bei48482_FM

The Solid State 351

Solution

The wire contains nfree electrons per unit volume. Each electron has the charge eand in the

time tit travels the distance dtalong the wire, as in Fig. 10.16. The number of free electrons

in the volume Adtis nAdt, and all of them pass through any cross section of the wire in the

time t. Thus the charge that passes through this cross section in tis QnAedt, and the

corresponding current is

InAed

The drift velocity of the electrons is therefore

d

From Example 9.8 we know that, in copper, n NV 8.5   1028 electrons /m^3 , and here

I1.0 A and A1.0 mm^2 1.0     10 ^6 m^2. Hence

d7.4     10 ^4 m/s

But if the free electrons have so small a drift velocity, why does an electric appliance go on as

soon as its switch is closed and not minutes or hours later? The answer is that applying a potential

difference across a circuit very rapidly creates an electric field in the circuit, and as a result all

the free electrons begin their drift almost simultaneously.

A potential difference Vacross the ends of a conductor of length Lproduces an elec-

tric field of magnitude EVLin the conductor. This field exerts a force of eEon a

free electron in the conductor, whose acceleration is

a (10.11)

When the electron undergoes a collision, it rebounds in an arbitrary direction and, on

the average, no longer has a component of velocity parallel to E. Imposing the field E

on the free electron gas in a metal superimposes a general drift on the faster but ran-

dom motions of the electron (Fig. 10.17). We can therefore ignore the electron’s mo-

tion at the Fermi velocity Fin calculating the drift velocity d.

eE



m

F



m

1.0 A



(8.5    1028 m^3 )(1.0     10 ^6 m^2 )(1.6    10 ^19 C)

I



nAe

Q



t

Area = A Volume = Avdt

vd

vdt

Figure 10.16The number of free electrons in a wire that drift past a cross-section of the wire in the

time tis nVnAdt, where nis the number of free electrons /m^3 in the wire.

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