Question

A current of $5$A is passing through a metallic wire of length $5$m and cross-sectional area of ${10}^{-5}{m}^2$. If the density of the electrons in the wire is $6.25\times {10}^{26}{m}^3$, find the time taken by the electrons to travel from one end of the wire to the other.

Answer 1

Current flowing through the wire, $I=5A$

Length of wire, $l=5m$

Cross-section area of wire, $A={10}^{-5}{m}^2$

Electron density, $n=6.25\times {10}^{26}$ per $m^3$

Using: $I=Ane{v}_d$ where $v_d=\frac{l}{t}$

We get: $I=Ane\frac{l}{t}$

Thus, time taken by electron to cross the wire, $t=\frac{Anel}{I}$

$\therefore$ $t=\frac{\left({10}^{-5}\right)(6.25\times {10}^{26})(1.6\times {10}^{-19})\left(5\right)}{5}=1000s$