A current of 10A exists in a wire of cross-sectional area of 5 mm2 with a drift velocity of 2 $\times$ 10$-$3 ms$-$1. The number of free electrons in each cubic meter of the wire is ___________.
Solution
$I = neA{V_d}$<br><br>$n = {I \over {eA{V_d}}}$<br><br>$$ = {{10} \over {1.6 \times {{10}^{ - 9}} \times 5 \times {{10}^{ - 6}} \times 2 \times {{10}^{ - 3}}}}$$<br><br>$= {{{{10}^{25}}} \over {16}} = 6.25 \times {10^{27}} = 625 \times {10^{25}}$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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