Current Electricity Charging Discharging Of Capacitors Question 353
Question: There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is 1 $ mm^{2} $ . If the number of free electrons per $ cm^{3} $ is $ 8.4\times 10^{22} $ , then the drift velocity would be
[CPMT 1990]
Options:
A) 1.0 $ mm/sec $
B) 1.0 $ m/sec $
C) 0.1 $ mm/sec $
D) 0.01 $ mm/sec $
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Answer:
Correct Answer: C
Solution:
$ v _{d}=\frac{i}{nAe}=\frac{1.344}{{{10}^{-6}}\times 1.6\times {{10}^{-19}}\times 8.4\times 10^{22}} $
$ =\frac{1.344}{10\times 1.6\times 8.4}=0.01cm/s=0.1mm/s $
BETA
