CBSE Class 11-science Answered
hello mam
When a potential difference of 1.5 V is applied across a wire of length 0.2 m and area of
cross-section 0.30 mm2
, a current of 2.4 A flows through the wire. If the number density of
free electrons in the wire is 8.4 X 1028 m-3
, calculate the average relaxation time.
please solve this sum
Asked by seeni2005 | 22 Jun, 2021, 04:06: PM
Expert Answer
Electric field set up across the conductor,
E = V/t = 1.5/0.2 = 7.5 V/m
Current density in the wire,
J = I/ A = 2.4/(0.3 x 10-6)
= 8 x 106 A/m2
Current density,
J = ne2τE/m
Thus,
Average relaxation time,
τ = mJ/ ne2E
= 9.1 x 10-31 x 8 x 106/ (8.4 x 1028) x (1.6 x 10-19)2 x 7.5
= 4.51 x 10-16 s
Answered by Shiwani Sawant | 22 Jun, 2021, 05:29: PM
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