# The number density of free electrons in a copper conductor estimated in Example

Question:

The number density of free electrons in a copper conductor estimated in Example $3.1$ is $8.5 \times 10^{28} \mathrm{~m}^{-3}$. How long does an electron take to drift from one end of a wire $3.0 \mathrm{~m}$ long to its other end? The area of cross-section of the wire is $2.0 \times 10^{-6} \mathrm{~m}^{2}$ and it is carrying a current of $3.0 \mathrm{~A}$.

Solution:

Number density of free electrons in a copper conductor, n = 8.5 × 1028 m−3 Length of the copper wire, l = 3.0 m

Area of cross-section of the wire, A = 2.0 × 10−6 m2

Current carried by the wire, I = 3.0 A, which is given by the relation,

nAeVd

Where,

e = Electric charge = 1.6 × 10−19 C

$V_{d}=$ Drift velocity $=\frac{\text { Length of the wire }(l)}{\text { Time taken to } \operatorname{cover} l(t)}$

$I=n A \mathrm{e} \frac{l}{t}$

$t=\frac{n A \mathrm{e} l}{I}$

$=\frac{3 \times 8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}$

$=2.7 \times 10^{4} \mathrm{~s}$

Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 × 104 s.