Question

oss-section of the wire? (e=1.60×10−19 C) A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cross-section of the wire? ( C) 1.5×1023 1.6×1018 6.3×1015 1.6×1017 3.7×1015

Answer 1

Answer:

(1.6 × 10¹⁸) /s

Explanation:

Current flowing in a wire is given by

I = (Q/t)

where Q = total charges on the electrons flowing in the wire

t = time

But Q = nq

where n = number of electrons flowing in the wire

q = charge on one electron = 1.602 × 10⁻¹⁹ C

So, I = nq/t

(n/t) = (I/q)

(n/t) = number of electrons per second, for any cross sectional Area.

(n/t) = (I/q)

I = 260 mA = 0.26 A

q = 1.6 × 10⁻¹⁹ C

(n/t) = (0.26/(1.602×10⁻¹⁹)) = (1.62 × 10¹⁸) /s = (1.60 × 10¹⁸) /s