question_answer

What is the drift velocity of electrons, if the current flowing through a copper wire of 1 mm diameter is 1.1A? Assume that each atom of copper contributes one electron. (Given, density of \[Cu=9\,\,g/c{{m}^{3}}\] and atomic weight of \[Cu=63\])

A)  0.3mm/s      

B)  0.Smm/s

C)  0.lmm/s      

D)  0.2mm/s

Correct Answer: C

Solution :

 Key Idea: 63 g of copper contains \[6.02\times {{10}^{23}}\] atoms. If charge carried by free electrons passing through a section of the wire in t seconds is q, then \[i=\frac{q}{t}\] Let area of wire is A and number of free electrons per unit volume of wire is n, then the number of electrons passing per second through a cross-section of wire will be \[nA\,{{v}_{d}}\], in t seconds \[A\,{{v}_{d}}\] electrons will pass. If change on electron is e, then the charge passing through any cross-section of wire in t seconds is \[q=(n\,A\,{{v}_{d}}\,t)\times e\] \[\therefore \] \[i=\frac{q}{t}=ne\,A{{v}_{d}}\] \[n=\frac{Avogadro\,s\,\,number}{Volume\text{ }of\text{ }63g\text{ }of\text{ }copper}\] \[=\frac{6.02\times {{10}^{23}}}{63/9}=\frac{6.02\times {{10}^{23}}}{7}/c{{m}^{3}}\] \[n=\frac{6.02\times {{10}^{29}}}{7}/{{m}^{3}}\] Area \[A=\pi \,{{r}^{2}}=\pi \,{{(0.5\times {{10}^{-3}})}^{2}}{{m}^{2}}\] Drift velocity\[{{V}_{d}}=\frac{i}{n\,eA}\] \[=\frac{1.1\times 7}{6.02\times {{10}^{29}}\times 1.6\times {{10}^{-19}}\times \pi \,{{(0.5\times {{10}^{-3}})}^{2}}}\] \[=\frac{7.7}{7.56}\times {{10}^{-4}}m/s\] \[\approx 0.1\,mm/s\].