Question

A gold wire has a 0.30 mm diameter cross section. Opposite ends of this wire are connected to the terminals of a 1.5 V battery. If the length of the wire is 5.5 cm, how much time, on average, is required for electrons leaving the negative terminal of the battery to reach the positive terminal? Assume the resistivity of gold is 2.44 10-8 Ω·m.

Answer 1

Answer :

The time is 0.46 sec.

Explanation:

Given that,

Diameter = 0.30 mm

Voltage = 1.5 V

Length = 5.5 cm

Resistivity of gold $\rho= 2.44\times10^{-8}\ \Omega m$

We know that,

The resistance is

$R=\dfrac{\rho L}{A}$...(I)

The current is

$I=\dfrac{V}{R}$...(II)

Put the value of R

$I=\dfrac{V}{\dfrac{\rho L}{A}}$

$I=\dfrac{VA}{\rho L}$

We need to calculate the drift velocity

Using formula of drift velocity

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

$v_{d}=\dfrac{\dfrac{VA}{\rho L}}{neA}$

$v_{d}=\dfrac{V}{ne\rho L}$

Put the value into the formula

$v_{d}=\dfrac{1.5}{5.90\times10^{28}\times1.6\times10^{-19}\times5.5\times10^{-2}\times2.44\times10^{-8}}$

$v_{d}=11.8\times10^{-2}\ cm/s$

We need to calculate the time

Using formula of time

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

Put the value into the formula

$t=\dfrac{5.5}{11.8\times10^{-2}}$

$t=0.46\ sec$

Hence, The time is 0.46 sec.