JEE Main 2019 — Electromagnetic Induction Question with Solution

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 

    $\mathrm{\epsilon }=Bvl$     
    $=\left(1\mathrm{T}\right)(1\mathrm{c}\mathrm{m} {\mathrm{s}}^{-1})\left(5\mathrm{c}\mathrm{m}\right)$
    $=1\times 1\times {10}^{-2}\times 5\times {10}^{-2}$
    $=5\times {10}^{-4}\mathrm{V}$
As given circuit forms balanced whetstone bridge, the current through $3\mathrm{\Omega }$ is zero.
Equivalent resistance $R_{\mathrm{e}\mathrm{q}}=\frac{4\times 2}{4+2}+1.7=3\mathrm{ }\mathrm{\Omega }$
Current in the circuit $i=\frac{\mathrm{\epsilon }}{R_{\mathrm{e}\mathrm{q}}}=\frac{5\times {10}^{-4}}{3}$
$=167\mathrm{ }\mathrm{\mu }\mathrm{A}\approx 170 \mathrm{\mu A}$