JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Electromagnetic Induction

Let $B_0$ be the field at the loop's near edge (side $AD$, at $x=0$) at some instant, with the gradient $\frac{dB}{dx}=-10^{-3}\text{T/cm}$ (field decreases along $+x$). At the far edge (side $BC$, $x=12$ cm):

Motional emfs in the sides $AB$ and $CD$ (parallel to the direction of motion) are zero since $\vec{v}$ is along their length. The motional emf in $AD$ is ${\epsilon }_1=B_0\ell v$ and in $BC$ is ${\epsilon }_2=B_{BC}\ell v$, where $\ell =5$ cm is the length of the sides perpendicular to motion. Their difference (net motional emf) is

In addition, since $B$ is decreasing with time at $10^{-3}$ T/s over the whole loop area $A=12\times 5=60{\text{cm}}^2=60\times 10^{-4}{\text{m}}^2$, there is an emf due to time variation:

Both contributions add (same sense), giving net emf ${\epsilon }_{net}=300\times 10^{-7}+60\times 10^{-7}=360\times 10^{-7}$ V. The power dissipated is

So the required value is $216$.