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CBSE Class 12 Physics 2014 Outside Delhi Set 2 Paper

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Question : 5 of 9
Marks: +1, -0
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5×10−7m22.5 \times 10^{-7} \mathrm{m}^2 carrying a current of 1.8A1.8 \mathrm{A}. Assume the density of conduction electrons to be 9×10289 \times 10^{28} m−3\mathrm{m}^{-3}.
Solution:  
Given:     \;\; Current (l)=1.8A(l)=1.8 \mathrm{A}
Charge density (n)=9×1028m−3(n)=9 \times 10^{28} \mathrm{m}^{-3}
Cross-section area (A)=2.5×10−7m2(A)=2.5 \times 10^{-7} \mathrm{m}^2
Charge of electron (q)=1.6×10−19C(q)=1.6 \times 10^{-19} \mathrm{C}
vd  =  InAq.v_d \;=\; \frac{I}{n A q} .
  =  1.89×1028×2.5×10−7×1.6×10−19\;=\; \frac{1.8}{9 \times 10^{28} \times 2.5 \times 10^{-7} \times 1.6 \times 10^{-19}}
  =0.0005m/s\;=0.0005 \text{m/s}
  =0.5mm s−1\;=0.5 \mathrm{mm}\,\mathrm{s}^{-1}
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