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

A)2.5×10-4 m/s

B)25 x 10-4 m/s

C)1.25 x 10-3 m/s

D)2.5 x 10-³ m/s

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Answer 1

Answer:

a is the correct answer 2.5×10 to the power-4

Answer 2

Answer:

The drift velocity of electrons is $2.5\times 10^-^4$m/s.i.e.option(A).

Explanation:

First, we need to understand the relationship between current and drift velocity.

The current in terms of drift velocity is given as,

$I=neAV_d$             (1)

Where,

I=current flowing through the cross-section

n=number of free electrons

e=charge on an electron

A=cross sectional area

Vd=drift velocity

From the question we have,

I=4A

A=1mm²=1×10⁻⁶m²

n=10²⁹

e=1.6 ×10⁻¹⁹C

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

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

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

Hence, the drift velocity of electrons is $2.5\times 10^-^4$ m/s.i.e.option(A).