Question

A steady current of 8 mA flows through a wire. The number of electrons passing through a cross-section of the wire in 10 s is ______.

Options

• 4.0 × 10¹⁶

• 5.0 × 10¹⁷

• 1.6 × 10¹⁶

• 1.0 × 10¹⁷

Answer 1

A steady current of 8 mA flows through a wire. The number of electrons passing through a cross-section of the wire in 10 s is 5.0 × 10¹⁷.

Explanation:

Given: I = 8 mA = 8 × 10^-3 A

t = 10 s

e = 1.6 × 10^-19

We know that,

Q = ne .......(i)

Q = It .......(ii)

Equating (i) and (ii), we get

ne = It

`n xx (1.6 xx 10^-19) = 8 xx 10^-3 xx 10`

n = `(8 xx 10^-2)/(1.6 xx 10^-19)`

n = `80/(16) xx 10^-2/10^-19`

n = `5 xx 10^17`