The figure shows a square loop L of side 5cm which is connected to...

If $B$ is magnetic field, $v$ is velocity of rod and $l$ is the length of rod, then induced EMF in moving road is given by

$ε = Bvl$
$= \left(1T\right) ( 1cm{s}^{- 1} ) ( 5cm )$
$= 1 \times 1 \times {10}^{- 2} \times 5 \times {10}^{- 2}$
$= 5 \times {10}^{- 4}V$
As given circuit forms balanced whetstone bridge, the current through $3Ω$ is zero.
Equivalent resistance ${R}_{eq} = \dfrac{4 \times 2}{4 + 2} + 1.7 = 3Ω$
Current in the circuit $i = \dfrac{ε}{{R}_{eq}} = \dfrac{5 \times {10}^{- 4}}{3}$
$= 167\muA \approx 170\mu A$