An electron takes 40 × 10 3   s to drift from one end of a metal wire of length 2   m to its other end. The area of cross-section of the wire is 4   mm 2 and it is carrying a current of 1 . 6   A . The number density of free electrons in the metal wire is

Question

An electron takes 40 &#215; 10 3 &#160; s to drift from one end of a metal wire of length 2 &#160; m to its other end. The area of cross-section of the wire is 4 &#160; mm 2 and it is carrying a current of 1 . 6 &#160; A . The number density of free electrons in the metal wire is

Aicharya · Accepted Answer

From the given data, drift velocity can be written as,
$v = \frac{2 m}{40 \times 10^{3} s} = \frac{1}{2 \times 10^{4}} m s^{- 1}$
Now for current we can write,
$I = n e A v \Rightarrow n = \frac{I}{e A v} \Rightarrow n = \frac{1.6}{(1.6 \times 10^{- 19}) (4 \times 10^{- 6}) (\frac{1}{2 \times 10^{4}})} = 5 \times 10^{28} m^{- 3}$

An electron takes 40 × 10 3   s to drift from one end of a metal wire of length 2   m to its…

Solution

An electron takes 40 &#215; 10 3 &#160; s to drift from one end of a metal wire of length 2 &#160; m to its…

Solution

An electron takes 40 × 10 3 s to drift from one end of a metal wire of length 2 m to its…