Question

The number of electrons per second flowing through any cross-section of the wire carrying current of $1A$ is

• A. $3.12\times {10}^{16}$
• B. $1.6\times {10}^{18}$
• C. $6.25\times {10}^{16}$
• D. $3.12\times {10}^{18}$
• E. $6.25\times {10}^{18}$

Answer 1

Answer: (C)

$6.25\times {10}^{16}$

Current $I=\frac{q}{t}$
$I=\frac{q}{1}$
$I=ne$
The number of electron
$n=\frac{I}{e}$
$=\frac{1}{1.6\times {10}^{-19}}$
$=6.25\times {10}^{18}$