Question

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? (e = 1.60 × 10-19 C)

Answer 1

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

$I=\dfrac{q}{t}$

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

$n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}$

n=162500000 t

$\dfrac{n}{t}=16250000$

The number of electron passe per second will be 162500000.