(b) When a potential difference 1.5 V is applied across a wire of length 0.2m and area of cross section 0.30 mm, a current of 2.4 A flows through the wire. If the number density of free electrons in the wire is 8.4 x 10" ma calculate the average relaxation time.
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Answer:
I=neAv
d
⇒v
d
=
neA
I
=
9×10
28
×1.6×10
−19
×5×10
−6
1.5
=0.02mm/s
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