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CGP EDU Academic Team
Published on: September 12, 2026
There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is 1 mm 2 . If the number of free electrons per cm 3 is $8.4 \times 10^{22}$ , then the drift velocity would be
Text Solution
Verified by ExpertsThe correct answer is:
C
$v_d = \frac{i}{nAe} = \frac{1.344}{10^{-6} \times 1.6 \times 10^{-19} \times 8.4 \times 10^{22}} = \frac{1.344}{10 \times 1.6 \times 8.4} = 0.01 \text{ cm/s} = 0.1 \text{ mm/s}$
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