Discussion Forum

When 5V  potential difference is applied across a wire of length 0.1m  the drift speed of electrons is  $2.5\times 10^{-4}ms^{-1}$. If the electron density in the wire is  $8\times 10^{28}m^{-3}$  the resistivity of the material is close to Ωm

Answer:

Explanation:

According to the question

 $v_{d}=2.5\times 10^{-4}m/s$

 $\Rightarrow$         $n=8\times 10^{28}/m^{3}$  

   we know that     $J=nev_{d}$   or   $I=nev_{d}A$

 where,    symbols have their usual  meaning

 $\Rightarrow$     $\frac{V}{R}=nev_{d}A$

      or        $\frac{V}{\frac{\rho L}{A}}=nev_{d}A$

     or     $\frac{V}{{\rho L}}=nev_{d}$

   or     $\rho  =\frac{V}{nev_{d}L}$

      =    $\frac{5}{8\times 10^{28}\times1.6\times 10^{-19}\times2.5 \times 10^{-4}\times0.1}$

          or    $\rho$    =$1.6\times 10^{-5}$  Ωm