Question

The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a copper wire. (This is a typical value for a simple circuit.) The diameter of the wire is 1.2 mm. In this case, about how many minutes would it take for a single electron in the electron sea to drift from one end to the other end of a wire 31 cm long?

Answer 1

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density $= 8.4\times10^{28}\ m^3$

Current $i = 4.6\times10^{18}\ electron/s$

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

$v_{d}=\dfrac{I}{neA}$

$v_{d}=\dfrac{Ne}{tne\times\pi r^2}$

Put the value into the formula

$v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}$

$v_{d}=4.842\times10^{-5}\ m/s$

We need to calculate the time

Using formula for time

$v_{d}=\dfrac{l}{t}$

$t=\dfrac{l}{v_{d}}$

Where, l = length

$v_{d}$ = drift velocity

Put the value into the formula

$t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}$

$t=6402.31\ sec$

$t=106.7\ minute$

Hence, The time is 106.7 minute.