Question

Current flowing through a cylindrical conductor of area of cross-section 1 mm2 is 4 A. If number of free electrons per m3 is 1029. Then drift velocity of electrons is

Answer 1

Answer:

The drift velocity of electrons is 2.5×10⁻⁴m/s.

Explanation:

This is the formula-based question.

First, let's see the relation between current and drift velocity,

$I=neAv_d$         (1)

Where,

I=the current flowing

n=density of electron

A=area of a cross-section through which the current is flowing

e=charge on an electron=1.6×10⁻¹⁹C

vd=drift velocity

From the question we have,

A=1mm²=1×10⁻⁶m²             (∴1mm=10⁻³)

I=4A

n=10²⁹

By substituting the value of I,n,e, and A in equation (1) we get;

$4=10^2^9\times 1.6\times 10^-^1^9\times 10^-^6\times v_d$

$v_d=\frac{4}{10^2^9\times \times 1.6\times 10^-^1^9\times 10^-^6}=2.5\times 10^-^4m/s$

Hence, the drift velocity of electrons is 2.5×10⁻⁴m/s.