Question

The number density of free electrons in a copper wire conductor is 8.5 x 10^8 m? How long does an electron take to drift from one end of a wire of 2 m long to its other end? The area of cross section of the wire is 2.0 x 10 m and it is carrying a current of 3.0A. ( e = 1.6 x 10").​

Answer 1

Explanation:

number of density of electrons,

$n = 8.5 \times {10}^{8}$

current carried by the wire,

$i = 3.0a$

area of cross-section of the wire,

$a = 2.0 \times 10m$

length of the wire,

$l = 2m$

charge on electron,

$e = 1.6 \times {10 }^{11}$

using the formula for drift velocity,

$vd = \frac{i}{nea}$

$= \frac{3.0}{8.5 \times {10}^{8} \times 1.6 \times {10}^{11} \times 2.0 \times 10}$

$= 1.1029 \times {10}^{ - 21}$

therefore,

time taken by an electron to drift from one end to another end is

$t = \frac{l}{vd}$

$= \frac{2}{1.1029 \times {10}^{ - 21} }$

$= 1.813 \times {10}^{21} sec$