In figure the cube is of $40\,\, cm$ edge. Four straight segment of wire $ab, bc, cd$ and $da$ form a closed loop that carries a current $I = 5\,A$. A uniform magnetic field $0.02\,\,T$ is in $+y\,-$ direction ratio of magnetic force on segement $ab$ and $bc$ is
Medium
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Inward magnetic field ( $\mathrm{x}$ ) increasing. Therefore, induced current in both the loop should be to the left side loop $\left(\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\mathrm{A} \cdot \frac{\mathrm{dB}}{\mathrm{dt}}\right) .$ Therefore net current in the complete loop will be in a direction shown below. Hence only option $(1)$ is correct.
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