Question

What happens to the drift velocity of the electrons and to the resistance if the length of the conductor is doubled keeping potential difference unchanged?

Answer 1

Drift velocity v is proportional to the current i (and electric field E) along the conductor length L.

  current i = V/R
 The electric field E = potential difference V / length of conductor L.
   drift speed = i /(n e A) = σ E/(ne) = V / (L ρ n e)

So drift velocity becomes half of its previous value.

====
Resistance = R = ρ L/A
     =>  Resistance becomes double, as L is twice.

Answer 2

"When length is doubled, the drift velocity is halved and the resistance is doubled.

We know that,

Drift velocity,$v_{d}=\frac{I}{A n e} \rightarrow(1)$

Where,

I is the current

A is the cross sectional area

n is the charge density

e is the charge of electron

We know that,

$I=J \times A=\sigma E A \rightarrow(2)$

Where,

J is current density

$\sigma$ is the conductivity

E is the electric field

On substituting, (2) in (1), we get,

$\Rightarrow v_{d}=\frac{\sigma E}{n e}$

We know that,

$E=\frac{V}{L}$

Where,

V is the potential difference

L is the length of the conductor

$\Rightarrow v_{ d }=\frac { \sigma V }{ { Ln }e }$

If V is constant, then

$V_{d} \propto\left(\frac{1}{L}\right)$

Thus, when the length of the conductor is doubled, the drift velocity is halved.

We know that,

$\frac{1}{\sigma}=\rho$

Where,

$\rho$ is resistivity

$\Rightarrow v_{d}=\frac{V}{L \rho n e}$

$\rho L=\frac{V}{v_{d} n e}$

Again, substituting equation (1), we get,

$\rho L=\frac{V \times A n e}{I \times n e}$

$\because$ $R=\frac{V}{I}$

$\rho L=R \times A$

$R=\frac{\rho L}{A}$

When length is doubled, the resistance is doubled."