Current Electricity Charging Discharging Of Capacitors Question 404
Question: There is a current of 40 ampere in a wire of $ {{10}^{-6}},m^{2} $ area of cross-section. If the number of free electron per $ m^{3} $ is $ 10^{29} $ , then the drift velocity will be
[Pb. PMT 2001]
Options:
A) $ 1.25\times ~10^{3}m/s $
B) $ 2.50\text{ }\times ~{{10}^{-3}}m/s $
C) $ 25.0\text{ }\times ~{{10}^{-3}}m/s $
D) $ 250\text{ }\times ~{{10}^{-3}}m/s $
Show Answer
Answer:
Correct Answer: B
Solution:
$ V _{d}=\frac{i}{neA}=\frac{40}{10^{29}\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}} $ = $ 2.5\times {{10}^{-3}}m/sec $ .
BETA
