Question

The number of electrons passing per second through a cross section of copper wire carrying 10${}^{-6}$ ampere current per one second is found to be:

• A. $1.6\times {10}^{19}$
• B. $6\times {10}^{35}$
• C. $6\times {10}^6$
• D. $6\times {10}^{12}$

Answer 1

Answer: (D)

$6\times {10}^{12}$

$Current=\frac{charge}{time}$

$\frac{\text{Number of electrons}\times 1.6\times {10}^{-19}}{1sec}={10}^{-6}$

$\therefore \text{Number of electrons}=\frac{{10}^{13}}{1.67}$

$\text{Number of electrons}=6\times {10}^{12}$

Hence, option D is correct.