Question

The drift velocity of electrons in silver wire with cross-sectional area $3.14\times {{10}^{-6}}{{m}^{2}}$ carrying a current of 20A is. Given the atomic weight of Ag=108, density of silver$=10.5\times {{10}^{3}}kg/{{m}^{3}}$.
A. $2.798\times {{10}^{-4}}m/\sec $
B. $67.98\times {{10}^{-4}}m/\sec $
C. $0.67\times {{10}^{-4}}m/\sec $
D. $6.798\times {{10}^{-4}}m/\sec $

Answer 1

Hint: As a first step, we could note down all the given values from the question. Then you could recall some expression for drift velocity in terms of these given quantities. Then, you could simply substitute all these values into the above expression and thus you could arrive at the answer.
Formula used:
Drift velocity,
${{v}_{d}}=\dfrac{I}{neA}$

Complete answer:
In the question, we are given area of cross section of a silver wire as $3.14\times {{10}^{-6}}{{m}^{2}}$ and the current passing through the wire would be 20A. The atomic weight of silver is given as 108g and its density is found to be$10.5\times {{10}^{3}}kg{{m}^{-3}}$. We are supposed to find the drift velocity of the electrons in this wire with the given information.
We know that number of electrons present in one kg of silver can be given by,
$N=\dfrac{6.022\times {{10}^{26}}}{108}$
Now, we are given the mass density of silver as $10.5\times {{10}^{3}}kg{{m}^{-3}}$ and its number density could be given by,
$n=\dfrac{6.022\times {{10}^{26}}}{108}\times 10.5\times {{10}^{3}}$
Now, we have relation for drift velocity in terms of current flowing through the wire, number density and charge of the electron as,
${{v}_{d}}=\dfrac{I}{neA}$
Substituting the given values we get,
$\Rightarrow {{v}_{d}}=\dfrac{20\times 108}{6.022\times {{10}^{26}}\times 10.5\times {{10}^{3}}\times 1.6\times {{10}^{-19}}\times 3.14\times {{10}^{-6}}}$
$\therefore {{v}_{d}}=6.798\times {{10}^{-4}}m{{s}^{-1}}$
Therefore, we found the drift velocity of the electrons to be, ${{v}_{d}}=6.798\times {{10}^{-4}}m{{s}^{-1}}$.

So, the correct answer is “Option D”.

Note: Though we have a number of physical quantities whose values are directly given in the question we do not have the ones that could be directly substituted. Thus we have found the number density from the given values. Later we have substituted the same so as to get the answer.