During the 4.0 min a 5.0 A current is set up in a wire, how many (a) coulombs and (b) electrons pass through any cross section across the wire’s width?
(a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s,
q = it = (5.0 A)(240 s) = 1.2 × 10³ C.
(b) The number of electrons N is given by q = Ne, where e is the magnitude of the charge on an electron. Thus,
N=q / e=(1200 \,C ) /\left(1.60 \times 10^{-19} \,C \right)=7.5 \times 10^{21} .