Solution :

The network behaves like a balanced wheatstone bridge.
The induced emf developed is given by
$$\eqalign{
& e = vB\ell = v \times 2 \times 0.1 = 0.2v\,......\left( {\text{i}} \right) \cr
& {\text{Now,}}\,\,e = IR \cr
& e = {10^{ - 3}} \times 4 = 4 \times {10^3}{\text{amp}}\,......\left( {{\text{ii}}} \right) \cr} $$
From (i) and (ii),
$$\eqalign{
& 0.2\,v = 4 \times {10^{ - 3}} \cr
& \therefore v = \frac{{4 \times {{10}^{ - 3}}}}{{0.2}} = 0.02\,m/s \cr} $$