Q. In a closed circuit, the current $I$(in ampere) at an instant of time $t$(in second) is given by $I = 4 - 0.08t.$ The number of electrons flowing in 50s through the cross-section of the conductor is
A) $1.25 \times 10^{19}$
B) $6.25 \times 10^{20}$
C) $5.25 \times 10^{19}$
D) $2.55 \times 10^{20}$
Explanation
(b) $I = 4 - 0.08t\ A$
Or $\frac{dq}{dt} = 4 - 0.08t\ A$
$or\ \ q = \int_{0}^{50}{(4 - 0.08t)dt\ C}\ $
$or\ Ne = \left\lbrack 4t - \frac{0.08t^{2}}{2} \right\rbrack_{0}^{50} = 100\ C$
Where N is number of electrons.
or $N = \frac{100}{e} = \frac{100}{1.6 \times 10^{- 19}}$
$\ \ \ \ \ \ \ \ \ \ \ = 6.25 \times 10^{20}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.