Question

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

Answer 1

Consider that a wire of length l and area of cross-section A be subjected to an electric field of strength E.

If V = Potential difference applied across the end of the wire,

∴ E = \(\frac{V}{l}\) or V = EI

Let n = number of free electrons per unit volume of the conductor,

vd = Drift velocity of electrons

∴ Charge flowing through the conductor wire,

q = n Ale

Time taken by the electrons to cross the conductor,

t = \(\frac{\text { Distance }}{\text { Velocity }} = \frac{l}{v_d}\)

Current I = \(\frac{\text {Charge} }{\text {time}} = \frac{nAle}{\frac{l}{v_d}}\)

I = n v_d Ae ........................(1)

This is the required relation between current and drift velocity.

Since n, A and e are constants,

∴ I ∝ v_d

Hence, the current flowing through a conductor is directly proportional to the drift velocity. The small value of drift velocity (is 10^-3 ms^-1) produces a large amount of electric current, due to the presence of extremely large number of free electrons in a conductor.

I = nv_d eA 

∴ Current density,

J = \(\frac{I}{A}\) = nev_d

or J ∝ v_d