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

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Consider a uniform metallic wire XY of length 1 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.

$E=lV $........(i)

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

Current in wire $I=tq $..........(i)

The distance traversed by each electron in time t=average velocity x time =vdt

If we consider two planes P and Q at a distance vd 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 x height

$Avdt$

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 $−c(c=1.6×10_{−19}C)$,then the total charge flowing through a cross section of the wire.

$q=(nA_{v}dt)(−e)=−neA_{v}dt$.........(iii)

$∴$ Current flowing in the wire,

$I=tq =t−v $

i.e , current$I=−neA_{v}d$.......(iv)

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

Numericaly $I=−neAτd$...............(v)

$∴$ Cureent dendity , $$J=\dfrac{t}P{l}{A-}

$$\therefgfore)$$ Curent density $$J=\dfrac{l}{A}.

$⇒JAI =d$

$⇒JαD$

This is current density of maetallic conductor is directly proportional to the drfyr velocity.

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