Question

A $1.5m$ long wire is made of a metal with the same electron density as copper. The wire is connected across the terminals of a $9.0\mathrm{V}$ battery. What conductivity would the metal need for the drift velocity of electrons in the wire to be $60\mathrm{mph}$? By what factor is this larger than the conductivity of copper?

Short answer

The factor the given metal is larger than the conductivity of copper is$6.07\times {10}^{10},1000times.$

Step-by-step solution

Given Information 

The given metal wire length$=1.5m$

Voltage$=9.0V$

Drift velocity$=60mph$

Explanation 

Find the conductance as

$J=\sigma E\Rightarrow \sigma =\frac{J}{E}$

Density of electrons, elementary charge, and drift speed are used in calculating current density,,

$J=e{n}_e{v}_d$

The electric field inside a wire of length $L$under potential difference $U$is,

$E=\frac{U}{L}$

We find that by substituting these two expressions at the conductance expression, we get

$\sigma =\frac{e{n}_e{v}_{d}L}{UJ}$

Here, $60mph=26.8m/s$

Substitute the values,

role="math" localid="1649139392068" $\sigma =\left(\frac{1.6\times {10}^{-19}C\times 8.5\times {10}^{28}e\times 26.8m/s\times 1.5m}{9V}\right)$

$\sigma =6.07\times {10}^{10}\mathrm{S}.$

Final Answer

This value is 1000 times greater than the value of copper,${\sigma }_{Cu}=6\times {10}^7\mathrm{s}.$