Q. $62.5 \times 10^{18}$ electrons per second are flowing through a wire of area of cross-section $0.1\ m^{2},$ the value of current flowing will be
A) 1 A
B) 0.1 A
C) 10 A
D) 0.11 A
Explanation
(c) $i = \frac{ne}{t} = \frac{62.5 \times 10^{18} \times 1.6 \times 10^{- 19}}{1} = 10\ ampere$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.