Question

A wire of area of cross-section A is made of metal P and carries current I. If density of the wire is p, mass of an atom of metal P is m and each atom of P contributes one free electron then the correct expression for drift speed of the free electrons is [e = charge of an electron]​

Answer 1

The expression is:

$\boxed{v_d=\frac{Im}{\rho g eA N_A}}$

Explanation:

We know that

$I=neAv_d$

Where

A is the area of cross section

n is electon density

I is current

$v_d$ is drift speed

Since each atom of P contributes one free electron

Therefore, n will be equal to number of atoms divided by volume

Thus,

$n=\frac{\text{mass of metal P}\times g}{\text{Atomic weight\times V}}\times N_A$              (where $N_A$ is Avogadro's number)

$\implies n=\frac{\rho g}{m}N_A$    

Therefore,

$v_d=\frac{I}{neA}$

$\implies v_d=\frac{Im}{\rho g eA N_A}$

Hope this answer is helpful.

Know More:

Q: Drift velocity of electron is of the order of:

Click Here: https://brainly.in/question/3729513