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`        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.`
one year ago

```							dear sandhyaas i=neavd  (vd=drift velocity) i/nea=vd  => 2/1029 *1.6*10-19*1*10-6*= vd =>1.2* 10-4      (1mm2=10-6m2) now time taken =distance travelled/velocity  t=l/vd   => t=4/1.2 *10-4=3.3* 104if u have any problem in it ask to me frequently
```
one year ago
```							As drift speed [(Vd)=i/nea], here i= 2A,n=10^29,e=1.6×10^-19,a=1×10^-6 meters, soVd=2/10^29×(1.6×10^-19)×10^-25So finally Vd=1.25×10^-4Now  time =distance ÷ velocityDistance =length = 4 metersSo time t=4/1.25×10^-4Which is equal to 3.2×10^4 seconds Now convert into hours by dividing the above value with 3600We get 8.8888....hours=8.9 hours By J SAI RAHULSRI CHAITANYA COLLEGE IPL CAMPUSKAKINADA
```
one year ago
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