Question
Establish a relation between electric current and drift velocity.

OR

Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.

Answer

Relation between electric current and drift velocity:

Consider a uniform metallic wire XY of length l and cross-sectional area A. A potential difference V is applied across the ends X and Y of the wire. This causes an electric field at each point of the wire of strength.

$\text{E}=\frac{\text{V}}{\text{l}}\dots\text{(i)}$

Due to this electric field, the electrons gain a drift velocity vd opposite to direction of electric field. If q be the charge passing through the cross-section of wire in t seconds, then

$\text{Current in write I}=\frac{\text{q}}{\text{r}}\dots\text{(ii)}$

The distance traversed by each electron in time t = average velocity × time = vd t

If we consider two planes P and Q at a distance vd t in a conductor, then the total charge flowing in time t will be equal to the total charge on the electrons present within the cylinder PQ.

The volume of this cylinder = cross sectional area × height = A vd t

If n is the number of free electrons in the wire per unit volume, then the number of free electrons

in the cylinder = n(Avd t)

If charge on each electron is -e (e = 1.6 × 10-19C), then the total charge flowing through a cross-section of the wire

q = (nAvd t) (-e) = –neAvd t ...(iii)

$\therefore$ Current flowing in the wire,

$\text{I}=\frac{\text{q}}{\text{t}}=\frac{-\text{v}}{\text{t}}$

i.e., current I = -neAvd ...(iv)

This is the relation between electric current and drift velocity. Negative sign shows that the direction of current is opposite to the drift velocity.

Numerically I = -neAvd ...(v)

$\therefore$ Current density, $\text{J}=\frac{\text{I}}{\text{A}}=\text{d}$

$\Rightarrow\text{J}\propto\text{vd.}$

That is, current density of a metallic conductor is directly proportional to the drift velocity.

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