A current i is passed through a silver strip of width d and area of...

The drift velocity of the free electrons is $v_d$.

The current $i$ flowing through the strip is given by:

$i = n e A v_d$

$\Rightarrow v_d = \dfrac{i}{n e A}$

When the magnetic field $B$ is applied perpendicular to the current, the electrons experience a magnetic force:

$F_m = e v_d B$

This force causes electrons to accumulate on one side of the strip, setting up a transverse electric field $E$. In the steady state, the electric force balances the magnetic force:

$e E = e v_d B$

$\Rightarrow E = v_d B$

The potential difference $V$ developed across the width $d$ of the conductor is:

$V = E d = v_d B d$

Substituting the value of $v_d$, we get:

$V = \left(\dfrac{i}{n e A}\right) B d = \dfrac{i B d}{A n e}$