Recall from Section 22.3 that a metal consists of a lattice of positively charged ions through which electrons can move relatively freely. Although the lattice positions are fixed (meaning there is no motion of ions through the metal), the ions vibrate around their lattice positions, as if they were connected to those po- sitions by springs. (The amplitude of the vibrations depends on temperature.) The ions consist of atomic nuclei surrounded by most of their electrons; the out- ermost one or more electrons of each atom are free to move through the entire lattice. Because of their thermal energy, these free electrons move through the lattice at very high speeds (105 m>s at room temperature, or about 0.1% of the speed of light), and they move in straight lines without any change in energy or momentum except in the instants when they collide with the ions.
In the absence of an electric field and over a time interval long enough for many collisions to take place, the displacement of an electron is very small in spite of its high speed (Figure 31.30a). This is true because each collision changes the direction of the electron’s motion, randomizing the direction of the electron’s velocity. Consequently, the average velocity of all the electrons is zero. However, when an electric field is applied, as in Figure 31.30b, the electric field causes the electrons to accelerate in the direction opposite to the electric field.
A quantitative description of the motion of charge carriers in a conductor was given by P. K. Drude in 1900, shortly after the discovery of the electron. His model (called the Drude model) applies remarkably well to metals. Let’s describe the motion of the electrons in this model. In the presence of a uniform electric field ES, the electrons are subject to a force -eES and therefore have an accelera- tion aS= -eES>me. For any one electron moving along a straight path between two consecutive collisions, the electron’s final velocity vSf just before the second collision is
vSf=vSi+aS∆t=vSi− eES
me∆t, (31.1)
where vSi is the electron’s initial velocity on that path (its velocity just after the first collision), ∆t is the time interval the electron spends on that path (in other words, the time interval between collisions), e is the elementary charge (Eq. 22.3), ES is the applied electric field, and me is the mass of the electron. The magnitude of vSi is roughly 105 m>s, as noted above, and the direction is determined by the first of the two collisions. Because of the high electron speed, the time interval
∆t between collisions is extremely short—on the order of 10-14 s.
To calculate the average velocity of all the electrons, we take the average of Eq. 31.1 for all of the electrons:
(vSf)av=(vSi)av− eES
me (∆t)av. (31.2) (I have assumed here that the electric field is either constant over time or takes a time interval much longer than ∆t to change significantly.)
Even though the magnitude of vSi is quite large, its average value for all electrons is zero because the collisions produce a random distribution of the directions of the initial velocities. The resulting average velocity, called the drift velocity vSd of the electrons, is thus
vS
d=-eES
met, (31.3)
where tK(∆t)av is the average time interval between collisions. (The value of t depends on the number density, size, and charge of the lattice ions, and on
Figure 31.30 The effect of an applied electric field on the motion of a free electron through a lattice of ions.
ES
S S
E = 0 (a) Motion in absence of an electric field
(b) Motion with applied electric field start
start end
end
Electron’s displacement is zero over long time interval.
Electron undergoes displacement in direction opposite to electric field.
ion
31.5 resistanCe and ohm’s law 843
Quantitative tools
temperature.) The magnitude of the drift velocity is called the drift speed. Equa- tion 31.3 shows that the drift velocity of the electrons is in a direction opposite that of the electric field, and the drift speed is proportional to the electric field magnitude.
31.11 (a) Does the electric field do work on the electrons of Figure 31.30b as they accelerate between collisions? (b) On average, does the kinetic energy of the electrons increase as they drift through the lattice? (c) What do your answers to parts a and b imply about the energy in the lattice?
In Chapter 27 we found that the current in a conductor can be expressed in terms of the speed v of the charge carriers (Eq. 27.16). Because the speed of the charge carriers is what we are now calling the drift speed, we can write Eq. 27.16 in the form
I=nAqvd, (31.4)
where A is the cross-sectional area of the conductor, q is the charge on each charge carrier in the conductor, and n is the number of carriers per unit vol- ume in the conductor. Because of the current’s dependence on cross-sectional area, it is convenient to introduce the current per unit area, called the current density, whose magnitude is given by the magnitude of the current per unit area:
JK I
A =nqvd. (31.5)
Because the drift velocity is a vector, the current density is a vector, too:
SJ
=nqvSd. (31.6)
The direction of the current density is the same as that of the drift velocity for positive charge carriers and opposite the direction of the drift velocity for nega- tive charge carriers. Therefore the current density is always in the same direction as the current. The SI unit of current density is A>m2.
Substituting the absolute value of the right side of Eq. 31.3 for vd and e for q in Eq. 31.5 yields
J=n(e) aeE
metb =ne2t
me E. (31.7)
Equation 31.7 shows that the current density is proportional to the applied electric field. The proportionality constant s is called the conductivity of the material of which the conductor is made:*
sK J
E. (31.8)
*Note that conductivity is not the same as surface charge density, which is represented by the same symbol.
Quantitative tools
The SI unit of conductivity is equal to (A>m2)>(V>m)=A>(V#m). The conduc- tivity is a measure of a material’s ability to conduct a current for a given applied electric field. The conductivity is a property of the material and is therefore the same for any piece of that material you might choose.
table 31.1 gives the conductivities of some common materials at room temperature. Note that the first four materials, all metals, have very similar conductivities. Nichrome is an alloy of nickel and chromium used in heating elements because of its relatively low conductivity. Silicon is a semiconductor;
its conductivity is intermediate between that of an electrical conductor and that of an electrical insulator. The conductivities of insulators (such as pure water and glass) are many orders of magnitude smaller than those of metals. Seawater has a greater conductivity than pure water because the ions dissolved in seawater serve as charge carriers for a current, just as the ions in the electrolyte of a bat- tery do.
Comparing Eqs. 31.7 and 31.8 gives us an expression for the conductivity of a metal in terms of the average time interval between collisions, the mass and charge of the electrons, and the number density of electrons present in the material:
s=ne2t
me . (31.9)
Because of the temperature dependence of t, the conductivity σ depends on temperature and, for some materials, on the magnitude of the current in the ma- terial, but it is independent of the shape of the piece of material in which the current density is measured.
We obtained Eq. 31.9 for a metal, in which the charge carriers are electrons, but it applies to any system in which the charge carriers in a conducting material move freely between collisions. To generalize Eq. 31.9 for charge carriers other than electrons, simply substitute q2 for e2, and mq for me, where q is the charge on the carriers and mq is the mass of one carrier.
The conductivity describes how large a current density and hence current are created by an external electric field. Such an electric field is produced by ap- plying a potential difference across a material. In Section 31.2, we looked at resistance as a way of relating the applied potential difference V across a circuit element to the resulting current in the element. In general, the resistance of any element is defined to be
RKV
I. (31.10)
table 31.1 Conductivities of various materials at room temperature A,(V~m)
Conductors Silver 6.3×107
Copper 5.9×107
Aluminum 3.6×107
Tungsten 1.8×107
Nichrome 6.7×105
Carbon 7.3×104
Semiconductors Silicon 4×10-4
Germanium 2
Poor conductors Seawater 4
Insulators Pure water 4.0×10-6
Glass 10-12
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Quantitative tools
The derived SI unit of resistance is the ohm (1 ΩK1 V>A). Resistance is always positive, so in Eq. 31.10 the direction in which V and I are measured must be such that they both have the same algebraic sign. The resistance of most circuit elements is typically in the range of 10 Ω to 100,000 Ω.
The concept of resistance is useful in describing conductors and other objects that provide a continuous conducting path for charge carriers, such as filaments, bulbs, and resistors. (It is not especially useful for other types of circuit elements, such as capacitors.) For some conducting materials, the resistance R of a piece of this material is fixed at a given temperature. For these materials, Eq. 31.10 indi- cates that the current in the conductor is proportional to the potential difference across it and inversely proportional to the resistance:
I=V
R. (31.11)
Such materials are said to be ohmic. If we plot the magnitude of the current in a piece of this material as a function of the potential difference across it, the result is a straight line whose slope is equal to the inverse of the resistance R of the piece (Figure 31.31).
Equation 31.11 is often referred to as Ohm’s law. It is important to keep two things in mind about Eq. 31.11. First, many materials and many circuit elements are not ohmic. For such materials and elements, a plot of current as a function of applied potential difference is not a straight line, but has a more complicated shape, indicating that R depends on the potential difference. Second, Eq. 31.11 is really a definition, not a law. It simply amounts to the observation that in certain materials, the current is proportional to the applied potential difference.
In this chapter, we concern ourselves only with ohmic materials and circuit elements.
Let’s now relate the resistance of a conductor to the conductivity of the ma- terial of which it is made. From Checkpoint 31.8 we know that the magnitude of the potential difference across a wire of length / is given by (Eq. 25.25 and Example 25.5):
V =E/. (31.12)
Combining Eqs. 31.5 and 31.8, we see that JK I>A=sE, so E= I>(sA).
Substituting this expression into Eq. 31.12 and dropping the absolute-value symbol, we obtain a relationship between V and I:
V= I
sA /=I /
sA. (31.13)
Substituting this expression into the definition of resistance, Eq. 31.10, we obtain for ohmic conductors,
R= /
sA. (31.14)
Equation 31.14 shows that the resistance of a conductor not only depends on the material from which it is made (through the conductivity s) but also is propor- tional to the length / of the conductor and inversely proportional to the cross- sectional area, as we had concluded in Section 31.4.
Figure 31.31 Current versus applied potential difference for an ohmic conductor. The current is proportional to the potential difference.
V slope = 1>R
I
00
Quantitative tools
example 31.7 Drifting electrons
Consider a piece of copper wire that is 10 m long and has a diameter of 1.0 mm. The number density of free electrons in copper is 8.4×1028 electrons>m3. If the wire carries a current of 2.0 A, what are (a) the magnitude of the potential difference across the wire, (b) the drift speed of the electrons in the wire, and (c) the average time interval between collisions for the electrons?
❶ GettinG Started Ohm’s law relates the potential differ- ence across the wire to the current in it and its resistance. Even though the wire is made of copper, it has a finite resistance, which depends on the conductivity of copper and on the length and cross-sectional area of the wire. This resistance is caused by collisions between the electrons and copper ions in the wire, and these collisions limit the drift speed of the electrons through the wire. The drift speed is related to the average time interval between collisions, which is one of the parameters I need to calculate and appears in the expression for conductivity in the Drude model, Eq. 31.3, vd=eEt>me.
❷ deviSe plan The potential difference across the wire is re- lated to the current by Eq. 31.11, but in order to use this equa- tion, I must determine the wire’s resistance, which is given by Eq. 31.14. I can look up the conductivity of copper in Table 31.1 (s=5.9×107 A>V#m). To obtain the drift speed of the elec- trons, I can use Eq. 31.4, which contains the charge on the electron (q=e=1.6×10-19 C), the current (given), the number den- sity of the electrons (given), and the cross-sectional area of the wire, which I can calculate from the given diameter. To obtain the average time interval between collisions, I can use Eq. 31.9, which contains the mass of the electron (me=9.11×10-31 kg).
❸ execute plan (a) I obtain the resistance of the wire from Eq. 31.14 and the cross-sectional area of the wire, A=pr2 = p(5.0×10-4 m)2=7.9×10-7 m2:
Rwire= /
sA= 10 m
(5.9×107 A>V#m)(7.9×10-7 m2)=0.21 Ω.
Now that I have Rwire, I can obtain Vwire from Eq. 31.11:
Vwire=IRwire=(2.0 A)(0.21 Ω)=0.42 V. ✔
(b) I first obtain the drift speed of the electrons from Eq. 31.4:
vd= I
neA= 2.0 C>s
(8.4×1028 m-3)(1.6×10-19 C)(7.9×10-7 m2)
=1.9×10-4 m>s. ✔
(c) Solving Eq. 31.9 for t, I get t=mes
ne2 =(9.11×10-31 kg)(5.9×107 A>V#m) (8.4×1028 m-3)(1.6×10-19 C)2 =2.5×10-14 s. ✔
❹ evaluate reSult Because the conductivity of copper is high, it makes sense that I obtain a small resistance and conse- quently a small potential difference across the wire even though it is 10 m long. My answer to part b shows that the drift speed of the electrons in the wire is very small. The drift speed indi- cates that it takes the electrons about 5 s to move just 1 mm, even though I know from experience that when I turn on a light switch, the light turns on instantly even if the bulb is meters away from the switch. The reason the light bulb comes on almost instantaneously, however, is that the current is the same through- out the circuit—all of the electrons throughout the circuit are set in motion almost simultaneously when I flip the switch.
How can my very small calculated value for drift speed be reasonable when the current is a significant 2.0 A=2.0 C>s at any given location in the wire? The reason is that the number density of electrons in the wire is very high. Although it takes 5 s for an electron to move 1 mm, there are nearly 1020 electrons per cubic millimeter of the wire. This means that on the order of 1020 electrons, or 10 coulombs of charge, pass a given loca- tion in 5 s—hence the current is 2 C>s, or 2 A.
The very high number density of the electrons is also responsible for the extremely short time interval between col- lisions. If I imagine the 1020 electrons per cubic millimeter arranged on a cubic lattice where each cube of the lattice has a side length of 1.0 mm, there are about 107 electrons along each side of the cube, and so the average distance between them is about 10-10 m. As the electrons move at about 105 m>s (see Section 31.5), they cover this average distance in about 10-15 s, which is within an order of magnitude of what I obtained.
31.12 If the temperature of a metal is raised, the amplitude of the vibrations of the metal-lattice ions increases. (a) What effect, if any, do you expect these greater vibrations to have on the resistance of a piece of that metal? (b) What effect does running a current in a metal have on the temperature of the metal? (c) Make a graph of current versus potential difference, taking into account the effect you described in part b.