Are the paths of electrons straight lines between successive collisions (with positive ions of the metal) in the (i) absence of electric field? (ii) presence of electric field? Establish a relation between drift velocity ‘v_d’ of an electron in a conductor of cross-section ‘A’ carrying current ‘I’ and concentration ‘n’ f free electrons per unit volume of conductor.

i) In the absence of electric field the path of electrons are straight lines between successive collisions.

ii) When an electric field is set up from positive to negative charge, the electrons are accelerated towards the negative charge. Therefore, in the presence of electric field the electrons follow a curved path in between successive collisions. 

Consider a conductor of length'l' and of uniform area of cross section 'A'. 



Therefore, 
Volume of conductor, V = Al
If, n is the number density of electrons, then
Total number of free electrons in the conductor = Aln
If, e is the total charge on each electron then,
Total charge on electrons in the conductor, q = Alne 

Let, a constant potential difference 'V' be applied across the ends of a conductor. Then,

Electric field set up across the conductor, E = V/l 

Because of this field, the free electrons present in the conductor begin to move with a drift velocity, v_d towards the left hand side.

Therefore,
Time taken by the free electrons to cross the conductor, t = I /V_d 

Hence, current, I = $\frac{\mathit{q}}{\mathit{t}}\mathit{=}\frac{\mathit{A}\mathit{l}\mathit{n}\mathit{e}}{\mathit{l}\mathit{/}{\mathit{v}}_{\mathit{d}}}$ 

$\Rightarrow$                  $\boxed{ \mathrm{I} = \mathrm{Anevd} }$ 

where n is the number of electrons per unit volume.