i=10A,A=5mm2=5×10−6m2 and vd=2×10−3ms−1
We know, i=neAvd
∴10=n×1.6×10−19×5×10−6×2×10−3
⇒n=0.625×1028=625×1025
JEE Main 2021 — Physics Electromagnetism
A current of 10A exists in a wire of cross-sectional area of 5mm2 with a drift velocity of 2×10−3ms−1. The number of free electrons in each cubic meter of the wire is
Held on 17 Mar 2021 · Verified 6 Jul 2026.
2×106
625×1025
2×1025
1×1023
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