A 1m long metal rod XY completes the circuit as shown in figure....

The force on the rod is given by

$F = ilB$

$⇒ F = \dfrac{Bvl}{R} \times lB$

$⇒ F = \dfrac{{B}^{2}{l}^{2}v}{R} . . . ( i )$

The data given is

$R = 5Ωl = 1mv = 4m{s}^{- 1}B = 0 . 15T$

Substituting the values in equation (i)

$F = \dfrac{{\left(0 . 15\right)}^{2} \times {\left(1\right)}^{2} \times 4}{5}$

$= 18 \times {10}^{- 3}N$