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 $-e (e = 1.6 \times 10^{-19}C)$, then the total charge flowing through a cross-section of the wire $q = (nA_vd t) (-e) = –neA_vd t ...(iii)$ $\therefore$ 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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