Question

$62.5\times {10}^{18}$ electrons per second are flowing through a wire of area of cross-section $0.1m^2$, the value of current flowing will be:

• A. $1A$
• B. $0.1A$
• C. $10A$
• D. $0.11A$

Answer 1

The correct option is C $10A$

As we know

$I=\frac{Q}{t}$

where $Q=ne$(number of electron$\times$ charge on each electron which is $1.6\times {10}^{-19}$)

Therefore,

$I=\frac{ne}{t}=\frac{62.5\times {10}^{18}\times 1.6\times {10}^{-19}}{1}=10A$