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Grade: 12th pass
        consider a wire of length 4 m and cross sectional area of 1 mm^2 carrying a current of 2A. If each cubic metre of the material contains 10^29 electrons, find the average time taken by an electron to cross the length of wire.
9 months ago

Answers : (2)

ashish
46 Points
							
dear sandhya
as i=neavd  (vd=drift velocity)
 i/nea=vd  => 2/1029 *1.6*10-19*1*10-6*= v=>1.2* 10-4      (1mm2=10-6m2
now time taken =distance travelled/velocity  
t=l/vd   => t=4/1.2 *10-4=3.3* 104
if u have any problem in it ask to me frequently
 
8 months ago
Rahul
15 Points
							
As drift speed [(Vd)=i/nea], here i= 2A,n=10^29,
e=1.6×10^-19,a=1×10^-6 meters, so
Vd=2/10^29×(1.6×10^-19)×10^-25
So finally Vd=1.25×10^-4
Now  time =distance ÷ velocity
Distance =length = 4 meters
So time t=4/1.25×10^-4
Which is equal to 3.2×10^4 seconds 
Now convert into hours by dividing the above value with 3600
We get 8.8888....hours=8.9 hours 
By J SAI RAHUL
SRI CHAITANYA COLLEGE IPL CAMPUS
KAKINADA
 
3 months ago
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