Table 28.1,

28.56. Visualize:

(8.5 x lOZs)n( 0.5 x l0-lo)'

= 1.5 x

m = 1.5 nm

@ dr

The current density in the beam increases with distance from the center. We consider a thin circular shell of width d r a t a distance r from the center to calculate the current density at the edge.

28-14

Chapter 28

Solve: (a) The beam current is 1.5 mA. This means the beam transports a charge of 1.5 x number of protons delivered in one second is

C in 1 s. The

1.5~10C -~ = 9.375 x 1015 1 . 6 ~ 1 0 - l 9c (b)From J = I / A ,the current in the ring of width dr at a distance r from the center is

2m2dr dI = JdA = Jage(i ) ( 2 m d r ) = Jdge R R The total current I = 1.5 mA is found by integrating this expression:

I-

= IdI = 1.5 x

a J~~~~= (is x 10-~A)

A =Ja8e j z m 2 d r = 2 d d g e R 2 R o 3 3 2z(2.5 x

=115A/m2

28.57.

Model: The currents in the two segments of the wire are the same. Visualize: The electric fields El and i2point in the direction of the current. Establish a cylindrical Gaussian surface with end area a that extends into both segments of the wire. Charge Q, = vu \

Area A

Gaussian cylinder

Solve: (a) Because current is conserved, I , = I2 = I. The cross-section areas of the two wires are the same, so the current densities are the same: J1 = J2 = Z/A. Thus the electric fields in the two segments have strengths

The electric field enters the Gaussian surface on the left (negative flux) and exits on the right. No flux passes through the wall of cylinder, so the net flux is @e = E# - E,a. The Gaussian cylinder encloses charge Q, = qa on the boundary between the segments. Gauss's law is

Thus the surface charge density on the boundary is

(b) From the expression obtained in part (a)

a Q = (5 A)(8.85 x 10-l2 C2 / Nm')

1 (1.0 x lo7 R-Im-'

6.0 x lo7 C2-Im-l

)

= 3.68 x

Assess: This charge corresponds to a deficit of a mere 23 electrons on the boundary between the metals.