question_answer

A uniform copper wire of length 1 m and cross-sectional area 8 x 10-7m2 carries a current of 1 A. Assuming that there are \[8\times {{10}^{28}}\]free electron/ m3 in copper, how long will an electron take to drift from one end of the wire to the other?

A)  \[0.8\times {{10}^{3}}s~\]

B)  \[1.6\times {{10}^{3}}s\]

C)  \[3.2\times {{10}^{3}}s\]

D)  \[6.4\times {{10}^{3}}s\]

Correct Answer: D

Solution :

Consider a conductor of length \[l\] and of uniform area of cross-section A. \[\therefore \] Volume of the conductor \[=Al\] If n is the number of free electrons per unit volume of the conductor, then total number of free electrons in the conductor \[=A\ln \]. Ifeis the charge on each electron, then total charge on all the free electrons in the conductor, \[q=A\operatorname{lne}\]. Let a constant potential difference V is applied across the ends of the conductor with the help of a battery. The electric field set up across the conductor is given by \[E=V/l\] Due to this field, the free electrons present in the conductor will begin to move with a drift velocity \[{{v}_{d}}\] towards the left hand side as shown in figure. Therefore, time taken by the free electrons to          cross the conductor, \[t=\frac{l}{{{v}_{d}}}\] Hence, current \[i=\frac{q}{t}=\frac{A\ln e}{l/{{v}_{d}}}\] or \[i=Ane{{v}_{d}}\] Here, \[i=1A,\,n=8\times {{10}^{28}}electron/{{m}^{3}}\] \[A=5\times {{10}^{-7}}{{m}^{2}}\] \[\Rightarrow \] \[1=8\times {{10}^{28}}\times 1.6\times {{10}^{-19}}\times 5\times {{10}^{-7}}\times {{v}_{d}}\] or \[{{v}_{d}}=\frac{1}{8\times {{10}^{28}}\times 1.6\times {{10}^{-19}}\times 5\times {{10}^{-7}}}\] Now, \[t=\frac{l}{{{v}_{d}}}\] \[=8\times {{10}^{28}}\times 1.6\times {{10}^{-19}}\times 5\times {{10}^{-7}}\] \[=64\times {{10}^{2}}\] \[=6.4\times {{10}^{3}}s\]